![10 CHAPTER 1 GENERAL PRINCIPLES
1
Rules for Use. Here are a few of the important rules that describe
the proper use of the various SI symbols:
• Quantities defined by several units which are multiples of one another
are separated by a dot to avoid confusion with prefix notation, as
indicated by . Also, (meter-second),
whereas ms (milli-second).
• The exponential power on a unit having a prefix refers to both the
unit and its prefix. For example, . Likewise,
mm2 represents .
• With the exception of the base unit the kilogram, in general avoid
the use of a prefix in the denominator of composite units. For
example, do not write , but rather ; also, should
be written as .
• When performing calculations, represent the numbers in terms of
their base or derived units by converting all prefixes to powers of 10.
The final result should then be expressed using a single prefix. Also,
after calculation, it is best to keep numerical values between 0.1 and
1000; otherwise, a suitable prefix should be chosen. For example,
1.5 Numerical Calculations
Numerical work in engineering practice is most often performed by using
handheld calculators and computers. It is important, however, that the
answers to any problem be reported with both justifiable accuracy and
appropriate significant figures. In this section we will discuss these topics
together with some other important aspects involved in all engineering
calculations.
Dimensional Homogeneity. The terms of any equation used to
describe a physical process must be dimensionally homogeneous; that is,
each term must be expressed in the same units. Provided this is the case,
all the terms of an equation can then be combined if numerical values
are substituted for the variables. Consider, for example, the equation
, where, in SI units, is the position in meters, m, is time in
seconds, s, is velocity in and is acceleration in . Regardless of
how this equation is evaluated, it maintains its dimensional homogeneity.
In the form stated, each of the three terms is expressed in meters
or solving for , , the terms are
each expressed in units of .
ms2
[ms2
, ms2
, (ms)s]
a = 2st2
- 2vt
a
[m,(ms) s, (ms2
) s2
,]
ms2
a
ms
v
t
s
s = vt + 1
2 at2
= 3000(10-6
) N # m = 3(10-3
) N # m = 3 mN # m
(50 kN)(60 nm) = [50(103
) N][60(10-9
) m]
Mmkg
mmg
kNm
Nmm
(mm)2
= mm # mm
mN2
= (mN)2
= mN # mN
m # s
N = kg # ms2
= kg # m # s-2
Computers are often used in engineering for
advanced design and analysis.](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-27-320.jpg)
![14 CHAPTER 1 GENERAL PRINCIPLES
1
Evaluate each of the following and express with SI units having an
appropriate prefix: (a) (50 mN)(6 GN), (b) (400 mm)(0.6 MN)2,
(c) .
SOLUTION
First convert each number to base units, perform the indicated
operations, then choose an appropriate prefix.
Part (a)
Ans.
NOTE: Keep in mind the convention .
Part (b)
Ans.
We can also write
Ans.
Part (c)
Ans.
= 50 kN3
kg
= 50(109
) N3
a
1 kN
103
N
b
3
1
kg
= 50(109
) N3
kg
45 MN3
900 Gg
=
45(106
N)3
900(106
) kg
= 0.144 m # MN2
144(109
) m # N2
= 144(109
) m # N2
a
1 MN
106
N
b a
1 MN
106
N
b
= 144 Gm # N2
= 144(109
) m # N2
= [400(10-3
) m][0.36(1012
) N2
]
(400 mm)(0.6 MN)2
= [400(10-3
) m][0.6(106
) N]2
kN2
= (kN)2
= 106
N2
= 300 kN2
= 300(106
) N2
a
1 kN
103
N
b a
1 kN
103
N
b
= 300(106
) N2
(50 mN)(6 GN) = [50(10-3
) N][6(109
) N]
45 MN3
900 Gg
EXAMPLE 1.3](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-31-320.jpg)
![58 CHAPTER 2 FORCE VECTORS
2
EXAMPLE 2.12
An elastic rubber band is attached to points A and B as shown in
Fig. 2–37a. Determine its length and its direction measured from A
toward B.
SOLUTION
We first establish a position vector from A to B, Fig. 2–37b. In
accordance with Eq.2–11,the coordinates of the tail A(1 m,0,–3 m) are
subtracted from the coordinates of the head B(–2 m, 2 m, 3 m), which
yields
These components of r can also be determined directly by realizing
that they represent the direction and distance one must travel along
each axis in order to move from A to B, i.e., along the x axis {–3i} m,
along the y axis {2j} m, and finally along the z axis {6k} m.
The length of the rubber band is therefore
Ans.
Formulating a unit vector in the direction of r, we have
The components of this unit vector give the coordinate direction
angles
Ans.
Ans.
Ans.
NOTE: These angles are measured from the positive axes of a localized
coordinate system placed at the tail of r, as shown in Fig. 2–37c.
g = cos-1
a
6
7
b = 31.0°
b = cos-1
a
2
7
b = 73.4°
a = cos-1
a-
3
7
b = 115°
u =
r
r
= -
3
7
i +
2
7
j +
6
7
k
r = 2(-3 m)2
+ (2 m)2
+ (6 m)2
= 7 m
= 5-3i + 2j + 6k6 m
r = [-2 m - 1 m]i + [2 m - 0]j + [3 m - (-3 m)]k
(a)
z
y
x
3 m
1 m
A
B
3 m
2 m
2 m
(b)
z
y
A
B
{6 k}
{2 j} m
{3 i} m
r
x
(c)
A
B
z¿
y¿
x¿
r 7 m
g 31.0
a 115
b 73.4
Fig. 2–37](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-75-320.jpg)
![EXAMPLE 3.3
92 CHAPTER 3 EQUILIBRIUM OF A PARTICLE
3
The 200-kg crate in Fig. 3-7a is suspended using the ropes and .
Each rope can withstand a maximum force of before it breaks. If
always remains horizontal, determine the smallest angle to which
the crate can be suspended before one of the ropes breaks.
u
AB
10 kN
AC
AB
SOLUTION
Free-Body Diagram. We will study the equilibrium of ring .There
are three forces acting on it, Fig. 3-7b.The magnitude of is equal to
the weight of the crate, i.e., .
Equations of Equilibrium. Applying the equations of equilibrium
along the x and y axes,
; (1)
(2)
From Eq. (1), is always greater than since .
Therefore, rope will reach the maximum tensile force of
before rope . Substituting into Eq. (2), we get
Ans.
The force developed in rope can be obtained by substituting the
values for and into Eq. (1).
FB = 9.81 kN
10(103
) N =
FB
cos11.31°
FC
u
AB
u = sin-1
(0.1962) = 11.31° = 11.3°
[10(103
)N] sin u - 1962 N = 0
FC = 10 kN
AB
10 kN
AC
cos u … 1
FB
FC
FC sin u - 1962 N = 0
+ c©Fy = 0;
FC =
FB
cos u
- FC cos u + FB = 0
:
+ ©Fx = 0;
FD = 200 (9.81) N = 1962 N 6 10 kN
FD
A
(a)
D
A B
C u
Fig. 3–7
FD 1962 N
y
x
(b)
A
FC
FB
u](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-109-320.jpg)
![3.4 THREE-DIMENSIONAL FORCE SYSTEMS 105
3
EXAMPLE 3.6
The 10-kg lamp in Fig. 3-11a is suspended from the three equal-length
cords. Determine its smallest vertical distance s from the ceiling if the
force developed in any cord is not allowed to exceed 50 N.
x
y
s
(a)
z
D
A
B
C
600 mm
120
120
Fig. 3–11
SOLUTION
Free-Body Diagram. Due to symmetry, Fig. 3-11b, the distance
. It follows that from and
, the tension T in each cord will be the same. Also, the angle
between each cord and the axis is .
Equation of Equilibrium. Applying the equilibrium equation along
the axis, with , we have
From the shaded triangle shown in Fig. 3-11b,
Ans.
s = 519 mm
tan 49.16° =
600 mm
s
g = cos-1 98.1
150
= 49.16°
3[(50 N) cos g] - 10(9.81) N = 0
gFz = 0;
T = 50 N
z
g
z
gFy = 0
gFx = 0
DA = DB = DC = 600 mm
x
y
s
600 mm
D
z
(b)
A
B
C
10(9.81) N
T
T
T
g](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-122-320.jpg)
![126 CHAPTER 4 FORCE SYSTEM RESULTANTS
4
EXAMPLE 4.3
Determine the moment produced by the force F in Fig. 4–14a about
point O. Express the result as a Cartesian vector.
SOLUTION
As shown in Fig. 4–14a, either or can be used to determine the
moment about point O.These position vectors are
Force F expressed as a Cartesian vector is
Thus
Ans.
or
Ans.
NOTE: As shown in Fig. 4–14b, acts perpendicular to the plane
that contains . Had this problem been worked using
notice the difficulty that would arise in obtaining the
moment arm d.
MO = Fd,
F, rA, and rB
MO
= 5-16.5i + 5.51j6 kN–m
+ [4(1.376) - 12(0.4588)]k
= [12(-1.376) - 0(1.376)]i - [4(-1.376) - 0(0.4588)]j
MO = rB * F = 3
i j k
4 12 0
0.4588 1.376 -1.376
3
= 5-16.5i + 5.51j6 kN–m
+ [0(1.376) - 0(0.4588)]k
= [0(-1.376) - 12(1.376)]i - [0(-1.376) - 12(0.4588)] j
MO = rA * F = 3
i j k
0 0 12
0.4588 1.376 -1.376
3
= 50.4588i + 1.376j - 1.376k6 kN
F = FuAB = 2 kNc
54i + 12j - 12k6 m
214 m22
+ 112 m22
+ 1-12 m22
d
rA = 512k6 m and rB = 54i + 12j6 m
rB
rA
12 m
4 m
12 m
A
B
O
x
y
z
(a)
F 2 kN
uAB
Fig. 4–14
(b)
A
B
O
x
y
z
F
rB
rA
MO](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-143-320.jpg)
![4.3 MOMENT OF A FORCE—VECTOR FORMULATION 127
4
EXAMPLE 4.4
x
z
O
5 ft
4 ft
2 ft
A
B
F2 {80i 40j 30k} lb
F1 {60i 40j 20k} lb
(a)
y
Fig. 4–15
x
y
z
O A
B
(b)
rA
rB
F1
F2
x
y
z
O
39.8
67.4
121
MRO
{30i 40j 60k} lb · ft
(c)
a
g
b
SOLUTION
Position vectors are directed from point O to each force as shown in
Fig. 4–15b.These vectors are
The resultant moment about O is therefore
rB = 54i + 5j - 2k6 ft
rA = 55j6 ft
NOTE: This result is shown in Fig. 4–15c. The coordinate direction
angles were determined from the unit vector for Realize that the
two forces tend to cause the rod to rotate about the moment axis in
the manner shown by the curl indicated on the moment vector.
MRO
.
Ans.
= 530i - 40j + 60k6 lb # ft
+ [51-302 - 1-221402]i - [41-302 - (-2)1802]j + [41402 - 51802]k
= [51202 - 01402]i - [0]j + [01402 - (5)1-602]k
= 3
i j k
0 5 0
-60 40 20
3 + 3
i j k
4 5 -2
80 40 -30
3
= rA * F1 + rB * F3
MRO
= ©1r * F2
Two forces act on the rod shown in Fig. 4–15a. Determine the
resultant moment they create about the flange at O. Express the result
as a Cartesian vector.](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-144-320.jpg)
![130 CHAPTER 4 FORCE SYSTEM RESULTANTS
4
EXAMPLE 4.6
Force F acts at the end of the angle bracket shown in Fig. 4–19a.
Determine the moment of the force about point O.
SOLUTION I (SCALAR ANALYSIS)
The force is resolved into its x and y components as shown in
Fig. 4–19b, then
a
b
or
Ans.
SOLUTION II (VECTOR ANALYSIS)
Using a Cartesian vector approach, the force and position vectors
shown in Fig. 4–19c are
The moment is therefore
Ans.
NOTE: It is seen that the scalar analysis (Solution I) provides a
more convenient method for analysis than Solution II since the
direction of the moment and the moment arm for each component
force are easy to establish. Hence, this method is generally
recommended for solving problems displayed in two dimensions,
whereas a Cartesian vector analysis is generally recommended only
for solving three-dimensional problems.
= 5-98.6k6 N # m
= 0i - 0j + [0.41-346.42 - 1-0.221200.02]k
MO = r * F = 3
i j k
0.4 -0.2 0
200.0 -346.4 0
3
= 5200.0i - 346.4j6 N
F = 5400 sin 30°i - 400 cos 30°j6 N
r = 50.4i - 0.2j6 m
MO = 5-98.6k6 N # m
= -98.6 N # m = 98.6 N # m
+MO = 400 sin 30° N10.2 m2 - 400 cos 30° N10.4 m2
0.4 m
0.2 m
30
O
F = 400 N
(a)
Fig. 4–19
0.4 m
0.2 m
(b)
x
400 cos 30 N
400 sin 30 N
O
y
y
x
0.4 m
0.2 m
30
O
F
(c)
r](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-147-320.jpg)
![140 CHAPTER 4 FORCE SYSTEM RESULTANTS
Vector Analysis. To find the moment of force F in Fig. 4–20b about
the y axis using a vector analysis, we must first determine the moment of
the force about any point O on the y axis by applying Eq. 4–7,
.The component along the y axis is the projection of
onto the y axis. It can be found using the dot product discussed in
Chapter 2, so that where j is the unit vector
for the y axis.
We can generalize this approach by letting be the unit vector that
specifies the direction of the axis shown in Fig. 4–21.Then the moment
of F about the axis is . This combination is referred to
as the scalar triple product. If the vectors are written in Cartesian form,
we have
This result can also be written in the form of a determinant, making it
easier to memorize.*
(4–11)
where
represent the x, y, z components of the unit
vector defining the direction of the axis
represent the x, y, z components of the
position vector extended from any point O on
the axis to any point A on the line of action
of the force
represent the x, y, z components of the force
vector.
When is evaluated from Eq. 4–11, it will yield a positive or negative
scalar.The sign of this scalar indicates the sense of direction of along
the axis. If it is positive, then will have the same sense as whereas
if it is negative, then will act opposite to
Once is determined, we can then express as a Cartesian vector,
namely,
(4–12)
The examples which follow illustrate numerical applications of the
above concepts.
Ma = Maua
Ma
Ma
ua.
Ma
ua,
Ma
a
Ma
Ma
Fz
Fy,
Fx,
a
rz
ry,
rx,
a
uax
, uay
, uaz
Ma = ua
# 1r * F2 = 3
uax
uay
uaz
rx ry rz
Fx Fy Fz
3
= uax
(ryFz - rzFy) - uay
(rxFz - rzFx) + uaz
(rxFy - ryFx)
Ma = [uax
i + uay
j + uaz
k] # 3
i j k
rx ry rz
Fx Fy Fz
3
Ma = ua
# (r * F)
a
ua
My = j # MO = j # (r * F),
MO
My
MO = r * F
*Take a moment to expand this determinant, to show that it will yield the above result.
r
O
MO r F
Ma
ua
a
Axis of projection
F
A
Fig. 4–21
x y
r
j
(b)
z
O
M0 r F
F
u
u My
4 Fig. 4–20](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-157-320.jpg)
![142 CHAPTER 4 FORCE SYSTEM RESULTANTS
4
EXAMPLE 4.8
Determine the moment produced by the force F in Fig. 4–23a,
which tends to rotate the rod about the AB axis.
SOLUTION
A vector analysis using will be considered for the
solution rather than trying to find the moment arm or perpendicular
distance from the line of action of F to the AB axis. Each of the terms
in the equation will now be identified.
Unit vector defines the direction of the AB axis of the rod,
Fig. 4–23b, where
Vector r is directed from any point on the AB axis to any point on the
line of action of the force. For example, position vectors and are
suitable, Fig. 4–23b. (Although not shown, or can also be
used.) For simplicity, we choose where
The force is
Substituting these vectors into the determinant form and expanding,
we have
This positive result indicates that the sense of is in the same
direction as
Expressing as a Cartesian vector yields
Ans.
The result is shown in Fig. 4–23b.
NOTE: If axis AB is defined using a unit vector directed from B toward
A,then in the above formulation would have to be used.This would
lead to Consequently, and
the same result would be obtained.
MAB = MAB1-uB2,
MAB = -80.50 N # m.
-uB
= 572.0i + 36.0j6 N # m
MAB = MABuB = 180.50 N # m210.8944i + 0.4472j2
MAB
uB.
MAB
= 80.50 N # m
+ 0[0.6102 - 0102]
= 0.8944[01-3002 - 0102] - 0.4472[0.61-3002 - 0102]
MAB = uB
# 1rD * F2 = 3
0.8944 0.4472 0
0.6 0 0
0 0 -300
3
F = 5-300k6 N
rD = 50.6i6 m
rD,
rBD
rBC
rD
rC
uB =
rB
rB
=
{0.4i + 0.2j} m
210.4 m2 2
+ 10.2 m2 2
= 0.8944i + 0.4472j
uB
MAB = uB
# 1r * F2
MAB
0.4 m
(a)
0.3 m
0.6 m
0.2 m
C
F = 300 N
B
x
y
z
A
Fig. 4–23
(b)
F
C
B
x
z
MAB
uB
rC
rD
A
D
y](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-159-320.jpg)
![4.5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS 143
4
EXAMPLE 4.9
Determine the magnitude of the moment of force F about segment
OA of the pipe assembly in Fig. 4–24a.
SOLUTION
The moment of F about the OA axis is determined from
where r is a position vector extending from any
point on the OA axis to any point on the line of action of F. As
indicated in Fig. 4–24b, either can be used;
however, will be considered since it will simplify the calculation.
The unit vector , which specifies the direction of the OA axis, is
and the position vector is
rOD = 50.5i + 0.5k6 m
rOD
uOA =
rOA
rOA
=
50.3i + 0.4j6 m
210.3 m22
+ 10.4 m22
= 0.6i + 0.8j
uOA
rOD
rOD, rOC, rAD, or rAC
MOA = uOA
# (r * F),
The force F expressed as a Cartesian vector is
= {200i - 200j + 100k} N
= (300 N)B
50.4i - 0.4j + 0.2k6 m
2(0.4 m)2
+ (-0.4 m)2
+ (0.2 m)2
R
F = Fa
rCD
rCD
b
Therefore,
Ans.
= 100 N#m
= 0.6[011002 - (0.5)1-2002] - 0.8[0.511002 - (0.5)12002] + 0
= 3
0.6 0.8 0
0.5 0 0.5
200 -200 100
3
MOA = uOA
# 1rOD * F2
0.1 m
0.3 m
0.2 m
0.4 m
0.5 m
0.5 m
(a)
x y
C
A
O
D
z
F 300 N
B
Fig. 4–24
x
y
z
F
(b)
D
A
C
O
rOD
rAD
rAC
rOC
uOA](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-160-320.jpg)
![4.6 MOMENT OF A COUPLE 153
4
EXAMPLE 4.13
Replace the two couples acting on the pipe column in Fig. 4–33a by a
resultant couple moment.
SOLUTION (VECTOR ANALYSIS)
The couple moment developed by the forces at A and B, can
easily be determined from a scalar formulation.
By the right-hand rule, acts in the direction, Fig. 4–33b. Hence,
Vector analysis will be used to determine caused by forces at C
and D. If moments are computed about point D, Fig. 4–33a,
then
Since and are free vectors, they may be moved to some
arbitrary point and added vectorially, Fig. 4–33c. The resultant couple
moment becomes
Ans.
MR = M1 + M2 = 560i + 22.5j + 30k6 N # m
M2
M1
= 522.5j + 30k6 N # m
= 10.3i2 * [100j - 75k] = 301i * j2 - 22.51i * k2
M2 = rDC * FC = 10.3i2 * C125A4
5 Bj - 125A3
5 BkD
M2 = rDC * FC,
M2,
M1 = 560i6 N # m
+i
M1
M1 = Fd = 150 N10.4 m2 = 60 N # m
M1,
0.3 m
150 N
125 N
125 N
3
4
5
D
z
y
5
3
4
C
0.4 m
150 N
A
B
x
(a)
Fig. 4–33
M2 37.5 N m
M1 60 N m
3
4
5
(b)
(c)
M1
M2
MR](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-170-320.jpg)
![174 CHAPTER 4 FORCE SYSTEM RESULTANTS
4
Replace the force and couple moment system acting on the beam in
Fig. 4–44a by an equivalent resultant force, and find where its line of
action intersects the beam, measured from point O.
EXAMPLE 4.17
SOLUTION
Force Summation. Summing the force components,
From Fig. 4–44b, the magnitude of is
Ans.
The angle is
Ans.
Moment Summation. We must equate the moment of about
point O in Fig. 4–44b to the sum of the moments of the force and
couple moment system about point O in Fig. 4–44a. Since the line of
action of acts through point O, only produces a moment
about this point.Thus,
a
Ans.
d = 2.25 m
-[8 kNA3
5 B] (0.5 m) + [8 kNA4
5 B](4.5 m)
2.40 kN(d) = -(4 kN)(1.5 m) - 15 kN#m
+(MR)O = ©MO;
(FR)y
(FR)x
FR
u = tan-1
a
2.40 kN
4.80 kN
b = 26.6°
u
FR = 214.80 kN22
+ 12.40 kN22
= 5.37 kN
FR
(FR)y = -4 kN + 8 kNA4
5 B = 2.40 kNc
+ c(FR)y = ©Fy;
(FR)x = 8 kNA3
5 B = 4.80 kN :
:
+ (FR)x = ©Fx;
(a)
O
4 kN
15 kNm
8 kN
3
4
5
1.5 m 1.5 m 1.5 m 1.5 m
0.5 m
y
x
Fig. 4–44
(b)
d
O
FR
(FR)x 4.80 kN
(FR)y 2.40 kN
u](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-191-320.jpg)
![4.8 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 177
4
EXAMPLE 4.20
Replace the force system in Fig. 4–47a by an equivalent resultant
force and specify its point of application on the pedestal.
SOLUTION
Force Summation. Here we will demonstrate a vector analysis.
Summing forces,
Ans.
Location. Moments will be summed about point O. The resultant
force is assumed to act through point P (x, y, 0), Fig. 4–47b.Thus
Equating the i and j components,
(1)
Ans.
(2)
Ans.
The negative sign indicates that the x coordinate of point P is
negative.
NOTE: It is also possible to establish Eq. 1 and 2 directly by summing
moments about the x and y axes. Using the right-hand rule, we have
700x = 300 lb14 in.2 - 500 lb14 in.)
(MR)y = ©My;
-700y = -100 lb(4 in.) - 500 lb(2 in.)
(MR)x = ©Mx;
x = -1.14 in.
700x = -800
y = 2 in.
-700y = -1400
700xj - 700yi = 1200j - 2000j - 1000i - 400i
- 1000(j * k) - 400(j * k)
-700x(i * k) - 700y(j * k) = -1200(i * k) + 2000(i * k)
+ [1-4i + 2j2 * 1-500k2] + [(-4j) * (100k)]
1xi + yj2 * 1-700k2 = [14i2 * 1-300k2]
rP * FR = (rA * FA) + (rB * FB) + (rC * FC)
(MR)O = ©MO;
FR
= 5-700k6 lb
= 5-300k6 lb + 5-500k6 lb + 5100k6 lb
FR = FA + FB + FC
FR = ©F;
x y
z
(a)
FB 500 lb
FA 300 lb
FC 100 lb 2 in.
4 in.
4 in.
4 in.
B
O
A
C
rB
rA
rC
Fig. 4–47
x y
z
(b)
FR {700k} lb
rP
O
P
y
x](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-194-320.jpg)
![5.2 FREE-BODY DIAGRAMS 203
Typical examples of actual supports are shown in the following sequence of photos. The numbers refer to the
connection types in Table 5–1.
5
The cable exerts a force on the bracket
in the direction of the cable. (1)
The rocker support for this bridge
girder allows horizontal movement
so the bridge is free to expand and
contract due to a change in
temperature. (5)
This concrete girder
rests on the ledge that
is assumed to act as
a smooth contacting
surface. (6)
This utility building is
pin supported at the top
of the column. (8)
The floor beams of this building
are welded together and thus
form fixed connections. (10)
Types of Connection Reaction Number of Unknowns
Two unknowns. The reactions are two components of
force, or the magnitude and direction of the resultant
force. Note that and are not necessarily equal [usually
not, unless the rod shown is a link as in (2)].
Three unknowns. The reactions are the couple moment
and the two force components, or the couple moment and
the magnitude and direction of the resultant force.
Two unknowns. The reactions are the couple moment
and the force which acts perpendicular to the rod.
F
Fy
M
or
Fx
F
fixed support
Fy
Fx
F
or
M M
f
f
f
u
TABLE 5–1 Continued
member fixed connected
to collar on smooth rod
smooth pin or hinge
(8)
(9)
(10)
u f
f](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-220-320.jpg)
 - [NB sin 30°](2 ft) - 750 lb(3 ft) = 0
+©MA = 0;
Ay
Ax](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-237-320.jpg)
![5.3 EQUATIONS OF EQUILIBRIUM 223
5
EXAMPLE 5.12
Determine the support reactions on the member in Fig. 5–19a. The
collar at A is fixed to the member and can slide vertically along the
vertical shaft.
A
B
(a)
1.5 m 1.5 m
1 m
45
900 N
500 N m
A
B
Ax
MA
900 N
NB
45
500 N m
1 m
1.5 m 1.5 m
y
x
(b)
Fig. 5–19
SOLUTION
Free-Body Diagram. The free-body diagram of the member is shown
in Fig. 5–19b.The collar exerts a horizontal force and moment
on the member. The reaction of the roller on the member is
vertical.
Equations of Equilibrium. The forces and can be determined
directly from the force equations of equilibrium.
Ans.
Ans.
The moment can be determined by summing moments either
about point A or point B.
a
b Ans.
or
a
b Ans.
The negative sign indicates that has the opposite sense of rotation
to that shown on the free-body diagram.
MA
MA = -1486 N # m = 1.49 kN # m
MA + 900 N [1.5 m + (1 m) cos 45°] - 500 N # m = 0
+©MB = 0;
MA = -1486 N # m = 1.49 kN # m
MA - 900 N(1.5 m) - 500 N # m + 900 N [3 m + (1 m) cos 45°] = 0
+ ©MA = 0;
MA
NB - 900 N
NB - 900 N = 0
+c©Fy = 0;
Ax = 0
:
+ © Fx = 0;
NB
Ax
NB
MA
Ax](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-240-320.jpg)
![5.4 TWO- AND THREE-FORCE MEMBERS 235
5
•5–61. If spring BC is unstretched with and the bell
crank achieves its equilibrium position when ,
determine the force F applied perpendicular to segment
AD and the horizontal and vertical components of reaction
at pin A. Spring BC remains in the horizontal postion at all
times due to the roller at C.
u = 15°
u = 0°
*5–60. The uniform rod has a length l and weight W. It is
supported at one end A by a smooth wall and the other end
by a cord of length s which is attached to the wall as
shown. Show that for equilibrium it is required that
.
h = [(s2
- l2
)3]12
5–59. A man stands out at the end of the diving board,
which is supported by two springs A and B, each having a
stiffness of . In the position shown the board
is horizontal. If the man has a mass of 40 kg, determine the
angle of tilt which the board makes with the horizontal after
he jumps off. Neglect the weight of the board and assume it
is rigid.
k = 15 kNm
5–62. The thin rod of length l is supported by the smooth
tube. Determine the distance a needed for equilibrium if
the applied load is P.
B
A
1 m 3 m
Prob. 5–59
Prob. 5–60
300 mm
400 mm
B
k 2 kN/m
D
C
A
150 F
u
Prob. 5–61
P
B
A
2r
a
l
Prob. 5–62
h
s
C
B
A
l](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-252-320.jpg)
![5.7 CONSTRAINTS AND STATICAL DETERMINACY 251
5
EXAMPLE 5.19
The bent rod in Fig. 5–32a is supported at A by a journal bearing, at
D by a ball-and-socket joint, and at B by means of cable BC. Using
only one equilibrium equation, obtain a direct solution for the
tension in cable BC. The bearing at A is capable of exerting force
components only in the z and y directions since it is properly aligned
on the shaft.
SOLUTION (VECTOR ANALYSIS)
Free-Body Diagram. As shown in Fig. 5–32b, there are six unknowns.
Equations of Equilibrium. The cable tension may be obtained
directly by summing moments about an axis that passes through
points D and A. Why? The direction of this axis is defined by the unit
vector u, where
Hence, the sum of the moments about this axis is zero provided
Here r represents a position vector drawn from any point on the axis
DA to any point on the line of action of force F (see Eq. 4–11). With
reference to Fig. 5–32b, we can therefore write
Ans.
Since the moment arms from the axis to and W are easy to obtain,
we can also determine this result using a scalar analysis. As shown in
Fig. 5–32b,
Ans.
T
B = 490.5 N
©MDA = 0; TB(1 m sin 45°) - 981 N(0.5 m sin 45°) = 0
TB
T
B = 490.5 N
- 0.70711-TB + 490.52 + 0 + 0 = 0
1-0.7071i - 0.7071j2 # [1-TB + 490.52i] = 0
+ 1-0.5j2 * 1-981k2D = 0
1-0.7071i - 0.7071j2 # C1-1j2 * 1TBk2
u # 1rB * TB + rE * W2 = 0
©MDA = u # ©1r * F2 = 0
= -0.7071i - 0.7071j
u =
rDA
rDA
= -
1
22
i -
1
22
j
TB
0.5 m
0.5 m
x
z
y
E
B
A
D
100 kg
C
(a)
1 m
TB
x
z
y
B
A
D
Az
Ay
Dy
Dz
Dx
rE
rB
W 981 N
u
(b)
45
0.5 m
0.5 m
Fig. 5–32](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-268-320.jpg)
 + T1(3 m) = 0
2T1(0.8 m) - [75(9.81) N](0.8 m)
+©MA = 0
E
T1
T2 = 512 N
Nm = 224 N
T1 = 256 N
T2
Nm
T1
T1(3 m) - Nm(0.8 m) - [40(9.81) N](1.5 m) = 0
+©MA = 0
Nm + 2T1 - 75(9.81) N = 0
+ c©Fy = 0
T2 = 2T1
2T1 - T2 = 0
+ c©Fy = 0;
NB = 0
T1
EXAMPLE 6.20
A B
C
D
H
E
F
2.2 m
(a)
0.8 m
Fig. 6–32
G
H
E
1.5 m
75 (9.81) N
40 (9.81) N
(b)
0.8 m 0.7 m
Ay NB 0
Ax
Nm
T1
T1
T1
T2 2 T1
T2
Nm
G
1.5 m
75 (9.81) N
40 (9.81) N
(c)
0.8 m 0.7 m
Ay NB 0
Ax
T1
T1
T1](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-325-320.jpg)
![7.3 RELATIONS BETWEEN DISTRIBUTED LOAD, SHEAR, AND MOMENT 355
7
If we rewrite the above equation in the form dV = w(x)dx and perform
an integration between any two points B and C on the beam, we see that
(7–2)
Relation Between the Shear and Moment. If we apply the
moment equation of equilibrium about point O on the free-body
diagram in Fig. 7–13b, we get
a
Dividing both sides of this equation by , and letting 0, yields
(7–3)
In particular, notice that the absolute maximum bending moment
occurs at the point where the slope , since this is
where the shear is equal to zero.
If Eq. 7–3 is rewritten in the form and integrated
between any two points B and C on the beam, we have
(7–4)
As stated previously, the above equations do not apply at points where
a concentrated force or couple moment acts. These two special cases
create discontinuities in the shear and moment diagrams, and as a result,
each deserves separate treatment.
Force. A free-body diagram of a small segment of the beam in
Fig. 7–13a, taken from under one of the forces, is shown in Fig. 7–14a.
Here force equilibrium requires
(7–5)
Since the change in shear is positive, the shear diagram will “jump”
upward when F acts upward on the beam. Likewise, the jump in shear
is downward when F acts downward.
1¢V2
¢V = F
+ c ©Fy = 0;
Change in
moment
=
Area under
shear diagram
¢M =
L
V dx
dM = 1 V dx
dMdx = 0
|M|max
Slope of
moment diagram
= Shear
dM
dx
= V
¢x :
¢x
¢M = V¢x + k w(x)¢x2
(M + ¢M) - [w(x)¢x] k¢x - V¢x - M = 0
+©M0 = 0;
Change in
shear
=
Area under
loading curve
¢V =
L
w1x2 dx
V
M
V
x
V
M M
(a)
F
Fig. 7–14](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-372-320.jpg)
![7.3 RELATIONS BETWEEN DISTRIBUTED LOAD, SHEAR, AND MOMENT 357
7
EXAMPLE 7.8
Draw the shear and moment diagrams for the cantilever beam in
Fig. 7–15a.
2 kN
1.5 kN/m
(a)
A
B
2 m 2 m
Fig. 7–15
SOLUTION
The support reactions at the fixed support B are shown in
Fig. 7–15b.
Shear Diagram. The shear at end A is –2 kN. This value is plotted
at x = 0, Fig. 7–15c. Notice how the shear diagram is constructed by
following the slopes defined by the loading w. The shear at x = 4 m is
–5 kN, the reaction on the beam. This value can be verified by finding
the area under the distributed loading; i.e.,
Moment Diagram. The moment of zero at x = 0 is plotted in
Fig. 7–15d. Construction of the moment diagram is based on knowing
its slope which is equal to the shear at each point. The change of
moment from x = 0 to x = 2 m is determined from the area under the
shear diagram. Hence, the moment at x = 2 m is
This same value can be determined from the method of sections,
Fig. 7–15e.
Mƒx=2 m = Mƒx=0 + ¢M = 0 + [-2 kN(2 m)] = -4 kN # m
V ƒx=4 m = Vƒx=2 m + ¢V = -2 kN - (1.5 kNm)(2m) = -5 kN
(d)
(c)
(b)
2 kN
2 m 2 m
2 4
5
2
By 5 kN
MB 11 kN m
x (m)
V (kN)
2
0
4
11
4
x (m)
M (kN m)
w 0
slope 0
w negative constant
slope negative constant
V negative constant
slope negative constant
V negative increasing
slope negative increasing
1.5 kN/m
(e)
2 m
V 2 kN
M 4 kN m
2 kN](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-374-320.jpg)
![358 CHAPTER 7 INTERNAL FORCES
7
Draw the shear and moment diagrams for the overhang beam in
Fig. 7–16a.
EXAMPLE 7.9
4 kN/m
4 m 2 m
(a)
A
B
Fig. 7–16
SOLUTION
The support reactions are shown in Fig. 7–16b.
Shear Diagram. The shear of –2 kN at end A of the beam is plotted
at x = 0, Fig. 7–16c. The slopes are determined from the loading and
from this the shear diagram is constructed, as indicated in the figure.
In particular, notice the positive jump of 10 kN at x = 4 m due to the
force , as indicated in the figure.
Moment Diagram. The moment of zero at x = 0 is plotted,
Fig. 7–16d, then following the behavior of the slope found from the
shear diagram, the moment diagram is constructed. The moment at
is found from the area under the shear diagram.
We can also obtain this value by using the method of sections, as
shown in Fig. 7–16e.
Mƒ x=4 m = Mƒx=0 + ¢M = 0 + [-2 kN(4 m)] = -8 kN # m
x = 4 m
By
4 m 2 m
Ay 2 kN
By 10 kN
A
2
8
4 6
4
0 x (m)
V (kN)
6
8
0 x (m)
M (kN m)
w 0
slope 0
V positive decreasing
slope positive decreasing
V negative constant
slope negative constant
w negative constant
slope negative constant
(d)
(c)
(b)
4 kN/m
4 m
2 kN
A
(e)
V 2 kN
M 8 kN m](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-375-320.jpg)
![7.3 RELATIONS BETWEEN DISTRIBUTED LOAD, SHEAR, AND MOMENT 359
7
EXAMPLE 7.10
The shaft in Fig. 7–17a is supported by a thrust bearing at A and a
journal bearing at B. Draw the shear and moment diagrams.
B
A
12 ft
(a)
120 lb/ft
Fig. 7–17
SOLUTION
The support reactions are shown in Fig. 7–17b.
Shear Diagram. As shown in Fig. 7–17c, the shear at x = 0 is +240.
Following the slope defined by the loading, the shear diagram is
constructed, where at B its value is –480 lb. Since the shear changes
sign, the point where V = 0 must be located.To do this we will use the
method of sections. The free-body diagram of the left segment of the
shaft, sectioned at an arbitrary position x within the region 0 x
9 ft, is shown in Fig. 7–17e. Notice that the intensity of the distributed
load at x is w = 10x, which has been found by proportional triangles,
i.e., .
Thus, for V = 0,
Moment Diagram. The moment diagram starts at 0 since there is
no moment at A, then it is constructed based on the slope as
determined from the shear diagram.The maximum moment occurs at
x = 6.93 ft, where the shear is equal to zero, since dM/dx = V = 0,
Fig. 7–17e,
x = 6.93 ft
240 lb -1
2(10x)x = 0
+ c©Fy = 0;
12012 = wx
…
x (ft)
B
120 lb/ft
A
12 ft
Ay 240 lb By 480 lb
12
6.93
6.93 12
240
480
V (lb)
x (ft)
0
0
M (lb ft)
w negative increasing
slope negative increasing
V negative increasing
slope negative increasing
(d)
(c)
(b)
linear
parabolic
cubic 1109
positive
decreasing
A
x
(e)
Ay 240 lb
x
3
[ ] x
1
2
10 x
10 x
V
M
Finally, notice how integration, first of the loading w which is linear,
produces a shear diagram which is parabolic, and then a moment
diagram which is cubic.
Mmax = 1109 lb # ft
a ; Mmax + 1
2 [(10)(6.93)] 6.93 A1
3 (6.93)B -240(6.93) = 0
+©M = 0](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-376-320.jpg)
![374 CHAPTER 7 INTERNAL FORCES
7
Determine the deflection curve, the length, and the maximum tension
in the uniform cable shown in Fig. 7–23.The cable has a weight per unit
length of
SOLUTION
For reasons of symmetry, the origin of coordinates is located at the
center of the cable.The deflection curve is expressed as We
can determine it by first applying Eq. 7–15, where
Integrating the term under the integral sign in the denominator,
we have
Substituting so that a
second integration yields
or
(1)
To evaluate the constants note that, from Eq. 7–14,
Since at then Thus,
(2)
The constant may be evaluated by using the condition at
in Eq. 1, in which case To obtain the deflection curve,
solve for s in Eq. 1, which yields
(3)
Now substitute into Eq. 2, in which case
dy
dx
= sinha
w0
FH
xb
s =
FH
w0
sinha
w0
FH
xb
C2 = 0.
x = 0
s = 0
C2
dy
dx
=
w0s
FH
C1 = 0.
s = 0,
dydx = 0
dy
dx
=
1
FH L
w0 ds or
dy
dx
=
1
FH
1w0s + C12
x =
FH
w0
esinh-1
c
1
FH
1w0s + C12d + C2 f
x =
FH
w0
1sinh-1
u + C22
du = 1w0FH2 ds,
u = 11FH21w0s + C12
x =
L
ds
[1 + 11FH
2
21w0s + C122
]12
x =
L
ds
c1 + 11FH
2
2a
L
w0 dsb
2
d
12
w1s2 = w0.
y = f1x2.
w0 = 5 Nm.
EXAMPLE 7.13
y
x
s
L 20 m
h 6 m
umax
Fig. 7–23](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-391-320.jpg)
![7.4 CABLES 375
7
Hence,
If the boundary condition at is applied, the constant
and therefore the deflection curve becomes
(4)
This equation defines the shape of a catenary curve. The constant
is obtained by using the boundary condition that at in
which case
(5)
Since and Eqs. 4 and 5 become
(6)
(7)
Equation 7 can be solved for by using a trial-and-error procedure.
The result is
and therefore the deflection curve, Eq. 6, becomes
Ans.
Using Eq. 3, with the half-length of the cable is
Hence,
Ans.
Since the maximum tension occurs when is
maximum, i.e., at Using Eq. 2 yields
And so,
Ans.
Tmax =
FH
cos umax
=
45.9 N
cos 52.8°
= 75.9 N
umax = 52.8°
dy
dx
`
s=12.1 m
= tan umax =
5 Nm112.1 m2
45.9 N
= 1.32
s = l2 = 12.1 m.
u
T = FHcos u,
l = 24.2 m
l
2
=
45.9 N
5 Nm
sinhc
5 Nm
45.9 N
110 m2d = 12.1 m
x = 10 m,
y = 9.19[cosh10.109x2 - 1] m
FH = 45.9 N
FH
6 m =
FH
5 Nm
ccosha
50 N
FH
b - 1d
y =
FH
5 Nm
ccosha
5 Nm
FH
xb - 1d
L = 20 m,
h = 6 m,
w0 = 5 Nm,
h =
FH
w0
ccosha
w0L
2FH
b - 1d
x = L2,
y = h
FH
y =
FH
w0
ccosha
w0
FH
xb - 1d
C3 = -FHw0,
x = 0
y = 0
y =
FH
w0
cosha
w0
FH
xb + C3](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-392-320.jpg)
 cos u = 0
+©M
A = 0;
u
NB = 29.43 N = 29.4 N
29.43 N - NB = 0
:
+ ©F
x = 0;
NB
F
A = 0.3(98.1 N) = 29.43 N
N
A = 98.1 N
N
A - 10(9.81) N = 0
+ c©F
y = 0;
NA
F
A = msN
A = 0.3N
A
FA](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-414-320.jpg)
![416 CHAPTER 8 FRICTION
8
The turnbuckle shown in Fig. 8–17 has a square thread with a mean
radius of 5 mm and a lead of 2 mm. If the coefficient of static friction
between the screw and the turnbuckle is determine the
moment M that must be applied to draw the end screws closer
together.
ms = 0.25,
EXAMPLE 8.7
Fig. 8–17
M
2 kN
2 kN
SOLUTION
The moment can be obtained by applying Eq. 8–3. Since friction at
two screws must be overcome, this requires
(1)
Here
and Substituting
these values into Eq. 1 and solving gives
Ans.
NOTE: When the moment is removed, the turnbuckle will be self-
locking; i.e., it will not unscrew since fs 7 u.
= 6374.7 N # mm = 6.37 N # m
M = 2[12000 N215 mm2 tan114.04° + 3.64°2]
u = tan-1
1l2pr2 = tan-1
12 mm[2p15 mm2]2 = 3.64°.
= 14.04°,
fs = tan-1
ms = tan-1
10.252
r = 5 mm,
W = 2000 N,
M = 2[Wr tan1u + f2]](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-433-320.jpg)
![422 CHAPTER 8 FRICTION
8
Integrating this equation between all the points of contact that the belt
makes with the drum, and noting that at and at
yields
Solving for we obtain
(8–6)
where
T2 = T1emb
T2,
ln
T2
T1
= mb
L
T2
T1
dT
T
= m
L
b
0
du
u = b,
T = T2
u = 0
T = T1
Motion or impending
motion of belt relative
to surface
r
T2
T1
u
Flat orV-belts are often used to transmit
the torque developed by a motor to a
wheel attached to a pump,fan or blower.
[
[
T
i
n
f
o
r
m
a
l
t
a
b
l
e
0
]
] T1 =
T2, belt tensions; opposes the direction of motion (or
impending motion) of the belt measured relative to the
surface, while acts in the direction of the relative belt
motion (or impending motion); because of friction,
T2 7 T1
T2
T1
m = coefficient of static or kinetic friction between the belt
and the surface of contact
b = angle of belt to surface contact, measured in radians
e = base of the natural logarithm
2.718 Á ,
Note that is independent of the radius of the drum, and instead it is
a function of the angle of belt to surface contact, As a result, this
equation is valid for flat belts passing over any curved contacting surface.
b.
T2](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-439-320.jpg)
![8.5 FRICTIONAL FORCES ON FLAT BELTS 423
8
EXAMPLE 8.8
The maximum tension that can be developed in the cord shown in
Fig.8–19a is 500 N. If the pulley at A is free to rotate and the coefficient
of static friction at the fixed drums B and C is determine the
largest mass of the cylinder that can be lifted by the cord.
ms = 0.25,
SOLUTION
Lifting the cylinder, which has a weight causes the cord to
move counterclockwise over the drums at B and C; hence, the
maximum tension in the cord occurs at D. Thus,
A section of the cord passing over the drum at B is shown in
Fig. 8–19b. Since the angle of contact between the drum
and the cord is Using Eq. 8–6, we have
Hence,
Since the pulley at A is free to rotate, equilibrium requires that the
tension in the cord remains the same on both sides of the pulley.
The section of the cord passing over the drum at C is shown in
Fig. 8–19c.The weight Why? Applying Eq. 8–6, we obtain
so that
Ans.
= 15.7 kg
m =
W
g
=
153.9 N
9.81 ms2
W = 153.9 N
277.4 N = We0.25[1342p]
T2 = T1emsb
;
W 6 277.4 N.
T1 =
500 N
e0.25[1342p]
=
500 N
1.80
= 277.4 N
500 N = T1e0.25[1342p]
T2 = T1emsb
;
b = 1135°180°2p = 3p4 rad.
180° = p rad
F = T2 = 500 N.
T2
W = mg,
T
A
(a)
C
B
D
45 45
135
Impending
motion
B
500 N
T1
(b)
W mg
277.4 N
135
Impending
motion
(c)
C
Fig. 8–19](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-440-320.jpg)
![P
3 in.
3 in.
45
440 CHAPTER 8 FRICTION
8
P
250 mm
B
A
30
Prob. 8–137
300 mm
P
30
30
Probs. 8–138/139
P
Prob. 8–141
W
P
r
A
B
Prob. 8–140
1.25 ft
P
1.25 ft
Prob. 8–142
Prob. 8–136
•8–141. The 1.2-Mg steel beam is moved over a level
surface using a series of 30-mm-diameter rollers for which
the coefficient of rolling resistance is 0.4 mm at the ground
and 0.2 mm at the bottom surface of the beam. Determine
the horizontal force P needed to push the beam forward at
a constant speed. Hint: Use the result of Prob. 8–140.
•8–137. The lawn roller has a mass of 80 kg. If the arm BA
is held at an angle of 30° from the horizontal and the
coefficient of rolling resistance for the roller is 25 mm,
determine the force P needed to push the roller at constant
speed. Neglect friction developed at the axle, A, and assume
that the resultant force P acting on the handle is applied
along arm BA.
8–138. Determine the force P required to overcome
rolling resistance and pull the 50-kg roller up the inclined
plane with constant velocity. The coefficient of rolling
resistance is .
8–139. Determine the force P required to overcome
rolling resistance and support the 50-kg roller if it rolls
down the inclined plane with constant velocity. The
coefficient of rolling resistance is .
a = 15 mm
a = 15 mm
*8–136. The wagon together with the load weighs 150 lb.
If the coefficient of rolling resistance is a = 0.03 in.,
determine the force P required to pull the wagon with
constant velocity.
*8–140. The cylinder is subjected to a load that has a
weight W. If the coefficients of rolling resistance for the
cylinder’s top and bottom surfaces are and ,
respectively, show that a horizontal force having a
magnitude of is required to move the
load and thereby roll the cylinder forward. Neglect the
weight of the cylinder.
P = [W(aA + aB)]2r
aB
aA
8–142. Determine the smallest horizontal force P that
must be exerted on the 200-lb block to move it forward.The
rollers each weigh 50 lb, and the coefficient of rolling
resistance at the top and bottom surfaces is .
a = 0.2 in](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-457-320.jpg)
![9.1 CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY 457
9
EXAMPLE 9.5
Locate the centroid of the area shown in Fig. 9–12a.
SOLUTION I
Differential Element. A differential element of thickness dx is
shown in Fig. 9–12a. The element intersects the curve at the arbitrary
point (x, y), and so it has a height y.
Area and Moment Arms. The area of the element is
and its centroid is located at
Integrations. Applying Eqs.9–4 and integrating with respect to x yields
y
'
= y2.
x
'
= x,
dA = y dx,
Ans.
Ans.
L
1m
0
1x2
22x2
dx
L
1m
0
x2
dx
=
0.100
0.333
= 0.3 m
y =
L
A
y
'
dA
L
A
dA
=
L
1m
0
1y22y dx
L
1m
0
y dx
=
x =
L
A
x
'
dA
L
A
dA
=
L
1m
0
xy dx
L
1m
0
y dx
=
L
1m
0
x3
dx
L
1m
0
x2
dx
=
0.250
0.333
= 0.75 m
SOLUTION II
Differential Element. The differential element of thickness dy is
shown in Fig. 9–12b. The element intersects the curve at the arbitrary
point (x, y), and so it has a length
Area and Moment Arms. The area of the element is
and its centroid is located at
Integrations. Applying Eqs. 9–4 and integrating with respect to y,
we obtain
y
'
= y
x
'
= x + a
1 - x
2
b =
1 + x
2
,
dA = 11 - x2 dy,
11 - x2.
Ans.
Ans.
=
0.100
0.333
= 0.3 m
y =
L
A
y
'
dA
L
A
dA
=
L
1m
0
y11 - x2 dy
L
1m
0
11 - x2 dy
=
L
1m
0
1y - y32
2 dy
L
1m
0
11 - 1y2 dy
1
2 L
1m
0
11 - y2 dy
L
1m
0
11 - 1y2 dy
=
0.250
0.333
= 0.75 m
x
'
=
L
A
x
'
dA
L
A
dA
=
L
1m
0
[11 + x22]11 - x2 dy
L
1m
0
11 - x2 dy
=
y x2
1 m
dx
1 m
y
x
y
(a)
(x, y)
~
~
(x, y)
x
Fig. 9–12
1 m
dy
1 m
y
x
y
(b)
(x, y)
~
~
(x, y)
x
y x2
(1 x)
NOTE: Plot these results and notice that they seem reasonable. Also,
for this problem, elements of thickness dx offer a simpler solution.](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-474-320.jpg)
![9.1 CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY 467
9
9–31. Locate the centroid of the area. Hint: Choose
elements of thickness dy and length .
[(2 - y) - y2
]
9–30. The steel plate is 0.3 m thick and has a density of
. Determine the location of its center of mass.
Also determine the horizontal and vertical reactions at the
pin and the reaction at the roller support. Hint: The normal
force at B is perpendicular to the tangent at B, which is
found from tan .
u = dydx
7850 kgm3
*9–28. Locate the centroid of the area.
•9–29. Locate the centroid of the area.
y
x
*9–32. Locate the centroid of the area.
•9–33. Locate the centroid of the area.
y
x
y
x
h
a
y xn
h
––
an
Probs. 9–28/29
y
A
B
x
2 m
2 m
2 m
y2
2x
Prob. 9–30
y
x
1 m
1 m
1 m
y x 2
y2
x
Prob. 9–31
y
x
1 ft
y 2x
2 ft
y2
4x
Probs. 9–32/33](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-484-320.jpg)
![9.3 THEOREMS OF PAPPUS AND GULDINUS 487
9
EXAMPLE 9.13
Determine the surface area and volume of the full solid in Fig. 9–22a.
1 in.
1 in.
2 in.
(a)
2.5 in.
z
Fig. 9–22
(b)
z
1 in.
3.5 in.
3 in.
2.5 in.
1 in.
2 in.
1 in.
2 in.
1 in.
(c)
z
3 in.
2.5 in. ( )(1 in.) 3.1667 in.
2
3
SOLUTION
Surface Area. The surface area is generated by revolving the four
line segments shown in Fig. 9–22b, radians about the z axis. The
distances from the centroid of each segment to the z axis are also
shown in the figure.Applying Eq. 9–7, yields
Ans.
Volume. The volume of the solid is generated by revolving the two
area segments shown in Fig. 9–22c, 2 radians about the z axis. The
distances from the centroid of each segment to the z axis are also
shown in the figure.Applying Eq. 9–10, we have
Ans.
= 47.6 in3
V=2p©rA=2pE(3.1667 in.)c
1
2
(1 in.)(1 in.) d +(3 in.)[(2 in.)(1 in.)F
p
= 143 in2
+ (3.5 in.)(3 in.) + (3 in.)(1 in.)]
A = 2p©rL = 2p[(2.5 in.)(2 in.) + (3 in.)¢ 3(1 in.)2
+ (1 in.)2
≤
2p](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-504-320.jpg)
![9.4 RESULTANT OF A GENERAL DISTRIBUTED LOADING 493
9
*9.4 Resultant of a General Distributed
Loading
In Sec. 4.9, we discussed the method used to simplify a two-dimensional
distributed loading to a single resultant force acting at a specific point. In
this section we will generalize this method to include flat surfaces that
have an arbitrary shape and are subjected to a variable load distribution.
Consider,for example,the flat plate shown in Fig.9–23a,which is subjected
to the loading defined by Pa, where 1
Knowing this function, we can determine the resultant force acting on
the plate and its location Fig. 9–23b.
Magnitude of Resultant Force. The force dF acting on the
differential area of the plate, located at the arbitrary point (x, y),
has a magnitude of
Notice that p(x, y) the colored differential volume element
shown in Fig. 9–23a. The magnitude of is the sum of the differential
forces acting over the plate’s entire surface area A.Thus:
(9–11)
This result indicates that the magnitude of the resultant force is equal to
the total volume under the distributed-loading diagram.
Location of Resultant Force. The location ( ) of is
determined by setting the moments of equal to the moments of all the
differential forces dF about the respective y and x axes: From Figs. 9–23a
and 9–23b, using Eq. 9–11, this results in
FR
FR
y
x,
FR =
L
A
p1x, y2 dA =
L
V
dV = V
FR = ©F;
FR
dA = dV,
dF = [p1x, y2 Nm2
]1dA m2
2 = [p1x, y2 dA] N.
dA m2
(x, y),
FR
Pa 1pascal2 = 1 Nm2
.
p1x, y2
p =
The resultant of a wind loading that is
distributed on the front or side walls of
this building must be calculated using
integration in order to design the
framework that holds the building
together.
x y
y
x
(a)
dF
p
p p(x, y)
dA dV
Fig. 9–23
x y
y
x
(b)
FR
(9–12)
x =
L
A
xp1x,y2dA
L
A
p1x,y2dA
=
L
V
x dV
L
V
dV
y =
L
A
yp1x,y2dA
L
A
p1x,y2dA
=
L
V
ydV
L
V
dV
Hence, the line of action of the resultant force passes through the
geometric center or centroid of the volume under the distributed-loading
diagram.](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-510-320.jpg)
![500 CHAPTER 9 CENTER OF GRAVITY AND CENTROID
9
1 m
1 m
(a)
E
Fig. 9–30
Determine the magnitude and location of the resultant force acting
on the triangular end plates of the water trough shown in Fig. 9–30a;
rw = 1000 kgm3
.
EXAMPLE 9.16
SOLUTION
The pressure distribution acting on the end plate E is shown in Fig.9–30b.
The magnitude of the resultant force is equal to the volume of this
loading distribution.We will solve the problem by integration. Choosing
the differential volume element shown in the figure, we have
The equation of line AB is
Hence, substituting and integrating with respect to z from to
yields
Ans.
This resultant passes through the centroid of the volume. Because of
symmetry,
Ans.
Since for the volume element, then
z
'
= z
x = 0
= 9810
L
1 m
0
1z - z2
2 dz = 1635 N = 1.64 kN
F = V =
L
V
dV =
L
1 m
0
119 6202z[0.511 - z2] dz
z = 1 m
z = 0
x = 0.511 - z2
dF = dV = p dA = rwgz12x dz2 = 19 620zx dz
0.5 m
y
x
z
1 m
z
dz
A
B
(b)
O
2x
dF
Ans.
NOTE: We can also determine the resultant force by applying Eq. 9–14,
FR = gzA = (9810 Nm3
)(1
3)(1 m)[1
2(1 m)(1 m)] = 1.64 kN.
= 0.5 m
9810
L
1 m
0
1z2
- z3
2 dz
1635
z =
L
V
z
'
dV
L
V
dV
=
L
1 m
0
z119 6202z[0.511 - z2] dz
1635
=](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-517-320.jpg)
![REVIEW PROBLEMS 509
9
9–127. Locate the centroid of the shaded area.
y
9–126. Determine the location ( , ) of the centroid for
the structural shape. Neglect the thickness of the member.
y
x
•9–125. Locate the centroid ( , ) of the area.
y
x *9–128. The load over the plate varies linearly along the
sides of the plate such that . Determine
the resultant force and its position ( , ) on the plate.
y
x
p = 2
3 [x(4 - y)] kPa
y
x
3 in.
1 in.
3 in.
6 in.
Prob. 9–125
1.5 in. 1.5 in. 1.5 in. 1.5 in.
1 in.
1 in.
3 in.
x
y
Prob. 9–126
x
y
a
—
2
a
—
2
a
a
a
Prob. 9–127
p
3 m
4 m
y
x
8 kPa
Prob. 9–128
p
x
y
6 m
5 m
100 Pa
300 Pa
Prob. 9–129
•9–129. The pressure loading on the plate is described by
the function . Determine
the magnitude of the resultant force and coordinates of the
point where the line of action of the force intersects
the plate.
p = 5-240(x + 1) + 3406 Pa](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-526-320.jpg)
![524 CHAPTER 10 MOMENTS OF INERTIA
10
Determine the moments of inertia for the cross-sectional area of the
member shown in Fig. 10–9a about the x and y centroidal axes.
SOLUTION
Composite Parts. The cross section can be subdivided into the three
rectangular areas A, B, and D shown in Fig. 10–9b. For the calculation,
the centroid of each of these rectangles is located in the figure.
Parallel-Axis Theorem. From the table on the inside back cover, or
Example 10.1, the moment of inertia of a rectangle about its
centroidal axis is Hence, using the parallel-axis theorem
for rectangles A and D, the calculations are as follows:
Rectangles A and D
Rectangle B
Summation. The moments of inertia for the entire cross section
are thus
Ans.
Ans.
= 5.601109
2 mm4
Iy = 2[1.901109
2] + 1.801109
2
= 2.901109
2 mm4
Ix = 2[1.4251109
2] + 0.051109
2
Iy =
1
12
11002160023
= 1.801109
2 mm4
Ix =
1
12
16002110023
= 0.051109
2 mm4
= 1.901109
2 mm4
Iy = Iy¿ + Adx
2
=
1
12
13002110023
+ 1100213002125022
= 1.4251109
2 mm4
Ix = Ix¿ + Ady
2
=
1
12
11002130023
+ 1100213002120022
I = 1
12 bh3
.
EXAMPLE 10.5
100 mm
400 mm
100 mm
100 mm
600 mm
400 mm
x
y
(a)
C
100 mm
100 mm
x
y
300 mm
300 mm
200 mm
250 mm
200 mm
(b)
A
B
D
250 mm
Fig. 10–9](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-541-320.jpg)
![536 CHAPTER 10 MOMENTS OF INERTIA
10
Determine the principal moments of inertia and the orientation of the
principal axes for the cross-sectional area of the member shown in
Fig. 10–18a with respect to an axis passing through the centroid.
SOLUTION
The moments and product of inertia of the cross section with respect
to the x, y axes have been determined in Examples 10.5 and 10.7. The
results are
Using Eq. 10–10, the angles of inclination of the principal axes u and
are
Thus, by inspection of Fig. 10–18b,
Ans.
The principal moments of inertia with respect to these axes are
determined from Eq. 10–11. Hence,
or
Ans.
NOTE: The maximum moment of inertia,
occurs with respect to the u axis since by inspection most of the cross-
sectional area is farthest away from this axis. Or, stated in another
manner, occurs about the u axis since this axis is located within
of the y axis, which has the larger value of Also, this
can be concluded by substituting the data with into the first
of Eqs. 10–9 and solving for .
Iu
u = 57.1°
I 1Iy 7 Ix2.
;45°
Imax
Imax = 7.541109
2 mm4
,
Imax = 7.541109
2 mm4
Imin = 0.9601109
2 mm4
Imax
min
= 4.251109
2 ; 3.291109
2
;
C
c
2.901109
2 - 5.601109
2
2
d
2
+ [-3.001109
2]2
=
2.901109
2 + 5.601109
2
2
Imax
min
=
Ix + Iy
2
;
C
a
Ix - Iy
2
b
2
+ Ixy
2
up2
= -32.9° and up1
= 57.1°
2up = -65.8° and 114.2°
tan 2up =
-Ixy
1Ix - Iy22
=
-[-3.001109
2]
[2.901109
2 - 5.601109
2]2
= -2.22
v
Ix = 2.901109
2 mm4
Iy = 5.601109
2 mm4
Ixy = -3.001109
2 mm4
EXAMPLE 10.8
100 mm
400 mm
100 mm
100 mm
600 mm
400 mm
x
y
(a)
C
x
y
(b)
C
u
v
up1
57.1
up2
32.9
Fig. 10–18](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-553-320.jpg)
![548 CHAPTER 10 MOMENTS OF INERTIA
10
A solid is formed by revolving the shaded area shown in Fig. 10–24a
about the y axis. If the density of the material is determine
the mass moment of inertia about the y axis.
5 slugft3
,
EXAMPLE 10.11
SOLUTION
Disk Element. The moment of inertia will be determined using this
disk element, as shown in Fig. 10–24b. Here the element intersects the
curve at the arbitrary point (x, y) and has a mass
Although all points on the element are not located at the same
distance from the y axis, it is still possible to determine the moment of
inertia of the element about the y axis. In the previous example it
was shown that the moment of inertia of a homogeneous cylinder
about its longitudinal axis is where m and R are the mass
and radius of the cylinder. Since the height of the cylinder is not
involved in this formula, we can also use this result for a disk.Thus, for
the disk element in Fig. 10–24b, we have
Substituting and integrating with respect to y,
from to yields the moment of inertia for the entire solid.
Ans.
Iy =
5p
2 L
1ft
0
x4
dy =
5p
2 L
1ft
0
y8
dy = 0.873 slug # ft2
y = 1 ft,
y = 0
r = 5 slugft3
,
x = y2
,
dIy =
1
2
1dm2x2
=
1
2
[r1px2
2 dy]x2
I = 1
2 mR2
,
dIy
dm = r dV = r1px2
2 dy
x
1 ft
y2
x
(a)
y
1 ft
y
1 ft
x
1 ft
y
dy
(x, y)
(b)
Fig. 10–24](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-565-320.jpg)
![10.8 MASS MOMENT OF INERTIA 549
10
Parallel-Axis Theorem. If the moment of inertia of the body
about an axis passing through the body’s mass center is known, then the
moment of inertia about any other parallel axis can be determined by
using the parallel-axis theorem.To derive this theorem, consider the body
shown in Fig. 10–25. The axis passes through the mass center G,
whereas the corresponding parallel z axis lies at a constant distance d
away. Selecting the differential element of mass dm, which is located at
point ( ), and using the Pythagorean theorem,
the moment of inertia of the body about the z axis is
Since the first integral represents The second
integral is equal to zero, since the axis passes through the body’s mass
center, i.e., since Finally, the third integral
is the total mass m of the body. Hence, the moment of inertia about the z
axis becomes
(10–15)
where
I = IG + md2
x = 0.
1x¿ dm = x1dm = 0
z¿
IG.
r¿2
= x¿2
+ y¿2
,
=
L
m
1x¿2
+ y¿2
2 dm + 2d
L
m
x¿ dm + d2
L
m
dm
I =
L
m
r2
dm =
L
m
[1d + x¿22
+ y¿2
] dm
r2
= 1d + x¿22
+ y¿2
,
y¿
x¿,
z¿
IG = moment of inertia about the axis passing through the mass
center G
z¿
m = mass of the body
d = distance between the parallel axes
y¿
x¿
z z¿
y¿
r¿
x¿
d
r
dm
A G
Fig. 10–25](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-566-320.jpg)
![10.8 MASS MOMENT OF INERTIA 551
10
O
0.25 m
0.125 m
G
(a)
Thickness 0.01 m
0.25 m
G G
– 0.125 m
(b)
Fig. 10–26
EXAMPLE 10.12
If the plate shown in Fig. 10–26a has a density of and a
thickness of 10 mm, determine its mass moment of inertia about an
axis perpendicular to the page and passing through the pin at O.
8000 kgm3
SOLUTION
The plate consists of two composite parts, the 250-mm-radius disk
minus a 125-mm-radius disk, Fig. 10–26b.The moment of inertia about
O can be determined by finding the moment of inertia of each of
these parts about O and then algebraically adding the results. The
computations are performed by using the parallel-axis theorem in
conjunction with the data listed in the table on the inside back cover.
Disk. The moment of inertia of a disk about an axis perpendicular
to the plane of the disk and passing through G is The mass
center of both disks is 0.25 m from point O.Thus,
Hole. For the smaller disk (hole), we have
The moment of inertia of the plate about the pin is therefore
Ans.
= 1.20 kg # m2
= 1.473 kg # m2
- 0.276 kg # m2
IO = 1IO2d - 1IO2h
= 0.276 kg # m2
= 1
213.93 kg210.125 m22
+ 13.93 kg210.25 m22
1IO2h = 1
2 mhrh
2
+ mhd2
mh = rhV
h = 8000 kgm3
[p10.125 m22
10.01 m2] = 3.93 kg
= 1.473 kg # m2
= 1
2115.71 kg210.25 m22
+ 115.71 kg210.25 m22
1IO2d = 1
2 mdrd
2
+ mdd2
md = rdV
d = 8000 kgm3
[p10.25 m22
10.01 m2] = 15.71 kg
IG = 1
2 mr2
.](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-568-320.jpg)
![570 CHAPTER 11 VIRTUAL WORK
11
Determine the required force P in Fig. 11–7a, needed to maintain
equilibrium of the scissors linkage when . The spring is
unstretched when . Neglect the mass of the links.
SOLUTION
Free-Body Diagram. Only and P do work when undergoes a
positive virtual displacement , Fig. 11–7b. For the arbitrary position
, the spring is stretched , so that
Virtual Displacements. The position coordinates, and ,
measured from the fixed point A, are used to locate and . These
coordinates are parallel to the line of action of their corresponding
forces. Expressing and in terms of the angle using
trigonometry,
Differentiating, we obtain the virtual displacements of points B and D.
(1)
(2)
Virtual-Work Equation. Force P does positive work since it acts in
the positive sense of its virtual displacement.The spring force does
negative work since it acts opposite to its positive virtual
displacement.Thus, the virtual-work equation becomes
Since , then this equation requires
When ,
Ans.
P = 500 sin 60° - 250 = 183 N
u = 60°
P = 500 sin u - 250
cos u du Z 0
[0.9P + 225 - 450 sin u] cos u du = 0
- [1500 sin u - 750] (0.3 cos u du) + P (0.9 cos u du) = 0
-FsdxB+PdxD = 0
dU = 0;
Fs
dxD = 0.9 cos u du
dxB = 0.3 cos u du
xD = 3[(0.3 m) sin u] = (0.9 m) sin u
xB = (0.3 m) sin u
u
xD
xB
P
Fs
xD
xB
= (1500 sin u - 750) N
Fs = ks = 5000 N/m [(0.3 m) sin u - (0.3 m) sin 30°]
(0.3 m) sin u - (0.3 m) sin 30°
u
du
u
Fs
u = 30°
u = 60°
EXAMPLE 11.2
A
B
k 5 kN/m
(a)
C
E
D
G
0.3m
0.3m
0.3m
0.3m
u
u
P
B
(b)
Gx
Ax
Ay
Fs
xD
xB
dxD
dxB
P
u
du
Fig. 11–7](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-587-320.jpg)
![11.3 PRINCIPLE OF VIRTUAL WORK FOR A SYSTEM OF CONNECTED RIGID BODIES 571
11
EXAMPLE 11.3
If the box in Fig. 11–8a has a mass of 10 kg, determine the couple
moment M needed to maintain equilibrium when . Neglect
the mass of the members.
u = 60°
D
C
A
B
M
0.45 m
0.45 m
(a)
0.2 m
0.4 m
u
u 0.45 m
C
b
A
M
(b)
yE
yE
Bx Dx
By Dy
10(9.81) N
u
d
u
d
u
u
d
Fig. 11–8
SOLUTION
Free-Body Diagram. When undergoes a positive virtual
displacement , only the couple moment M and the weight of the box
do work, Fig. 11–8b.
Virtual Displacements. The position coordinate , measured from
the fixed point B, locates the weight, . Here,
where b is a constant distance. Differentiating this equation, we obtain
(1)
Virtual-Work Equation. The virtual-work equation becomes
Substituting Eq. 1 into this equation
Since , then
Since it is required that , then
Ans.
M = 44.145 cos 60° = 22.1 N # m
u = 60°
M - 44.145 cos u = 0
du Z 0
du(M - 44.145 cos u) = 0
Mdu - 10(9.81) N(0.45 m cos u du) = 0
Mdu - [10(9.81) N]dyE = 0
dU = 0;
dyE = 0.45 m cos u du
yE = (0.45 m) sin u + b
10(9.81) N
yE
du
u](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-588-320.jpg)
![572 CHAPTER 11 VIRTUAL WORK
11
The mechanism in Fig. 11–9a supports the 50-lb cylinder. Determine
the angle for equilibrium if the spring has an unstretched length of
2 ft when . Neglect the mass of the members.
SOLUTION
Free-Body Diagram. When the mechanism undergoes a positive
virtual displacement , Fig. 11–9b, only and the 50-lb force do work.
Since the final length of the spring is , then
Virtual Displacements. The position coordinates and are
established from the fixed point A to locate at and at .
The coordinate also measured from A, specifies the position of the
50-lb force at B. The coordinates can be expressed in terms of using
trigonometry.
Differentiating, we obtain the virtual displacements of points , ,
and as
(1)
(2)
(3)
Virtual-Work Equation. The virtual-work equation is written as if
all virtual displacements are positive, thus
Since , then
Solving by trial and error,
Ans.
u = 34.9°
800 sin u cos u - 800 sin u + 100 cos u = 0
du Z 0
du(800 sin u cos u - 800 sin u + 100 cos u) = 0
-(400 - 400 cos u)(-1 sin u du) = 0
(400 - 400 cos u)(-3 sin u du) + 50(2 cos u du)
FsdxE + 50dyB - FsdxD
= 0
dU = 0;
dyB = 2 cos u du
dxE = -3 sin u du
dxD = -1 sin u du
B
E
D
yB = (2 ft) sin u
xE = 3[(1 ft) cos u] = (3 ft) cos u
xD = (1 ft) cos u
u
yB,
E
D
Fs
xE
xD
Fs = ks = (200 lb/ft)(2 ft - 2 ft cos u) = (400 - 400 cos u) lb
2(1 ft cos u)
Fs
du
u = 0°
u
EXAMPLE 11.4
C
D E
k 200 lb/ft
B
A
1 ft
1 ft 1 ft
(a)
1 ft
u u
E
D
50 lb
(b)
Ax
Ay Cy
Fs Fs
xD
yB
xE
dxE
dxD
dyB
u
du
Fig. 11–9](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-589-320.jpg)
![11.7 STABILITY OF EQUILIBRIUM CONFIGURATION 587
11
EXAMPLE 11.6
If the spring AD in Fig. 11–18a has a stiffness of 18 kN/m and is
unstretched when , determine the angle for equilibrium.The
load has a mass of 1.5 Mg. Investigate the stability at the equilibrium
position.
SOLUTION
Potential Energy. The gravitational potential energy for the load
with respect to the fixed datum, shown in Fig. 11–18b, is
u
u = 60°
Vg = mgy = 1500(9.81) N[(4 m) sin u + h] = 58 860 sin u + 14 715h
where h is a constant distance. From the geometry of the system, the
elongation of the spring when the load is on the platform is
.
Thus, the elastic potential energy of the system is
The potential energy function for the system is therefore
(1)
Equilibrium. When the system is in equilibrium,
Since
Solving by trial and error,
Ans.
Stability. Taking the second derivative of Eq. 1,
Substituting yields
Ans.
And for ,
Ans.
d2
V
du2
= 64 073 7 0 Stable
u = 45.51°
d2
V
du2
= -60 409 6 0 Unstable
u = 28.18°
d2
V
du2
= -58 860 sin u - 288 000 cos 2u + 144 000 cos u
u = 45.51°
u = 28.18° and
58 860 cos u - 144 000 sin 2u + 144 000 sin u = 0
sin 2u = 2 sin u cos u,
58 860 cos u - 288 000 sin u cos u + 144 000 sin u = 0
dV
du
= 58 860 cos u + 18 000(4 cos u - 2)(-4 sin u) = 0
V = Vg + V
e = 58 860 sin u + 14 715h + 9000(4 cos u - 2)2
Ve = 1
2 ks2
= 1
2(18 000 N/m)(4 m cos u - 2 m)2
= 9000(4 cos u - 2)2
s = (4 m) cos u - (4 m) cos 60° = (4 m) cos u - 2 m
(a)
2 m
2 m
A
C
E
B
D
G
k 18 kN/m
u
u
2 m
2 m
A
C
E
B
D
G
k 18 kN/m
(b)
4 m cosu
(4 m)sinu
h
y
Datum
u
u
Fig 11–18](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-604-320.jpg)
![604 PARTIAL SOLUTIONS AND ANSWERS
F2–14.
Require .
Ans.
F2–15.
Ans.
F2–16.
Ans.
F2–17.
Ans.
F2–18.
Ans.
F2–19. Ans.
Ans.
Ans.
F2–20. Ans.
Ans.
Ans.
F2–21.
Ans.
= 5180i + 270j - 540k6 N
= (630 N)A2
7 i + 3
7 j - 6
7 kB
FB = FBuB
rB = 52i + 3j - 6k6 m
u = 180° - 131.8° = 48.2°
a = cos-1
A -4 ft
6 ft B = 131.8°
rAB = 2(-4 ft)2
+ (2 ft)2
+ (4 ft)2
= 6 ft
rAB = 5-4i + 2j + 4k6 ft
a = 132°, b = 48.2°, g = 70.5°
rAB = 2(-6 m)2
+ (6 m)2
+ (3 m)2
= 9 m
rAB = 5-6i + 6j + 3k6 m
FR = F1 + F2 = 5490i + 683j - 266k6 lb
= 5489.90i + 282.84j - 565.69k6 lb
+ (800 lb) sin 45° (-k)
+ [(800 lb) cos 45°] sin 30°j
F2 = [(800 lb) cos 45°] cos 30° i
= 5400j + 300k6 lb
F1 = A4
5 B(500 lb) j + A3
5 B (500 lb)k
F2 = 5265i - 459j + 530k6 N
Fy = (530.33 N) sin 60° = 459.3 N
Fx = (530.33 N) cos 60° = 265.1 N
F¿ = (750 N) cos 45° = 530.33 N
Fz = (750 N) sin 45° = 530.33 N
F = 5-21.2i + 28.3j + 35.4k6 lb
Fy = A4
5 B(35.36 lb) = 28.28 lb
Fx = A3
5 B (35.36 lb) = 21.21 lb
F¿ = (50 lb) cos 45° = 35.36 lb
Fz = (50 lb) sin 45° = 35.36 lb
= 5250i - 354j - 250k6 N
F = FuF = (500 N)(0.5i - 0.7071j - 0.5k)
a = 60°
cos2
a + cos2
135° + cos2
120° = 1
= 5-250i - 354j + 250k6 N
F = FuF = (500 N)(-0.5i - 0.7071j + 0.5k)
b = 135°
cos b = 21 - cos2
120° - cos2
60° = ;0.7071 F2–22.
F2–23.
Ans.
F2–24.
Ans.
F2–25.
Ans.
F2–26.
Ans.
F2–27.
Ans.
F2–28.
FOA = FOA uOA = 5231i + 96.2j6 N
FOA = F # uOA = 250 N
F = FuF = [650j] N
uOA = 12
13 i + 5
13 j
cos u = 5
13; u = 67.4°
uOA
# j = uOA (1) cos u
uOA = 12
13 i + 5
13 j
u = cos-1
(uAB
# uF) = 68.9°
uF = 4
5 i - 3
5 j
uAB = -3
5 j + 4
5 k
u = cos-1
(uAO
# uF) = 57.7°
uF = -0.5345i + 0.8018j + 0.2673k
uAO = -1
3 i + 2
3 j - 2
3 k
FR = FB + FC = 5-620i + 610j -540k6 lb
= 5-420i + 210j - 140k6 lb
= (490 lb)A -6
7 i + 3
7 j - 2
7 kB
FC = FCuC
= 5-200i + 400j - 400k6 lb
= (600 lb)A -1
3 i + 2
3 j - 2
3 kB
FB = FBuB
= 1.18 kN
FR = 2(480 N)2
+ (-60 N)2
+ (-1080 N)2
= 5120i + 180j - 360k6 N
= (420 N) A2
7 i + 3
7 j - 6
7 kB
FC = FCuC
= 5360i - 240j - 720k6 N
= (840 N)A3
7 i - 2
7 j - 6
7 kB
FB = FBuB
= 5-400i + 700j - 400k6 N
F = FuAB = 900NA -4
9 i + 7
9 j - 4
9 kB](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-621-320.jpg)
![FUNDAMENTAL PROBLEMS 605
F2–29.
Ans.
F2–30.
Ans.
Ans.
Chapter 3
F3–1.
Ans.
Ans.
F3–2.
Ans.
F3–3.
Ans.
F3–4.
Ans.
F3–5.
Ans.
mA = 20 kg
+ c ©Fy = 0; (392.4 N)sin 30° - mA(9.81) = 0
l0 = 0.283 m
Fsp = k(l - l0); 43.35 = 200(0.5 - l0)
Fsp = 43.35 N
+Q©Fx = 0; 4
5 (Fsp) - 5(9.81) sin 45° = 0
T = 40.9 N
u = tan-1
A0.15 m
0.2 m B = 36.87°
+ c©Fy = 0; 2T sin u - 49.05 N = 0
f = u
:
+
©Fx = 0; T cos u - T cos f = 0
LABC = 2A 5 ft
cos 13.5° B = 10.3 ft
u = 13.5°
+ c©Fy = 0; -2(1500) sin u + 700 = 0
FAC = 518 lb
FAB = 478 lb
+ c©Fy = 0; 3
5 FAC + FAB sin 30° - 550 = 0
+
: ©F
x = 0; 4
5F
AC - F
AB cos 30° = 0
= 401 lb
(FA)per = 2(600 lb)2
- (446.41 lb)2
(FA)proj = F # uA = 446.41 lb = 446 lb
uA = -2
3 i + 2
3 j + 1
3 k
= 5-150i + 259.81j + 519.62k6 lb
+ [(600 lb) sin 60°] k
+ [(600 lb) cos 60°] cos 30° j
F = [(-600 lb) cos 60°] sin 30°i
(FAO)proj = F # uAO = 244 N
= - 0.5547j - 0.8321k
uAO =
5-4 j- 6 k6m
2(-4 m)2
+ (-6 m)2
= 5219.78i + 54.94j - 329.67k6 N
F = (400 N)
54 i+1 j-6 k6m
2(4 m)2
+ (1 m)2
+ (-6 m)2
F3–6.
Ans.
Ans.
Ans.
Ans.
F3–7. (1)
(2)
(3)
Ans.
Ans.
Ans.
F3–8.
Ans.
Ans.
Ans.
F3–9.
Ans.
Ans.
Ans.
F3–10.
Ans.
Ans.
Ans.
FAB = 138.60 lb = 139 lb
©F
x = 0; F
AB - 0.25(202.92) - 0.5(175.74) = 0
FAC = 202.92 lb = 203 lb
FAD = 175.74 lb = 176 lb
©Fz = 0; 0.8660F
AC + 0.7071F
AD - 300 = 0
©F
y = 0; 0.4330F
AC - 0.5F
AD = 0
= -0.5FAD i - 0.5FAD j + 0.7071FAD k
FAD = FAD5cos 120° i + cos 120° j + cos 45°k6
= -0.25F
AC i + 0.4330F
AC j + 0.8660F
AC k
+ cos 60° cos 30° j + sin 60° k6
FAC = FAC 5-cos 60° sin 30° i
FAC = 646.41 N = 646 N
1
3 (900) + 692.82 sin 30° - FAC = 0
©Fx = 0;
FAB = 692.82 N = 693 N
FAB cos 30° - 2
3 (900) = 0
©Fy = 0;
FAD = 900 N
2
3FAD - 600 = 0
©Fz = 0;
FAD = FADa
rAD
rAD
b = 1
3 FADi - 2
3 FAD j + 2
3 F
AD k
FAB = 506.25 N = 506 N
©Fx = 0; FAB - 843.75A3
5 B = 0
FAC = 843.75 N = 844 N
©Fy = 0; FACA4
5 B - 1125A3
5 B = 0
FAD = 1125 N = 1.125 kN
©Fz = 0; FADA4
5 B - 900 = 0
F2 = 879 N
F1 = 466 N
F3 = 776 N
©Fz = 0; A4
5 BF3 + A3
5 BF1 - 900 N = 0
©Fy = 0; A4
5 BF1 - C A3
5 BF3D A4
5 B = 0
©Fx = 0; CA3
5 BF3D A3
5B + 600 N - F2 = 0
u = 21.9°
TCD = 395 N
+ c©Fy = 0; TCD sin u - 15(9.81) N = 0
+
:©Fx = 0; TCD cos u - 366.11 N = 0
TBC = 366.11 N = 366 N
+
:©Fx = 0; TBC - 379.03 N cos 15° = 0
TAB = 379.03 N = 379 N
+ c©Fy = 0; TAB sin 15° - 10(9.81) N = 0](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-622-320.jpg)
- [(600 N) cos 60°](0.25 m)
- C A4
5 B500 ND(0.25 m)
(MR)O = C A3
5 B500 ND(0.425 m)
+(MR)O = ©Fd;
F3–11.
(1)
(2)
(3)
Ans.
Ans.
Ans.
Chapter 4
F4–1. a
Ans.
F4–2. a
b Ans.
F4–3. a
Ans.
F4–4. a
Ans.
F4–5. c
Ans.
F4–6. a
Ans.
F4–7. a
Ans.
= 1254 N # m = 1.25 kN # m
- (300N)[(2.5 m) sin 45°]
+ (500 N)[3 m + (2.5 m) cos 45°]
(MR)O = -(600 N)(1 m)
+(MR)O = ©Fd;
= 1.06 kN # m
- 500 cos 45° (3 sin 45°)
+MO = 500 sin 45° (3 + 3 cos 45°)
= 11.2 N # m
- 50 cos 60°(0.2 sin 45°)
+MO = 50 sin 60° (0.1 + 0.2 cos 45° + 0.1)
= 3.07 kip # ft
+MO = (600 lb)(4 ft + (3 ft)cos 45° - 1 ft)
= 36.7 N # m
- [(300 N) cos 30°][(0.3 m) sin 45°]
+MO = [(300 N) sin 30°][0.4m + (0.3 m) cos 45°]
= -460 N # m = 460 N # m
+MO = - A4
5 B(100 N)(2 m)- A3
5 B(100 N)(5 m)
= 2.49 kip # ft
+MO = 600 sin 50° (5) + 600 cos 50° (0.5)
FD = 346.15 lb = 346 lb
FC = 1.5(162 lb) = 242 lb
FB = 162 lb
©Fz = 0; 2
7 FB + 3
7 FC - 150 = 0
©Fy = 0; 3
7 FB - 2
7 FC = 0
©Fx = 0; -6
7 FB - 6
7 FC + FD = 0
W = 5-150k6 lb
FD = FDi
= -6
7 FCi - 2
7FCj + 3
7 FC k
= FCB
5-6i - 2j + 3k6 ft
2(-6 ft)2
+ (-2 ft)2
+ (3ft)2
R
FC = FCa
rAC
rAC
b
= -6
7FBi + 3
7 FBj + 2
7 FBk
= FBB
5-6i + 3j + 2k6 ft
2(-6 ft)2
+ (3 ft)2
+ (2 ft)2
R
FB = FBa
rAB
rAB
b](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-623-320.jpg)
![FUNDAMENTAL PROBLEMS 607
F4–13.
Ans.
F4–14.
Ans.
F4–15.
F4–16.
Ans.
F4–17.
Ans.
F4–18.
Ans.
Ans.
Ans.
= -160 N#m
Mz = 240 N(2 m) - 320 N(2 m)
= -120 N#m
My = 300 N(2 m) - 240 N(3 m)
= -360 N # m
Mx = 300 N(2 m) - 320 N(3 m)
Fz = (500 N)A3
5 B = 300 N
Fy = C A4
5 B500 ND A4
5 B = 320 N
Fx = C A4
5 B500 ND A3
5 B = 240 N
MAB = MABuAB = 53.2i - 2.4j6 lb#ft
= -4 lb#ft
= 4
i j k
-0.8 0.6 0
0 0 2
50 -40 20
4
MAB = uAB
#(rAC * F)
uAB =
rAB
rAB
=
5-4i + 3j6ft
2(-4 ft)2
+ (3 ft)2
= -0.8i + 0.6j
= 210 N # m
Mp = j#(rA * F) = 3
0 1 0
-3 -4 2
30 -20 50
3
= 17.4 N # m
MO = i#(rA * F) = 3
1 0 0
0 0.3 0.25
-100 100 141.42
3
= 5-100i + 100j + 141.42k6 N
+ (200 N) cos 60° j + (200 N) cos 45° k
F = (200 N) cos 120° i
= -72 N # m
MOA = uOA
#(rAB * F) = 3
0.6 0.8 0
0 0 -0.2
300 -200 150
3
uOA =
rA
rA
=
50.3i + 0.4j6 m
2(0.3 m)2
+ (0.4 m)2
= 20 N # m
Mx = i#(rOB*F) = 3
1 0 0
0.3 0.4 -0.2
300 -200 150
3
F4–19. c
Ans.
Also,
c
Ans.
F4–20. a
Ans.
F4–21. a
Ans.
F4–22. a
b
F4–23.
Ans.
F4–24.
Ans.
Also,
Ans.
= {108j + 144k} N#m
= 3
i j k
0 0 0.3
0 -360 270
3 + 3
i j k
0.4 0 0.3
0 360 -270
3
Mc = (rA * FA) + (rB * FB)
= {108j + 144k} N#m
Mc = rAB * FB = 3
i j k
0.4 0 0
0 360 -270
3
= 5360j - 270k6 N
FB = A4
5 B(450 N)j - A3
5 B(450 N) k
(Mc)R = {-20i - 40j + 100k} lb#ft
(Mc)R = ©Mc;
= 5180i - 240j6 lb#ft
(Mc)3 = (Mc)3 u3 = (300 lb#ft)A1.5
2.5 i - 2
2.5 jB
= 5-250k6 lb#ft
(Mc)2 = (Mc)2u2 = (250 lb#ft)(-k)
= 5-200i + 200j + 350k6 lb#ft
= (450 lb#ft)A- 2
4.5 i + 2
45 j + 3.5
4.5 kB
(Mc)1 = (Mc)1u1
u3 = 1.5
2.5 i - 2
2.5 j
u2 = -k
= - 2
4.5 i + 2
4.5 j + 3.5
4.5 k
u1 =
r1
r1
=
[-2i + 2j + 3.5k] ft
2(-2 ft)2
+ (2 ft)2
+ (3.5 ft)2
= 20 kN # m
+MC = 10A3
5 B(2) - 10A4
5 B(4) = -20 kN # m
F = 2.33 kN
-1.5 kN#m = (2 kN)(0.3 m) -F(0.9 m)
+(MB)R = ©MB
= 2600 lb # ft
+MCR
= 300(4) + 200(4) + 150(4)
= 740 N # m
+MCR
= 300(5) - 400(2) + 200(0.2)
+ 200(0.2) = 740 N # m
+MCR
= ©MA = 400(3) - 400(5) + 300(5)](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-624-320.jpg)
 + 500(9)
MAR
= ©MA;
+
= 1550 lb
FR = 1
2 (50)(6) + 150(6) + 500
+ TFR = ©Fy;
d = 1 m
-27(d) = 1
2 (6)(3)(1) - 1
2 (6)(6)(2)
+(MR)A = ©MA;
FR = 27 kN T
-FR = - 1
2 (6)(3) - 1
2 (6)(6)
+ cFR = ©Fy;
d = 8.36 ft
1650d = C1
2 (6)(150)D(4) + [8(150)](10)
MAR
= ©MA;
+
FR = 1
2 (6)(150) + 8(150) = 1650 lb
d = 1.25 m
- 9(3)(1.5) - 3(1.5)(3.75)
-40.5(d) = 6(1.5)(0.75)
+(MR)A = ©MA;
FR = 40.5 kN T
-FR = -6(1.5) - 9(3) - 3(1.5)
+ cFR = ©Fy;](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-626-320.jpg)
 = 0
-NA sin 30°(0.15 m) - 433.0 N(0.2 m)
+©MB = 0;
NC = 433.0 N = 433 N
NC sin 30° - (250 N) sin 60° = 0
+
:©Fx = 0;
FA = 93.5 N
+ (151.71 N) sin 60° - 25(9.81) N = 0
FA + (78.53 N) sin 15°
+ c©Fy = 0;
TAB = 78.53 N = 78.5 N
TAB cos 15° - (151.71 N) cos 60° = 0
+
:©Fx = 0;
NC = 151.71 N = 152 N
NC(0.7 m) - [25(9.81) N] (0.5 m) cos 30° = 0
+©MA = 0;
Chapter 5
F5–1.
Ans.
c
Ans.
Ans.
F5–2. a
Ans.
Ans.
Ans.
F5–3. a
Ans.
Ans.
Ans.
F5–4.
Ans.
Ans.
a
Ans.
MA = 3.90 kN # m
- 400 sin 30°(4.5) - 400 cos 30°(3 sin 60°) = 0
MA - 200(2.5) - 200(3.5) - 200(4.5)
+©MA = 0;
Ay = 800 N
Ay - 200 - 200 - 200 - 400 sin 30° = 0
+ c©Fy = 0;
Ax = 346 N
-Ax + 400 cos 30° = 0
:
+
©Fx = 0;
Ay = 5.49 kN
Ay + 8.047 kN - (5 kN) sin 45° - 10 kN = 0
+ c©Fy = 0;
Ax = 3.54 kN
(5 kN) cos 45° - Ax = 0
+
:©Fx = 0;
NB = 8.047 kN = 8.05 kN
- 5 kN(4 m) = 0
- 10 kN[2 m + (6 m) cos 45°]
NB[6 m + (6 m) cos 45°]
+©MA = 0;
Ay = -4 kN = 4 kN T
Ay + (11.31 kN) sin 45° - 4 kN = 0
+ c©Fy = 0;
Ax = -8 kN = 8 kN ;
+
:©Fx = 0; Ax + (11.31 kN) cos 45° = 0
FCD = 11.31 kN = 11.3 kN
FCD sin 45°(1.5 m) - 4 kN(3 m) = 0
+©MA = 0;
Ay = 140 lb
+ c©Fy = 0; Ay + 260 - 500A4
5 B = 0
By = 260 lb
©MA = 0; By(10) - 500A4
5 B(5) - 600 = 0
+
Ax = 300 lb
:
+
©Fx = 0; -Ax + 500A3
5 B = 0](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-627-320.jpg)
![616 PARTIAL SOLUTIONS AND ANSWERS
F9–9.
Ans.
F9–10.
Ans.
Ans.
F9–11.
Ans.
Ans.
Ans.
F9–12.
Ans.
Ans.
Ans.
F9–13.
Ans.
Ans.
F9–14.
Ans.
Ans.
= 22.6 m3
= 2pC1.8A1
2 B(0.9)(1.2) + 1.95(0.9)(1.5)D
V = 2p©rA
= 77.5 m2
= 2pC1.952(0.9)2
+ (1.2)2
+ 2.4(1.5) + 1.95(0.9) + 1.5(2.7)D
A = 2p©rL
= 18.8 m3
= 2pC0.75(1.5)(2) + 0.5A1
2 B(1.5)(2)D
V = 2p©rA
= 37.7 m2
= 2pC0.75(1.5) + 1.5(2) + 0.752(1.5)2
+ (2)2
D
A = 2p©rL
z =
©z V
©V
=
2.835
3.6
= 0.7875 m
y =
©y V
©V
=
5.00625
3.6
= 1.39 m
= 0.391 m
=
0.25[0.5(2.5)(1.8)] + 0.25B
1
2
(1.5)(1.8)(0.5)R + B
1
2
(1.5)(1.8)(0.5)R
0.5(2.5)(1.8) +
1
2
(1.5)(1.8)(0.5) +
1
2
(1.5)(1.8)(0.5)
x =
©x V
©V
= 2.67 ft
z =
©z V
©V
=
3[2(7)(6)] + 1.5[4(2)(3)]
2(7)(6) + 4(2)(3)
= 2.94 ft
y =
©y V
©V
=
3.5[2(7)(6)] + 1[4(2)(3)]
2(7)(6) + 4(2)(3)
= 1.67 ft
x =
©x V
©V
=
1[2(7)(6)] + 4[4(2)(3)]
2(7)(6) + 4(2)(3)
= 1.33 in.
y =
©y A
©A
=
2[4(0.5)] + 0.25[(0.5)(2.5)]
4(0.5) + (0.5)(2.5)
= 0.827 in.
x =
©x A
©A
=
0.25[4(0.5)] + 1.75[0.5(2.5)]
4(0.5) + 0.5(2.5)
= 162.5 mm
y =
©y A
©A
=
100[2(200)(50)] + 225[50(400)]
2(200)(50) + 50(400)
F9–3.
Ans.
F9–4.
Ans.
F9–5.
Ans.
F9–6.
Ans.
F9–7.
Ans.
Ans.
Ans.
F9–8.
Ans.
= 237.5 mm
y =
©y A
©A
=
150[300(50)] + 325[50(300)]
300(50) + 50(300)
= -61.5 mm
=
0(300) + 0(600) + (-200)(400)
300 + 600 + 400
z =
©z L
©L
= 323 mm
=
0(300) + 300(600) + 600(400)
300 + 600 + 400
y =
©y L
©L
= 265 mm
=
150(300) + 300(600) + 300(400)
300 + 600 + 400
x =
©x L
©L
= 0.786 ft
z =
L
V
z dV
L
V
dV
=
L
2 ft
0
zc
9p
64
(4 - z)2
dzd
L
2 ft
0
9p
64
(4 - z)2
dz
= 0.667 m
y =
L
V
y dV
L
V
dV
=
L
1 m
0
y¢
p
4
ydy≤
L
1 m
0
p
4
y dy
=
9
16
L
x =
L
m
x dm
L
m
dm
=
L
L
0
xBm0¢1 +
x2
L2 ≤dxR
L
L
0
m0¢1 +
x2
L2 ≤dx
= 1.2 m
y =
L
A
y dA
L
A
dA
=
L
2 m
0
ya2a
y1/2
22
b bdy
L
2 m
0
2a
y1/2
22
bdy](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-633-320.jpg)
![618 PARTIAL SOLUTIONS AND ANSWERS
Ans.
Ans.
F11–5.
Ans.
F11–6.
Ans.
u = 20.9°
+ 5400 sin u cos u)du = 0
(135 cos u - 5400 sin u
dU = 0; PdxC + FspdyB = 0
yB = 2[0.3 cos u] dyB = -0.6 sin u du
xC = 3[0.3 sin u] dxC = 0.9 cos u du
Fsp = 15 000 (0.6 - 0.6 cos u)
u = 56.33° = 56.3°
- 1226.25 cos u)du = 0
(15 000 sin u cos u - 7500 sin u
dU = 0; -WdyG + A -FspdxAB = 0
xA = 5 cos u dxC = -5 sin u du
yG = 2.5 sin u dyG = 2.5 cos u du
u = 54.31° = 54.3°
64800 cos u - 37 800 = 0
sin u = 0 u = 0°
sin u (64 800 cos u - 37 800)du = 0
F11–2.
Ans.
F11–3.
Ans.
Ans.
F11–4.
- 36A103
B(cos u - 0.5)(-1.8 sin u du) = 0
6A103
B(-0.9 sin u du)
dU = 0; PdxB + A -Fsp dxCB = 0
xC = 2(0.9 cos u) dxC = -1.8 sin udu
xB = 0.9 cos u dxB = -0.9 sin udu
u = 77.16° = 77.2°
-5400 cos u + 1200 = 0
sin u = 0 u = 0°
- 2000(-0.6 sin u du) = 0
-9A103
B sin u (0.6 cos u du)
dU = 0; -FspdxB + (-PdyC) = 0
yC = 0.6 cos u dyC = -0.6 sin u du
xB = 0.6 sin u dxB = 0.6 cos u du
P = 245.25 cot u|u=60° = 142 N
(5P sin u - 1226.25 cos u)du = 0
dU = 0; -PdxA + (-WdyG) = 0
yG = 2.5 sin u dyG = 2.5 cos u du
xA = 5 cos u dxA = -5 sin u du](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-635-320.jpg)
![ANSWERS TO SELECTED PROBLEMS 623
2–97.
2–98.
2–99.
2–101.
2–102.
2–103.
2–105.
2–106.
2–107.
2–109.
2–110. F = {143i + 248j - 201k} lb
g = 2.81°
b = 92.6°
a = 88.8°
FR = 18.5 kN
FB = {0.970i - 1.68j + 7.76k} kN
+ (0 - 0.75 cos 30°)j + (3 - 0)k
rB = [0 - (-0.75 sin 30°)]i
FC = {0.857 i + 0.857 j + 4.85 k} kN
+ [0 - (-0.75 cos 45°)]j + (3 - 0)k
rC = [0 - (-0.75 sin 45°)]i
FA = {-1.46 i + 5.82 k} kN
= {-0.75 i + 0j + 3 k} m
rA = (0 - 0.75)i + (0 - 0)j + (3 - 0)k
F = {-6.61i - 3.73j + 9.29k} lb
F = 105 lb
g = 180°
b = 90°
a = 90°
FR = 240 lb
FD = 5-30i - 20j - 60k6 lb
FC = 5-30i + 20j - 60k6 lb
FB = 530i + 20j - 60k6 lb
FA = 530i - 20j - 60k6 lb
FA = FB = FC = 326 lb
g = 180°
b = 90°
a = 90°
FR = 1.24 kip
z = 16 ft
y = 18 ft
x = 24 ft
u = F
F = -120
170 i - 90
170 j - 80
170 k
FR = 3.46 kN
FB = 2.42 kN
FC = 1.62 kN
FB = {53.2i - 79.8j - 146k} N
FA = {-43.5i + 174j - 174k} N
FC = {159i + 183j - 59.7k} N
FA = {285j - 93.0k} N
+ (0 - 0.750) k} m
rCD = {[-0.5 - (-2.5)] i + [0 - (-2.299)] j
+ (0 - 0.750) k} m
rAB = {(0 - 0)i + [0 - (-2.299)] j 2–113.
2–114.
2–115.
2–117.
2–118.
2–119.
2–121.
2–122.
2–123.
2–125.
2–126.
2–127.
2–129.
2–130.
2–131.
2–133.
2–134.
2–135.
2–137.
2–138.
2–139.
u = 52.7°
FAB = 215 lb
u = 85.2°
Fr = 178 N
u = 143°
rBC = {6i + 4j - 2k} ft
rBA = {-3i} ft
u = 52.7°
FR = 215 lb
u = 100°
FR = 178 N
(F1)F2
= 5.44 lb
uF2
= cos 135°i + cos 60°j + cos 60°k
uF1
= cos 30° sin 30°i + cos 30° cos 30°j - sin 30°k
Fy = 260 N
Fx = -75 N
FAC = 45.5 lb
Fx = 47.8 lb
u = 34.2°
rAC = {-15i - 8j + 12k} ft
rAB = {-15i + 3j + 8k} ft
u = 97.3°
(F1)F2
= 50.6 N
f = 65.8°
uOA = 1
3 i + 2
3 j - 2
3 k
uOD = -sin 30°i + cos 30°j
(FBC)⬜ = 316 N
(FBC)|| = 245 N
FAC = 5293j + 219k6 lb
FAC = 366 lb
(FAC)z = -569 lb
uAC = 0.1581i + 0.2739j - 0.9487k
F2 = 373 N
F1 = 333 N
FBC = {32i - 32j} N
FBC = 45.2 N
|Proj F2| = 71.6 N
u1 = cos 120° i + cos 60° j + cos 45° k
(FED)⬜ = 498 N
(FED)|| = 334 N
rBC = 5.39 m
(FAO)⬜ = 2(56)2
- (46.86)2
= 30.7 N
+ 16A-2
7 B = 46.9 N
(F
AO)|| = (24)A3
7 B + (-48)A -6
7 B](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-640-320.jpg)
![636 ANSWERS TO SELECTED PROBLEMS
6–127.
6–129.
6–130.
6–131.
6–133. Member AC:
Member AC:
Member CB:
6–134.
6–135.
Ey = 5.69 kip
Ex = 8.31 kip
Ay = 0.308 kip
Ax = 8.31 kip
P = kL
2 tan u sin u (2 - csc u)
By = 97.4 N
Bx = 97.4 N
Ay = 397 N
Ax = 117 N
Cy = 97.4 N
Cx = 402.6 N
FDF = 424 lb (T)
FCF = 300 lb (C)
FCD = 500 lb (C)
FEF = 300 lb (C)
FDE = 0
FAE = 367 lb (C)
FAD = 0
FAC = 972 lb (T)
FAB = 300 lb (C)
FBE = 500 lb (T)
FBC = 0
FBF = 0
FDF = 424 lb (T)
FCF = 300 lb (C)
FCD = 300 lb (C)
FEF = 300 lb (C)
FDE = 0
FAE = 667 lb (C)
FAD = 333 lb (T)
FAC = 583 lb (T)
FAB = 300 lb (C)
FBE = 500 lb (T)
FBC = 0
FBF = 0
Joint A: FAE = 8.00 kN (T)
Joint B: FBA = 17.9 kN (C)
Joint D: FDE = 8.00 kN (T)
FCD = 8.00 kN (T)
Joint C: FCB = 17.9 kN (C)
FDB = FBE = 0
FB = 133 lb
MCz = 0
MCy = -429 N # m
Cz = 125 N
Cy = 61.9 N
6–113.
6–114.
6–115.
6–117.
6–118.
6–119.
6–121.
6–122.
6–123.
6–125.
6–126.
Cx = 47.3 N
Ay = 115 N
Ax = 172 N
Az = 0
By = -13.3 lb
Bx = -30 lb
Bz = 0
Bz + 6
9(270) - 180 = 0
FDE = 270 lb
-6
9 FDE(3) + 180(3) = 0
Bz = Dz = 283 N
By = Dy = 283 N
Bx = Dx = 42.5 N
P = 283 N
W3 = 75 lb
W2 = 21 lb
W1 = 3 lb
M = 4PL sin2
u
sin f [cos(f - u)]
Nc = 4P sin2
u
sin f
Cy = 53.3 lb
Cx = 413 lb
Bx = 333 lb
By = 133 lb
Ay = 80 lb
Ax = 80 lb
NC = 15.0 lb
Ay = 0
Ax = 120 lb
MD = 2.66 kN # m
Dy = 1.96 kN
Dx = 0
Cy = 7.01 kN
Cx = 2.17 kN
FAB = 9.23 kN
lAB = 861.21 mm, LCAB = 76.41°,
NE = 187 N
FBC = 15.4 kN (C)
FIJ = 9.06 kN (T)
W1 = b
aW
Wx
12b + 3c(4b) + W¢1 -
x
3b + 3
4c
≤(b) - W1(a) = 0
©MA = 0; FCD(c) -
Wx
A3b + 3
4cB
A1
4cB = 0
©ME = 0; W(x) - NBA3b + 3
4cB = 0](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-653-320.jpg)
![640 ANSWERS TO SELECTED PROBLEMS
7–118.
7–119.
7–121.
7–122.
7–123.
7–125.
7–126.
7–127.
Chapter 8
8–1.
8–2.
8–3.
8–5.
u = 52.0°
180(10 cos u) - 0.4(180)(10 sin u) - 180(3) = 0
ms = 0.256
P = 474 N
N = 494.94 N
P = 140 N
P cos 30° + 0.25N - 50(9.81) sin 30° = 0
h = 93.75 ft
l = 238 ft
s = 18.2 ft
ME = 86.6 lb # ft
V
E = 28.9 lb
NE = 0
ND = FCD = -86.6 lb
V
D = MD = 0
FCD = 86.6 lb
M = -300 - 200y
N = -150 lb
V = 200 lb
0 … y … 2 ft
M = 150 cos u + 200 sin u - 150
N = 150 cos u + 200 sin u
V = 150 sin u - 200 cos u
0° … u … 180°
M = {16.0 - 2.71x - 0.0981x2
} kN # m
V = { - 0.196x - 2.71} kN
2 m 6 x … 5 m
M = {5.29x - 0.0981x2
} kN # m
V = {5.29 - 0.196x} kN
0 … x 6 2 m
M = {27.0 - 4.50x} kN # m
V = -4.50 kN
3 m 6 x … 6 m
M = {1.50x} kN # m
V = 1.50 kN
0 … x 6 3 m
Ay = 1.50 kN
FCD = 6.364 kN
Tmax = 76.7 lb
a = 0.366L
MD = -54.9 N # m
V
D = -220 N
ND = -220 N
Segment CD
ME = 112.5 N # m
7–97.
7–98.
7–99.
7–101.
7–102.
7–103.
7–105.
7–106.
7–107.
7–109.
7–111.
7–113.
7–114.
7–115.
7–117.
V
E = 0
NE = 80.4 N
Segment CE
FBC = 310.58 N
x = 5+
, V = -1.14, M = 2.29
x = 2-
, V = 4.86, M = 9.71
h = 10.6 ft
Total length = 55.6 ft
h = 1.47 m
y = 135.92Ccosh 7.3575(10-3
)x - 1D
dy
dx = sinh 7.3575(10-3
)x
L = 15.5 m
Tmax = 1.60 kN
y = 23.5[cosh 0.0425x - 1] m
FH = 1153.41 N
L = 45 = 2e
FH
49.05
sinha
49.05
FH
(20)b f
y =
FH
49.05
ccosha
49.05
FH
xb - 1d m
L = 51.3 m
Tmax = 5.36 kN
Tmax = 48.7 kip
y = 46.0(10-6
)x3
+ 0.176x
w0 = 77.8 kNm
y = 150 m at x = -(1000 - x0)
y = 75 m at x = x0
y =
w0
4FH
x2
dy
dx
=
w0
2FH
x
L = 13.4 ft
h = 2.68 ft
4.42 kip
w0 = 264 lb/ft
10 =
w0
2 FH
(25 - x)2
15 =
w0
2 FH
x2
w0 = 0.846 kN/m
P = 71.4 lb
xB = 4.36 ft
Joint C:
30 - 2xB
2(xB - 3)2
+ 64
TBC = 102
Joint B:
13xB - 15
2(xB - 3)2
+ 64
TBC = 200](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-657-320.jpg)
![644 ANSWERS TO SELECTED PROBLEMS
9–43.
9–45.
9–46.
9–47.
9–49.
9–50.
9–51.
9–53.
9–54.
9–55.
9–57.
9–58.
9–59.
9–61.
9–62.
9–63.
9–65.
x =
2.4971(10-3
)
16.347(10-3
)
= 153 mm
©m = 16.4 kg
y = 293 mm
y =
b(W2 - W1) 2b2
- c2
cW
x =
W1
W
b
y =
441.2(104
)
81(104
)
= 544 mm
x = 0
y = 2.56 in.
x = 4.83 in.
x =
4Aro
3
- ri
3
B
3pAro
2
- ri
2
B
y = 9.648
6.84 = 1.41 m
x = 15.192
6.84 = 2.22 m
y = 2.57 in.
y = 2.00 in.
= 5.125 in.
y = 3[2(6)(1)] + 5.5(6)(1) + 9(6)(1)
2(6)(1) + 6(1) + 6(1)
y = 12 in.
x = 2.64 in.
Ax = 0
Ay = 1.32 kN
Ey = 342 N
y = 9.24 m
x = 1.65 m
f = 30° - 10.89° = 19.1°
u = tan-1 50
400 sin 60° - 88.60 = 10.89°
y = 88.6 mm
x = -50 mm
z = 0.157 in.
y = 0.0370 in.
x = 0.0740 in.
z = 2.14 in.
y = 1.07 in.
x = -0.590 in.
z = 169.44(103
)
1361.37 = 124 mm
y = 60(103
)
1361.37 = 44.1 mm
x = 164.72(103
)
1361.37 = 121 mm
z = 8
15 r
m = pkr4
4
9–18.
9–19.
9–21.
9–22.
9–23.
9–25.
9–26.
9–27.
9–29.
9–30.
9–31.
9–33.
9–34.
9–35.
9–37.
9–38.
9–39.
9–41.
9–42.
z = a
p
y = 3
4 h
V = pa 2
h
6
y = 23
55 a
y
'
= y
dm = pr0Aa2
- y2
+ ay - y3
a B dy
y = 4.36 ft
z = 2
9 h
y = 3.2 m
y
'
= y
dV = p
16 y3
dy
y = a
2(10 - 3p)
x = 5
9 a
m = 3
2 r0 abt
y = 1 ft
y
'
= y
dA = Ay
2 - y2
4 B dy
y = 0.357 m
x = 0.914 m
Ay = 73.9 kN
Ax = 24.6 kN
NB = 55.1 kN
y = 0
x = 1.20 m
y = n + 1
2(2n + 1) h
y = y
2
dA = y dx
y = 0.45 m
x = 0.45 m
y = 1.14 ft
x = 1.6 ft
A = 2.25 ft2
y
'
= 1
2 Ax + x3
9 B
x
'
= x
dA = Ax - x3
9 B dx
y = 0.541 in.
x = 1.08 in.
x = 5a
8
x
'
= x
dA = 2kAx - x2
2a B dx
x = - 0.833a
Ay = 1.98 kN
Ax = 0
FBC = 2.64 kN](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-661-320.jpg)
![ANSWERS TO SELECTED PROBLEMS 645
9–66.
9–67.
9–69.
9–70.
9–71.
9–73.
9–74.
9–75.
9–77.
9–78. or
9–79.
9–81.
9–82.
9–83.
9–85.
9–86.
9–87.
9–89.
9–90.
9–91.
9–93.
9–94.
9–95.
9–97.
= 536 m3
V = 2pC A4(4)
3p B A1
4p (4)2
B + (2)(8)(4)D
2.26 gallons
R = 29.3 kip
V
c = 20.5 m3
V
h = 2p[0.75(6) + 0.6333(0.780) + 0.1(0.240)]
V = 0.0376 m3
A = 1.06 m2
V = 50.6 in3
A = 116 in2
= 0.0486 m3
V = 2p[(112.5)(75)(375) + (187.5)(325)(75)]
A = 141 in2
A = 1365 m2
= 77.0 m3
+ 1.667A2(1.5)
2 B D
V = 2pC A4(3)
3p Ap(32
)
4 B + 0.5(1.5)(1)
V = 101 ft3
V = 3485 ft3
A = 2p(184) = 1156 ft2
z = 122 mm
h = 48 mm
h = 80 mm
x =
11.02(106
)p
172(103
)p
= 64.1 mm
z = 1.67 in.
y = 2.79 in.
x = 2.19 in.
z = 754 mm
z = 1.0333p
9.3333p = 111 mm
y = 11.0 ft
x = 19.0 ft
Ay = 5.99 kN
By = 4.66 kN
y = 3.07 m
x = 4.56 m
u = 30.2°
z = 371 433.63
16 485.84 = 22.5 mm
x = 216 000
16 485.84 = 13.1 mm
x = L + (n - 1)d
2
y = 3.80 ft
x = 5.07 ft
z =
1.8221(10-3
)
16.347(10-3
)
= 111 mm
y = -15 mm 9–98.
9–99.
9–101.
9–102.
9–103.
9–105.
9–106.
9–107.
9–109.
9–110.
9–111.
9–113.
9–114.
9–115.
9–117.
9–118.
9–119.
9–121.
9–122.
9–123.
9–125.
9–126.
9–127.
9–129.
y = 3.00 m
x = 2.74 m
FR = 7.62 kN
dFR = 6A- 240
x + 1 + 340B dx
y = -0.262a
y = 1.63 in.
x = 0
y = 39.833
27.998 = 1.42 in.
x = 76.50
27.998 = 2.73 in.
z = 2
3 a
x = y = 0
y = 87.5 mm
y = 1.33 in.
y
'
= x2
2
dA = x2
dx
FR = 170 kN
x = 1.51 m
F.S. = 2.66
(Wcon)r = 282.53 kN
(Wcon)p = 188.35 kN
Fh = 176.58 kN
Fv = 39.24 kN
mA = 5.89 Mg
L = 2.31 m
NC = 13.1 kN
wC = 58.86 kN
wB = 39.24 kN
FR = 450 lb
FR = 225 lb
FR = 41.7 kN
dFR = A26.556721 - y2
- 6.9367y21 - y2
B dy
h = 2.7071 - 0.7071y
FABDC = 1800 lb
FCDEF = 750 lb
d = 3.65 m
d = 2.68 m
-176 580(2) + 73 575dA2
3dB = 0
h = 106 mm
V = 22.1(103
) ft3
A = 2p[7.5(2241) + 15(30)] = 3.56(103
) ft2
14.4 liters
A = 43.18 m2
V = 25.5 m3](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-662-320.jpg)
![646 ANSWERS TO SELECTED PROBLEMS
10–34.
10–35.
10–37.
10–38.
10–39.
10–41. Consider a large rectangle and a hole.
10–42.
10–43. ,
10–45. Consider three segments.
10–46.
10–47.
10–49. Consider three segments.
10–50.
10–51.
10–53.
10–54.
10–55.
10–57. Consider rectangular segments,
, , and
10–58.
10–59.
10–61.
10–62.
10–63.
10–65.
Ixy = 3.12 m4
y
'
= y
2
x
'
= x
dA = 1
8(x3
+ 2x2
+ 4x) dx
Ixy = 48 in4
Ixy = a2
b2
8
Ixy = 0.667 in4
dA = x dy
y = y
x = x
2
Ix¿ = 30.2(106
) mm4
Iy = 153(106
) mm4
Ix = 115(106
) mm4
150 mm * 12 mm
100 mm * 12 mm
226 mm * 12 mm
Ix = 22.9(106
) mm4
Ix = 388 in4
Ix¿ = 15.896 + 36.375 = 52.3 in4
y = 61.75
13 = 4.75 in.
Ix = 2.51(106
) mm4
Iy¿ = 1.21(109
) mm4
Ix¿ = 124(106
) mm4
Iy = 914(106
) mm4
Iy = 548(106
) mm4
Ix = 548(106
) mm4
Ix¿ = 64.0 in4
y = 2.00 in.
Iy = 2.51(106
) mm4
Ix = 52.7(106
) mm4
Ix = 2.17(10-3
) m4
Ix¿ = 722(10)6
mm4
y = 170 mm
= 74 in4
Iy = C 1
12(2)(63
)D + 2C 1
12(3)(13
) + 3(1)(2.5)2
D
Iy¿ = 122(106
) mm4
Ix¿ = 34.4(106
) mm4
y = 22.5 mm
= 10.3(109
) mm4
+ C - p
4(75)4
+ (-p(75)2
(450)2
D
+ C 1
12(200)(300)3
+ 200(300)(450)2
Chapter 10
10–1.
10–2.
10–3.
10–5.
10–6.
10–7.
10–9.
10–10.
10–11.
10–13.
10–14.
10–15.
10–17.
10–18.
10–19.
10–21.
10–22.
10–23.
10–25.
10–26.
10–27.
10–29.
10–30.
10–31.
10–33.
+ 1
2(200)(300)(200)2
D
(Iy)triangle = C 1
36(200)(3003
)
Iy = 45.5(106
) mm4
Ix = 76.6(106
) mm4
= 54.7 in4
Iy = 1
12(2)(6)3
+ 2C 1
12(4)(1)3
+ 1(4)(1.5)2
D
Ix¿ = 57.9 in4
y = 2.20 in.
J0 =
pr0
4
4
Iy =
pr0
4
8
dA = (rdu) dr
Iy = 30.9 in4
Ix = 9.05 in4
Iy = 307 in4
dA = x1/3
dx
Iy = 2
15hb3
Ix = 2
7bh3
Iy = 1
12hb3
dA = Ah - h
bxB dx
Iy = 1.07 in4
Ix = 19.5 in4
Iy = 0.333 in4
dA = (2 - 2x3
) dx
Iy = 10.7 in4
Ix = 307 in4
JO = 0.491 m4
Iy = 0.2857 m4
dA = 2x4
dx
Ix = 0.2051 m4
dA = C1 - Ay
2 B1/4
D dy
Ix = 0.205 m4
Iy = 4.57 m4
Ix = 2.13 m4
dA = A2 - y2
2 B dy
Ix = 0.0606 m4
Iy = 2.67 m4
Ix = 0.533 m4
dA = [2 - (4y)1/3
] dy](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-663-320.jpg)
![ANSWERS TO SELECTED PROBLEMS 647
10–66.
10–67.
10–69.
10–70.
10–71.
10–73. Consider three segments.
10–74.
10–75.
10–77. Consider three segments.
10–78.
10–79.
10–81.
10–82.
10–83.
10–85.
10–86.
Imax = 31.7 in4
y = 1.68 in.
x = 1.68 in.
Iuv = 111 in4
Iv = 238 in4
Iu = 109 in4
R = 128.72 in4
Iavg = 173.72 in4
y = 8.25 in.
Iuv = -126(106
) mm4
Iv = 258(106
) mm4
Iu = 112(106
) mm4
x = 48.2 mm
Iuv = -3.08(106
) mm4
Iv = 47.0(106
) mm4
Iu = 43.4(106
) mm4
y = 82.5 mm
Imin = 5.03(106
) mm4
, (up)2 = -77.7°
Imax = 113(106
) mm4
, (up)1 = 12.3°
Ixy = -22.4(106
) mm4
Iy = 9.907(106
) mm4
Ix = 107.83(106
) mm4
Iuv = 111 in4
Iv = 238 in4
Iu = 109 in4
y = 8.25 in.
Iuv = 17.5 in4
Iv = 23.6 in4
Iu = 43.9 in4
Ixy = -13.05(106
) mm4
Iuv = -126(106
) mm4
Iv = 258(106
) mm4
Iu = 112(106
) mm4
x = 48.2 mm
Ixy = -110 in4
Ixy = 17.1(106
) mm4
Ixy = 36.0 in4
Ixy = 35.7 in4
Ixy = 10.7 in4
dA = x1/2
dx, x
'
= x, y
'
= y
2
Ixy = 3
16b2
h2
Ixy = 0.333 m4
d
b
10–87.
d
b
10–89.
10–90.
10–91.
10–93.
10–94.
10–95.
10–97.
10–98.
10–99.
10–101.
10–102.
10–103.
10–105.
10–106.
10–107.
10–109.
10–110.
10–111.
10–113. Consider four triangles and a rectangle.
10–114.
10–115.
10–117.
10–118. Ix = 0.610 ft4
Iy = 2.13 ft4
dA = 1
4(4 - x2
) dx
y = 0.875 in., Ix¿ = 2.27 in4
Ix = 1
12a4
Iy = 0.187d4
IO = 0.113 kg # m2
IO = 0.276 kg # m2
IA = 222 slug # ft2
IO = 84.94 slug # ft2
Ix = 3.25 g # m2
Iz = 2.25 kg # m2
IG = 4.45 kg # m2
y
'
= 1(3) + 2.25(5)
3 + 5 = 1.78 m
Iy = 0.144 kg # m2
Iz = 0.150 kg # m2
IO = 53.2 kg # m2
L = 6.39 m
0.5 = 1.5(6) + 0.65[1.3(2)] + 0[L(2)]
6 + 1.3(2) + L(2)
Iy = 1.71(103
) kg # m2
Iz = 63.2 slug # ft2
Iz = 87.7(103
) kg # m2
dIz = rp
8192z8
dz
Ix = 93
10 mb2
Iy = 2
5 mb2
kx = 57.7 mm
dIx = rp
2 (2500 x2
) dx
dm = rp(50x) dx
Iz = 7
18 ml 2
Ix = 3
10 mr 2
Iz = 3
10 mr0
2
dIz = 1
2 rpar0 -
r0
h
zb
4
dz
dm = rpar0 -
r0
h
zb
2
dz
(up)2 = 77.7°
(up)1 = 12.3°
Imin = 5.03(106
) mm4
Imax = 113(106
) mm4
(u)p)2 = 45°
(up)1 = 45°
Imin = 8.07 in4](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-664-320.jpg)
![12.2 RECTILINEAR KINEMATICS: CONTINUOUS MOTION 17
12
•12–21. Two particles A and B start from rest at the origin
and move along a straight line such that
and , where t is in
seconds. Determine the distance between them when
and the total distance each has traveled in .
12–22. A particle moving along a straight line is subjected
to a deceleration , where is in . If it
has a velocity and a position when
, determine its velocity and position when .
12–23. A particle is moving along a straight line such that
its acceleration is defined as , where is in
meters per second. If when and ,
determine the particle’s position, velocity, and acceleration
as functions of time.
t = 0
s = 0
v = 20 ms
v
a = (-2v) ms2
t = 4 s
t = 0
s = 10 m
v = 8 ms
ms
v
a = (-2v3
) ms2
t = 4 s
t = 4 s
aB = (12t2
- 8) fts2
aA = (6t - 3) fts2
s = 0
*12–16. As a train accelerates uniformly it passes successive
kilometer marks while traveling at velocities of and
then . Determine the train’s velocity when it passes
the next kilometer mark and the time it takes to travel the
2-km distance.
•12–17. A ball is thrown with an upward velocity of
from the top of a 10-m high building. One second later
another ball is thrown vertically from the ground with a
velocity of . Determine the height from the ground
where the two balls pass each other.
12–18. A car starts from rest and moves with a constant
acceleration of until it achieves a velocity of .
It then travels with constant velocity for 60 seconds.
Determine the average speed and the total distance
traveled.
12–19. A car is to be hoisted by elevator to the fourth floor
of a parking garage, which is 48 ft above the ground. If the
elevator can accelerate at decelerate at
and reach a maximum speed of determine the shortest
time to make the lift, starting from rest and ending at rest.
*12–20. A particle is moving along a straight line such that
its speed is defined as , where s is in meters.
If when , determine the velocity and
acceleration as functions of time.
t = 0
s = 2 m
v = (-4s2
) ms
8 fts,
0.3 fts2
,
0.6 fts2
,
25 ms
1.5 ms2
10 ms
5 ms
10 ms
2 ms
12–15. Tests reveal that a normal driver takes about
before he or she can react to a situation to avoid a collision.It
takes about 3 s for a driver having 0.1% alcohol in his system
to do the same. If such drivers are traveling on a straight road
at 30 mph (44 ) and their cars can decelerate at ,
determine the shortest stopping distance d for each from the
moment they see the pedestrians. Moral: If you must drink,
please don’t drive!
2 fts2
fts
0.75 s
d
v1 44 ft/s
Prob. 12–15
*12–24. A particle starts from rest and travels along a
straight line with an acceleration ,
a = (30 - 0.2v) fts2
where is in . Determine the time when the velocity of
the particle is .
•12–25. When a particle is projected vertically upwards
with an initial velocity of , it experiences an acceleration
, where g is the acceleration due to gravity,
a = -(g + kv2
)
v0
v = 30 fts
fts
v
k is a constant and is the velocity of the particle.
Determine the maximum height reached by the particle.
12–26. The acceleration of a particle traveling along a
straight line is , where t is in seconds. If
, when , determine the velocity and
acceleration of the particle at .
12–27. A particle moves along a straight line with an
acceleration of , where s is in
meters. Determine the particle’s velocity when , if it
starts from rest when . Use Simpson’s rule to
evaluate the integral.
*12–28. If the effects of atmospheric resistance are
accounted for, a falling body has an acceleration defined by
the equation , where is in
and the positive direction is downward. If the body is
released from rest at a very high altitude, determine (a) the
velocity when , and (b) the body’s terminal or
maximum attainable velocity (as ).
t : q
t = 5 s
ms
v
a = 9.81[1 - v2
(10-4
)] ms2
s = 1 m
s = 2 m
a = 5(3s13
+ s52
) ms2
s = 4 m
t = 0
s = 0
v = 0
a = (0.02et
) ms2
v](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-688-320.jpg)
![18 CHAPTER 12 KINEMATICS OF A PARTICLE
12
*12–36. The acceleration of a particle traveling along a
straight line is , where s is in meters. If
at , determine the velocity of the particle at
, and the position of the particle when the velocity
is maximum.
•12–37. Ball A is thrown vertically upwards with a velocity
of . Ball B is thrown upwards from the same point with
the same velocity t seconds later. Determine the elapsed
time from the instant ball A is thrown to when the
balls pass each other, and find the velocity of each ball at
this instant.
12–38. As a body is projected to a high altitude above the
earth’s surface, the variation of the acceleration of gravity
with respect to altitude y must be taken into account.
Neglecting air resistance, this acceleration is determined
from the formula , where is the
constant gravitational acceleration at sea level, R is the
radius of the earth, and the positive direction is measured
upward. If and , determine the
minimum initial velocity (escape velocity) at which a
projectile should be shot vertically from the earth’s surface
so that it does not fall back to the earth. Hint: This requires
that as
12–39. Accounting for the variation of gravitational
acceleration a with respect to altitude y (see Prob. 12–38),
derive an equation that relates the velocity of a freely
falling particle to its altitude. Assume that the particle is
released from rest at an altitude from the earth’s surface.
With what velocity does the particle strike the earth if it is
released from rest at an altitude ? Use the
numerical data in Prob. 12–38.
*12–40. When a particle falls through the air, its initial
acceleration diminishes until it is zero, and
thereafter it falls at a constant or terminal velocity If
this variation of the acceleration can be expressed as
determine the time needed for the
velocity to become Initially the particle falls
from rest.
•12–41. A particle is moving along a straight line such that
its position from a fixed point is ,
where t is in seconds. Determine the total distance traveled
by the particle from to .Also, find the average
speed of the particle during this time interval.
t = 3s
t = 1s
s = (12 - 15t2
+ 5t3
) m
v = vf2.
a = 1gv2
f21v2
f - v2
2,
vf.
a = g
y0 = 500 km
y0
y : q.
v = 0
R = 6356 km
g0 = 9.81 ms2
g0
a = -g0[R2
(R + y)2
]
t 6 2v0g
v0
s = 2 m
s = 0
v = 0
a = (8 - 2s) ms2
•12–29. The position of a particle along a straight line is
given by , where t is in
seconds. Determine the position of the particle when
and the total distance it travels during the 6-s time
interval. Hint: Plot the path to determine the total distance
traveled.
12–30. The velocity of a particle traveling along a straight
line is , where k is constant. If when ,
determine the position and acceleration of the particle as a
function of time.
12–31. The acceleration of a particle as it moves along a
straight line is given by where t is in
seconds. If and when determine
the particle’s velocity and position when Also,
determine the total distance the particle travels during this
time period.
*12–32. Ball A is thrown vertically upward from the top
of a 30-m-high-building with an initial velocity of 5 . At
the same instant another ball B is thrown upward from the
ground with an initial velocity of 20 . Determine the
height from the ground and the time at which they pass.
•12–33. A motorcycle starts from rest at and travels
along a straight road with a constant acceleration of
until it reaches a speed of 50 . Afterwards it maintains
this speed. Also, when , a car located 6000 ft down the
road is traveling toward the motorcycle at a constant speed
of 30 . Determine the time and the distance traveled by
the motorcycle when they pass each other.
12–34. A particle moves along a straight line with a
velocity , where s is in millimeters.
Determine the acceleration of the particle at .
How long does the particle take to reach this position if
when ?
쐍12–35. A particle has an initial speed of If it
experiences a deceleration of where t is in
seconds, determine its velocity, after it has traveled 10 m.
How much time does this take?
a = 1-6t2 ms2
,
27 ms.
t = 0
s = 500 mm
s = 2000 mm
v = (200s) mms
fts
t = 0
fts
6 fts2
t = 0
ms
ms
t = 6 s.
t = 0,
v = 2 ms
s = 1 m
a = 12t - 12 ms2
,
t = 0
s = 0
v = v0 - ks
t = 6 s
s = (1.5t3
- 13.5t2
+ 22.5t) ft](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-689-320.jpg)
![12.3 RECTILINEAR KINEMATICS: ERRATIC MOTION 23
12
EXAMPLE 12.7
The car in Fig. 12–14a starts from rest and travels along a straight
track such that it accelerates at for 10 s, and then decelerates
at . Draw the and s–t graphs and determine the time
needed to stop the car. How far has the car traveled?
SOLUTION
v–t Graph. Since the graph is determined by integrating
the straight-line segments of the a–t graph. Using the initial condition
when we have
When Using this as the initial
condition for the next time period, we have
v = 101102 = 100 ms.
t = 10 s,
v = 10t
L
v
0
dv =
L
t
0
10 dt,
a = (10) ms2
;
0 … t 6 10 s;
t = 0,
v = 0
v–t
dv = a dt,
t¿
v–t
2 ms2
10 ms2
t (s)
a (m/s2
)
(a)
10
2
10
A1
A2
t¿
t (s)
v (m/s)
(b)
100
10
v 10t
v 2t 120
t¿ 60
t (s)
s (m)
(c)
10 60
500
3000
s 5t2
s t2
120t 600
Fig. 12–14
The s–t graph is shown in Fig. 12–14c.
NOTE: A direct solution for s is possible when since the
triangular area under the graph would yield the displacement
from to Hence,
t¿ = 60 s.
t = 0
¢s = s - 0
v–t
t¿ = 60 s,
Ans.
¢s = 1
2160 s21100 ms2 = 3000 m
v = (-2t + 120) ms
L
v
100 ms
dv =
L
t
10 s
-2 dt,
a = (-2) ms2
;
10 s 6 t … t¿;
When we require This yields, Fig. 12–14b,
Ans.
A more direct solution for is possible by realizing that the area
under the a–t graph is equal to the change in the car’s velocity. We
require Fig. 12–14a.Thus
Ans.
s–t Graph. Since integrating the equations of the
graph yields the corresponding equations of the s–t graph. Using the
initial condition when we have
When Using this initial condition,
s = 511022
= 500 m.
t = 10 s,
s = (5t2
) m
L
s
0
ds =
L
t
0
10t dt,
v = (10t) ms;
0 … t … 10 s;
t = 0,
s = 0
v–t
ds = v dt,
t¿ = 60 s
0 = 10 ms2
110 s2 + 1-2 ms2
21t¿ - 10 s2
¢v = 0 = A1 + A2,
t¿
t¿ = 60 s
v = 0.
t = t¿
L
s
500 m
ds =
L
t
10 s
1-2t + 1202 dt
v = (-2t + 120) ms;
10 s … t … 60 s;
When the position is
Ans.
s = -16022
+ 1201602 - 600 = 3000 m
t¿ = 60 s,
s = (-t2
+ 120t - 600) m
s - 500 = -t2
+ 120t - [-11022
+ 1201102]](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-694-320.jpg)
![24 CHAPTER 12 KINEMATICS OF A PARTICLE
12 EXAMPLE 12.8
The –s graph describing the motion of a motorcycle is shown in
Fig. 12–15a. Construct the a–s graph of the motion and determine the
time needed for the motorcycle to reach the position
SOLUTION
a–s Graph. Since the equations for segments of the –s graph are
given, the a–s graph can be determined using
The results are plotted in Fig. 12–15b.
Time. The time can be obtained using the –s graph and
because this equation relates s, and t. For the first segment of
motion, when so
At Therefore,
using these initial conditions for the second segment of motion,
Therefore, at
Ans.
NOTE: The graphical results can be checked in part by calculating slopes.
For example, at
Also, the results can be checked in part by inspection. The –s graph
indicates the initial increase in velocity (acceleration) followed by
constant velocity 1a = 02.
v
a = v1dvds2 = 10150 - 102200 = 2 ms2
.
s = 0,
t =
400
50
+ 4.05 = 12.0 s
s = 400ft,
t = a
s
50
+ 4.05b s
t - 8.05 =
s
50
- 4;
L
t
8.05 s
dt =
L
s
200 m
ds
50
;
dt =
ds
v
=
ds
50
v = 50 fts;
200 ft 6 s … 400 ft;
t = 5 ln[0.212002 + 10] - 5 ln 10 = 8.05 s.
s = 200 ft,
t = (5 ln10.2s + 102 - 5 ln 10) s
L
t
0
dt =
L
s
0
ds
0.2s + 10
dt =
ds
v
=
ds
0.2s + 10
v = (0.2s + 10) fts;
0 … s 6 200 ft;
t = 0,
s = 0
v,
v = dsdt,
v
a = v
dv
ds
= 1502
d
ds
1502 = 0
v = 50 fts
200 ft 6 s … 400 ft;
a = v
dv
ds
= 10.2s + 102
d
ds
10.2s + 102 = 0.04s + 2
v = (0.2s + 10) fts
0 … s 6 200 ft;
a ds = v dv.
v
s = 400 ft.
v
(a)
v (ft/s)
s (ft)
10
50
200 400
v 0.2s 10
v 50
(b)
200 400
s (ft)
a (ft/s2
)
10
2
a 0.04s 2
a 0
Fig. 12–15](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-695-320.jpg)
![26 CHAPTER 12 KINEMATICS OF A PARTICLE
12 PROBLEMS
12–46. A train starts from station A and for the first
kilometer, it travels with a uniform acceleration. Then, for
the next two kilometers, it travels with a uniform speed.
Finally, the train decelerates uniformly for another
kilometer before coming to rest at station B. If the time for
the whole journey is six minutes, draw the graph and
determine the maximum speed of the train.
12–47. The particle travels along a straight line with the
velocity described by the graph. Construct the graph.
a-s
v–t
*12–44. A freight train starts from rest and travels with a
constant acceleration of . After a time it
maintains a constant speed so that when it has
traveled 2000 ft. Determine the time and draw the –t
graph for the motion.
•12–45. If the position of a particle is defined by
, where t is in seconds, construct
the , , and graphs for .
0 … t … 10 s
a-t
v-t
s-t
s = [2 sin (p5)t + 4] m
v
t¿
t = 160 s
t¿
0.5 fts2
12–42. The speed of a train during the first minute has
been recorded as follows:
*12–48. The a–s graph for a jeep traveling along a straight
road is given for the first 300 m of its motion. Construct the
–s graph.At , .
v = 0
s = 0
v
t (s) 0 20 40 60
( )
ms
v 0 16 21 24
Plot the graph, approximating the curve as straight-line
segments between the given points. Determine the total
distance traveled.
12–43. A two-stage missile is fired vertically from rest with
the acceleration shown. In 15 s the first stage A burns out
and the second stage B ignites. Plot the and graphs
which describe the two-stage motion of the missile for
.
0 … t … 20 s
s-t
v-t
v-t
Prob. 12–43
a (m/s2
)
t (s)
15
18
25
20
A
B
s (m)
a (m/s2
)
2
200 300
Prob. 12–48
s (m)
v (m/s)
3 6
4
10
13
v 2s 4
v s 7
Prob. 12–47](https://image.slidesharecdn.com/engineeringmechanics-combinedstaticsdynamics12theditionbyrussellc-220608233255-66a65017/85/Engineering-Mechanics-Combined-Statics-Dynamics-12th-Edition-by-Russell-C-Hibbeler-pdf-697-320.jpg)
Engineering Mechanics--Combined Statics Dynamics, 12th Edition by Russell C. Hibbeler.pdf
- 2. STATICS AND DYNAMICS TWELFTH EDITION ENGINEERING MECHANICS R. C. HIBBELER PRENTICE HALL Upper Saddle River, NJ 07458
- 3. Library of Congress Cataloging-in-Publication Data on File Vice President and Editorial Director, ECS: Marcia Horton Acquisitions Editor: Tacy Quinn Associate Editor: Dee Bernhard Editorial Assistant: Bernadette Marciniak Managing Editor: Scott Disanno Production Editor: Rose Kernan Art Director, Interior and Cover Designer: Kenny Beck Art Editor: Gregory Dulles Media Editor: Daniel Sandin Operations Specialist: Lisa McDowell Manufacturing Manager: Alexis Heydt-Long Senior Marketing Manager: Tim Galligan About the Cover: Background image: Pulleys on Elevator Traction Wheels/Getty/The Image Bank/Sandra Baker Inset image: Elevator in a Business Center/Shutterstock/Khafizor Ivan Harisovich © 2010 by R.C. Hibbeler Published by Pearson Prentice Hall Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, without permission in writing from the publisher. Pearson Prentice Hall™ is a trademark of Pearson Education, Inc. The author and publisher of this book have used their best efforts in preparing this book.These efforts include the development, research, and testing of the theories and programs to determine their effectiveness.The author and publisher shall not be liable in any event for incidental or consequential damages with, or arising out of, the furnishing, performance, or use of these programs. Pearson Education Ltd., London Pearson Education Australia Pty. Ltd., Sydney Pearson Education Singapore, Pte. Ltd. Pearson Education North Asia Ltd., Hong Kong Pearson Education Canada, Inc., Toronto Pearson Educación de Mexico, S.A. de C.V. Pearson Education—Japan, Tokyo Pearson Education Malaysia, Pte. Ltd. Pearson Education, Inc., Upper Saddle River, New Jersey Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 ISBN-10: 0-13-814929-1 ISBN-13: 978-0-13-814929-1
- 4. To the Student With the hope that this work will stimulate an interest in Engineering Mechanics and provide an acceptable guide to its understanding.
- 5. The main purpose of this book is to provide the student with a clear and thorough presentation of the theory and application of engineering mechanics. To achieve this objective, this work has been shaped by the comments and suggestions of hundreds of reviewers in the teaching profession, as well as many of the author’s students. The twelfth edition of this book has been significantly enhanced from the previous edition and it is hoped that both the instructor and student will benefit greatly from these improvements. New Features Fundamental Problems. These problem sets are located just after the example problems. They offer students simple applications of the concepts and, therefore, provide them with the chance to develop their problem-solving skills before attempting to solve any of the standard problems that follow.You may consider these problems as extended examples since they all have partial solutions and answers that are given in the back of the book. Additionally, the fundamental problems offer students an excellent means of studying for exams; and they can be used at a later time as a preparation for the Fundamentals in Engineering Exam. Content Revisions. Each section of the text was carefully reviewed and, in many areas, the material has been redeveloped to better explain the concepts. This has included adding or changing several of the examples in order to provide more emphasis on the applications of the important concepts. Conceptual Problems. Throughout the text, usually at the end of each chapter, there is a set of problems that involve conceptual situations related to the application of the mechanics principles contained in the chapter. These analysis and design problems are intended to engage the students in thinking through a real-life situation as depicted in a photo. They can be assigned after the students have developed some expertise in the subject matter. Additional Photos. The relevance of knowing the subject matter is reflected by the real world applications depicted in over 120 new and updated photos placed throughout the book. These photos are generally used to explain how the principles of mechanics apply to real-world situations. In some sections, photographs have been used to show how engineers must first make an idealized model for analysis and then proceed to draw a free-body diagram of this model in order to apply the theory. New Problems. There are approximately 50%, or about 1600, new problems added to this edition including aerospace and petroleum engineering, and biomechanics applications. Also, this new edition now has approximately 17% more problems than in the previous edition. Hallmark Features Besides the new features mentioned above, other outstanding features that define the contents of the text include the following. Organization and Approach. Each chapter is organized into well-defined sections that contain an explanation of specific topics, illustrative example problems, and a set of homework problems. The topics within each section are placed into subgroups defined by boldface titles.The purpose of this is to present a structured method for introducing each new definition or concept and to make the book convenient for later reference and review. PREFACE
- 6. Chapter Contents. Each chapter begins with an illustration demonstrating a broad- range application of the material within the chapter. A bulleted list of the chapter contents is provided to give a general overview of the material that will be covered. Emphasis on Free-Body Diagrams. Drawing a free-body diagram is particularly important when solving problems, and for this reason this step is strongly emphasized throughout the book. In particular, special sections and examples are devoted to show how to draw free-body diagrams. Specific homework problems have also been added to develop this practice. Procedures for Analysis. A general procedure for analyzing any mechanical problem is presented at the end of the first chapter. Then this procedure is customized to relate to specific types of problems that are covered throughout the book.This unique feature provides the student with a logical and orderly method to follow when applying the theory. The example problems are solved using this outlined method in order to clarify its numerical application. Realize, however, that once the relevant principles have been mastered and enough confidence and judgment have been obtained, the student can then develop his or her own procedures for solving problems. Important Points. This feature provides a review or summary of the most important concepts in a section and highlights the most significant points that should be realized when applying the theory to solve problems. Conceptual Understanding. Through the use of photographs placed throughout the book, theory is applied in a simplified way in order to illustrate some of its more important conceptual features and instill the physical meaning of many of the terms used in the equations.These simplified applications increase interest in the subject matter and better prepare the student to understand the examples and solve problems. Homework Problems. Apart from the Fundamental and Conceptual type problems mentioned previously, other types of problems contained in the book include the following: • Free-Body Diagram Problems. Some sections of the book contain introductory problems that only require drawing the free-body diagram for the specific problems within a problem set. These assignments will impress upon the student the importance of mastering this skill as a requirement for a complete solution of any equilibrium problem. • General Analysis and Design Problems. The majority of problems in the book depict realistic situations encountered in engineering practice. Some of these problems come from actual products used in industry. It is hoped that this realism will both stimulate the student’s interest in engineering mechanics and provide a means for developing the skill to reduce any such problem from its physical description to a model or symbolic representation to which the principles of mechanics may be applied. Throughout the book, there is an approximate balance of problems using either SI or FPS units. Furthermore, in any set, an attempt has been made to arrange the problems in order of increasing difficulty except for the end of chapter review problems, which are presented in random order. • Computer Problems. An effort has been made to include some problems that may be solved using a numerical procedure executed on either a desktop computer or a programmable pocket calculator.The intent here is to broaden the student’s capacity for using other forms of mathematical analysis without sacrificing the time needed to focus on the application of the principles of mechanics. Problems of this type, which either can or must be solved using numerical procedures, are identified by a “square” symbol ( ) preceding the problem number. . PREFACE V
- 7. With so many homework problems in this new edition, they have now been placed in three different categories. Problems that are simply indicated by a problem number have an answer given in the back of the book. If a bullet (•) proceeds the problem number, then a suggestion, key equation, or additional numerical result is given along with the answer. Finally, an asterisk (*) before every fourth problem number indicates a problem without an answer. Accuracy. As with the previous editions, apart from the author, the accuracy of the text and problem solutions has been thoroughly checked by four other parties: Scott Hendricks, Virginia Polytechnic Institute and State University; Karim Nohra, University of South Florida; Kurt Norlin, Laurel Tech Integrated Publishing Services; and finally Kai Beng, a practicing engineer, who in addition to accuracy review provided content development suggestions. Contents Statics The subject of Statics is covered in the first 11 chapters, in which the principles are first applied to simple, then to more complicated situations. In a general sense, each principle is applied first to a particle, then a rigid body subjected to a coplanar system of forces, and finally to three-dimensional force systems acting on a rigid body. Chapter 1 begins with an introduction to mechanics and a discussion of units. The vector properties of a concurrent force system are introduced in Chapter 2.This theory is then applied to the equilibrium of a particle in Chapter 3. Chapter 4 contains a general discussion of both concentrated and distributed force systems and the methods used to simplify them.The principles of rigid-body equilibrium are developed in Chapter 5 and then applied to specific problems involving the equilibrium of trusses, frames, and machines in Chapter 6, and to the analysis of internal forces in beams and cables in Chapter 7.Applications to problems involving frictional forces are discussed in Chapter 8, and topics related to the center of gravity and centroid are treated in Chapter 9. If time permits, sections involving more advanced topics, indicated by stars (★), may be covered. Most of these topics are included in Chapter 10 (area and mass moments of inertia) and Chapter 11 (virtual work and potential energy). Note that this material also provides a suitable reference for basic principles when it is discussed in more advanced courses. Finally, Appendix A provides a review and list of mathematical formulas needed to solve the problems in the book. Alternative Coverage. At the discretion of the instructor, some of the material may be presented in a different sequence with no loss of continuity. For example, it is possible to introduce the concept of a force and all the necessary methods of vector analysis by first covering Chapter 2 and Section 4.2 (the cross product). Then after covering the rest of Chapter 4 (force and moment systems), the equilibrium methods of Chapters 3 and 5 can be discussed. Dynamics The kinematics of a particle is discussed in Chapter 12, followed by a discussion of particle kinetics in Chapter 13 (Equation of Motion), Chapter 14 (Work and Energy), and Chapter 15 (Impulse and Momentum). The concepts of particle dynamics contained in these four chapters are then summarized in a “review” section, and the student is given the chance to identify and solve a variety of problems. A similar sequence of presentation is given for the planar motion of a rigid body: Chapter 16 (Planar Kinematics), Chapter 17 (Equations of Motion), Chapter 18 (Work and Energy), and Chapter 19 (Impulse and Momentum), followed by a summary and review set of problems for these chapters. VI PREFACE
- 8. If time permits, some of the material involving three-dimensional rigid-body motion may be included in the course.The kinematics and kinetics of this motion are discussed in Chapters 20 and 21, respectively. Chapter 22 (Vibrations) may be included if the student has the necessary mathematical background. Sections of the book that are considered to be beyond the scope of the basic dynamics course are indicated by a star (★) and may be omitted.Note that this material also provides a suitable reference for basic principles when it is discussed in more advanced courses. Finally, Appendix A provides a list of mathematical formulas needed to solve the problems in the book, Appendix B provides a brief review of vector analysis, and Appendix C reviews application of the chain rule. Alternative Coverage. At the discretion of the instructor, it is possible to cover Chapters 12 through 19 in the following order with no loss in continuity: Chapters 12 and 16 (Kinematics), Chapters 13 and 17 (Equations of Motion), Chapter 14 and 18 (Work and Energy), and Chapters 15 and 19 (Impulse and Momentum). Acknowledgments The author has endeavored to write this book so that it will appeal to both the student and instructor.Through the years, many people have helped in its development, and I will always be grateful for their valued suggestions and comments.Specifically,I wish to thank the following individuals who have contributed their comments relative to preparing the twelfth edition of this work. Yesh P. Singh, University of Texas–San Antonio Manoj Chopra, University of Central Florida Kathryn McWilliams, University of Saskatchewan Daniel Linzell, Penn State University Larry Banta, West Virginia University Manohar L.Arora, Colorado School of Mines Robert Rennaker, University of Oklahoma Ahmad M. Itani, University of Nevada Per Reinhall, University of Washington Faissal A. Moslehy, University of Central Florida Richard R. Neptune, University of Texas at Austin Robert Rennaker, University of Oklahoma There are a few people that I feel deserve particular recognition. Vince O’Brien, Director of Team-Based Project Management, and Rose Kernan, my production editor for many years, have both provided me with their encouragement and support. Frankly, without their help, this totally revised and enhanced edition would not be possible.Furthermore a long-time friend and associate, Kai Beng Yap, was of great help to me in checking the entire manuscript and helping to prepare the problem solutions.A special note of thanks also goes to Kurt Norlin of Laurel Tech Integrated Publishing Services in this regard. During the production process I am thankful for the assistance of my wife, Conny, and daughter, Mary Ann, with the proofreading and typing needed to prepare the manuscript for publication. Lastly, many thanks are extended to all my students and to members of the teaching profession who have freely taken the time to e-mail me their suggestions and comments. Since this list is too long to mention, it is hoped that those who have given help in this manner will accept this anonymous recognition. I would greatly appreciate hearing from you if at any time you have any comments, suggestions, or problems related to any matters regarding this edition. Russell Charles Hibbeler [email protected] PREFACE VII
- 9. Immediate and specific feedback on wrong answers coach students individually. Specific feedback on common errors helps explain why a particular answer is not correct. Hints provide individualized coaching. Skip the hints you don’t need and access only the ones that you need, for the most efficient path to the correct solution. Resources to Accompany Engineering Mechanics: Statics, Twelfth Edition MasteringEngineering is the most technologically advanced online tutorial and homework system. It tutors students individually while providing instructors with rich teaching diagnostics. MasteringEngineering is built upon the same platform as MasteringPhysics, the only online physics homework system with research showing that it improves student learning.A wide variety of published papers based on NSF-sponsored research and tests illustrate the benefits of MasteringEngineering.To read these papers, please visit www.masteringengineering.com. MasteringEngineering for Students MasteringEngineering improves understanding. As an Instructor-assigned homework and tutorial system, MasteringEngineering is designed to provide students with customized coaching and individualized feedback to help improve problem-solving skills. Students complete homework efficiently and effectively with tutorials that provide targeted help.
- 10. Contact your Pearson Prentice Hall representative for more information. One click compiles all your favorite teaching diagnostics— the hardest problem, class grade distribution, even which students are spending the most or the least time on homework. Color-coded gradebook instantly identifies students in difficulty, or challenging areas for your class. MasteringEngineering for Instructors Incorporate dynamic homework into your course with automatic grading and adaptive tutoring. Choose from a wide variety of stimulating problems, including free-body diagram drawing capabilities, algorithmically-generated problem sets, and more. MasteringEngineering emulates the instructor’s office-hour environment, coaching students on problem-solving techniques by asking students simpler sub-questions.
- 11. X RESOURCES FOR INSTRUCTORS Resources for Instructors • Instructor’s Solutions Manuals. These supplements provide complete solutions supported by problem statements and problem figures.The twelfth edition manuals were revised to improve readability and was triple accuracy checked. • Instructor’s Resource CD-ROM. Visual resources to accompany the text are located on this CD as well as on the Pearson Higher Education website:www.pearsonhighered.com.If you are in need of a login and password for this site,please contact your local Pearson representative.Visual resources include all art from the text, available in PowerPoint slide and JPEG format. • Video Solutions. Developed by Professor Edward Berger, University of Virginia, video solutions are located on the Companion Website for the text and offer step-by-step solution walkthroughs of representative homework problems from each section of the text. Make efficient use of class time and office hours by showing students the complete and concise problem-solving approaches that they can access any time and view at their own pace. The videos are designed to be a flexible resource to be used however each instructor and student prefers. A valuable tutorial resource, the videos are also helpful for student self-evaluation as students can pause the videos to check their understanding and work alongside the video. Access the videos at www.prenhall.com/hibbeler and follow the links for the Engineering Mechanics: Statics and Dynamics,Twelfth Edition text. Resources for Students • Statics and Dynamics Study Packs. These supplements contains chapter-by-chapter study materials, a Free-Body Diagram Workbook and access to the Companion Website where additional tutorial resources are located. • Companion Website. The Companion Website, located at www.prenhall.com/hibbeler, includes opportunities for practice and review including: • Video Solutions—Complete, step-by-step solution walkthroughs of representative homework problems from each section.Videos offer: • Fully worked Solutions—Showing every step of representative homework problems, to help students make vital connections between concepts. • Self-paced Instruction—Students can navigate each problem and select, play, rewind, fast-forward, stop, and jump-to- sections within each problem’s solution. • 24/7 Access—Help whenever students need it with over 20 hours of helpful review. An access code for the Engineering Mechanics: Statics and Dynamics,Twelfth Edition website is included inside the Statics Study Pack or the Dynamics Study Pack. To redeem the code and gain access to the site, go to www.prenhall.com/hibbeler and follow the directions on the access code card.Access can also be purchased directly from the site. • Statics and Dynamics Practice Problems Workbooks. These workbooks contain additional worked problems. The problems are partially solved and are designed to help guide students through difficult topics. Ordering Options The Statics Study Pack and MasteringEngineering resources are available as stand-alone items for student purchase and are also available packaged with the texts.The ISBN for each valuepack is as follows: • Engineering Mechanics: Statics with Study Pack: 0-13-246200-1 • Engineering Mechanics: Statics with Study Pack and MasteringEngineering Student Access Card: 0-13-701629-8 The Dynamics Study Pack and MasteringEngineering resources are available as stand-alone items for student purchase and are also available packaged with the texts.The ISBN for each valuepack is as follows: • Engineering Mechanics: Dynamics with Study Pack: 0-13-700239-4 • Engineering Mechanics: Dynamics with Study Pack and MasteringEngineering Student Access Card: 0-13-701630-1 Custom Solutions New options for textbook customization are now available for Engineering Mechanics,Twelfth Edition.Please contact your local Pearson/Prentice Hall representative for details.
- 12. 1 General Principles 3 Chapter Objectives 3 1.1 Mechanics 3 1.2 Fundamental Concepts 4 1.3 Units of Measurement 7 1.4 The International System of Units 9 1.5 Numerical Calculations 10 1.6 General Procedure for Analysis 12 2 Force Vectors 17 Chapter Objectives 17 2.1 Scalars and Vectors 17 2.2 Vector Operations 18 2.3 Vector Addition of Forces 20 2.4 Addition of a System of Coplanar Forces 32 2.5 Cartesian Vectors 43 2.6 Addition of Cartesian Vectors 46 2.7 Position Vectors 56 2.8 Force Vector Directed Along a Line 59 2.9 Dot Product 69 3 Equilibrium of a Particle 85 Chapter Objectives 85 3.1 Condition for the Equilibrium of a Particle 85 3.2 The Free-Body Diagram 86 3.3 Coplanar Force Systems 89 3.4 Three-Dimensional Force Systems 103 4 Force System Resultants 117 Chapter Objectives 117 4.1 Moment of a Force—Scalar Formulation 117 4.2 Cross Product 121 4.3 Moment of a Force—Vector Formulation 124 4.4 Principle of Moments 128 4.5 Moment of a Force about a Specified Axis 139 4.6 Moment of a Couple 148 4.7 Simplification of a Force and Couple System 160 4.8 Further Simplification of a Force and Couple System 170 4.9 Reduction of a Simple Distributed Loading 183 5 Equilibrium of a Rigid Body 199 Chapter Objectives 199 5.1 Conditions for Rigid-Body Equilibrium 199 5.2 Free-Body Diagrams 201 5.3 Equations of Equilibrium 214 5.4 Two- and Three-Force Members 224 5.5 Free-Body Diagrams 237 5.6 Equations of Equilibrium 242 5.7 Constraints and Statical Determinacy 243 CONTENTS xi
- 13. 6 Structural Analysis 263 Chapter Objectives 263 6.1 Simple Trusses 263 6.2 The Method of Joints 266 6.3 Zero-Force Members 272 6.4 The Method of Sections 280 6.5 Space Trusses 290 6.6 Frames and Machines 294 7 Internal Forces 329 Chapter Objectives 329 7.1 Internal Forces Developed in Structural Members 329 7.2 Shear and Moment Equations and Diagrams 345 7.3 Relations between Distributed Load, Shear, and Moment 354 7.4 Cables 365 8 Friction 387 Chapter Objectives 387 8.1 Characteristics of Dry Friction 387 8.2 Problems Involving Dry Friction 392 8.3 Wedges 412 8.4 Frictional Forces on Screws 414 8.5 Frictional Forces on Flat Belts 421 8.6 Frictional Forces on Collar Bearings, Pivot Bearings, and Disks 429 8.7 Frictional Forces on Journal Bearings 432 8.8 Rolling Resistance 434 9 Center of Gravity and Centroid 447 Chapter Objectives 447 9.1 Center of Gravity, Center of Mass, and the Centroid of a Body 447 9.2 Composite Bodies 470 9.3 Theorems of Pappus and Guldinus 484 9.4 Resultant of a General Distributed Loading 493 9.5 Fluid Pressure 494 10 Moments of Inertia 511 Chapter Objectives 511 10.1 Definition of Moments of Inertia for Areas 511 10.2 Parallel-Axis Theorem for an Area 512 10.3 Radius of Gyration of an Area 513 10.4 Moments of Inertia for Composite Areas 522 10.5 Product of Inertia for an Area 530 10.6 Moments of Inertia for an Area about Inclined Axes 534 10.7 Mohr’s Circle for Moments of Inertia 537 10.8 Mass Moment of Inertia 545 XII CONTENTS
- 14. 11 Virtual Work 563 Chapter Objectives 563 11.1 Definition of Work 563 11.2 Principle of Virtual Work 565 11.3 Principle of Virtual Work for a System of Connected Rigid Bodies 567 11.4 Conservative Forces 579 11.5 Potential Energy 580 11.6 Potential-Energy Criterion for Equilibrium 582 11.7 Stability of Equilibrium Configuration 583 Appendix A. Mathematical Review and Expressions 598 Fundamental Problems Partial Solutions and Answers 603 Answers to Selected Problems 620 Index 650 12 Kinematics of a Particle 3 Chapter Objectives 3 12.1 Introduction 3 12.2 Rectilinear Kinematics: Continuous Motion 5 12.3 Rectilinear Kinematics: Erratic Motion 19 12.4 General Curvilinear Motion 32 12.5 Curvilinear Motion: Rectangular Components 34 12.6 Motion of a Projectile 39 12.7 Curvilinear Motion: Normal and Tangential Components 53 12.8 Curvilinear Motion: Cylindrical Components 67 12.9 Absolute Dependent Motion Analysis of Two Particles 81 12.10 Relative-Motion of Two Particles Using Translating Axes 87 13 Kinetics of a Particle: Force and Acceleration 107 Chapter Objectives 107 13.1 Newton’s Second Law of Motion 107 13.2 The Equation of Motion 110 13.3 Equation of Motion for a System of Particles 112 13.4 Equations of Motion: Rectangular Coordinates 114 13.5 Equations of Motion: Normal and Tangential Coordinates 131 13.6 Equations of Motion: Cylindrical Coordinates 144 *13.7 Central-Force Motion and Space Mechanics 155 14 Kinetics of a Particle: Work and Energy 169 Chapter Objectives 169 14.1 The Work of a Force 169 14.2 Principle of Work and Energy 174 14.3 Principle of Work and Energy for a System of Particles 176 CONTENTS XIII
- 15. XIV CONTENTS 16.3 Rotation about a Fixed Axis 314 16.4 Absolute Motion Analysis 329 16.5 Relative-Motion Analysis: Velocity 337 16.6 Instantaneous Center of Zero Velocity 351 16.7 Relative-Motion Analysis: Acceleration 363 16.8 Relative-Motion Analysis using Rotating Axes 377 17 Planar Kinetics of a Rigid Body: Force and Acceleration 395 Chapter Objectives 395 17.1 Moment of Inertia 395 17.2 Planar Kinetic Equations of Motion 409 17.3 Equations of Motion: Translation 412 17.4 Equations of Motion: Rotation about a Fixed Axis 425 17.5 Equations of Motion: General Plane Motion 440 18 Planar Kinetics of a Rigid Body: Work and Energy 455 Chapter Objectives 455 18.1 Kinetic Energy 455 18.2 The Work of a Force 458 18.3 The Work of a Couple 460 18.4 Principle of Work and Energy 462 18.5 Conservation of Energy 477 14.4 Power and Efficiency 192 14.5 Conservative Forces and Potential Energy 201 14.6 Conservation of Energy 205 15 Kinetics of a Particle: Impulse and Momentum 221 Chapter Objectives 221 15.1 Principle of Linear Impulse and Momentum 221 15.2 Principle of Linear Impulse and Momentum for a System of Particles 228 15.3 Conservation of Linear Momentum for a System of Particles 236 15.4 Impact 248 15.5 Angular Momentum 262 15.6 Relation Between Moment of a Force and Angular Momentum 263 15.7 Principle of Angular Impulse and Momentum 266 15.8 Steady Flow of a Fluid Stream 277 *15.9 Propulsion with Variable Mass 282 Review 1. Kinematics and Kinetics of a Particle 298 16 Planar Kinematics of a Rigid Body 311 Chapter Objectives 311 16.1 Planar Rigid-Body Motion 311 16.2 Translation 313
- 16. CONTENTS XV 19 Planar Kinetics of a Rigid Body: Impulse and Momentum 495 Chapter Objectives 495 19.1 Linear and Angular Momentum 495 19.2 Principle of Impulse and Momentum 501 19.3 Conservation of Momentum 517 *19.4 Eccentric Impact 521 Review 2. Planar Kinematics and Kinetics of a Rigid Body 534 20 Three-Dimensional Kinematics of a Rigid Body 549 Chapter Objectives 549 20.1 Rotation About a Fixed Point 549 *20.2 The Time Derivative of a Vector Measured from Either a Fixed or Translating-Rotating System 552 20.3 General Motion 557 *20.4 Relative-Motion Analysis Using Translating and Rotating Axes 566 21 Three-Dimensional Kinetics of a Rigid Body 579 Chapter Objectives 579 *21.1 Moments and Products of Inertia 579 21.2 Angular Momentum 589 21.3 Kinetic Energy 592 *21.4 Equations of Motion 600 *21.5 Gyroscopic Motion 614 21.6 Torque-Free Motion 620 22 Vibrations 631 Chapter Objectives 631 *22.1 Undamped Free Vibration 631 *22.2 Energy Methods 645 *22.3 Undamped Forced Vibration 651 *22.4 Viscous Damped Free Vibration 655 *22.5 Viscous Damped Forced Vibration 658 *22.6 Electrical Circuit Analogs 661 Appendix A. Mathematical Expressions 670 B. Vector Analysis 672 C. The Chain Rule 677 Fundamental Problems Partial Solutions and Answers 680 Answers to Selected Problems 699 Index 725
- 17. Credits Chapter 1, Space Shuttle Discovery lifts off from launch pad 39-1 at the Kennedy Space Center, May 31, 2008 in Cape Canaveral, Florida.The Space Shuttle is carrying the main unit of Japan’s Kibo science lab to the International Space Station. Getty Images. Chapter 1 text, Astronaut floating in space.Alamy Images Royalty Free. Chapter 2, Erasmus suspension bridge, Rotterdam, Holland. Alamy Images. Chapter 3, Prefabricated section of a building being hoisted into place by a large crane.Alamy Images. Chapter 4, Engineer turns screws with spanner, close-up of hands. Getty Images/ Digital Vision. Chapter 5, Lifeboat being lifted by a mobile hydraulic crane, Grimsby, Humberside, North Lincolnshire, England, UK.Alamy Images. Chapter 6, Fog lifting off the river running under a Pratt Truss steel bridge over the St. John River, New Brunswick, Canada at Perth Andover.Alamy Images. Chapter 7, Reinforcing rods encased in concrete. Russ C. Hibbeler. Chapter 8, Calliper brake on bicycle.Alamy Images. Chapter 9, Water tower, Harmony, Bluff County, Minnesota.Alamy Images. Chapter 10, Steel framework at construction site. Corbis Royalty Free. Chapter 11, Crane boom. Getty Images Inc.—Stone Allstock. Chapter 12, The United States Navy Blue Angels perform in an air show as part of San Francisco’s Fleet Week celebration. ©Roger Ressmeyer/CORBIS.All Rights Reserved. Chapter 13, Orange juice factory, elevated view. Getty Images. Chapter 14, White roller coaster of the Mukogaokayuen ground, Kanagawa. ©Yoshio Kosaka/CORBIS.All Rights Reserved. Chapter 15, Close-up of golf club hitting a golf ball off the tee.Alamy Images Royalty Free. Chapter 16, Windmills in Livermore, part of an extensive wind farm, an alternative source of electrical power, California, United States. Brent Winebrenner/Lonely Planet Images/ Photo 20-20. Chapter 17, Burnout drag racing car at Santa Pod Raceway, England.Alamy Images. Chapter 18, Drilling rig. Getty Images. Chapter 19, NASA shuttle docking with the International Space Station. Dennis Hallinan/Alamy Images. Chapter 20, Man watching robotic welding. ©Ted Horowitz/CORBIS.All Rights Reserved. Chapter 21, A spinning Calypso ride provides a blur of bright colors at Waldameer Park and Water World in Erie, Pennsylvania. Jim Cole/Alamy Images. Chapter 22, A train track and train wheel give great perspective to the size and power of railway transportation. Joe Belanger/Alamy Images
- 19. The design of this rocket and gantry structure requires a basic knowledge of both statics and dynamics, which form the subject matter of engineering mechanics.
- 20. General Principles CHAPTER OBJECTIVES • To provide an introduction to the basic quantities and idealizations of mechanics. • To give a statement of Newton’s Laws of Motion and Gravitation. • To review the principles for applying the SI system of units. • To examine the standard procedures for performing numerical calculations. • To present a general guide for solving problems. 1.1 Mechanics Mechanics is a branch of the physical sciences that is concerned with the state of rest or motion of bodies that are subjected to the action of forces. In general, this subject can be subdivided into three branches: rigid-body mechanics, deformable-body mechanics, and fluid mechanics. In this book we will study rigid-body mechanics since it is a basic requirement for the study of the mechanics of deformable bodies and the mechanics of fluids. Furthermore, rigid-body mechanics is essential for the design and analysis of many types of structural members,mechanical components,or electrical devices encountered in engineering. Rigid-body mechanics is divided into two areas: statics and dynamics. Statics deals with the equilibrium of bodies, that is, those that are either at rest or move with a constant velocity; whereas dynamics is concerned with the accelerated motion of bodies. We can consider statics as a special case of dynamics, in which the acceleration is zero; however, statics deserves separate treatment in engineering education since many objects are designed with the intention that they remain in equilibrium. 1
- 21. 4 CHAPTER 1 GENERAL PRINCIPLES 1 Historical Development. The subject of statics developed very early in history because its principles can be formulated simply from measurements of geometry and force. For example, the writings of Archimedes (287–212 B.C.) deal with the principle of the lever. Studies of the pulley, inclined plane, and wrench are also recorded in ancient writings—at times when the requirements for engineering were limited primarily to building construction. Since the principles of dynamics depend on an accurate measurement of time, this subject developed much later. Galileo Galilei (1564–1642) was one of the first major contributors to this field. His work consisted of experiments using pendulums and falling bodies. The most significant contributions in dynamics, however, were made by Isaac Newton (1642–1727), who is noted for his formulation of the three fundamental laws of motion and the law of universal gravitational attraction. Shortly after these laws were postulated, important techniques for their application were developed by such notables as Euler, D’Alembert, Lagrange, and others. 1.2 Fundamental Concepts Before we begin our study of engineering mechanics, it is important to understand the meaning of certain fundamental concepts and principles. Basic Quantities. The following four quantities are used throughout mechanics. Length. Length is used to locate the position of a point in space and thereby describe the size of a physical system. Once a standard unit of length is defined, one can then use it to define distances and geometric properties of a body as multiples of this unit. Time. Time is conceived as a succession of events. Although the principles of statics are time independent, this quantity plays an important role in the study of dynamics. Mass. Mass is a measure of a quantity of matter that is used to compare the action of one body with that of another.This property manifests itself as a gravitational attraction between two bodies and provides a measure of the resistance of matter to a change in velocity. Force. In general, force is considered as a “push” or “pull” exerted by one body on another. This interaction can occur when there is direct contact between the bodies, such as a person pushing on a wall, or it can occur through a distance when the bodies are physically separated. Examples of the latter type include gravitational, electrical, and magnetic forces. In any case, a force is completely characterized by its magnitude, direction, and point of application.
- 22. 1.2 FUNDAMENTAL CONCEPTS 5 1 Idealizations. Models or idealizations are used in mechanics in order to simplify application of the theory. Here we will consider three important idealizations. Particle. A particle has a mass, but a size that can be neglected. For example, the size of the earth is insignificant compared to the size of its orbit, and therefore the earth can be modeled as a particle when studying its orbital motion.When a body is idealized as a particle, the principles of mechanics reduce to a rather simplified form since the geometry of the body will not be involved in the analysis of the problem. Rigid Body. A rigid body can be considered as a combination of a large number of particles in which all the particles remain at a fixed distance from one another, both before and after applying a load. This model is important because the material properties of any body that is assumed to be rigid will not have to be considered when studying the effects of forces acting on the body. In most cases the actual deformations occurring in structures, machines, mechanisms, and the like are relatively small, and the rigid-body assumption is suitable for analysis. Concentrated Force. A concentrated force represents the effect of a loading which is assumed to act at a point on a body. We can represent a load by a concentrated force, provided the area over which the load is applied is very small compared to the overall size of the body.An example would be the contact force between a wheel and the ground. A Three forces act on the hook at A. Since these forces all meet at a point, then for any force analysis, we can assume the hook to be represented as a particle. Steel is a common engineering material that does not deform very much under load.Therefore,we can consider this railroad wheel to be a rigid body acted upon by the concentrated force of the rail.
- 23. 6 CHAPTER 1 GENERAL PRINCIPLES 1 Newton’s Three Laws of Motion. Engineering mechanics is formulated on the basis of Newton’s three laws of motion, the validity of which is based on experimental observation. These laws apply to the motion of a particle as measured from a nonaccelerating reference frame.They may be briefly stated as follows. First Law. A particle originally at rest, or moving in a straight line with constant velocity, tends to remain in this state provided the particle is not subjected to an unbalanced force, Fig. 1–1a. Second Law. A particle acted upon by an unbalanced force F experiences an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force, Fig. 1–1b.* If F is applied to a particle of mass m, this law may be expressed mathematically as (1–1) F = ma Third Law. The mutual forces of action and reaction between two particles are equal, opposite, and collinear, Fig. 1–1c. *Stated another way, the unbalanced force acting on the particle is proportional to the time rate of change of the particle’s linear momentum. Equilibrium v F2 F1 F3 (a) Accelerated motion a F (b) Action – reaction force of A on B force of B on A F F A B (c) Fig. 1–1
- 24. 1.3 UNITS OF MEASUREMENT 7 1 Weight. According to Eq. 1–2, any two particles or bodies have a mutual attractive (gravitational) force acting between them. In the case of a particle located at or near the surface of the earth, however, the only gravitational force having any sizable magnitude is that between the earth and the particle. Consequently, this force, termed the weight, will be the only gravitational force considered in our study of mechanics. From Eq.1–2,we can develop an approximate expression for finding the weight W of a particle having a mass . If we assume the earth to be a nonrotating sphere of constant density and having a mass ,then if r is the distance between the earth’s center and the particle, we have Letting yields (1–3) By comparison with , we can see that g is the acceleration due to gravity. Since it depends on r, then the weight of a body is not an absolute quantity.Instead,its magnitude is determined from where the measurement was made. For most engineering calculations, however, g is determined at sea level and at a latitude of 45°,which is considered the“standard location.” 1.3 Units of Measurement The four basic quantities—length, time, mass, and force—are not all independent from one another;in fact,they are related by Newton’s second law of motion, . Because of this, the units used to measure these quantities cannot all be selected arbitrarily. The equality is maintained only if three of the four units, called base units, are defined and the fourth unit is then derived from the equation. F = ma F = ma F = ma W = mg g = GMer2 W = G mMe r2 m2 = Me m1 = m F = force of gravitation between the two particles G = universal constant of gravitation; according to experimental evidence, G = 66.73(10-12 ) m3 (kg # s2 ) m1, m2 = mass of each of the two particles r = distance between the two particles Newton’s Law of Gravitational Attraction. Shortly after formulating his three laws of motion,Newton postulated a law governing the gravitational attraction between any two particles.Stated mathematically, (1–2) where F = G m1m2 r2 The astronaut is weightless, for all practical purposes, since she is far removed from the gravitational field of the earth.
- 25. 8 CHAPTER 1 GENERAL PRINCIPLES 1 SI Units. The International System of units, abbreviated SI after the French “Système International d’Unités,” is a modern version of the metric system which has received worldwide recognition. As shown in Table 1–1, the SI system defines length in meters (m), time in seconds (s), and mass in kilograms (kg). The unit of force, called a newton (N), is derived from .Thus, 1 newton is equal to a force required to give 1 kilogram of mass an acceleration of . If the weight of a body located at the “standard location” is to be determined in newtons, then Eq. 1–3 must be applied. Here measurements give ; however, for calculations, the value will be used.Thus, (1–4) Therefore, a body of mass 1 kg has a weight of 9.81 N, a 2-kg body weighs 19.62 N, and so on, Fig. 1–2a. U.S. Customary. In the U.S.Customary system of units (FPS) length is measured in feet (ft), time in seconds (s), and force in pounds (lb), Table 1–1.The unit of mass, called a slug, is derived from . Hence, 1 slug is equal to the amount of matter accelerated at when acted upon by a force of . Therefore, if the measurements are made at the “standard location,” where , then from Eq. 1–3, (1–5) And so a body weighing 32.2 lb has a mass of 1 slug, a 64.4-lb body has a mass of 2 slugs, and so on, Fig. 1–2b. m = W g (g = 32.2 fts2 ) g = 32.2 fts2 1 lb (slug = lb # s2 ft) 1 fts2 F = ma W = mg (g = 9.81 ms2 ) g = 9.81 ms2 g = 9.806 65 ms2 1 ms2 (N = kg # ms2 ) F = ma TABLE 1–1 Systems of Units Name Length Time Mass Force International System of Units SI meter m second s kilogram kg a kg # m s2 b N newton* U.S. Customary FPS foot ft second s a lb # s2 ft b slug* pound lb *Derived unit. 9.81 N 1 kg (a) 32.2 lb 1 slug (b) Fig. 1–2
- 26. 1.4 THE INTERNATIONAL SYSTEM OF UNITS 9 1 Conversion of Units. Table 1–2 provides a set of direct conversion factors between FPS and SI units for the basic quantities. Also, in the FPS system, recall that 1 ft = 12 in. (inches), 5280 ft = 1 mi (mile), 1000 lb = 1 kip (kilo-pound), and 2000 lb = 1 ton. 1.4 The International System of Units The SI system of units is used extensively in this book since it is intended to become the worldwide standard for measurement. Therefore, we will now present some of the rules for its use and some of its terminology relevant to engineering mechanics. Prefixes. When a numerical quantity is either very large or very small, the units used to define its size may be modified by using a prefix. Some of the prefixes used in the SI system are shown in Table 1–3. Each represents a multiple or submultiple of a unit which, if applied successively, moves the decimal point of a numerical quantity to every third place.* For example, 4 000 000 N = 4 000 kN (kilo-newton) = 4 MN (mega-newton), or 0.005 m = 5 mm (milli-meter). Notice that the SI system does not include the multiple deca (10) or the submultiple centi (0.01), which form part of the metric system. Except for some volume and area measurements, the use of these prefixes is to be avoided in science and engineering. * The kilogram is the only base unit that is defined with a prefix. TABLE 1–2 Conversion Factors Quantity Unit of Measurement (FPS) Equals Unit of Measurement (SI) Force lb 4.448 N Mass slug 14.59 kg Length ft 0.304 8 m TABLE 1–3 Prefixes Exponential Form Prefix SI Symbol Multiple 1 000 000 000 109 giga G 1 000 000 106 mega M 1 000 103 kilo k Submultiple 0.001 10–3 milli m 0.000 001 10–6 micro m 0.000 000 001 10–9 nano n
- 27. 10 CHAPTER 1 GENERAL PRINCIPLES 1 Rules for Use. Here are a few of the important rules that describe the proper use of the various SI symbols: • Quantities defined by several units which are multiples of one another are separated by a dot to avoid confusion with prefix notation, as indicated by . Also, (meter-second), whereas ms (milli-second). • The exponential power on a unit having a prefix refers to both the unit and its prefix. For example, . Likewise, mm2 represents . • With the exception of the base unit the kilogram, in general avoid the use of a prefix in the denominator of composite units. For example, do not write , but rather ; also, should be written as . • When performing calculations, represent the numbers in terms of their base or derived units by converting all prefixes to powers of 10. The final result should then be expressed using a single prefix. Also, after calculation, it is best to keep numerical values between 0.1 and 1000; otherwise, a suitable prefix should be chosen. For example, 1.5 Numerical Calculations Numerical work in engineering practice is most often performed by using handheld calculators and computers. It is important, however, that the answers to any problem be reported with both justifiable accuracy and appropriate significant figures. In this section we will discuss these topics together with some other important aspects involved in all engineering calculations. Dimensional Homogeneity. The terms of any equation used to describe a physical process must be dimensionally homogeneous; that is, each term must be expressed in the same units. Provided this is the case, all the terms of an equation can then be combined if numerical values are substituted for the variables. Consider, for example, the equation , where, in SI units, is the position in meters, m, is time in seconds, s, is velocity in and is acceleration in . Regardless of how this equation is evaluated, it maintains its dimensional homogeneity. In the form stated, each of the three terms is expressed in meters or solving for , , the terms are each expressed in units of . ms2 [ms2 , ms2 , (ms)s] a = 2st2 - 2vt a [m,(ms) s, (ms2 ) s2 ,] ms2 a ms v t s s = vt + 1 2 at2 = 3000(10-6 ) N # m = 3(10-3 ) N # m = 3 mN # m (50 kN)(60 nm) = [50(103 ) N][60(10-9 ) m] Mmkg mmg kNm Nmm (mm)2 = mm # mm mN2 = (mN)2 = mN # mN m # s N = kg # ms2 = kg # m # s-2 Computers are often used in engineering for advanced design and analysis.
- 28. 1.5 NUMERICAL CALCULATIONS 11 1 Keep in mind that problems in mechanics always involve the solution of dimensionally homogeneous equations, and so this fact can then be used as a partial check for algebraic manipulations of an equation. Significant Figures. The number of significant figures contained in any number determines the accuracy of the number. For instance, the number 4981 contains four significant figures. However, if zeros occur at the end of a whole number, it may be unclear as to how many significant figures the number represents.For example,23 400 might have three (234), four (2340), or five (23 400) significant figures.To avoid these ambiguities, we will use engineering notation to report a result. This requires that numbers be rounded off to the appropriate number of significant digits and then expressed in multiples of (103), such as (103), (106), or (10–9). For instance,if 23 400 has five significant figures,it is written as 23.400(103),but if it has only three significant figures, it is written as 23.4(103). If zeros occur at the beginning of a number that is less than one, then the zeros are not significant. For example, 0.00821 has three significant figures. Using engineering notation, this number is expressed as 8.21(10–3). Likewise, 0.000582 can be expressed as 0.582(10–3) or 582(10–6). Rounding Off Numbers. Rounding off a number is necessary so that the accuracy of the result will be the same as that of the problem data. As a general rule, any numerical figure ending in five or greater is rounded up and a number less than five is rounded down. The rules for rounding off numbers are best illustrated by examples. Suppose the number 3.5587 is to be rounded off to three significant figures. Because the fourth digit (8) is greater than 5, the third number is rounded up to 3.56. Likewise 0.5896 becomes 0.590 and 9.3866 becomes 9.39. If we round off 1.341 to three significant figures, because the fourth digit (1) is less than 5, then we get 1.34. Likewise 0.3762 becomes 0.376 and 9.871 becomes 9.87. There is a special case for any number that has a 5 with zeroes following it. As a general rule, if the digit preceding the 5 is an even number, then this digit is not rounded up. If the digit preceding the 5 is an odd number, then it is rounded up. For example, 75.25 rounded off to three significant digits becomes 75.2, 0.1275 becomes 0.128, and 0.2555 becomes 0.256. Calculations. When a sequence of calculations is performed, it is best to store the intermediate results in the calculator. In other words, do not round off calculations until expressing the final result. This procedure maintains precision throughout the series of steps to the final solution. In this text we will generally round off the answers to three significant figures since most of the data in engineering mechanics, such as geometry and loads, may be reliably measured to this accuracy.
- 29. 12 CHAPTER 1 GENERAL PRINCIPLES 1 1.6 General Procedure for Analysis The most effective way of learning the principles of engineering mechanics is to solve problems. To be successful at this, it is important to always present the work in a logical and orderly manner, as suggested by the following sequence of steps: • Read the problem carefully and try to correlate the actual physical situation with the theory studied. • Tabulate the problem data and draw any necessary diagrams. • Apply the relevant principles, generally in mathematical form.When writing any equations, be sure they are dimensionally homogeneous. • Solve the necessary equations, and report the answer with no more than three significant figures. • Study the answer with technical judgment and common sense to determine whether or not it seems reasonable. • Statics is the study of bodies that are at rest or move with constant velocity. • A particle has a mass but a size that can be neglected. • A rigid body does not deform under load. • Concentrated forces are assumed to act at a point on a body. • Newton’s three laws of motion should be memorized. • Mass is measure of a quantity of matter that does not change from one location to another. • Weight refers to the gravitational attraction of the earth on a body or quantity of mass. Its magnitude depends upon the elevation at which the mass is located. • In the SI system the unit of force, the newton, is a derived unit. The meter, second, and kilogram are base units. • Prefixes G, M, k, m, , and n are used to represent large and small numerical quantities. Their exponential size should be known, along with the rules for using the SI units. • Perform numerical calculations with several significant figures, and then report the final answer to three significant figures. • Algebraic manipulations of an equation can be checked in part by verifying that the equation remains dimensionally homogeneous. • Know the rules for rounding off numbers. m When solving problems, do the work as neatly as possible. Being neat will stimulate clear and orderly thinking, and vice versa. Important Points
- 30. 1.6 GENERAL PROCEDURE FOR ANALYSIS 13 1 Convert to How many is this? SOLUTION Since and , the factors of conversion are arranged in the following order, so that a cancellation of the units can be applied: Ans. From Table 1–2, 1 ft = 0.3048 m.Thus, Ans. NOTE: Remember to round off the final answer to three significant figures. = 1.82 fts 0.556 ms = a 0.556 m s b a 1 ft 0.3048 m b = 2000 m 3600 s = 0.556 ms 2 kmh = 2 km h a 1000 m km b a 1 h 3600 s b 1 h = 3600 s 1 km = 1000 m fts ms 2 kmh Convert the quantities and to appropriate SI units. SOLUTION Using Table 1–2, . Ans. Since and , then Ans. = 26.8 Mgm3 = 26.8(103 ) kgm3 52 slugft3 = 52 slug ft3 a 14.59 kg 1 slug b a 1 ft 0.304 8 m b 3 1 ft = 0.304 8 m 1 slug = 14.593 8 kg = 1334.5 N # s = 1.33 kN # s 300 lb # s = 300 lb # sa 4.448 N 1 lb b 1 lb = 4.448 2 N 52 slugft3 300 lb # s EXAMPLE 1.1 EXAMPLE 1.2
- 31. 14 CHAPTER 1 GENERAL PRINCIPLES 1 Evaluate each of the following and express with SI units having an appropriate prefix: (a) (50 mN)(6 GN), (b) (400 mm)(0.6 MN)2, (c) . SOLUTION First convert each number to base units, perform the indicated operations, then choose an appropriate prefix. Part (a) Ans. NOTE: Keep in mind the convention . Part (b) Ans. We can also write Ans. Part (c) Ans. = 50 kN3 kg = 50(109 ) N3 a 1 kN 103 N b 3 1 kg = 50(109 ) N3 kg 45 MN3 900 Gg = 45(106 N)3 900(106 ) kg = 0.144 m # MN2 144(109 ) m # N2 = 144(109 ) m # N2 a 1 MN 106 N b a 1 MN 106 N b = 144 Gm # N2 = 144(109 ) m # N2 = [400(10-3 ) m][0.36(1012 ) N2 ] (400 mm)(0.6 MN)2 = [400(10-3 ) m][0.6(106 ) N]2 kN2 = (kN)2 = 106 N2 = 300 kN2 = 300(106 ) N2 a 1 kN 103 N b a 1 kN 103 N b = 300(106 ) N2 (50 mN)(6 GN) = [50(10-3 ) N][6(109 ) N] 45 MN3 900 Gg EXAMPLE 1.3
- 32. PROBLEMS 15 1 PROBLEMS 1–1. Round off the following numbers to three significant figures: (a) 4.65735 m, (b) 55.578 s, (c) 4555 N, and (d) 2768 kg. 1–2. Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) , (b) , (c) , and (d) . 1–3. Represent each of the following quantities in the correct SI form using an appropriate prefix: (a) 0.000431 kg, (b) , and (c) 0.00532 km. *1–4. Represent each of the following combinations of units in the correct SI form: (a) , (b) , and (c) . 1–5. Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) ,(b) ,and (c) . 1–6. Represent each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) 45 320 kN, (b) , and (c) 0.005 63 mg. 1–7. A rocket has a mass of slugs on earth. Specify (a) its mass in SI units and (b) its weight in SI units. If the rocket is on the moon, where the acceleration due to gravity is , determine to three significant figures (c) its weight in SI units and (d) its mass in SI units. *1–8. If a car is traveling at , determine its speed in kilometers per hour and meters per second. 1–9. The pascal (Pa) is actually a very small unit of pressure. To show this, convert to . Atmospheric pressure at sea level is in2. How many pascals is this? 1–10. What is the weight in newtons of an object that has a mass of: (a) 10 kg, (b) 0.5 g, and (c) 4.50 Mg? Express the result to three significant figures. Use an appropriate prefix. 1–11. Evaluate each of the following to three significant figures and express each answer in Sl units using an appropriate prefix: (a) 354 mg(45 km) (0.0356 kN), (b) (0.004 53 Mg)(201 ms), and (c) 435 MN 23.2 mm. 14.7 lb lbft2 1 Pa = 1 Nm2 55 mih gm = 5.30 fts2 250(103 ) 568(105 ) mm MN(kg # ms) MgmN kNms mN(kg # ms) Nmm Mgms 35.3(103 ) N kNms MNks2 Nmm mMN *1–12. The specific weight (wt. vol.) of brass is . Determine its density (mass vol.) in SI units. Use an appropriate prefix. 1–13. Convert each of the following to three significant figures: (a) to , (b) to , and (c) 15 ft h to mm s. 1–14. The density (mass volume) of aluminum is . Determine its density in SI units. Use an appropriate prefix. 1–15. Water has a density of . What is the density expressed in SI units? Express the answer to three significant figures. *1–16. Two particles have a mass of 8 kg and 12 kg, respectively. If they are 800 mm apart, determine the force of gravity acting between them. Compare this result with the weight of each particle. 1–17. Determine the mass in kilograms of an object that has a weight of (a) 20 mN, (b) 150 kN, and (c) 60 MN. Express the answer to three significant figures. 1–18. Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) , (b) , and (c) . 1–19. Using the base units of the SI system, show that Eq. 1–2 is a dimensionally homogeneous equation which gives F in newtons. Determine to three significant figures the gravitational force acting between two spheres that are touching each other.The mass of each sphere is 200 kg and the radius is 300 mm. *1–20. Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) , and (b) . 1–21. Evaluate (204 mm)(0.00457 kg) (34.6 N) to three significant figures and express the answer in SI units using an appropriate prefix. (35 mm)2 (48 kg)3 (0.631 Mm)(8.60 kg)2 (400 m)3 (0.005 mm)2 (200 kN)2 1.94 slugft3 5.26 slugft3 kNm3 450 lbft3 N # m 20 lb # ft 520 lbft3
- 33. This bridge tower is stabilized by cables that exert forces at the points of connection. In this chapter we will show how to express these forces as Cartesian vectors and then determine the resultant force.
- 34. Force Vectors 2 Sense Magnitude Direction A u Fig. 2–1 CHAPTER OBJECTIVES • To show how to add forces and resolve them into components using the Parallelogram Law. • To express force and position in Cartesian vector form and explain how to determine the vector’s magnitude and direction. • To introduce the dot product in order to determine the angle between two vectors or the projection of one vector onto another. 2.1 Scalars and Vectors All physical quantities in engineering mechanics are measured using either scalars or vectors. Scalar. A scalar is any positive or negative physical quantity that can be completely specified by its magnitude. Examples of scalar quantities include length, mass, and time. Vector. A vector is any physical quantity that requires both a magnitude and a direction for its complete description. Examples of vectors encountered in statics are force, position, and moment. A vector is shown graphically by an arrow.The length of the arrow represents the magnitude of the vector, and the angle between the vector and a fixed axis defines the direction of its line of action.The head or tip of the arrow indicates the sense of direction of the vector, Fig. 2–1. In print, vector quantities are represented by bold face letters such as A, and its magnitude of the vector is italicized, A. For handwritten work, it is often convenient to denote a vector quantity by simply drawing an arrow on top of it, A : . u
- 35. 18 CHAPTER 2 FORCE VECTORS A A 2A 0.5 Scalar multiplication and division A Fig. 2–2 A A B B R (a) (c) (b) R A B A B Parallelogram law P Fig. 2–3 2 2.2 Vector Operations Multiplication and Division of a Vector by a Scalar. If a vector is multiplied by a positive scalar, its magnitude is increased by that amount. When multiplied by a negative scalar it will also change the directional sense of the vector. Graphic examples of these operations are shown in Fig. 2–2. Vector Addition. All vector quantities obey the parallelogram law of addition. To illustrate, the two “component” vectors A and B in Fig. 2–3a are added to form a “resultant” vector R A B using the following procedure: • First join the tails of the components at a point so that it makes them concurrent, Fig. 2–3b. • From the head of B, draw a line parallel to A. Draw another line from the head of A that is parallel to B.These two lines intersect at point P to form the adjacent sides of a parallelogram. • The diagonal of this parallelogram that extends to P forms R, which then represents the resultant vector R A B, Fig. 2–3c. + = + = We can also add B to A, Fig. 2–4a, using the triangle rule, which is a special case of the parallelogram law, whereby vector B is added to vector A in a “head-to-tail” fashion, i.e., by connecting the head of A to the tail of B, Fig. 2–4b. The resultant R extends from the tail of A to the head of B. In a similar manner, R can also be obtained by adding A to B, Fig. 2–4c. By comparison, it is seen that vector addition is commutative; in other words, the vectors can be added in either order, i.e., . R = A + B = B + A
- 36. 2.2 VECTOR OPERATIONS 19 A A B B R R R A B R B A (b) Triangle rule Triangle rule (c) A B (a) Fig. 2–4 A B R Addition of collinear vectors R A B Fig. 2–5 2 As a special case, if the two vectors A and B are collinear, i.e., both have the same line of action, the parallelogram law reduces to an algebraic or scalar addition R A B, as shown in Fig. 2–5. + = Vector Subtraction. The resultant of the difference between two vectors A and B of the same type may be expressed as This vector sum is shown graphically in Fig. 2–6. Subtraction is therefore defined as a special case of addition, so the rules of vector addition also apply to vector subtraction. R' = A - B = A + (–B) R¿ A ⴚB B A ⴚB A R¿ or Parallelogram law Triangle construction Vector subtraction Fig. 2–6
- 37. 20 CHAPTER 2 FORCE VECTORS 2 FR F1 F2 FR FR F1 F1 F1 F2 F2 F2 (c) (b) (a) v Fig. 2–7 2.3 Vector Addition of Forces Experimental evidence has shown that a force is a vector quantity since it has a specified magnitude, direction, and sense and it adds according to the parallelogram law. Two common problems in statics involve either finding the resultant force, knowing its components, or resolving a known force into two components. We will now describe how each of these problems is solved using the parallelogram law. Finding a Resultant Force. The two component forces F1 and F2 acting on the pin in Fig. 2–7a can be added together to form the resultant force FR = F1 + F2, as shown in Fig. 2–7b. From this construction, or using the triangle rule, Fig. 2–7c, we can apply the law of cosines or the law of sines to the triangle in order to obtain the magnitude of the resultant force and its direction. Finding the Components of a Force. Sometimes it is necessary to resolve a force into two components in order to study its pulling or pushing effect in two specific directions. For example, in Fig. 2–8a, F is to be resolved into two components along the two members, defined by the u and axes. In order to determine the magnitude of each component, a parallelogram is constructed first, by drawing lines starting from the tip of F, one line parallel to u, and the other line parallel to . These lines then intersect with the and u axes, forming a parallelogram. The force components Fu and F are then established by simply joining the tail of F to the intersection points on the u and axes, Fig. 2–8b. This parallelogram can then be reduced to a triangle, which represents the triangle rule, Fig. 2–8c. From this, the law of sines can then be applied to determine the unknown magnitudes of the components. v v v v v Fu u v Fv F FR F2 F1 Using the parallelogram law force F caused by the vertical member can be resolved into components acting along the suspension cables a and b. The parallelogram law must be used to determine the resultant of the two forces acting on the hook.
- 38. 2.3 VECTOR ADDITION OF FORCES 21 2 F u (b) F Fu Fu (c) F u (a) v v Fv Fv Fig. 2–8 Addition of Several Forces. If more than two forces are to be added, successive applications of the parallelogram law can be carried out in order to obtain the resultant force. For example, if three forces F1, F2, F3 act at a point O, Fig. 2–9, the resultant of any two of the forces is found, say, F1 + F2—and then this resultant is added to the third force, yielding the resultant of all three forces; i.e., FR = (F1 + F2)+F3. Using the parallelogram law to add more than two forces, as shown here, often requires extensive geometric and trigonometric calculation to determine the numerical values for the magnitude and direction of the resultant. Instead, problems of this type are easily solved by using the “rectangular- component method,” which is explained in Sec. 2.4. F1 FR F1 F2 F3 3 F1 F3 F2 F1 F2 F1 F2 FR F3 O Fig. 2–9 The resultant force on the hook requires the addition of ,then this resultant is added to . F3 F1 + F2 FR
- 39. 22 CHAPTER 2 FORCE VECTORS 2 A C B b (c) c a Sine law: sin a sin b sin c A B C Cosine law: C A2 B2 2AB cos c FR F1 F2 F Fu u (b) (a) v Fv Fig. 2–10 Procedure for Analysis Problems that involve the addition of two forces can be solved as follows: Parallelogram Law. • Two “component” forces F1 and F2 in Fig. 2–10a add according to the parallelogram law, yielding a resultant force FR that forms the diagonal of the parallelogram. • If a force F is to be resolved into components along two axes u and , Fig. 2–10b, then start at the head of force F and construct lines parallel to the axes, thereby forming the parallelogram. The sides of the parallelogram represent the components, F and F . • Label all the known and unknown force magnitudes and the angles on the sketch and identify the two unknowns as the magnitude and direction of FR, or the magnitudes of its components. Trigonometry. • Redraw a half portion of the parallelogram to illustrate the triangular head-to-tail addition of the components. • From this triangle, the magnitude of the resultant force can be determined using the law of cosines, and its direction is determined from the law of sines. The magnitudes of two force components are determined from the law of sines. The formulas are given in Fig. 2–10c. v u v Important Points • A scalar is a positive or negative number. • A vector is a quantity that has a magnitude, direction, and sense. • Multiplication or division of a vector by a scalar will change the magnitude of the vector.The sense of the vector will change if the scalar is negative. • As a special case, if the vectors are collinear, the resultant is formed by an algebraic or scalar addition.
- 40. The screw eye in Fig. 2–11a is subjected to two forces, F1 and F2. Determine the magnitude and direction of the resultant force. SOLUTION Parallelogram Law. The parallelogram is formed by drawing a line from the head of F1 that is parallel to F2, and another line from the head of F2 that is parallel to F1.The resultant force FR extends to where these lines intersect at point A, Fig. 2–11b. The two unknowns are the magnitude of FR and the angle (theta). Trigonometry. From the parallelogram, the vector triangle is constructed, Fig. 2–11c. Using the law of cosines Ans. Applying the law of sines to determine , Thus, the direction (phi) of FR, measured from the horizontal, is Ans. NOTE: The results seem reasonable, since Fig. 2–11b shows FR to have a magnitude larger than its components and a direction that is between them. f = 39.8° + 15.0° = 54.8° f u = 39.8° sin u = 150 N 212.6 N (sin 115º) 150 N sin u = 212.6 N sin 115° u = 213 N = 210 000 + 22 500 - 30 000(-0.4226) = 212.6 N FR = 2(100 N)2 + (150 N)2 - 2(100 N)(150 N) cos 115° u EXAMPLE 2.1 2.3 VECTOR ADDITION OF FORCES 23 F1 100 N F2 150 N 10 15 (a) Fig. 2–11 (c) FR 150 N 100 N 15 115 u f 2 FR 90 25 65 10 15 100 N A 65 115 150 N (b) 115 360 2(65) 2 u
- 41. Resolve the horizontal 600-lb force in Fig. 2–12a into components acting along the u and axes and determine the magnitudes of these components. v EXAMPLE 2.2 24 CHAPTER 2 FORCE VECTORS u 30 30 30 30 30 120 120 120 30 30 600 lb (a) u C B A 600 lb (b) Fu F (c) 600 lb Fu F v v v v Fig. 2–12 SOLUTION The parallelogram is constructed by extending a line from the head of the 600-lb force parallel to the axis until it intersects the u axis at point B, Fig. 2–12b. The arrow from A to B represents Fu. Similarly, the line extended from the head of the 600-lb force drawn parallel to the u axis intersects the axis at point C, which gives F . The vector addition using the triangle rule is shown in Fig. 2–12c. The two unknowns are the magnitudes of Fu and F . Applying the law of sines, Ans. Ans. NOTE: The result for Fu shows that sometimes a component can have a greater magnitude than the resultant. Fv = 600 lb Fv sin 30° = 600 lb sin 30° Fu = 1039 lb Fu sin 120° = 600 lb sin 30° v v v v 2
- 42. Determine the magnitude of the component force F in Fig. 2–13a and the magnitude of the resultant force FR if FR is directed along the positive y axis. EXAMPLE 2.3 2.3 VECTOR ADDITION OF FORCES 25 SOLUTION The parallelogram law of addition is shown in Fig. 2–13b, and the triangle rule is shown in Fig. 2–13c.The magnitudes of FR and F are the two unknowns.They can be determined by applying the law of sines. Ans. Ans. FR = 273 lb FR sin 75° = 200 lb sin 45° F = 245 lb F sin 60° = 200 lb sin 45° y 45 45 45 45 200 lb 30 30 30 (a) F y 45 200 lb (b) F FR 75 60 60 200 lb (c) F FR Fig. 2–13 2
- 43. It is required that the resultant force acting on the eyebolt in Fig. 2–14a be directed along the positive x axis and that F2 have a minimum magnitude. Determine this magnitude, the angle q, and the corresponding resultant force. EXAMPLE 2.4 26 CHAPTER 2 FORCE VECTORS x x x (a) (b) (c) FR FR F2 F2 F2 F1 800 N F1 800 N F1 800 N u 90 u u 60 60 60 Fig. 2–14 SOLUTION The triangle rule for is shown in Fig. 2–14b. Since the magnitudes (lengths) of FR and F2 are not specified,then F2 can actually be any vector that has its head touching the line of action of FR, Fig.2–14c.However,as shown,the magnitude of F2 is a minimum or the shortest length when its line of action is perpendicular to the line of action of FR, that is, when Ans. Since the vector addition now forms a right triangle, the two unknown magnitudes can be obtained by trigonometry. Ans. Ans. F2 = (800 N)sin 60° = 693 N FR = (800 N)cos 60° = 400 N u = 90° FR = F1 + F2 2
- 44. 30 40 500 N 200 N F2–2 FUNDAMENTAL PROBLEMS* 2.3 VECTOR ADDITION OF FORCES 27 x 2 kN 6 kN 45 60 F2–1 y x 800 N 600 N 30 F2–3 F2–3. Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis. F2–6. If force F is to have a component along the u axis of , determine the magnitude of F and the magnitude of its component F along the axis. v v Fu = 6 kN 30 lb u v 30 15 F2–4 A C B 450 lb 45 30 F2–5 u v F 45 105 F2–6 F2–1. Determine the magnitude of the resultant force acting on the screw eye and its direction measured clockwise from the x axis. F2–4. Resolve the 30-lb force into components along the u and axes, and determine the magnitude of each of these components. v F2–2. Two forces act on the hook. Determine the magnitude of the resultant force. F2–5. The force acts on the frame. Resolve this force into components acting along members AB and AC, and determine the magnitude of each component. F = 450 lb 2 * Partial solutions and answers to all Fundamental Problems are given in the back of the book.
- 45. PROBLEMS 28 CHAPTER 2 FORCE VECTORS 8 kN T x y u 45 Probs. 2–1/2/3 u F1 200 lb F2 150 lb v 30 30 45 Probs. 2–4/5/6 •2–1. If and , determine the magnitude of the resultant force acting on the eyebolt and its direction measured clockwise from the positive x axis. 2–2. If and , determine the magnitude of the resultant force acting on the eyebolt and its direction measured clockwise from the positive x axis. 2–3. If the magnitude of the resultant force is to be 9 kN directed along the positive x axis,determine the magnitude of force T acting on the eyebolt and its angle . u T = 5 kN u = 60° T = 6 kN u = 30° *2–4. Determine the magnitude of the resultant force acting on the bracket and its direction measured counterclockwise from the positive u axis. •2–5. Resolve F1 into components along the u and axes, and determine the magnitudes of these components. 2–6. Resolve F2 into components along the u and axes, and determine the magnitudes of these components. v v •2–9. The plate is subjected to the two forces at A and B as shown. If , determine the magnitude of the resultant of these two forces and its direction measured clockwise from the horizontal. 2–10. Determine the angle of for connecting member A to the plate so that the resultant force of FA and FB is directed horizontally to the right.Also, what is the magnitude of the resultant force? u u = 60° y x u B FA 3 kN FB A u 30 Probs. 2–7/8 A B FA 8 kN FB 6 kN 40 u Probs. 2–9/10 2–7. If and the resultant force acts along the positive u axis, determine the magnitude of the resultant force and the angle . *2–8. If the resultant force is required to act along the positive u axis and have a magnitude of 5 kN, determine the required magnitude of FB and its direction . u u FB = 2 kN 2
- 46. 2.3 VECTOR ADDITION OF FORCES 29 2–11. If the tension in the cable is 400 N, determine the magnitude and direction of the resultant force acting on the pulley. This angle is the same angle of line AB on the tailboard block. u *2–12. The device is used for surgical replacement of the knee joint. If the force acting along the leg is 360 N, determine its components along the x and y axes. •2–13. The device is used for surgical replacement of the knee joint. If the force acting along the leg is 360 N, determine its components along the x and y axes. ¿ ¿ *2–16. Resolve F1 into components along the u and axes and determine the magnitudes of these components. •2–17. Resolve F2 into components along the u and axes and determine the magnitudes of these components. v v 2–14. Determine the design angle for strut AB so that the 400-lb horizontal force has a component of 500 lb directed from A towards C.What is the component of force acting along member AB? Take . 2–15. Determine the design angle between struts AB and AC so that the 400-lb horizontal force has a component of 600 lb which acts up to the left, in the same direction as from B towards A.Take . u = 30° f (0° … f … 90°) f = 40° u (0° … u … 90°) 400 N 30 y A B x 400 N u Prob. 2–11 60 360 N 10 y x y¿ x¿ Probs. 2–12/13 A C B 400 lb u f Probs. 2–14/15 F1 250 N F2 150 N u v 30 30 105 Probs. 2–16/17 2
- 47. 30 CHAPTER 2 FORCE VECTORS •2–25. Two forces F1 and F2 act on the screw eye. If their lines of action are at an angle apart and the magnitude of each force is determine the magnitude of the resultant force FR and the angle between FR and F1. F1 = F2 = F, u y x F3 5 kN F1 4 kN F2 u Probs. 2–23/24 F2 F1 u Prob. 2–25 y 20° x A B FA FB u Prob. 2–18/19 F1 F2 x y u f 60 Probs. 2–20/21/22 2–18. The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop a resultant force of 950 N directed along the positive x axis. Set . 2–19. The truck is to be towed using two ropes. If the resultant force is to be 950 N, directed along the positive x axis, determine the magnitudes of forces FA and FB acting on each rope and the angle of FB so that the magnitude of FB is a minimum. FA acts at 20° from the x axis as shown. u u = 50° *2–20. If , , and the resultant force is 6 kN directed along the positive y axis,determine the required magnitude of F2 and its direction . •2–21. If and the resultant force is to be 6 kN directed along the positive y axis, determine the magnitudes of F1 and F2 and the angle if F2 is required to be a minimum. 2–22. If , , and the resultant force is to be directed along the positive y axis, determine the magnitude of the resultant force if F2 is to be a minimum. Also, what is F2 and the angle ? u F1 = 5 kN f = 30° u f = 30° u F1 = 5 kN f = 45° 2–23. If and , determine the magnitude of the resultant force acting on the plate and its direction measured clockwise from the positive x axis. *2–24. If the resultant force FR is directed along a line measured 75° clockwise from the positive x axis and the magnitude of F2 is to be a minimum, determine the magnitudes of FR and F2 and the angle . u … 90° F2 = 6 kN u = 30°
- 48. 2.3 VECTOR ADDITION OF FORCES 31 2 x y B A 30 FA FB u Probs. 2–26/27 FB FA y x 30 u Probs. 2–28/29 45 30 y x 400 lb 600 lb F u Prob. 2–31 300 lb 200 lb x y F 30 u Prob. 2–30 2–26. The log is being towed by two tractors A and B. Determine the magnitudes of the two towing forces FA and FB if it is required that the resultant force have a magnitude and be directed along the x axis. Set . 2–27. The resultant FR of the two forces acting on the log is to be directed along the positive x axis and have a magnitude of 10 kN,determine the angle of the cable,attached to B such that the magnitude of force FB in this cable is a minimum. What is the magnitude of the force in each cable for this situation? u u = 15° FR = 10 kN *2–28. The beam is to be hoisted using two chains. Deter- mine the magnitudes of forces FA and FB acting on each chain in order to develop a resultant force of 600 N directed along the positive y axis. Set . •2–29. The beam is to be hoisted using two chains. If the resultant force is to be 600 N directed along the positive y axis, determine the magnitudes of forces FA and FB acting on each chain and the angle of FB so that the magnitude of FB is a minimum. FA acts at 30° from the y axis, as shown. u u = 45° 2–31. Three cables pull on the pipe such that they create a resultant force having a magnitude of 900 lb. If two of the cables are subjected to known forces, as shown in the figure, determine the angle of the third cable so that the magnitude of force F in this cable is a minimum. All forces lie in the x–y plane. What is the magnitude of F? Hint: First find the resultant of the two known forces. u 2–30. Three chains act on the bracket such that they create a resultant force having a magnitude of 500 lb. If two of the chains are subjected to known forces, as shown, determine the angle of the third chain measured clockwise from the positive x axis, so that the magnitude of force F in this chain is a minimum. All forces lie in the x–y plane. What is the magnitude of F? Hint: First find the resultant of the two known forces. Force F acts in this direction. u
- 49. 32 CHAPTER 2 FORCE VECTORS 2 2.4 Addition of a System of Coplanar Forces When a force is resolved into two components along the x and y axes, the components are then called rectangular components. For analytical work we can represent these components in one of two ways, using either scalar notation or Cartesian vector notation. Scalar Notation. The rectangular components of force F shown in Fig. 2–15a are found using the parallelogram law, so that . Because these components form a right triangle, their magnitudes can be determined from Instead of using the angle , however, the direction of F can also be defined using a small “slope” triangle, such as shown in Fig. 2–15b. Since this triangle and the larger shaded triangle are similar, the proportional length of the sides gives or and or Here the y component is a negative scalar since Fy is directed along the negative y axis. It is important to keep in mind that this positive and negative scalar notation is to be used only for computational purposes, not for graphical representations in figures. Throughout the book, the head of a vector arrow in any figure indicates the sense of the vector graphically; algebraic signs are not used for this purpose. Thus, the vectors in Figs. 2–15a and 2–15b are designated by using boldface (vector) notation.* Whenever italic symbols are written near vector arrows in figures, they indicate the magnitude of the vector, which is always a positive quantity. Fy = -Fa b c b Fy F = b c Fx = F a a c b Fx F = a c u Fx = F cos u and Fy = F sin u F = Fx + Fy (a) F y x Fx u Fy Fx (b) F y x a b c Fig. 2–15 *Negative signs are used only in figures with boldface notation when showing equal but opposite pairs of vectors, as in Fig. 2–2.
- 50. 2.4 ADDITION OF A SYSTEM OF COPLANAR FORCES 33 2 Cartesian Vector Notation. It is also possible to represent the x and y components of a force in terms of Cartesian unit vectors i and j. Each of these unit vectors has a dimensionless magnitude of one, and so they can be used to designate the directions of the x and y axes, respectively, Fig. 2–16. * Since the magnitude of each component of F is always a positive quantity, which is represented by the (positive) scalars Fx and Fy, then we can express F as a Cartesian vector, Coplanar Force Resultants. We can use either of the two methods just described to determine the resultant of several coplanar forces.To do this, each force is first resolved into its x and y components, and then the respective components are added using scalar algebra since they are collinear. The resultant force is then formed by adding the resultant components using the parallelogram law. For example, consider the three concurrent forces in Fig. 2–17a, which have x and y components shown in Fig. 2–17b. Using Cartesian vector notation, each force is first represented as a Cartesian vector, i.e., The vector resultant is therefore If scalar notation is used, then we have These are the same results as the i and j components of FR determined above. (+ c) FRy = F1y + F2y - F3y ( : + ) FRx = F1x - F2x + F3x = (FRx)i + (FRy)j = (F1x - F2x + F3x)i + (F1y + F2y - F3y)j = F1xi + F1y j - F2x i + F2y j + F3x i-F3y j FR = F1 + F2 + F3 F1 = F1x i + F1y j F2 = -F2x i + F2y j F3 = F3x i - F3y j F = Fx i + Fy j F Fx Fy y x i j Fig. 2–16 F3 F1 F2 (a) x y (b) x y F2x F2y F1y F1x F3x F3y Fig. 2–17 *For handwritten work, unit vectors are usually indicated using a circumflex, e.g., and . These vectors have a dimensionless magnitude of unity, and their sense (or arrowhead) will be described analytically by a plus or minus sign, depending on whether they are pointing along the positive or negative x or y axis. j ¿ i ¿
- 51. 34 CHAPTER 2 FORCE VECTORS 2 We can represent the components of the resultant force of any number of coplanar forces symbolically by the algebraic sum of the x and y components of all the forces, i.e., (2–1) Once these components are determined, they may be sketched along the x and y axes with their proper sense of direction, and the resultant force can be determined from vector addition, as shown in Fig. 2–17. From this sketch, the magnitude of FR is then found from the Pythagorean theorem; that is, Also, the angle , which specifies the direction of the resultant force, is determined from trigonometry: The above concepts are illustrated numerically in the examples which follow. u = tan-1 2 FRy FRx 2 u FR = 2F2 Rx + F2 Ry FRx = ©Fx FRy = ©Fy The resultant force of the four cable forces acting on the supporting bracket can be determined by adding algebraically the separate x and y components of each cable force. This resultant FR produces the same pulling effect on the bracket as all four cables. Important Points • The resultant of several coplanar forces can easily be determined if an x, y coordinate system is established and the forces are resolved along the axes. • The direction of each force is specified by the angle its line of action makes with one of the axes, or by a sloped triangle. • The orientation of the x and y axes is arbitrary, and their positive direction can be specified by the Cartesian unit vectors i and j. • The x and y components of the resultant force are simply the algebraic addition of the components of all the coplanar forces. • The magnitude of the resultant force is determined from the Pythagorean theorem, and when the components are sketched on the x and y axes, the direction can be determined from trigonometry. F1 F2 F3 F4 x y (c) x y FR FRy FRx u Fig. 2–17
- 52. 2.4 ADDITION OF A SYSTEM OF COPLANAR FORCES 35 2 EXAMPLE 2.5 Determine the x and y components of F1 and F2 acting on the boom shown in Fig. 2–18a. Express each force as a Cartesian vector. SOLUTION Scalar Notation. By the parallelogram law, F1 is resolved into x and y components, Fig. 2–18b. Since F1x acts in the –x direction, and F1y acts in the +y direction, we have Ans. Ans. The force F2 is resolved into its x and y components as shown in Fig. 2–17c. Here the slope of the line of action for the force is indicated. From this “slope triangle” we could obtain the angle , e.g., , and then proceed to determine the magnitudes of the components in the same manner as for F1. The easier method, how- ever, consists of using proportional parts of similar triangles, i.e., Similarly, Notice how the magnitude of the horizontal component, F2x, was obtained by multiplying the force magnitude by the ratio of the horizontal leg of the slope triangle divided by the hypotenuse; whereas the magnitude of the vertical component, F2y, was obtained by multiplying the force magnitude by the ratio of the vertical leg divided by the hypotenuse. Hence, Ans. Ans. Cartesian Vector Notation. Having determined the magnitudes and directions of the components of each force, we can express each force as a Cartesian vector. Ans. Ans. F2 = 5240i - 100j6 N F1 = 5-100i + 173j6 N F2y = -100 N = 100 NT F2x = 240 N = 240 N : F2y = 260 Na 5 13 b = 100 N F2x = 260 Na 12 13 b = 240 N F2x 260 N = 12 13 u = tan-1 ( 5 12) u F1y = 200 cos 30° N = 173 N = 173 Nc F1x = -200 sin 30° N = -100 N = 100 N ; y x F1 200 N F2 260 N 30 (a) 5 12 13 y x F1 200 N F1x 200 sin 30 N 30 F1y 200 cos 30 N (b) y x F2 260 N (c) 5 12 13 F2x 260 12 — 13 ( (N F2y 260 5 — 13 ( (N Fig. 2–18
- 53. 36 CHAPTER 2 FORCE VECTORS 2 EXAMPLE 2.6 The link in Fig. 2–19a is subjected to two forces F1 and F2. Determine the magnitude and direction of the resultant force. SOLUTION I Scalar Notation. First we resolve each force into its x and y components, Fig. 2–19b, then we sum these components algebraically. The resultant force, shown in Fig. 2–18c, has a magnitude of Ans. From the vector addition, Ans. SOLUTION II Cartesian Vector Notation. From Fig. 2–19b, each force is first expressed as a Cartesian vector. Then, The magnitude and direction of FR are determined in the same manner as before. NOTE: Comparing the two methods of solution, notice that the use of scalar notation is more efficient since the components can be found directly, without first having to express each force as a Cartesian vector before adding the components. Later, however, we will show that Cartesian vector analysis is very beneficial for solving three-dimensional problems. = 5236.8i + 582.8j6 N + (600 sin 30° N + 400 cos 45° N)j FR = F1 + F2 = (600 cos 30° N - 400 sin 45° N)i F2 = 5-400 sin 45°i + 400 cos 45°j6 N F1 = 5600 cos 30°i + 600 sin 30°j6 N u = tan-1 a 582.8 N 236.8 N b = 67.9° = 629 N FR = 2(236.8 N)2 + (582.8 N)2 = 582.8 Nc + cFRy = ©Fy; FRy = 600 sin 30° N + 400 cos 45° N = 236.8 N : : + FRx = ©Fx; FRx = 600 cos 30° N - 400 sin 45° N y F1 600 N x F2 400 N 45 30 (a) y F1 600 N x F2 400 N 30 (b) 45 y FR x (c) 582.8 N 236.8 N u Fig. 2–19
- 54. 2.4 ADDITION OF A SYSTEM OF COPLANAR FORCES 37 2 EXAMPLE 2.7 F3 200 N (a) y x F1 400 N F2 250 N 3 5 4 45 250 N (b) y 45 400 N 4 x 200 N 3 5 FR 296.8 N 383.2 N (c) y x u Fig. 2–20 The end of the boom O in Fig. 2–20a is subjected to three concurrent and coplanar forces. Determine the magnitude and direction of the resultant force. SOLUTION Each force is resolved into its x and y components,Fig.2–20b.Summing the x components, we have The negative sign indicates that FRx acts to the left, i.e., in the negative x direction, as noted by the small arrow. Obviously, this occurs because F1 and F3 in Fig. 2–20b contribute a greater pull to the left than F2 which pulls to the right. Summing the y components yields The resultant force, shown in Fig. 2–20c, has a magnitude of Ans. From the vector addition in Fig. 2–20c, the direction angle is Ans. NOTE: Application of this method is more convenient, compared to using two applications of the parallelogram law, first to add F1 and F2 then adding F3 to this resultant. u = tan-1 a 296.8 383.2 b = 37.8° u = 485 N FR = 2(-383.2 N)2 + (296.8 N)2 = 296.8 Nc + cFRy = ©Fy; FRy = 250 cos 45° N + 200A3 5 B N = -383.2 N = 383.2 N ; : + FRx = ©Fx; FRx = -400 N + 250 sin 45° N - 200A4 5 B N
- 55. 38 CHAPTER 2 FORCE VECTORS 2 FUNDAMENTAL PROBLEMS F2–10. If the resultant force acting on the bracket is to be 750 N directed along the positive x axis, determine the magnitude of F and its direction . u F2–11. If the magnitude of the resultant force acting on the bracket is to be 80 lb directed along the u axis, determine the magnitude of F and its direction . u 3 4 5 y x F2 450 N F1 300 N F3 600 N 45 F2–7 F 600 N 325 N 12 5 13 y x u 45 F2–10 90 lb 50 lb F 3 4 5 x u y 45 u F2–11 F3 15 kN F2 20 kN F1 15 kN y x 4 4 3 3 5 5 F2–12 y x 300 N 400 N 250 N 3 4 5 30 F2–8 3 4 5 F2 400 lb F1 700 lb y x F3 600 lb 30 F2–9 F2–7. Resolve each force acting on the post into its x and y components. F2–8. Determine the magnitude and direction of the resultant force. F2–9. Determine the magnitude of the resultant force acting on the corbel and its direction measured counterclockwise from the x axis. u F2–12. Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis. u
- 56. 2.4 ADDITION OF A SYSTEM OF COPLANAR FORCES 39 2 •2–33. If and , determine the magnitude of the resultant force acting on the eyebolt and its direction measured clockwise from the positive x axis. 2–34. If the magnitude of the resultant force acting on the eyebolt is 600 N and its direction measured clockwise from the positive x axis is , determine the magni- tude of F1 and the angle . f u = 30° f = 30° F1 = 600 N PROBLEMS x A 175 lb 12 5 13 y Prob. 2–35 x y F2 5 4 3 F1 4 kN F3 5 kN f 30 Probs. 2–36/37/38 x y F1 30 lb F2 40 lb F3 25 lb 15 15 45 Prob. 2–32 y x 3 4 5 F2 500 N F1 F3 450 N f 60 Probs. 2–33/34 *2–32. Determine the magnitude of the resultant force acting on the pin and its direction measured clockwise from the positive x axis. 2–35. The contact point between the femur and tibia bones of the leg is at A. If a vertical force of 175 lb is applied at this point, determine the components along the x and y axes. Note that the y component represents the normal force on the load-bearing region of the bones. Both the x and y components of this force cause synovial fluid to be squeezed out of the bearing space. *2–36. If and ,determine the magnitude of the resultant force acting on the plate and its direction measured clockwise from the positive x axis. •2–37. If the magnitude for the resultant force acting on the plate is required to be 6 kN and its direction measured clockwise from the positive x axis is , determine the magnitude of F2 and its direction . 2–38. If and the resultant force acting on the gusset plate is directed along the positive x axis, determine the magnitudes of F2 and the resultant force. f = 30° f u = 30° u F2 = 3 kN f = 30°
- 57. 40 CHAPTER 2 FORCE VECTORS 2 2–46. The three concurrent forces acting on the screw eye produce a resultant force . If and F1 is to be 90° from F2 as shown, determine the required magnitude of F3 expressed in terms of F1 and the angle . u F2 = 2 3 F1 FR = 0 F3 260 lb F2 300 lb 5 12 13 3 4 5 x y F1 f Probs. 2–43/44/45 y x 60 30 F2 F3 F1 u Prob. 2–46 A x y F1 400 N 600 N 3 4 5 30 u Probs. 2–39/40 FB x y B A 30 FA 700 N u Probs. 2–41/42 •2–41. Determine the magnitude and direction of FB so that the resultant force is directed along the positive y axis and has a magnitude of 1500 N. 2–42. Determine the magnitude and angle measured counterclockwise from the positive y axis of the resultant force acting on the bracket if and . u = 20° FB = 600 N u 2–43. If and , determine the magnitude of the resultant force acting on the bracket and its direction measured clockwise from the positive x axis. *2–44. If the magnitude of the resultant force acting on the bracket is 400 lb directed along the positive x axis, determine the magnitude of F1 and its direction . •2–45. If the resultant force acting on the bracket is to be directed along the positive x axis and the magnitude of F1 is required to be a minimum, determine the magnitudes of the resultant force and F1. f F1 = 250 lb f = 30° 2–39. Determine the magnitude of F1 and its direction so that the resultant force is directed vertically upward and has a magnitude of 800 N. *2–40. Determine the magnitude and direction measured counterclockwise from the positive x axis of the resultant force of the three forces acting on the ring A. Take and . u = 20° F1 = 500 N u
- 58. 2.4 ADDITION OF A SYSTEM OF COPLANAR FORCES 41 2 •2–49. Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis. 2–47. Determine the magnitude of FA and its direction so that the resultant force is directed along the positive x axis and has a magnitude of 1250 N. *2–48. Determine the magnitude and direction measured counterclockwise from the positive x axis of the resultant force acting on the ring at O if and . u = 45° FA = 750 N u 2–51. If and , determine the magnitude of the resultant force acting on the bracket and its direction measured clockwise from the positive x axis. *2–52. If the magnitude of the resultant force acting on the bracket is to be 450 N directed along the positive u axis, determine the magnitude of F1 and its direction . •2–53. If the resultant force acting on the bracket is required to be a minimum, determine the magnitudes of F1 and the resultant force. Set . f = 30° f f = 30° F1 = 150 N 2–50. The three forces are applied to the bracket. Determine the range of values for the magnitude of force P so that the resultant of the three forces does not exceed 2400 N. 3000 N 800 N P 90 60 Prob. 2–50 5 12 13 y x u F3 260 N F2 200 N F1 f 30 Probs. 2–51/52/53 30 y x O B A FA FB 800 N u Probs. 2–47/48 F1 = 60 lb F2 70 lb F3 50 lb y x 60 45 1 2 1 Prob. 2–49
- 59. 42 CHAPTER 2 FORCE VECTORS 2 *2–56. The three concurrent forces acting on the post produce a resultant force . If , and F1 is to be 90° from F2 as shown, determine the required magnitude of F3 expressed in terms of F1 and the angle . u F2 = 1 2 F1 FR = 0 2–58. Express each of the three forces acting on the bracket in Cartesian vector form with respect to the x and y axes. Determine the magnitude and direction of F1 so that the resultant force is directed along the positive axis and has a magnitude of . FR = 600 N x¿ u •2–57. Determine the magnitude of force F so that the resultant force of the three forces is as small as possible. What is the magnitude of this smallest resultant force? F2 350 N F1 F3 100 N y x x¿ 30 30 u Prob. 2–58 x y u 12 5 13 F2 25 F3 52 lb F1 80 lb u Probs. 2–54/55 x y F1 F2 F3 u Prob. 2–56 2–54. Three forces act on the bracket. Determine the magnitude and direction of F2 so that the resultant force is directed along the positive u axis and has a magnitude of 50 lb. 2–55. If and , determine the magnitude and direction measured clockwise from the positive x axis of the resultant force of the three forces acting on the bracket. u = 55° F2 = 150 lb u F 8 kN 14 kN 45 30 Prob. 2–57
- 60. 2.5 CARTESIAN VECTORS 43 2 2.5 Cartesian Vectors The operations of vector algebra, when applied to solving problems in three dimensions, are greatly simplified if the vectors are first represented in Cartesian vector form. In this section we will present a general method for doing this; then in the next section we will use this method for finding the resultant force of a system of concurrent forces. Right-Handed Coordinate System. We will use a right- handed coordinate system to develop the theory of vector algebra that follows.A rectangular coordinate system is said to be right-handed if the thumb of the right hand points in the direction of the positive z axis when the right-hand fingers are curled about this axis and directed from the positive x towards the positive y axis, Fig. 2–21. Rectangular Components of a Vector. A vector A may have one, two, or three rectangular components along the x, y, z coordinate axes, depending on how the vector is oriented relative to the axes. In general, though, when A is directed within an octant of the x, y, z frame, Fig. 2–22, then by two successive applications of the parallelogram law, we may resolve the vector into components as and then . Combining these equations, to eliminate , A is represented by the vector sum of its three rectangular components, (2–2) Cartesian Unit Vectors. In three dimensions, the set of Cartesian unit vectors, i, j, k, is used to designate the directions of the x, y, z axes, respectively. As stated in Sec. 2.4, the sense (or arrowhead) of these vectors will be represented analytically by a plus or minus sign, depending on whether they are directed along the positive or negative x, y, or z axes.The positive Cartesian unit vectors are shown in Fig. 2–23. A = Ax + Ay + Az A¿ A¿ = Ax + Ay A = A¿ + Az z y x Fig. 2–21 A Ax z y x Ay Az A¿ Fig. 2–22 k j i z y x Fig. 2–23
- 61. 44 CHAPTER 2 FORCE VECTORS 2 Cartesian Vector Representation. Since the three components of A in Eq. 2–2 act in the positive i, j, and k directions, Fig. 2–24, we can write A in Cartesian vector form as (2–3) There is a distinct advantage to writing vectors in this manner. Separating the magnitude and direction of each component vector will simplify the operations of vector algebra, particularly in three dimensions. Magnitude of a Cartesian Vector. It is always possible to obtain the magnitude of A provided it is expressed in Cartesian vector form. As shown in Fig. 2–25, from the blue right triangle, , and from the gray right triangle, . Combining these equations to eliminate , yields (2–4) Hence, the magnitude of A is equal to the positive square root of the sum of the squares of its components. Direction of a Cartesian Vector. We will define the direction of A by the coordinate direction angles a (alpha), b (beta), and g (gamma), measured between the tail of A and the positive x, y, z axes provided they are located at the tail of A, Fig. 2–26. Note that regardless of where A is directed, each of these angles will be between 0° and 180°. To determine a, b, and g, consider the projection of A onto the x, y, z axes, Fig. 2–27. Referring to the blue colored right triangles shown in each figure, we have (2–5) These numbers are known as the direction cosines of A. Once they have been obtained, the coordinate direction angles a, b, g can then be determined from the inverse cosines. cos a = Ax A cos b = Ay A cos g = Az A A = 2A2 x + A2 y + A2 z A¿ A¿ = 2A2 x + Ay 2 A = 2A¿2 + A2 z A = Axi + Ay j + Azk A Ax i z y x Ay j Az k k i j Fig. 2–24 A Axi z y x Ayj Azk A A¿ Ay Ax Az Fig. 2–25
- 62. An easy way of obtaining these direction cosines is to form a unit vector uA in the direction of A, Fig. 2–26. If A is expressed in Cartesian vector form, , then uA will have a magnitude of one and be dimensionless provided A is divided by its magnitude, i.e., (2–6) where .By comparison with Eqs.2–7,it is seen that the i, j, k components of uA represent the direction cosines of A, i.e., (2–7) Since the magnitude of a vector is equal to the positive square root of the sum of the squares of the magnitudes of its components, and uA has a magnitude of one, then from the above equation an important relation between the direction cosines can be formulated as (2–8) Here we can see that if only two of the coordinate angles are known, the third angle can be found using this equation. Finally, if the magnitude and coordinate direction angles of A are known, then A may be expressed in Cartesian vector form as (2–9) = Axi + Ayj + Azk = A cos ai + A cos b j + A cos gk A = AuA cos2 a + cos2 b + cos2 g = 1 uA = cos ai + cos b j + cos gk A = 2Ax 2 + A2 y + A2 z uA = A A = Ax A i + Ay A j + Az A k A = Axi + Ay j + Azk 2.5 CARTESIAN VECTORS 45 2 z y x 90 A Ax a z y x 90 A Ay b z y x Az 90 A g Fig 2–27 A Axi z y x Ayj Azk uA g a b Fig. 2–26
- 63. 46 CHAPTER 2 FORCE VECTORS 2 Sometimes, the direction of A can be specified using two angles, and (phi), such as shown in Fig. 2–28. The components of A can then be determined by applying trigonometry first to the blue right triangle, which yields and Now applying trigonometry to the other shaded right triangle, Therefore A written in Cartesian vector form becomes You should not memorize this equation, rather it is important to understand how the components were determined using trigonometry. 2.6 Addition of Cartesian Vectors The addition (or subtraction) of two or more vectors are greatly simplified if the vectors are expressed in terms of their Cartesian components. For example, if and , Fig. 2–29, then the resultant vector, R, has components which are the scalar sums of the i, j, k components of A and B, i.e., If this is generalized and applied to a system of several concurrent forces, then the force resultant is the vector sum of all the forces in the system and can be written as (2–10) Here ΣFx, ΣFy, and ΣFz represent the algebraic sums of the respective x, y, z or i, j, k components of each force in the system. FR = ©F = ©Fxi + ©Fy j + ©Fzk R = A + B = (Ax + Bx)i + (Ay + By)j + (Az + Bz)k B = Bxi + Byj + Bzk A = Axi + Ay j + Azk A = A sin f cos u i + A sin f sin u j + A cos f k Ay = A¿ sin u = A sin f sin u Ax = A¿ cos u = A sin f cos u A¿ = A sin f Az = A cos f f u z y x R B A (Az Bz)k (Ax Bx)i (Ay By)j Fig. 2–29 y x Ay Az Ax A¿ A z O u f Fig. 2–28
- 64. 2.6 ADDITION OF CARTESIAN VECTORS 47 2 Important Points • Cartesian vector analysis is often used to solve problems in three dimensions. • The positive directions of the x, y, z axes are defined by the Cartesian unit vectors i, j, k, respectively. • The magnitude of a Cartesian vector is . • The direction of a Cartesian vector is specified using coordinate direction angles which the tail of the vector makes with the positive x, y, z axes, respectively. The components of the unit vector represent the direction cosines of . Only two of the angles have to be specified. The third angle is determined from the relationship . • Sometimes the direction of a vector is defined using the two angles q and as in Fig. 2–28. In this case the vector components are obtained by vector resolution using trigonometry. • To find the resultant of a concurrent force system, express each force as a Cartesian vector and add the i, j, k components of all the forces in the system. f cos2 a + cos2 b + cos2 g = 1 a, b, g a, b, g uA = AA a, b, g A = 2A2 x + A2 y + A2 z EXAMPLE 2.8 Express the force F shown in Fig. 2–30 as a Cartesian vector. SOLUTION Since only two coordinate direction angles are specified,the third angle α must be determined from Eq. 2–8; i.e., Hence, two possibilities exist, namely, By inspection it is necessary that , since Fx must be in the +x direction. Using Eq. 2–9, with , we have Ans. Show that indeed the magnitude of . F = 200 N = 5100.0i + 100.0j + 141.4k6 N = (200 cos 60° N)i + (200 cos 60° N)j + (200 cos 45° N)k F = F cos ai + F cos bj + F cos gk F = 200 N a = 60° a = cos-1 (0.5) = 60° or a = cos-1 (-0.5) = 120° cos a = 21 - (0.5)2 - (0.707)2 = ;0.5 cos2 a + cos2 60° + cos2 45° = 1 cos2 a + cos2 b + cos2 g = 1 z y x 45 F 200 N 60 a Fig. 2–30 The resultant force acting on the bow the ship can be determined by first representing each rope force as a Cartesian vector and then summing the i, j, and k components.
- 65. 48 CHAPTER 2 FORCE VECTORS 2 EXAMPLE 2.9 Determine the magnitude and the coordinate direction angles of the resultant force acting on the ring in Fig. 2–31a. SOLUTION Since each force is represented in Cartesian vector form, the resultant force, shown in Fig. 2–31b, is The magnitude of FR is Ans. The coordinate direction angles a, b, g are determined from the components of the unit vector acting in the direction of FR. so that Ans. Ans. Ans. These angles are shown in Fig. 2–31b. NOTE: In particular, notice that since the j component of uF R is negative. This seems reasonable considering how F1 and F2 add according to the parallelogram law. b 7 90° g = 19.6° cos g = 0.9422 b = 102° cos b = -0.2094 a = 74.8° cos a = 0.2617 = 0.2617i - 0.2094j + 0.9422k uFR = FR FR = 50 191.0 i - 40 191.0 j + 180 191.0 k = 191 lb FR = 2(50 lb)2 + (-40 lb)2 + (180 lb)2 = 191.0 lb = 550i - 40j + 180k6 lb FR = ©F = F1 + F2 = 560j + 80k6 lb + 550i - 100j + 100k6 lb (a) z y x F2 {50i 100j 100k} lb F1 {60j 80k} lb (b) z y x F2 FR {50i 40j 180k} lb F1 102 74.8 19.6 a g b Fig. 2–31
- 66. 2.6 ADDITION OF CARTESIAN VECTORS 49 2 EXAMPLE 2.10 Express the force F shown in Fig. 2–32a as a Cartesian vector. SOLUTION The angles of 60° and 45° defining the direction of F are not coordinate direction angles.Two successive applications of the parallelogram law are needed to resolve F into its x, y, z components First , then , Fig. 2–32b. By trigonometry, the magnitudes of the components are Realizing that Fy has a direction defined by –j, we have Ans. To show that the magnitude of this vector is indeed 100 lb, apply Eq. 2–4, If needed, the coordinate direction angles of F can be determined from the components of the unit vector acting in the direction of F. Hence, so that These results are shown in Fig. 2–31c. g = cos-1 (0.866) = 30.0° b = cos-1 (-0.354) = 111° a = cos-1 (0.354) = 69.3° = 0.354i - 0.354j + 0.866k = 35.4 100 i - 35.4 100 j + 86.6 100 k u = F F = Fx F i + Fy F j + Fz F k = 2(35.4)2 + (-35.4)2 + (86.6)2 = 100 lb F = 2F2 x + F2 y + F2 z F = 535.4i - 35.4j + 86.6k6 lb Fy = F¿ sin 45º = 50 sin 45° lb = 35.4 lb Fx = F¿ cos 45º = 50 cos 45° lb = 35.4 lb F¿ = 100 cos 60° lb = 50 lb Fz = 100 sin 60° lb = 86.6 lb F¿ = Fx + Fy F = F¿ + Fz (a) z y x F 100 lb 60 45 z F¿ Fx Fz y x F 100 lb 60 45 Fy (c) z y x F 100 lb 69.3 111 30.0 Fig. 2–32
- 67. 50 CHAPTER 2 FORCE VECTORS 2 EXAMPLE 2.11 Two forces act on the hook shown in Fig. 2–32a. Specify the magnitude of and its coordinate direction angles of that the resultant force FR acts along the positive y axis and has a magnitude of 800 N. SOLUTION To solve this problem, the resultant force FR and its two components, F1 and F2, will each be expressed in Cartesian vector form. Then, as shown in Fig. 2–33a, it is necessary that . Applying Eq. 2–9, Since FR has a magnitude of 800 N and acts in the +j direction, We require To satisfy this equation the i, j, k components of FR must be equal to the corresponding i, j, k components of (F1 + F2). Hence, The magnitude of F2 is thus Ans. We can use Eq. 2–9 to determine 2, 2, 2. Ans. Ans. Ans. These results are shown in Fig. 2–32b. g2 = 77.6° cos g2 = 150 700 ; b2 = 21.8° cos b2 = 650 700 ; a2 = 108° cos a2 = -212.1 700 ; g b a = 700 N F2 = 2(-212.1 N)2 + (650 N)2 + (150 N)2 0 = 212.1 + F2x 800 = 150 + F2y 0 = -150 + F2z F2x = -212.1 N F2y = 650 N F2z = 150 N 800j = (212.1 + F2x)i + (150 + F2y)j + (-150 + F2z)k 800j = 212.1i + 150j - 150k + F2xi + F2yj + F2zk FR = F1 + F2 FR = (800 N)(+j) = 5800j6 N F2 = F2xi + F2y j + F2zk = 5212.1i + 150j - 150k6 N = 300 cos 45° i + 300 cos 60° j + 300 cos 120° k F1 = F1 cos a1i + F1 cos b1j + F1 cos g1k FR = F1 + F2 F2 F2 z F2 F1 300 N (a) x y 60 45 120 z (b) F1 300 N F2 700 N FR 800 N x y g2 77.6 b2 21.8 a2 108 Fig. 2–33
- 68. 2.6 ADDITION OF CARTESIAN VECTORS 51 2 F2–14. Express the force as a Cartesian vector. F2–15. Express the force as a Cartesian vector. F2–17. Express the force as a Cartesian vector. F2–18. Determine the resultant force acting on the hook. FUNDAMENTAL PROBLEMS F 500 N z y x 60 60 F2–14 F 500 N z y x 45 60 F2–15 y z x 30 F 75 lb 45 F2–13 z y x 3 4 5 F 50 lb 45 F2–16 F 750 N z y x 45 60 F2–17 F2 800 lb F1 500 lb 3 4 5 y z x 30 45 F2–18 F2–13. Determine its coordinate direction angles of the force. F2–16. Express the force as a Cartesian vector.
- 69. 52 CHAPTER 2 FORCE VECTORS 2 2–59. Determine the coordinate angle for F2 and then express each force acting on the bracket as a Cartesian vector. *2–60. Determine the magnitude and coordinate direction angles of the resultant force acting on the bracket. g •2–61. Express each force acting on the pipe assembly in Cartesian vector form. 2–62. Determine the magnitude and direction of the resultant force acting on the pipe assembly. •2–65. The two forces F1 and F2 acting at A have a resultant force of . Determine the magnitude and coordinate direction angles of F2. 2–66. Determine the coordinate direction angles of the force F1 and indicate them on the figure. FR = 5-100k6 lb PROBLEMS y z F2 600 N F1 450 N 45 30 45 60 x Probs. 2–59/60 z y x 5 3 4 F2 400 lb F1 600 lb 120 60 Probs. 2–61/62 F y z x a b g Probs. 2–63/64 y x F2 A 30 50 F1 60 lb z B Probs. 2–65/66 2–63. The force F acts on the bracket within the octant shown. If , , and , determine the x, y, z components of F. *2–64. The force F acts on the bracket within the octant shown. If the magnitudes of the x and z components of F are and , respectively, and , determine the magnitude of F and its y component. Also, find the coordinate direction angles and . g a b = 60° Fz = 600 N Fx = 300 N g = 45° b = 60° F = 400 N
- 70. 2.6 ADDITION OF CARTESIAN VECTORS 53 2 2–67. The spur gear is subjected to the two forces caused by contact with other gears. Express each force as a Cartesian vector. *2–68. The spur gear is subjected to the two forces caused by contact with other gears. Determine the resultant of the two forces and express the result as a Cartesian vector. •2–69. If the resultant force acting on the bracket is , determine the magnitude and coordinate direction angles of F. 2–70. If the resultant force acting on the bracket is to be , determine the magnitude and coordinate direction angles of F. FR = 5800j6 N FR = 5-300i + 650j + 250k6 N •2–73. The shaft S exerts three force components on the die D. Find the magnitude and coordinate direction angles of the resultant force. Force F2 acts within the octant shown. 2–71. If , , , and , determine the magnitude and coordinate direction angles of the resultant force acting on the hook. *2–72. If the resultant force acting on the hook is , determine the magnitude and coordinate direction angles of F. FR = 5-200i + 800j + 150k6 lb F = 400 lb g = 60° b 6 90° a = 120° 135 F1 50 lb F2 180 lb 24 7 25 60 60 z y x Probs. 2–67/68 F F1 750 N y z x a b g 30 45 Probs. 2–69/70 F1 600 lb F z x y 4 3 5 a b g 30 Probs. 2–71/72 S D z y x 3 4 5 F1 400 N F3 200 N F2 300 N g2 60 a2 60 Prob. 2–73
- 71. 54 CHAPTER 2 FORCE VECTORS 2 2–74. The mast is subjected to the three forces shown. Determine the coordinate direction angles of F1 so that the resultant force acting on the mast is . 2–75. The mast is subjected to the three forces shown. Determine the coordinate direction angles of F1 so that the resultant force acting on the mast is zero. a1, b1, g1 FR = 5350i6 N a1, b1, g1 *2–76. Determine the magnitude and coordinate direction angles of F2 so that the resultant of the two forces acts along the positive x axis and has a magnitude of 500 N. •2–77. Determine the magnitude and coordinate direction angles of F2 so that the resultant of the two forces is zero. 2–79. Specify the magnitude of F3 and its coordinate direction angles so that the resultant force . FR = 59j6 kN a3, b3, g3 2–78. If the resultant force acting on the bracket is directed along the positive y axis, determine the magnitude of the resultant force and the coordinate direction angles of F so that . b 6 90° F3 300 N F2 200 N x z F1 y b1 a1 g1 Probs. 2–74/75 y x z F1 180 N F2 60 15 b2 a2 g2 Probs. 2–76/77 x y z F 500 N F1 600 N a b g 30 30 Prob. 2–78 x z 5 12 13 y F3 30 F2 10 kN F1 12 kN g3 b3 a3 Prob. 2–79
- 72. 2.6 ADDITION OF CARTESIAN VECTORS 55 2 *2–80. If , , and = 45°, determine the magnitude and coordinate direction angles of the resultant force acting on the ball-and-socket joint. f u = 30° F3 = 9 kN •2–85. Two forces F1 and F2 act on the bolt. If the resultant force FR has a magnitude of 50 lb and coordinate direction angles and , as shown, determine the magnitude of F2 and its coordinate direction angles. b = 80° a = 110° 4 3 5 F3 F2 8 kN F1 10 kN z y x u f 30 60 Prob. 2–80 z Fz Fy Fx F y x a b g Probs. 2–81/82 x y z 3 4 5 F3 45 30 F1 80 N F2 110 N FR 120 N Probs. 2–83/84 F2 80 110 x y z g F1 20 lb FR 50 lb Prob. 2–85 2–83. Three forces act on the ring. If the resultant force FR has a magnitude and direction as shown, determine the magnitude and the coordinate direction angles of force F3. *2–84. Determine the coordinate direction angles of F1 and FR. •2–81. The pole is subjected to the force F, which has components acting along the x, y, z axes as shown. If the magnitude of F is 3 kN, , and , determine the magnitudes of its three components. 2–82. The pole is subjected to the force F which has components and . If , determine the magnitudes of F and Fy. b = 75° Fz = 1.25 kN Fx = 1.5 kN g = 75° b = 30°
- 73. 56 CHAPTER 2 FORCE VECTORS 2 2.7 Position Vectors In this section we will introduce the concept of a position vector. It will be shown that this vector is of importance in formulating a Cartesian force vector directed between two points in space. x, y, z Coordinates. Throughout the book we will use a right- handed coordinate system to reference the location of points in space.We will also use the convention followed in many technical books, which requires the positive z axis to be directed upward (the zenith direction) so that it measures the height of an object or the altitude of a point.The x, y axes then lie in the horizontal plane, Fig. 2–34. Points in space are located relative to the origin of coordinates, O, by successive measurements along the x, y, z axes. For example, the coordinates of point A are obtained by starting at O and measuring xA = +4 m along the x axis, then yA = +2 m along the y axis, and finally zA = –6 m along the z axis. Thus, A(4 m, 2 m, –6 m). In a similar manner, measurements along the x, y, z axes from O to B yield the coordinates of B, i.e., B(6 m, –1 m, 4 m). Position Vector. A position vector r is defined as a fixed vector which locates a point in space relative to another point. For example, if r extends from the origin of coordinates, O, to point P(x, y, z), Fig. 2–35a, then r can be expressed in Cartesian vector form as Note how the head-to-tail vector addition of the three components yields vector r, Fig. 2–35b. Starting at the origin O, one “travels” x in the +i direction, then y in the +j direction, and finally z in the +k direction to arrive at point P(x, y, z). r = xi + yj + zk z y x 4 m 1 m 2 m O B A 2 m 4 m 6 m Fig. 2–34 z y x y j r x i O z k (a) P(x, y, z) z y x z k r x i O (b) P(x, y, z) y j Fig. 2–35
- 74. 2.7 POSITION VECTORS 57 2 In the more general case, the position vector may be directed from point A to point B in space, Fig. 2–36a. This vector is also designated by the symbol r. As a matter of convention, we will sometimes refer to this vector with two subscripts to indicate from and to the point where it is directed.Thus, r can also be designated as rAB.Also, note that rA and rB in Fig. 2–36a are referenced with only one subscript since they extend from the origin of coordinates. From Fig. 2–36a, by the head-to-tail vector addition, using the triangle rule, we require Solving for r and expressing rA and rB in Cartesian vector form yields or (2–11) Thus, the i, j, k components of the position vector r may be formed by taking the coordinates of the tail of the vector and subtracting them from the corresponding coordinates of the head .We can also form these components directly, Fig. 2–36b, by starting at A and moving through a distance of (xB – xA) along the positive x axis (+i), then (yB – yA) along the positive y axis (+j), and finally (zB – zA) along the positive z axis (+k) to get to B. B(xB, yB, zB) A(xA, yA, zA) r = (xB - xA)i + (yB - yA)j + (zB - zA)k r = rB - rA = (xBi + yB j + zBk) - (xAi + yA j + zAk) rA + r = rB z y x (a) B(xB, yB, zB) A(xA, yA, zA) rA rB r (b) z y x (xB xA)i r B A (yB yA)j (zB zA)k Fig. 2–36 A r B u If an x,y,z coordinate system is established,then the coordinates of points A and B can be determinded. From this the position vector r acting along the cable can be formulated. Its magnitude represents the length of the cable, and its unit vector, u = r/r, gives the direction defined by . a, b, g
- 75. 58 CHAPTER 2 FORCE VECTORS 2 EXAMPLE 2.12 An elastic rubber band is attached to points A and B as shown in Fig. 2–37a. Determine its length and its direction measured from A toward B. SOLUTION We first establish a position vector from A to B, Fig. 2–37b. In accordance with Eq.2–11,the coordinates of the tail A(1 m,0,–3 m) are subtracted from the coordinates of the head B(–2 m, 2 m, 3 m), which yields These components of r can also be determined directly by realizing that they represent the direction and distance one must travel along each axis in order to move from A to B, i.e., along the x axis {–3i} m, along the y axis {2j} m, and finally along the z axis {6k} m. The length of the rubber band is therefore Ans. Formulating a unit vector in the direction of r, we have The components of this unit vector give the coordinate direction angles Ans. Ans. Ans. NOTE: These angles are measured from the positive axes of a localized coordinate system placed at the tail of r, as shown in Fig. 2–37c. g = cos-1 a 6 7 b = 31.0° b = cos-1 a 2 7 b = 73.4° a = cos-1 a- 3 7 b = 115° u = r r = - 3 7 i + 2 7 j + 6 7 k r = 2(-3 m)2 + (2 m)2 + (6 m)2 = 7 m = 5-3i + 2j + 6k6 m r = [-2 m - 1 m]i + [2 m - 0]j + [3 m - (-3 m)]k (a) z y x 3 m 1 m A B 3 m 2 m 2 m (b) z y A B {6 k} {2 j} m {3 i} m r x (c) A B z¿ y¿ x¿ r 7 m g 31.0 a 115 b 73.4 Fig. 2–37
- 76. 2.8 FORCE VECTOR DIRECTED ALONG A LINE 59 2 2.8 Force Vector Directed Along a Line Quite often in three-dimensional statics problems, the direction of a force is specified by two points through which its line of action passes. Such a situation is shown in Fig. 2–38, where the force F is directed along the cord AB.We can formulate F as a Cartesian vector by realizing that it has the same direction and sense as the position vector r directed from point A to point B on the cord.This common direction is specified by the unit vector . Hence, Although we have represented F symbolically in Fig. 2–38, note that it has units of force, unlike r, which has units of length. F = Fu = Fa r r b = Fa (xB - xA)i + (yB - yA)j + (zB - zA)k 2(xB - xA)2 + (yB - yA)2 + (zB - zA)2 b u = rr z y x r u B F A Fig. 2–38 r F u Important Points • A position vector locates one point in space relative to another point. • The easiest way to formulate the components of a position vector is to determine the distance and direction that must be traveled along the x, y, z directions—going from the tail to the head of the vector. • A force F acting in the direction of a position vector r can be represented in Cartesian form if the unit vector u of the position vector is determined and it is multiplied by the magnitude of the force, i.e., F = Fu = F(r/r). The force F acting along the chain can be represented as a Cartesian vector by establishing x, y, z axes and first forming a position vector r along the length of the chain. Then the corresponding unit vector u = r/r that defines the direction of both the chain and the force can be determined. Finally, the magnitude of the force is combined with its direction, . F = Fu
- 77. 60 CHAPTER 2 FORCE VECTORS 2 EXAMPLE 2.13 The man shown in Fig. 2–39a pulls on the cord with a force of 70 lb. Represent this force acting on the support A as a Cartesian vector and determine its direction. SOLUTION Force F is shown in Fig. 2–39b. The direction of this vector, u, is determined from the position vector r, which extends from A to B. Rather than using the coordinates of the end points of the cord, r can be determined directly by noting in Fig.2–39a that one must travel from A {–24k} ft, then {–8j} ft, and finally {12i} ft to get to B.Thus, The magnitude of r, which represents the length of cord AB, is Forming the unit vector that defines the direction and sense of both r and F, we have Since F, has a magnitude of 70 lb and a direction specified by u, then Ans. The coordinate direction angles are measured between r (or F) and the positive axes of a localized coordinate system with origin placed at A, Fig. 2–39b. From the components of the unit vector: Ans. Ans. Ans. NOTE: These results make sense when compared with the angles identified in Fig. 2–39b. g = cos-1 a -24 28 b = 149° b = cos-1 a -8 28 b = 107° a = cos-1 a 12 28 b = 64.6° = 530i - 20j - 60k6 lb F = Fu = 70 lba 12 28 i - 8 28 j - 24 28 kb u = r r = 12 28 i - 8 28 j - 24 28 k r = 2(12 ft)2 + (-8 ft)2 + (-24 ft)2 = 28 ft r = 512i - 8j - 24k6 ft y x z A 30 ft 8 ft 6 ft 12 ft B (a) F 70 lb (b) x¿ y¿ z¿ A u r B g b a Fig. 2–39
- 78. 2.8 FORCE VECTOR DIRECTED ALONG A LINE 61 2 EXAMPLE 2.14 The force in Fig. 2–40a acts on the hook. Express it as a Cartesian vector. SOLUTION As shown in Fig. 2–40b, the coordinates for points A and B are and or Therefore, to go from A to B, one must travel {4i} m, then {3.464 j} m, and finally {1 k} m.Thus, Force FB expressed as a Cartesian vector becomes Ans. = 5-557i + 482j + 139k6 N FB = FB uB = (750 N)(-0.74281i + 0.6433j + 0.1857k) = -0.7428i + 0.6433j + 0.1857k uB = a rB rB b = 5-4i + 3.464j + 1k6 m 2(-4 m)2 + (3.464 m)2 + (1 m)2 B(-2 m, 3.464 m, 3 m) Bc - a 4 5 b5 sin 30° m, a 4 5 b5 cos 30° m, a 3 5 b 5 md A(2 m, 0, 2 m) 2 m (a) 2 m y x A B z 5 m 30° FB 750 N (b) y x z rB FB uB A(2 m, 0 , 2 m) B(–2 m, 3.464 m, 3 m) 3 4 5 )(5 m) 3 5 ( )(5 m) 4 5 ( Fig. 2–40
- 79. 62 CHAPTER 2 FORCE VECTORS 2 EXAMPLE 2.15 The roof is supported by cables as shown in the photo. If the cables exert forces and on the wall hook at A as shown in Fig. 2–40a, determine the resultant force acting at A. Express the result as a Cartesian vector. SOLUTION The resultant force FR is shown graphically in Fig.2–41b.We can express this force as a Cartesian vector by first formulating FAB and FAC as Cartesian vectors and then adding their components.The directions of FAB and FAC are specified by forming unit vectors uAB and uAC along the cables.These unit vectors are obtained from the associated position vectors rAB and rAC.With reference to Fig. 2–41a, to go from A to B, we must travel and, then .Thus, To go from A to C, we must travel , then , and finally .Thus, The resultant force is therefore Ans. = 5151i + 40j - 151k6 N FR = FAB + FAC = 570.7i - 70.7k6 N + 580i + 40j - 80k6 N = 580i + 40j - 80k6 N FAC = FAC a rAC rAC b = (120 N) a 4 6 i + 2 6 j - 4 6 kb rAC = 2(4 m)2 + (2 m)2 + (-4 m)2 = 6 m rAC = 54i + 2j - 4k6 m 54j6 52j6 m 5-4k6 m FAB = 570.7i - 70.7k6 N FAB = FAB a rAB rAB b = (100 N) a 4 5.66 i - 4 5.66 kb rAB = 2(4 m)2 + (-4 m)2 = 5.66 m rAB = 54i - 4k6 m 5-4i6 m 5-4k6 m FAC = 120 N FAB = 100 N y x B C A FAB FAC rAB rAC FR (b) z Fig. 2–41 (a) y x 2 m 4 m B 4 m A C FAB 100 N FAC 120 N z
- 80. 2.8 FORCE VECTOR DIRECTED ALONG A LINE 63 2 FUNDAMENTAL PROBLEMS F2–23. Determine the magnitude of the resultant force at . A F2–24. Determine the resultant force at . A F2–19. Express the position vector in Cartesian vector form, then determine its magnitude and coordinate direction angles. rAB F2–22. Express the force as a Cartesian vector. 4 m 2 m 7 m 2 m z y A B x F 900 N F2–22 z y x 6 m FB 840 N FC 420 N 3 m 3 m 2 m 2 m B C A F2–23 3 m 2 m 2 m 4 m 4 m y x A B z F 630 N F2–21 z B A y x 4 m 2 m 3 m 3 m 3 m rAB F2–19 4 ft z A y x 4 ft 2 ft B O u F2–20 4 ft 6 ft 4 ft 3 ft 4 ft 2 ft z y x FC 490 lb FB 600 lb 2 ft C B A F2–24 F2–20. Determine the length of the rod and the position vector directed from What is the angle ? u A to B. F2–21. Express the force as a Cartesian vector.
- 81. 64 CHAPTER 2 FORCE VECTORS 2 2–86. Determine the position vector r directed from point A to point B and the length of cord AB.Take . 2–87. If the cord AB is 7.5 m long, determine the coordinate position +z of point B z = 4 m *2–88. Determine the distance between the end points A and B on the wire by first formulating a position vector from A to B and then determining its magnitude. •2–89. Determine the magnitude and coordinate direction angles of the resultant force acting at A. 2–90. Determine the magnitude and coordinate direction angles of the resultant force. PROBLEMS 2 ft 4 ft 3 ft 3 ft 4 ft 2.5 ft B A x C z FC 750 lb FB 600 lb Prob. 2–89 z x B A y 1 in. 3 in. 8 in. 2 in. 30 60 Prob. 2–88 3 m 2 m 6 m z y z B x A Probs. 2–86/87 x z y C B A 600 N 500 N 8 m 4 m 4 m 2 m Prob. 2–90
- 82. 11 2.8 FORCE VECTOR DIRECTED ALONG A LINE 65 2 2–91. Determine the magnitude and coordinate direction angles of the resultant force acting at A. *2–92. Determine the magnitude and coordinate direction angles of the resultant force. 2–95. Express force F as a Cartesian vector; then determine its coordinate direction angles. 120 z y 120 4 ft A B C 6 ft O FA FB FC x 120 Probs. 2–93/94 A C B 4 ft 7 ft 3 ft x y z F2 81 lb F1 100 lb 40 4 ft Prob. 2–92 y x B C A 6 m 3 m 45 4.5 m 6 m FB 900 N FC 600 N z Prob. 2–91 y x z B A 10 ft 70 30 7 ft 5 ft F 135 lb Prob. 2–95 •2–93. The chandelier is supported by three chains which are concurrent at point O. If the force in each chain has a magnitude of 60 lb, express each force as a Cartesian vector and determine the magnitude and coordinate direction angles of the resultant force. 2–94. The chandelier is supported by three chains which are concurrent at point O. If the resultant force at O has a magnitude of 130 lb and is directed along the negative z axis, determine the force in each chain.
- 83. 66 CHAPTER 2 FORCE VECTORS 2 y B C D A x z 4 m 4 m 1.5 m 1 m 3 m 2 m FA 250 N FB 175 N Prob. 2–98 x y z 2.5 m 1.5 m 0.5 m 1 m 30 A C B D FA 300 N FC 250 N Prob. 2–97 x z y x y 6 m 4 m 18 m C A D 400 N 800 N 600 N 24 m O 16 m B Prob. 2–96 *2–96. The tower is held in place by three cables. If the force of each cable acting on the tower is shown, determine the magnitude and coordinate direction angles of the resultant force.Take , . y = 15 m x = 20 m a, b, g •2–97. The door is held opened by means of two chains. If the tension in AB and CD is and , respectively, express each of these forces in Cartesian vector form. FC = 250 N FA = 300 N 2–98. The guy wires are used to support the telephone pole. Represent the force in each wire in Cartesian vector form. Neglect the diameter of the pole. z A x y 6 m 1500 N 3 m FB FC B C 2 m x z Probs. 2–99/100 2–99. Two cables are used to secure the overhang boom in position and support the 1500-N load. If the resultant force is directed along the boom from point A towards O, determine the magnitudes of the resultant force and forces FB and FC. Set and . *2–100. Two cables are used to secure the overhang boom in position and support the 1500-N load. If the resultant force is directed along the boom from point A towards O, determine the values of x and z for the coordinates of point C and the magnitude of the resultant force. Set and . FC = 2400 N FB = 1610 N z = 2 m x = 3 m
- 84. 2.8 FORCE VECTOR DIRECTED ALONG A LINE 67 2 •2–101. The cable AO exerts a force on the top of the pole of . If the cable has a length of 34 ft, determine the height z of the pole and the location (x, y) of its base. F = 5-120i - 90j - 80k6 lb 2–102. If the force in each chain has a magnitude of 450 lb, determine the magnitude and coordinate direction angles of the resultant force. 2–103. If the resultant of the three forces is , determine the magnitude of the force in each chain. FR = 5-900k6 lb *2–104. The antenna tower is supported by three cables. If the forces of these cables acting on the antenna are , , and , determine the magnitude and coordinate direction angles of the resultant force acting at A. FD = 560 N FC = 680 N FB = 520 N •2–105. If the force in each cable tied to the bin is 70 lb, determine the magnitude and coordinate direction angles of the resultant force. 2–106. If the resultant of the four forces is , determine the tension developed in each cable. Due to symmetry, the tension in the four cables is the same. FR = 5-360k6 lb 24 m 10 m 18 m 8 m 16 m 12 m 18 m z x y A O C B D FB FC FD Prob. 2–104 120 120 3 ft 7 ft 120 FA FB FC z C A D B y x Probs. 2–102/103 y z A z x F x y O Prob. 2–101 z B C E D A x y 6 ft 3 ft 3 ft 2 ft 2 ft FC FD FB FA Probs. 2–105/106
- 85. 68 CHAPTER 2 FORCE VECTORS 2 x z y D C A B 3 m 30 0.75 m 45 FB 8 kN FC 5 kN FA 6 kN Prob. 2–109 2 m 1 m 30 120 120 B A z y x F 200 N Prob. 2–108 3 ft 20 y x A B z 5 ft 6 ft F 12 lb Prob. 2–107 2–107. The pipe is supported at its end by a cord AB. If the cord exerts a force of on the pipe at A, express this force as a Cartesian vector. F = 12 lb *2–108. The load at A creates a force of 200 N in wire AB. Express this force as a Cartesian vector, acting on A and directed towards B. •2–109. The cylindrical plate is subjected to the three cable forces which are concurrent at point D. Express each force which the cables exert on the plate as a Cartesian vector, and determine the magnitude and coordinate direction angles of the resultant force. 2–110. The cable attached to the shear-leg derrick exerts a force on the derrick of . Express this force as a Cartesian vector. F = 350 lb 30 50 ft 35 ft x y z A B F 350 lb Prob. 2–110
- 86. 2.9 DOT PRODUCT 69 2 2.9 Dot Product Occasionally in statics one has to find the angle between two lines or the components of a force parallel and perpendicular to a line.In two dimensions, these problems can readily be solved by trigonometry since the geometry is easy to visualize. In three dimensions, however, this is often difficult, and consequently vector methods should be employed for the solution.The dot product, which defines a particular method for “multiplying” two vectors, will be is used to solve the above-mentioned problems. The dot product of vectors A and B, written A · B, and read “A dot B” is defined as the product of the magnitudes of A and B and the cosine of the angle between their tails, Fig. 2–41. Expressed in equation form, (2–12) where . The dot product is often referred to as the scalar product of vectors since the result is a scalar and not a vector. Laws of Operation. 1. Commutative law: 2. Multiplication by a scalar: 3. Distributive law: It is easy to prove the first and second laws by using Eq. 2–12. The proof of the distributive law is left as an exercise (see Prob. 2–111). Cartesian Vector Formulation. Equation 2–12 must be used to find the dot product for any two Cartesian unit vectors. For example, and . If we want to find the dot product of two general vectors A and B that are expressed in Cartesian vector form, then we have Carrying out the dot-product operations, the final result becomes (2–13) Thus, to determine the dot product of two Cartesian vectors, multiply their corresponding x, y, z components and sum these products algebraically. Note that the result will be either a positive or negative scalar. A # B = AxBx + AyBy + AzBz + AzBx(k # i) + AzBy(k # j) + AzBz(k # k) + AyBx(j # i) + (AyBy(j # j) + AyBz(j # k) = AxBx(i # i) + AxBy(i # j) + AxBz(i # k) A # B = (Axi + Ayj + Azk) # (Bxi + Byj + Bzk) i # j = (1)(1) cos 90° = 0 i # i = (1)(1) cos 0° = 1 A # (B + D) = (A # B) + (A # D) a(A # B) = (aA) # B = A # (aB) A # B = B # A 0° … u … 180° A # B = AB cos u u A B u Fig. 2–41
- 87. 70 CHAPTER 2 FORCE VECTORS 2 Applications. The dot product has two important applications in mechanics. • The angle formed between two vectors or intersecting lines. The angle between the tails of vectors A and B in Fig. 2–41 can be determined from Eq. 2–12 and written as Here is found from Eq. 2–13. In particular, notice that if , so that A will be perpendicular to B. • The components of a vector parallel and perpendicular to a line. The component of vector A parallel to or collinear with the line in Fig. 2–43 is defined by Aa where .This component is sometimes referred to as the projection of A onto the line, since a right angle is formed in the construction. If the direction of the line is specified by the unit vector ua, then since ua = 1, we can determine the magnitude of Aa directly from the dot product (Eq. 2–12); i.e., Hence, the scalar projection of A along a line is determined from the dot product of A and the unit vector ua which defines the direction of the line. Notice that if this result is positive, then Aa has a directional sense which is the same as ua, whereas if Aa is a negative scalar, then Aa has the opposite sense of direction to ua The component Aa represented as a vector is therefore The component of A that is perpendicular to line aa can also be obtained, Fig. 2–43. Since , then . There are two possible ways of obtaining . One way would be to determine from the dot product, , then .Alternatively, if Aa is known, then by Pythagorean’s theorem we can also write . A⬜ = 2A2 - Aa 2 A⬜ = A sin u u = cos-1 (A # uA A) u A⬜ A⬜ = A - Aa A = Aa + A⬜ Aa = Aa ua Aa = A cos u = A # ua Aa = A cos u aa¿ u = cos-1 0 = 90° A # B = 0 A # B u = cos-1 a A # B AB b 0° … u … 180° u The angle between the rope and the connecting beam can be determined by formulating unit vectors along the beam and rope and then using the dot product . ub # ur = (1)(1) cos u u ur uA A u The projection of the cable force F along the beam can be determined by first finding the unit vector that defines this direction.Then apply the dot product, . Fb = F # ub ub F Fb ub A⬜ a a ua Aa A cos u ua A u Fig. 2–43
- 88. 2.9 DOT PRODUCT 71 2 Important Points • The dot product is used to determine the angle between two vectors or the projection of a vector in a specified direction. • If vectors A and B are expressed in Cartesian vector form, the dot product is determined by multiplying the respective x, y, z scalar components and algebraically adding the results, i.e., . • From the definition of the dot product, the angle formed between the tails of vectors A and B is . • The magnitude of the projection of vector A along a line whose direction is specified by ua is determined from the dot product . Aa = A # ua aa u = cos-1 (A # BAB) A # B = AxBx + AyBy + AzBz EXAMPLE 2.16 Determine the magnitudes of the projection of the force F in Fig. 2–44 onto the u and axes. v SOLUTION Projections of Force. The graphical representation of the projections is shown in Fig.2–44.From this figure,the magnitudes of the projections of F onto the u and v axes can be obtained by trigonometry: Ans. Ans. NOTE: These projections are not equal to the magnitudes of the components of force F along the u and axes found from the parallelogram law. They will only be equal if the u and axes are perpendicular to one another. v v (Fv)proj = (100 N)cos 15° = 96.6 N (Fu)proj = (100 N)cos 45° = 70.7 N F 100 N u (Fu)proj v 15 45 (F )proj v Fig. 2–44
- 89. 72 CHAPTER 2 FORCE VECTORS 2 EXAMPLE 2.17 (a) z y x 6 m 2 m 3 m A B F {300 j} N (b) F F FAB z y x A B uB Fig 2–45 The frame shown in Fig. 2–45a is subjected to a horizontal force F = {300j}. Determine the magnitude of the components of this force parallel and perpendicular to member AB. SOLUTION The magnitude of the component of F along AB is equal to the dot product of F and the unit vector uB, which defines the direction of AB, Fig. 2–44b. Since then Ans. Since the result is a positive scalar, FAB has the same sense of direction as uB, Fig. 2–45b. Expressing FAB in Cartesian vector form, we have Ans. The perpendicular component, Fig. 2–45b, is therefore Its magnitude can be determined either from this vector or by using the Pythagorean theorem, Fig. 2–45b: Ans. = 155 N = 2(300 N)2 - (257.1 N)2 F⬜ = 2F2 - F2 AB = 5-73.5i + 80j - 110k6 N F⬜ = F - FAB = 300j - (73.5i + 220j + 110k) = 573.5i + 220j + 110k6N = (257.1 N)(0.286i + 0.857j + 0.429k) FAB = FABuB = 257.1 N = (0)(0.286) + (300)(0.857) + (0)(0.429) FAB = F cos u = F # uB = (300j) # (0.286i + 0.857j + 0.429k) uB = rB rB = 2i + 6j + 3k 2(2)2 + (6)2 + (3)2 = 0.286i + 0.857j + 0.429k
- 90. 2.9 DOT PRODUCT 73 2 EXAMPLE 2.18 F 80 lb 2 ft 2 ft 1 ft B 1 ft y x z (a) C A u (c) x F 80 lb F z y A B FBA u B y x z (b) C A u rBC rBA Fig. 2–46 The pipe in Fig. 2–46a is subjected to the force of F = 80 lb. Determine the angle between F and the pipe segment BA and the projection of F along this segment. u SOLUTION Angle . First we will establish position vectors from B to A and B to C; Fig. 2–46b.Then we will determine the angle between the tails of these two vectors. Thus, Ans. Components of F. The component of F along BA is shown in Fig. 2–46b.We must first formulate the unit vector along BA and force F as Cartesian vectors. Thus, Ans. = 59.0 lb = 0 a- 2 3 b + (-75.89)a- 2 3 b + (25.30) a 1 3 b FBA = F # uBA = (-75.89j + 25.30k) # a - 2 3 i - 2 3 j + 1 3 kb F = 80 lba rBC rBC b = 80a -3j + 1k 210 b = -75.89j + 25.30k uBA = rBA rBA = (-2i - 2j + 1k) 3 = - 2 3 i - 2 3 j + 1 3 k u = 42.5° = 0.7379 cos u = rBA # rBC rBArBC = (-2)(0) + (-2)(-3) + (1)(1) 3210 rBC = 5-3j + 1k6 ft, rBC = 210ft rBA = 5-2i - 2j + 1k6 ft, rBA = 3 ft u u NOTE: Since is known, then also, . FBA = F cos u = 80 lb cos 42.5º = 59.0 lb u
- 91. 74 CHAPTER 2 FORCE VECTORS 2 2 m 2 m 1 m z y A O x F {6 i 9 j 3 k} kN u F2–25 FUNDAMENTAL PROBLEMS F2–26. Determine the angle between the force and the line . AB u F2–27. Determine the angle between the force and the line . F2–28. Determine the component of projection of the force along the line . OA OA u F2–30. Determine the components of the force acting parallel and perpendicular to the axis of the pole. F2–25. Determine the angle between the force and the line AO. u F2–29. Find the magnitude of the projected component of the force along the pipe. F 650 N x A O y 13 12 5 u F2–27/28 O z y x 4 m 6 m 5 m B A F 400 N 4 m F2–29 y x z A F 600 N C B 4 m 4 m 3 m u F2–26 z x y A F 600 lb 60 30 4 ft 2 ft 4 ft O F2–30
- 92. 2.9 DOT PRODUCT 75 2 z x y C B A 3 m 1.5 m 1 m 3 m FAB 560 N 1.5 m Prob. 2–112 2–111. Given the three vectors A, B, and D, show that . *2–112. Determine the projected component of the force acting along cable AC. Express the result as a Cartesian vector. FAB = 560 N A # (B + D) = (A # B) + (A # D) •2–113. Determine the magnitudes of the components of force acting along and perpendicular to line AO. F = 56 N 2–115. Determine the magnitudes of the components of acting along and perpendicular to segment DE of the pipe assembly. F = 600 N PROBLEMS y x A C B z 1 m 4 m 3 m 3 m 1 m 5 m u Prob. 2–114 y x z C O D A B 3 m 1.5 m 1 m 1 m F 56 N Prob. 2–113 x y E D C B A z 2 m 2 m 2 m 2 m 3 m F 600 N Prob. 2–115 2–114. Determine the length of side BC of the triangular plate. Solve the problem by finding the magnitude of rBC; then check the result by first finding q , rAB, and rAC and then using the cosine law.
- 93. 76 CHAPTER 2 FORCE VECTORS 2 z O x y 40 mm 40 mm 20 mm F {500 k} N A Prob. 2–119 F 80 N A E B y F C x D z 2 m 2 m 1.5 m 1.5 m 2 m 2 m Prob. 2–118 x z y 45 60 120 F1 600 N F2 {120i + 90j – 80k}N u Probs. 2–116/117 *2–116. Two forces act on the hook. Determine the angle between them.Also, what are the projections of F1 and F2 along the y axis? •2–117. Two forces act on the hook. Determine the magnitude of the projection of F2 along F1. u 2–118. Determine the projection of force along line BC. Express the result as a Cartesian vector. F = 80 N 2–119. The clamp is used on a jig. If the vertical force acting on the bolt is , determine the magnitudes of its components F1 and F2 which act along the OA axis and perpendicular to it. F = {-500k} N *2–120. Determine the magnitude of the projected component of force FAB acting along the z axis. •2–121. Determine the magnitude of the projected component of force FAC acting along the z axis. 12 ft 18 ft 12 ft x B D C A O y z 12 ft 36 ft FAB 700 lb FAC 600 lb 30 Probs. 2–120/121
- 94. 2.9 DOT PRODUCT 77 2 2–122. Determine the projection of force acting along line AC of the pipe assembly. Express the result as a Cartesian vector. 2–123. Determine the magnitudes of the components of force acting parallel and perpendicular to segment BC of the pipe assembly. F = 400 N F = 400 N *2–124. Cable OA is used to support column OB. Determine the angle it makes with beam OC. •2–125. Cable OA is used to support column OB. Determine the angle it makes with beam OD. f u 2–126. The cables each exert a force of 400 N on the post. Determine the magnitude of the projected component of F1 along the line of action of F2. 2–127. Determine the angle between the two cables attached to the post. u *2–128. A force of is applied to the handle of the wrench. Determine the angle between the tail of the force and the handle AB. u F = 80 N x z y 20 35 45 60 120 F1 400 N F2 400 N u Probs. 2–126/127 z x C B O D y 4 m 30 8 m 8 m A u f Probs. 2–124/125 x A B C y z 4 m 3 m F 400 N 30 45 Probs. 2–122/123 x z B A y 300 mm 500 mm F 80 N 30 45 u Prob. 2–128
- 95. 78 CHAPTER 2 FORCE VECTORS 2 z A O x y 300 mm 300 mm 300 mm F 300 N 30 30 Prob. 2–132 60 y z 60 30 30 x F2 25 lb F1 30 lb u Probs. 2–133/134 z A O x y 300 mm 300 mm 300 mm F 300 N 30 30 Prob. 2–131 y z x 8 ft 3 ft 12 ft 8 ft 15 ft A C B F u Probs. 2–129/130 •2–129. Determine the angle between cables AB and AC. 2–130. If F has a magnitude of 55 lb, determine the magnitude of its projected components acting along the x axis and along cable AC. u 2–131. Determine the magnitudes of the projected components of the force acting along the x and y axes. F = 300 N *2–132. Determine the magnitude of the projected component of the force acting along line OA. F = 300 N •2–133. Two cables exert forces on the pipe. Determine the magnitude of the projected component of F1 along the line of action of F2. 2–134. Determine the angle between the two cables attached to the pipe. u
- 96. CHAPTER REVIEW 79 2 CHAPTER REVIEW A scalar is a positive or negative number; e.g., mass and temperature. A vector has a magnitude and direction, where the arrowhead represents the sense of the vector. Multiplication or division of a vector by a scalar will change only the magnitude of the vector. If the scalar is negative, the sense of the vector will change so that it acts in the opposite sense. If vectors are collinear, the resultant is simply the algebraic or scalar addition. Parallelogram Law Two forces add according to the parallelogram law. The components form the sides of the parallelogram and the resultant is the diagonal. To find the components of a force along any two axes, extend lines from the head of the force, parallel to the axes, to form the components. To obtain the components or the resultant, show how the forces add by tip-to-tail using the triangle rule, and then use the law of cosines and the law of sines to calculate their values. F1 sin u1 = F2 sin u2 = FR sin uR FR = 2F1 2 + F2 2 - 2 F1F2 cos uR R = A + B A 2 A 0.5 A 1.5 A A a b Components Resultant FR F1 F2 u1 u2 uR FR F1 F2 A B R
- 97. 80 CHAPTER 2 FORCE VECTORS 2 Rectangular Components: Two Dimensions Vectors Fx and Fy are rectangular components of F. The resultant force is determined from the algebraic sum of its components. u = tan-1 2 FRy FRx 2 FR = 2(FRx)2 + (FRy)2 FRy = ©Fy FRx = ©Fx Cartesian Vectors The unit vector u has a length of one, no units, and it points in the direction of the vector F. u = F F A force can be resolved into its Cartesian components along the x, y, z axes so that . The magnitude of F is determined from the positive square root of the sum of the squares of its components. F = Fxi + Fyj + Fzk The coordinate direction angles are determined by formulating a unit vector in the direction of F. The x, y, z components of u represent cos , cos , cos . g b a a, b, g u = cos a i + cos b j + cos g k u = F F = Fx F i + Fy F j + Fz F k F = 2Fx 2 + F2 y + F2 z F Fy y x Fx Fx i x F z Fz k y Fy j a b u g u 1 F F x y FRy FR FRx x y F2x F2y F1y F1x F3x F3y
- 98. CHAPTER REVIEW 81 2 The coordinate direction angles are related so that only two of the three angles are independent of one another. cos2 a + cos2 b + cos2 g = 1 To find the resultant of a concurrent force system, express each force as a Cartesian vector and add the i, j, k components of all the forces in the system. FR = ©F = ©Fxi + ©Fyj + ©Fzk Position and Force Vectors A position vector locates one point in space relative to another.The easiest way to formulate the components of a position vector is to determine the distance and direction that one must travel along the x, y, and z directions—going from the tail to the head of the vector. If the line of action of a force passes through points A and B, then the force acts in the same direction as the position vector r, which is defined by the unit vector u. The force can then be expressed as a Cartesian vector. Dot Product The dot product between two vectors A and B yields a scalar. If A and B are expressed in Cartesian vector form, then the dot product is the sum of the products of their x, y, and z components = AxBx + AyBy + AzBz A # B = AB cos u The dot product can be used to determine the angle between A and B. The dot product is also used to determine the projected component of a vector A onto an axis defined by its unit vector ua. aa Aa = A cos u ua = (A # ua)ua u = cos-1 a A # B AB b F = Fu = Fa r r b + (zB - zA)k + (yB - yA)j r = (xB - xA)i y r B A x (xB xA)i (yB yA)j z (zB zA)k z y x u B r F A A a u A u Aa A cos ua u a A B u
- 99. 82 CHAPTER 2 FORCE VECTORS 2 REVIEW PROBLEMS 2–138. Determine the magnitude and direction of the resultant of the three forces by first finding the resultant and then forming . Specify its direction measured counter- clockwise from the positive x axis. FR = F¿ + F2 F¿ = F1 + F3 FR = F1 + F2 + F3 2–135. Determine the x and y components of the 700-lb force. 2–139. Determine the design angle ( 90°) between the two struts so that the 500-lb horizontal force has a component of 600 lb directed from A toward C.What is the component of force acting along member BA? u u *2–136. Determine the magnitude of the projected component of the 100-lb force acting along the axis BC of the pipe. •2–137. Determine the angle between pipe segments BA and BC. u y x 700 lb 30 60 Prob. 2–135 y C B A D z 8 ft 3 ft 6 ft 2 ft 4 ft x F 100 lb u Probs. 2–136/137 x y 30 30 45 F1 80 N F2 75 N F3 50 N Prob. 2–138 500 lb 20 A B C u Prob. 2–139
- 100. REVIEW PROBLEMS 83 2 2–142. Cable AB exerts a force of 80 N on the end of the 3-m-long boom OA. Determine the magnitude of the projection of this force along the boom. *2–140. Determine the magnitude and direction of the smallest force F3 so that the resultant force of all three forces has a magnitude of 20 lb. 2–143. The three supporting cables exert the forces shown on the sign. Represent each force as a Cartesian vector. •2–141. Resolve the 250-N force into components acting along the u and axes and determine the magnitudes of these components. v F2 10 lb F3 4 3 5 F1 5 lb u Prob. 2–140 u v 40 20 250 N Prob. 2–141 O A 80 N 3 m B z y x 4 m 60 Prob. 2–142 2 m z C 2 m y x A D E B 3 m 3 m 2 m FB 400 N FC 400 N FE 350 N Prob. 2–143
- 101. Whenever cables are used for hoisting loads, they must be selected so that they do not fail when they are placed at their points of attachment. In this chapter, we will show how to calculate cable loadings for such cases.
- 102. Equilibrium of a Particle CHAPTER OBJECTIVES • To introduce the concept of the free-body diagram for a particle. • To show how to solve particle equilibrium problems using the equations of equilibrium. 3.1 Condition for the Equilibrium of a Particle A particle is said to be in equilibrium if it remains at rest if originally at rest, or has a constant velocity if originally in motion. Most often, however, the term “equilibrium” or, more specifically, “static equilibrium” is used to describe an object at rest.To maintain equilibrium, it is necessary to satisfy Newton’s first law of motion, which requires the resultant force acting on a particle to be equal to zero.This condition may be stated mathematically as (3–1) where is the vector sum of all the forces acting on the particle. Not only is Eq. 3–1 a necessary condition for equilibrium, it is also a sufficient condition. This follows from Newton’s second law of motion, which can be written as Since the force system satisfies Eq. 3–1, then and therefore the particle’s acceleration Consequently, the particle indeed moves with constant velocity or remains at rest. a = 0. ma = 0, ©F = ma. ©F ©F = 0 3
- 103. 86 CHAPTER 3 EQUILIBRIUM OF A PARTICLE 3.2 The Free-Body Diagram To apply the equation of equilibrium, we must account for all the known and unknown forces which act on the particle. The best way to do this is to think of the particle as isolated and “free” from its surroundings. A drawing that shows the particle with all the forces that act on it is called a free-body diagram (FBD). Before presenting a formal procedure as to how to draw a free-body diagram, we will first consider two types of connections often encountered in particle equilibrium problems. Springs. If a linearly elastic spring (or cord) of undeformed length lo is used to support a particle, the length of the spring will change in direct proportion to the force F acting on it, Fig. 3–1. A characteristic that defines the “elasticity” of a spring is the spring constant or stiffness k. The magnitude of force exerted on a linearly elastic spring which has a stiffness k and is deformed (elongated or compressed) a distance , measured from its unloaded position, is (3–2) If s is positive, causing an elongation, then F must pull on the spring; whereas if s is negative, causing a shortening, then F must push on it. For example, if the spring in Fig. 3–1 has an unstretched length of 0.8 m and a stiffness and it is stretched to a length of 1 m, so that then a force is needed. Cables and Pulleys. Unless otherwise stated, throughout this book, except in Sec. 7.4, all cables (or cords) will be assumed to have negligible weight and they cannot stretch.Also, a cable can support only a tension or “pulling” force, and this force always acts in the direction of the cable. In Chapter 5, it will be shown that the tension force developed in a continuous cable which passes over a frictionless pulley must have a constant magnitude to keep the cable in equilibrium. Hence, for any angle shown in Fig. 3–2, the cable is subjected to a constant tension T throughout its length. u, 500 Nm(0.2 m) = 100 N F = ks = s = l - lo = 1 m - 0.8 m = 0.2 m, k = 500 Nm F = ks s = l - lo 1©F2 3 F s o Fig. 3–1 T T Cable is in tension u Fig. 3–2
- 104. 3.2 THE FREE-BODY DIAGRAM 87 3 The bucket is held in equilibrium by the cable, and instinctively we know that the force in the cable must equal the weight of the bucket. By drawing a free-body diagram of the bucket we can understand why this is so. This diagram shows that there are only two forces acting on the bucket, namely, its weight W and the force T of the cable. For equilibrium, the resultant of these forces must be equal to zero, and so T = W. W T The spool has a weight W and is suspended from the crane boom. If we wish to obtain the forces in cables AB and AC, then we should consider the free-body diagram of the ring at A.Here the cables AD exert a resultant force of W on the ring and the condition of equilibrium is used to obtain and TC. TB TC TB W D C A A B Procedure for Drawing a Free-Body Diagram Since we must account for all the forces acting on the particle when applying the equations of equilibrium,the importance of first drawing a free-body diagram cannot be overemphasized.To construct a free- body diagram, the following three steps are necessary. Draw Outlined Shape. Imagine the particle to be isolated or cut “free” from its surroundings by drawing its outlined shape. Show All Forces. Indicate on this sketch all the forces that act on the particle. These forces can be active forces, which tend to set the particle in motion, or they can be reactive forces which are the result of the constraints or supports that tend to prevent motion. To account for all these forces, it may be helpful to trace around the particle’s boundary, carefully noting each force acting on it. Identify Each Force. The forces that are known should be labeled with their proper magnitudes and directions. Letters are used to represent the magnitudes and directions of forces that are unknown.
- 105. 88 CHAPTER 3 EQUILIBRIUM OF A PARTICLE 3 EXAMPLE 3.1 45 60 C E B A (a) D k Fig. 3–3 FCE (Force of cord CE acting on sphere) 58.9 N (Weight or gravity acting on sphere) (b) FCE (Force of sphere acting on cord CE) FEC (Force of knot acting on cord CE) (c) C FCBA (Force of cord CBA acting on knot) FCD (Force of spring acting on knot) FCE (Force of cord CE acting on knot) 60 (d) The sphere in Fig. 3–3a has a mass of 6 kg and is supported as shown. Draw a free-body diagram of the sphere,the cord CE,and the knot at C. SOLUTION Sphere. By inspection, there are only two forces acting on the sphere, namely, its weight, 6 kg , and the force of cord CE.The free-body diagram is shown in Fig. 3–3b. Cord CE. When the cord CE is isolated from its surroundings, its free-body diagram shows only two forces acting on it, namely, the force of the sphere and the force of the knot, Fig. 3–3c. Notice that shown here is equal but opposite to that shown in Fig. 3–3b, a consequence of Newton’s third law of action–reaction. Also, and pull on the cord and keep it in tension so that it doesn’t collapse. For equilibrium, Knot. The knot at C is subjected to three forces, Fig. 3–3d. They are caused by the cords CBA and CE and the spring CD. As required, the free-body diagram shows all these forces labeled with their magnitudes and directions. It is important to recognize that the weight of the sphere does not directly act on the knot. Instead, the cord CE subjects the knot to this force. FCE = FEC. FEC FCE FCE 19.81 ms2 2 = 58.9 N
- 106. 3.3 COPLANAR FORCE SYSTEMS 89 3 3.3 Coplanar Force Systems If a particle is subjected to a system of coplanar forces that lie in the x–y plane as in Fig. 3–4, then each force can be resolved into its i and j components. For equilibrium, these forces must sum to produce a zero force resultant, i.e., For this vector equation to be satisfied, the force’s x and y components must both be equal to zero. Hence, (3–3) These two equations can be solved for at most two unknowns, generally represented as angles and magnitudes of forces shown on the particle’s free-body diagram. When applying each of the two equations of equilibrium, we must account for the sense of direction of any component by using an algebraic sign which corresponds to the arrowhead direction of the component along the x or y axis. It is important to note that if a force has an unknown magnitude, then the arrowhead sense of the force on the free-body diagram can be assumed. Then if the solution yields a negative scalar, this indicates that the sense of the force is opposite to that which was assumed. For example, consider the free-body diagram of the particle subjected to the two forces shown in Fig. 3–5. Here it is assumed that the unknown force F acts to the right to maintain equilibrium. Applying the equation of equilibrium along the x axis, we have Both terms are “positive” since both forces act in the positive x direction. When this equation is solved, Here the negative sign indicates that F must act to the left to hold the particle in equilibrium, Fig. 3–5. Notice that if the axis in Fig. 3–5 were directed to the left, both terms in the above equation would be negative, but again, after solving, indicating that F would be directed to the left. F = -10 N, +x F = -10 N. +F + 10 N = 0 : + ©Fx = 0; ©Fx = 0 ©Fy = 0 ©Fxi + ©Fyj = 0 ©F = 0 y F2 F1 F3 F4 x Fig. 3–4 F x 10 N Fig. 3–5
- 107. 90 CHAPTER 3 EQUILIBRIUM OF A PARTICLE 3 Procedure for Analysis Coplanar force equilibrium problems for a particle can be solved using the following procedure. Free-Body Diagram. • Establish the x, y axes in any suitable orientation. • Label all the known and unknown force magnitudes and directions on the diagram. • The sense of a force having an unknown magnitude can be assumed. Equations of Equilibrium. • Apply the equations of equilibrium, and • Components are positive if they are directed along a positive axis, and negative if they are directed along a negative axis. • If more than two unknowns exist and the problem involves a spring, apply to relate the spring force to the deformation s of the spring. • Since the magnitude of a force is always a positive quantity, then if the solution for a force yields a negative result, this indicates its sense is the reverse of that shown on the free-body diagram. F = ks ©Fy = 0. ©Fx = 0 The chains exert three forces on the ring at A, as shown on its free-body diagram. The ring will not move, or will move with constant velocity, provided the summation of these forces along the x and along the y axis is zero. If one of the three forces is known, the magnitudes of the other two forces can be obtained from the two equations of equilibrium. TC TB TD y x B D A A C
- 108. 3.3 COPLANAR FORCE SYSTEMS 91 3 Determine the tension in cables BA and BC necessary to support the 60-kg cylinder in Fig. 3-6a. SOLUTION Free-Body Diagram. Due to equilibrium, the weight of the cylinder causes the tension in cable to be , Fig. 3-6b. The forces in cables and BC can be determined by investigating the equilibrium of ring .Its free-body diagram is shown in Fig.3-6c.The magnitudes of and are unknown, but their directions are known. Equations of Equilibrium. Applying the equations of equilibrium along the x and y axes, we have (1) (2) Equation (1) can be written as . Substituting this into Eq. (2) yields So that Ans. Substituting this result into either Eq. (1) or Eq. (2), we get Ans. NOTE: The accuracy of these results, of course, depends on the accuracy of the data, i.e., measurements of geometry and loads. For most engineering work involving a problem such as this, the data as measured to three significant figures would be sufficient. TA = 420 N TC = 475.66 N = 476 N TC sin 45° + A3 5 B(0.8839TC) - 60(9.81) N = 0 TA = 0.8839TC TC sin 45° + A3 5 BTA - 60(9.81) N = 0 + c©Fy = 0; TC cos 45° - A4 5 BTA = 0 : + ©Fx = 0; TC TA B BA TBD = 60(9.81) N BD EXAMPLE 3.2 (a) B 3 4 5 A D C 45 Fig. 3–6 60 (9.81) N TBD 60 (9.81) N (b) TBD 60 (9.81) N TA TC y x (c) B 3 4 5 45
- 109. EXAMPLE 3.3 92 CHAPTER 3 EQUILIBRIUM OF A PARTICLE 3 The 200-kg crate in Fig. 3-7a is suspended using the ropes and . Each rope can withstand a maximum force of before it breaks. If always remains horizontal, determine the smallest angle to which the crate can be suspended before one of the ropes breaks. u AB 10 kN AC AB SOLUTION Free-Body Diagram. We will study the equilibrium of ring .There are three forces acting on it, Fig. 3-7b.The magnitude of is equal to the weight of the crate, i.e., . Equations of Equilibrium. Applying the equations of equilibrium along the x and y axes, ; (1) (2) From Eq. (1), is always greater than since . Therefore, rope will reach the maximum tensile force of before rope . Substituting into Eq. (2), we get Ans. The force developed in rope can be obtained by substituting the values for and into Eq. (1). FB = 9.81 kN 10(103 ) N = FB cos11.31° FC u AB u = sin-1 (0.1962) = 11.31° = 11.3° [10(103 )N] sin u - 1962 N = 0 FC = 10 kN AB 10 kN AC cos u … 1 FB FC FC sin u - 1962 N = 0 + c©Fy = 0; FC = FB cos u - FC cos u + FB = 0 : + ©Fx = 0; FD = 200 (9.81) N = 1962 N 6 10 kN FD A (a) D A B C u Fig. 3–7 FD 1962 N y x (b) A FC FB u
- 110. EXAMPLE 3.4 3.3 COPLANAR FORCE SYSTEMS 93 3 Determine the required length of cord AC in Fig. 3–8a so that the 8-kg lamp can be suspended in the position shown. The undeformed length of spring AB is and the spring has a stiffness of kAB = 300 Nm. l¿ AB = 0.4 m, (a) A B 300 N/m 30 2 m C kAB Fig. 3–8 SOLUTION If the force in spring AB is known, the stretch of the spring can be found using From the problem geometry, it is then possible to calculate the required length of AC. Free-Body Diagram. The lamp has a weight and so the free-body diagram of the ring at A is shown in Fig. 3–8b. Equations of Equilibrium. Using the x, y axes, Solving, we obtain The stretch of spring AB is therefore so the stretched length is The horizontal distance from C to B, Fig. 3–8a, requires Ans. lAC = 1.32 m 2 m = lAC cos 30° + 0.853 m lAB = 0.4 m + 0.453 m = 0.853 m lAB = l¿ AB + sAB sAB = 0.453 m 135.9 N = 300 Nm1sAB2 TAB = kABsAB; TAB = 135.9 N TAC = 157.0 N TAC sin 30° - 78.5 N = 0 + c©Fy = 0; TAB - TAC cos 30° = 0 : + ©Fx = 0; W = 819.812 = 78.5 N F = ks. y x W 78.5 N A (b) 30 TAC TAB
- 111. 94 CHAPTER 3 EQUILIBRIUM OF A PARTICLE 3 FUNDAMENTAL PROBLEMS All problem solutions must include an FBD. F3–1. The crate has a weight of 550 lb. Determine the force in each supporting cable. F3–2. The beam has a weight of 700 lb. Determine the shortest cable ABC that can be used to lift it if the maximum force the cable can sustain is 1500 lb. F3–3. If the 5-kg block is suspended from the pulley B and the sag of the cord is d = 0.15 m, determine the force in cord ABC. Neglect the size of the pulley. F3–4. The block has a mass of 5 kg and rests on the smooth plane. Determine the unstretched length of the spring. F3–5. If the mass of cylinder C is 40 kg, determine the mass of cylinder A in order to hold the assembly in the position shown. F3–6. Determine the tension in cables AB, BC, and CD, necessary to support the 10-kg and 15-kg traffic lights at B and C, respectively.Also, find the angle . u 30 4 3 5 A B C D 10 ft A C B u u d 0.15m D A C B 0.4 m 45 0.4 m 0.3 m k 200 N/m 40 kg D A C E B 30 B A C D u 15 F3–4 F3–1 F3–2 F3–3 F3–6 F3–5
- 112. 3.3 COPLANAR FORCE SYSTEMS 95 3 All problem solutions must include an FBD. •3–1. Determine the force in each cord for equilibrium of the 200-kg crate. Cord remains horizontal due to the roller at , and has a length of . Set . 3–2. If the 1.5-m-long cord can withstand a maximum force of , determine the force in cord and the distance y so that the 200-kg crate can be supported. BC 3500 N AB y = 0.75 m 1.5 m AB C BC •3–5. The members of a truss are connected to the gusset plate. If the forces are concurrent at point O, determine the magnitudes of F and T for equilibrium.Take . 3–6. The gusset plate is subjected to the forces of four members. Determine the force in member B and its proper orientation for equilibrium. The forces are concurrent at point O.Take . F = 12 kN u u = 30° 3–7. The towing pendant AB is subjected to the force of 50 kN exerted by a tugboat. Determine the force in each of the bridles, BC and BD, if the ship is moving forward with constant velocity. PROBLEMS C B A 2 m y Probs. 3–1/2 FAB A B C D G 30 45 Probs. 3–3/4 5 kN A B C D T O 45 u F 8 kN Probs. 3–5/6 30 A B C D 50 kN 20 Prob. 3–7 3–3. If the mass of the girder is and its center of mass is located at point G, determine the tension developed in cables , , and for equilibrium. *3–4. If cables and can withstand a maximum tensile force of , determine the maximum mass of the girder that can be suspended from cable so that neither cable will fail. The center of mass of the girder is located at point . G AB 20 kN BC BD BD BC AB 3 Mg
- 113. 96 CHAPTER 3 EQUILIBRIUM OF A PARTICLE 3 *3–8. Members and support the 300-lb crate. Determine the tensile force developed in each member. •3–9. If members and can support a maximum tension of and , respectively, determine the largest weight of the crate that can be safely supported. 250 lb 300 lb AB AC AB AC 3–10. The members of a truss are connected to the gusset plate. If the forces are concurrent at point O, determine the magnitudes of F and T for equilibrium.Take . 3–11. The gusset plate is subjected to the forces of three members. Determine the tension force in member C and its angle for equilibrium.The forces are concurrent at point O. Take . F = 8 kN u u = 90° *3–12. If block weighs and block weighs , determine the required weight of block and the angle for equilibrium. •3–13. If block weighs 300 lb and block weighs 275 lb, determine the required weight of block and the angle for equilibrium. u C B D u D 100 lb C 200 lb B A B C 4 ft 4 ft 3 ft Probs. 3–8/9 x y A O F T B 9 kN C 4 5 3 u Probs. 3–10/11 A B D C u 30 Probs. 3–12/13 3 m 3 m 4 m kAC 20 N/m kAB 30 N/m C B A D Probs. 3–14/15 3–14. Determine the stretch in springs AC and AB for equilibrium of the 2-kg block. The springs are shown in the equilibrium position. 3–15. The unstretched length of spring AB is 3 m. If the block is held in the equilibrium position shown, determine the mass of the block at D.
- 114. 3.3 COPLANAR FORCE SYSTEMS 97 3 *3–16. Determine the tension developed in wires and required for equilibrium of the 10-kg cylinder. Take . •3–17. If cable is subjected to a tension that is twice that of cable , determine the angle for equilibrium of the 10-kg cylinder. Also, what are the tensions in wires and ? CB CA u CA CB u = 40° CB CA 3–18. Determine the forces in cables AC and AB needed to hold the 20-kg ball D in equilibrium. Take and . 3–19. The ball D has a mass of 20 kg.If a force of is applied horizontally to the ring at A, determine the dimension d so that the force in cable AC is zero. F = 100 N d = 1 m F = 300 N *3–20. Determine the tension developed in each wire used to support the 50-kg chandelier. •3–21. If the tension developed in each of the four wires is not allowed to exceed , determine the maximum mass of the chandelier that can be supported. 600 N 쐍3–22. A vertical force is applied to the ends of the 2-ft cord AB and spring AC. If the spring has an unstretched length of 2 ft, determine the angle for equilibrium.Take 3–23. Determine the unstretched length of spring AC if a force causes the angle for equilibrium. Cord AB is 2 ft long.Take k = 50 lbft. u = 60° P = 80 lb k = 15 lbft. u P = 10 lb 30° A B C u Probs. 3–16/17 A C B F D 2 m 1.5 m d Probs. 3–18/19 A B D C 30 30 45 Prob. 3–20/21 2 ft k 2 ft A B C P u Probs. 3–22/23
- 115. 98 CHAPTER 3 EQUILIBRIUM OF A PARTICLE 3 *3–24. If the bucket weighs 50 lb, determine the tension developed in each of the wires. •3–25. Determine the maximum weight of the bucket that the wire system can support so that no single wire develops a tension exceeding 100 lb. 3–26. Determine the tensions developed in wires , , and and the angle required for equilibrium of the 30-lb cylinder and the 60-lb cylinder . 3–27. If cylinder weighs 30 lb and , determine the weight of cylinder . F u = 15° E F E u BA CB CD *3–28. Two spheres A and B have an equal mass and are electrostatically charged such that the repulsive force acting between them has a magnitude of 20 mN and is directed along line AB. Determine the angle the tension in cords AC and BC, and the mass m of each sphere. u, A B E C D 4 3 5 30 30 Probs. 3–24/25 D A C F E B u 30 45 Probs. 3–26/27 C 30 20 mN 20 mN 30 B u A Prob. 3–28 12 5 13 B A C D u Prob. 3–29 •3–29. The cords BCA and CD can each support a maximum load of 100 lb. Determine the maximum weight of the crate that can be hoisted at constant velocity and the angle for equilibrium. Neglect the size of the smooth pulley at C. u
- 116. 3.3 COPLANAR FORCE SYSTEMS 99 3 *3–32. Determine the magnitude and direction of the equilibrium force exerted along link AB by the tractive apparatus shown. The suspended mass is 10 kg. Neglect the size of the pulley at A. FAB u •3–33. The wire forms a loop and passes over the small pulleys at A, B, C, and D. If its end is subjected to a force of , determine the force in the wire and the magnitude of the resultant force that the wire exerts on each of the pulleys. 3–34. The wire forms a loop and passes over the small pulleys at A, B, C, and D. If the maximum resultant force that the wire can exert on each pulley is 120 N, determine the greatest force P that can be applied to the wire as shown. P = 50 N 3–35. The picture has a weight of 10 lb and is to be hung over the smooth pin B. If a string is attached to the frame at points A and C, and the maximum force the string can support is 15 lb, determine the shortest string that can be safely used. F A B C E D 2 ft 2 ft k 30 lb/ft k 30 lb/ft θ θ Probs. 3–30/31 45 A B 75 FAB u Prob. 3–32 P A B D C 30 30 Probs. 3–33/34 C A 9 in. 9 in. B Prob. 3–35 •3–30. The springs on the rope assembly are originally unstretched when . Determine the tension in each rope when . Neglect the size of the pulleys at B and D. 3–31. The springs on the rope assembly are originally stretched 1 ft when . Determine the vertical force F that must be applied so that . u = 30° u = 0° F = 90 lb u = 0°
- 117. 100 CHAPTER 3 EQUILIBRIUM OF A PARTICLE 3 A O C 1 ft B 2 ft F D 2 ft 2 ft Prob. 3–36 d A C B 12 in. k u Probs. 3–37/38 C D B A 1 ft 1.5 ft Prob. 3–39 A B k 800 N/m D 500 mm 400 mm 400 mm 300 mm C Prob. 3–40 •3–37. The 10-lb weight is supported by the cord AC and roller and by the spring that has a stiffness of . and an unstretched length of 12 in. Determine the distance d to where the weight is located when it is in equilibrium. 3–38. The 10-lb weight is supported by the cord AC and roller and by a spring. If the spring has an unstretched length of 8 in. and the weight is in equilibrium when ., determine the stiffness k of the spring. d = 4 in k = 10 lbin •3–39. A “scale” is constructed with a 4-ft-long cord and the 10-lb block D.The cord is fixed to a pin at A and passes over two small pulleys at B and C. Determine the weight of the suspended block at B if the system is in equilibrium. •*3–40. The spring has a stiffness of and an unstretched length of 200 mm. Determine the force in cables BC and BD when the spring is held in the position shown. k = 800 Nm *3–36. The 200-lb uniform tank is suspended by means of a 6-ft-long cable, which is attached to the sides of the tank and passes over the small pulley located at O. If the cable can be attached at either points A and B or C and D, determine which attachment produces the least amount of tension in the cable.What is this tension?
- 118. 3.3 COPLANAR FORCE SYSTEMS 101 3 3–42. Determine the mass of each of the two cylinders if they cause a sag of when suspended from the rings at A and B. Note that when the cylinders are removed. s = 0 s = 0.5 m •3–43. The pail and its contents have a mass of 60 kg. If the cable BAL is 15 m long, determine the distance y of the pulley at A for equilibrium. Neglect the size of the pulley. •*3–44. A scale is constructed using the 10-kg mass, the 2-kg pan P, and the pulley and cord arrangement. Cord BCA is 2 m long. If , determine the mass D in the pan. Neglect the size of the pulley. s = 0.75 m •3–41. A continuous cable of total length 4 m is wrapped around the small pulleys at A, B, C, and D. If each spring is stretched 300 mm, determine the mass m of each block. Neglect the weight of the pulleys and cords.The springs are unstretched when d = 2 m. B C A k 500 N/m k 500 N/m d D Prob. 3–41 1 m 2 m 2 m 1.5 m s B A C D k 100 N/m k 100 N/m Prob. 3–42 2 m y C B A 10 m Prob. 3–43 1.5 m 0 s P D A C B 1.5 m Prob. 3–44
- 119. 102 CHAPTER 3 EQUILIBRIUM OF A PARTICLE 3 A B C B A C A C B D F B A B¿ C CONCEPTUAL PROBLEMS P3–1. The concrete wall panel is hoisted into position using the two cables AB and AC of equal length. Establish appropriate dimensions and use an equilibrium analysis to show that the longer the cables the less the force in each cable. P3–4. The two chains AB and AC have equal lengths and are subjected to the vertical force F. If AB is replaced by a shorter chain , show that this chain would have to support a larger tensile force than in order to maintain equilibrium. AB AB¿ P3–2. The truss is hoisted using cable ABC that passes through a very small pulley at B. If the truss is placed in a tipped position, show that it will always return to the horizontal position to maintain equilibrium. P3–3. The device DB is used to pull on the chain ABC so as to hold a door closed on the bin. If the angle between AB and the horizontal segment BC is 30º, determine the angle between DB and the horizontal for equilibrium.
- 120. 3.4 THREE-DIMENSIONAL FORCE SYSTEMS 103 3 3.4 Three-Dimensional Force Systems In Section 3.1 we stated that the necessary and sufficient condition for particle equilibrium is (3–4) In the case of a three-dimensional force system, as in Fig. 3–9, we can resolve the forces into their respective i, j, k components, so that .To satisfy this equation we require (3–5) These three equations state that the algebraic sum of the components of all the forces acting on the particle along each of the coordinate axes must be zero. Using them we can solve for at most three unknowns, generally represented as coordinate direction angles or magnitudes of forces shown on the particle’s free-body diagram. ©Fz = 0 ©Fy = 0 ©Fx = 0 ©Fxi + ©Fy j + ©Fzk = 0 ©F = 0 Procedure for Analysis Three-dimensional force equilibrium problems for a particle can be solved using the following procedure. Free-Body Diagram. • Establish the x, y, z axes in any suitable orientation. • Label all the known and unknown force magnitudes and directions on the diagram. • The sense of a force having an unknown magnitude can be assumed. Equations of Equilibrium. • Use the scalar equations of equilibrium, in cases where it is easy to resolve each force into its x, y, z components. • If the three-dimensional geometry appears difficult, then first express each force on the free-body diagram as a Cartesian vector, substitute these vectors into and then set the i, j, k components equal to zero. • If the solution for a force yields a negative result, this indicates that its sense is the reverse of that shown on the free-body diagram. ©F = 0, ©Fz = 0, ©Fy = 0, ©Fx = 0, F3 F2 F1 x y z Fig. 3–9 The ring at A is subjected to the force from the hook as well as forces from each of the three chains.If the electromagnet and its load have a weight W, then the force at the hook will be W, and the three scalar equations of equilibrium can be applied to the free-body diagram of the ring in order to determine the chain forces, , and FD. FC FB, A D C B FC FD FB W
- 121. 104 CHAPTER 3 EQUILIBRIUM OF A PARTICLE 3 EXAMPLE 3.5 A 90-lb load is suspended from the hook shown in Fig. 3–10a. If the load is supported by two cables and a spring having a stiffness , determine the force in the cables and the stretch of the spring for equilibrium. Cable AD lies in the x–y plane and cable AC lies in the x–z plane. SOLUTION The stretch of the spring can be determined once the force in the spring is determined. Free-Body Diagram. The connection at A is chosen for the equilibrium analysis since the cable forces are concurrent at this point.The free-body diagram is shown in Fig. 3–10b. Equations of Equilibrium. By inspection, each force can easily be resolved into its x, y, z components, and therefore the three scalar equations of equilibrium can be used. Considering components directed along each positive axis as “positive,” we have (1) (2) (3) Solving Eq. (3) for then Eq. (1) for and finally Eq. (2) for yields Ans. Ans. Ans. The stretch of the spring is therefore Ans. NOTE: Since the results for all the cable forces are positive, each cable is in tension; that is, it pulls on point A as expected, Fig. 3–10b. sAB = 0.416 ft 207.8 lb = (500 lbft)1sAB2 FB = ksAB FB = 207.8 lb FD = 240 lb FC = 150 lb FB, FD, FC, A3 5 BFC - 90 lb = 0 ©Fz = 0; -FD cos 30° + FB = 0 ©Fy = 0; FD sin 30° - A4 5 BFC = 0 ©Fx = 0; k = 500 lbft x y z (a) 30 C 90 lb A 5 3 4 k = 500 lb/ft B D Fig. 3–10 y x z (b) 30 90 lb A 5 3 4 FC FB FD
- 122. 3.4 THREE-DIMENSIONAL FORCE SYSTEMS 105 3 EXAMPLE 3.6 The 10-kg lamp in Fig. 3-11a is suspended from the three equal-length cords. Determine its smallest vertical distance s from the ceiling if the force developed in any cord is not allowed to exceed 50 N. x y s (a) z D A B C 600 mm 120 120 Fig. 3–11 SOLUTION Free-Body Diagram. Due to symmetry, Fig. 3-11b, the distance . It follows that from and , the tension T in each cord will be the same. Also, the angle between each cord and the axis is . Equation of Equilibrium. Applying the equilibrium equation along the axis, with , we have From the shaded triangle shown in Fig. 3-11b, Ans. s = 519 mm tan 49.16° = 600 mm s g = cos-1 98.1 150 = 49.16° 3[(50 N) cos g] - 10(9.81) N = 0 gFz = 0; T = 50 N z g z gFy = 0 gFx = 0 DA = DB = DC = 600 mm x y s 600 mm D z (b) A B C 10(9.81) N T T T g
- 123. 106 CHAPTER 3 EQUILIBRIUM OF A PARTICLE 3 EXAMPLE 3.7 y x z (a) 8 ft 3 ft 4 ft 4 ft C B D A Fig. 3–12 y x z W 40 lb (b) FB A FC FD Determine the force in each cable used to support the 40-lb crate shown in Fig. 3–12a. SOLUTION Free-Body Diagram. As shown in Fig. 3–12b, the free-body diagram of point A is considered in order to “expose” the three unknown forces in the cables. Equations of Equilibrium. First we will express each force in Cartesian vector form. Since the coordinates of points B and C are and C( 4 ft, 8 ft), we have Equilibrium requires Equating the respective i, j, k components to zero yields (1) (2) (3) Equation (2) states that Thus, solving Eq. (3) for and and substituting the result into Eq. (1) to obtain we have Ans. Ans. FD = 15.0 lb FB = FC = 23.6 lb FD, FC FB FB = FC. 0.848FB + 0.848FC - 40 = 0 ©Fz = 0; -0.424FB + 0.424FC = 0 ©Fy = 0; -0.318FB - 0.318FC + FD = 0 ©Fx = 0; - 0.318FCi + 0.424FCj + 0.848FCk + FDi - 40k = 0 -0.318FBi - 0.424FBj + 0.848FBk FB + FC + FD + W = 0 ©F = 0; W = 5-40k6 lb FD = FDi = -0.318FCi + 0.424FCj + 0.848FCk FC = FCc -3i + 4j + 8k 2 1-322 + 1422 + 1822 d = -0.318FBi - 0.424FBj + 0.848FBk FB = FBc -3i - 4j + 8k 2 1-322 + 1-422 + 1822 d -3 ft, B1-3 ft, -4 ft, 8 ft2
- 124. 3.4 THREE-DIMENSIONAL FORCE SYSTEMS 107 3 EXAMPLE 3.8 y 1 m 2 m z 60 135 2 m D 120 x (a) B A k 1.5 kN/m C Fig. 3–13 y x z W 981 N A FC (b) FD FB Determine the tension in each cord used to support the 100-kg crate shown in Fig. 3–13a. SOLUTION Free-Body Diagram. The force in each of the cords can be determined by investigating the equilibrium of point A.The free-body diagram is shown in Fig. 3–13b. The weight of the crate is Equations of Equilibrium. Each force on the free-body diagram is first expressed in Cartesian vector form. Using Eq. 2–9 for and noting point for we have Equilibrium requires W = 5-981k6 N = -0.333FDi + 0.667FDj + 0.667FDk FD = FDc -1i + 2j + 2k 2 1-122 + 1222 + 1222 d = -0.5FCi - 0.707FCj + 0.5FCk FC = FC cos 120°i + FC cos 135°j + FC cos 60°k FB = FBi FD, D1-1 m, 2 m, 2 m2 FC W = 10019.812 = 981 N. Equating the respective i, j, k components to zero, (1) (2) (3) Solving Eq. (2) for in terms of and substituting this into Eq. (3) yields is then determined from Eq. (2). Finally, substituting the results into Eq. (1) gives Hence, Ans. Ans. Ans. FB = 694 N FD = 862 N FC = 813 N FB. FD FC. FC FD 0.5FC + 0.667FD - 981 = 0 ©Fz = 0; -0.707FC + 0.667FD = 0 ©Fy = 0; FB - 0.5FC - 0.333FD = 0 ©Fx = 0; - 0.333FD i + 0.667FD j + 0.667FDk - 981k = 0 FBi - 0.5FCi - 0.707FCj + 0.5FCk FB + FC + FD + W = 0 ©F = 0;
- 125. 108 CHAPTER 3 EQUILIBRIUM OF A PARTICLE 3 FUNDAMENTAL PROBLEMS All problem solutions must include an FBD. F3–7. Determine the magnitude of forces so that the particle is held in equilibrium. F3, F2, F1, F3–10. Determine the tension developed in cables AB, AC, and AD. F3–8. Determine the tension developed in cables AB, AC, and AD. F3–9. Determine the tension developed in cables AB, AC, and AD. F3–11. The 150-lb crate is supported by cables AB, AC, and AD. Determine the tension in these wires. 900 N 600 N z x y 4 4 4 3 3 3 5 5 F1 F2 F3 5 A C z y x B D 3 3 4 4 5 5 900 lb 2 m 1 m 2 m A C z y x B D 600 N 30 A C z y x B 60º 300 lb 30 45 120 60 D A D E B C 2 ft 3 ft 3 ft 2 ft 6 ft F3–9 F3–11 F3–8 F3–7 F3–10
- 126. 3.4 THREE-DIMENSIONAL FORCE SYSTEMS 109 3 All problem solutions must include an FBD. •3–45. Determine the tension in the cables in order to support the 100-kg crate in the equilibrium position shown. 3–46. Determine the maximum mass of the crate so that the tension developed in any cable does not exceeded 3 kN. *3–48. Determine the tension developed in cables , , and required for equilibrium of the 300-lb crate. •3–49. Determine the maximum weight of the crate so that the tension developed in any cable does not exceed 450 lb. AD AC AB 3–47. The shear leg derrick is used to haul the 200-kg net of fish onto the dock. Determine the compressive force along each of the legs AB and CB and the tension in the winch cable DB.Assume the force in each leg acts along its axis. 2.5 m 2 m 2 m 2 m 1 m A z D y x B C Probs. 3–45/46 4 m 4 m 2 m 2 m 5.6 m D B C A x y z Prob. 3–47 A D C x 1 ft 3 ft 2 ft 1 ft 2 ft 2 ft y z 2 ft B Probs. 3–48/49 3 ft d y x C D B A 3500 lb 4 ft 3 ft 10 ft 4 ft 2 ft z Probs. 3–50/51 PROBLEMS 3–50. Determine the force in each cable needed to support the 3500-lb platform. Set . 3–51. Determine the force in each cable needed to support the 3500-lb platform. Set . d = 4 ft d = 2 ft
- 127. 110 CHAPTER 3 EQUILIBRIUM OF A PARTICLE 3 *3–52. Determine the force in each of the three cables needed to lift the tractor which has a mass of 8 Mg. •3–53. Determine the force acting along the axis of each of the three struts needed to support the 500-kg block. *3–56. The ends of the three cables are attached to a ring at A and to the edge of a uniform 150-kg plate. Determine the tension in each of the cables for equilibrium. •3–57. The ends of the three cables are attached to a ring at A and to the edge of the uniform plate. Determine the largest mass the plate can have if each cable can support a maximum tension of 15 kN. 2 m 1.25 m 1.25 m 1 m 3 m A D C B y x z Prob. 3–52 0.75 m 1.25 m 3 m 2.5 m z A B C D x y 2 m Prob. 3–53 x x A B C y z z 6 m 3 m 2 m D Probs. 3–54/55 z A B x y D C 10 m 6 m 6 m 6 m 4 m 2 m 2 m 12 m 2 m Probs. 3–56/57 3–54. If the mass of the flowerpot is 50 kg, determine the tension developed in each wire for equilibrium. Set and . 3–55. If the mass of the flowerpot is 50 kg, determine the tension developed in each wire for equilibrium. Set and . z = 1.5 m x = 2 m z = 2 m x = 1.5 m
- 128. 3.4 THREE-DIMENSIONAL FORCE SYSTEMS 111 3 3–58. Determine the tension developed in cables , , and required for equilibrium of the 75-kg cylinder. 3–59. If each cable can withstand a maximum tension of 1000 N, determine the largest mass of the cylinder for equilibrium. AD AC AB *3–64. The thin ring can be adjusted vertically between three equally long cables from which the 100-kg chandelier is suspended. If the ring remains in the horizontal plane and , determine the tension in each cable. •3–65. The thin ring can be adjusted vertically between three equally long cables from which the 100-kg chandelier is suspended. If the ring remains in the horizontal plane and the tension in each cable is not allowed to exceed , determine the smallest allowable distance required for equilibrium. z 1 kN z = 600 mm 1 m 3 m 3 m 4 m 1.5 m 2 m 2 m 1 m A C z y x B D Probs. 3–58/59 A z y x B d 2 m 2 m 3 m 6 m 6 m D C Probs. 3–60/61 B C A F y z x 4 ft 5 ft 5 ft z y Probs. 3–62/63 x y z z 0.5 m 120 120 120 A B C D Probs. 3–64/65 3–62. A force of holds the 400-lb crate in equilibrium. Determine the coordinates (0, y, z) of point A if the tension in cords AC and AB is 700 lb each. 3–63. If the maximum allowable tension in cables AB and AC is 500 lb, determine the maximum height z to which the 200-lb crate can be lifted. What horizontal force F must be applied? Take . y = 8 ft F = 100 lb *3–60. The 50-kg pot is supported from A by the three cables. Determine the force acting in each cable for equilibrium.Take . •3–61. Determine the height d of cable AB so that the force in cables AD and AC is one-half as great as the force in cable AB. What is the force in each cable for this case? The flower pot has a mass of 50 kg. d = 2.5 m
- 129. 112 CHAPTER 3 EQUILIBRIUM OF A PARTICLE 3 •3–69. Determine the angle such that an equal force is developed in legs OB and OC.What is the force in each leg if the force is directed along the axis of each leg? The force F lies in the plane. The supports at A, B, C can exert forces in either direction along the attached legs. x-y u *3–68. The three outer blocks each have a mass of 2 kg, and the central block E has a mass of 3 kg. Determine the sag s for equilibrium of the system. 120 1.5 ft 80 lb d C A B D 120 120 Prob. 3–66 A B C D F 120 120 120 3 ft y z x 4 ft Prob. 3–67 s 60 30 30 1 m 1 m A D E B C Prob. 3–68 120 5 ft 10 ft 120 120 y x z O B C A F = 100 lb u Prob. 3–69 3–66. The bucket has a weight of 80 lb and is being hoisted using three springs, each having an unstretched length of and stiffness of . Determine the vertical distance d from the rim to point A for equilibrium. k = 50 lbft l0 = 1.5 ft 3–67. Three cables are used to support a 900-lb ring. Determine the tension in each cable for equilibrium.
- 130. CHAPTER REVIEW 113 3 CHAPTER REVIEW Particle Equilibrium When a particle is at rest or moves with constant velocity, it is in equilibrium. This requires that all the forces acting on the particle form a zero resultant force. In order to account for all the forces that act on a particle, it is necessary to draw its free-body diagram.This diagram is an outlined shape of the particle that shows all the forces listed with their known or unknown magnitudes and directions. Two Dimensions The two scalar equations of force equilibrium can be applied with reference to an established x, y coordinate system. The tensile force developed in a continuous cable that passes over a frictionless pulley must have a constant magnitude throughout the cable to keep the cable in equilibrium. If the problem involves a linearly elastic spring, then the stretch or compression s of the spring can be related to the force applied to it. FR = πF = 0 Cable is in tension Three Dimensions If the three-dimensional geometry is difficult to visualize, then the equilibrium equation should be applied using a Cartesian vector analysis. This requires first expressing each force on the free- body diagram as a Cartesian vector. When the forces are summed and set equal to zero, then the i, j, and k components are also zero. ©Fz = 0 ©Fy = 0 ©Fx = 0 ©F = 0 ©Fy = 0 ©Fx = 0 F = ks T T u F4 F3 F1 F2 F3 F2 F1 x y z
- 131. 114 CHAPTER 3 EQUILIBRIUM OF A PARTICLE 3 REVIEW PROBLEMS •3–73. Two electrically charged pith balls, each having a mass of 0.15 g, are suspended from light threads of equal length. Determine the magnitude of the horizontal repulsive force, F, acting on each ball if the measured distance between them is . r = 200 mm B A C u F Prob. 3–70 x O y 70 30 5 kN 7 kN 3 4 5 F2 F1 u Prob. 3–71/72 A B 50 mm 150 mm 150 mm r 200 mm F –F Prob. 3–73 x 1.5 m 1.5 m 2 m 4 m A z B y 6 m O C Prob. 3–74 3–70. The 500-lb crate is hoisted using the ropes AB and AC. Each rope can withstand a maximum tension of 2500 lb before it breaks. If AB always remains horizontal, determine the smallest angle to which the crate can be hoisted. u 3–74. The lamp has a mass of 15 kg and is supported by a pole AO and cables AB and AC. If the force in the pole acts along its axis, determine the forces in AO, AB, and AC for equilibrium. 3–71. The members of a truss are pin connected at joint O. Determine the magnitude of and its angle for equilibrium. Set . *3–72. The members of a truss are pin connected at joint O. Determine the magnitudes of and for equilibrium. Set . u = 60° F2 F1 F2 = 6 kN u F1
- 132. REVIEW PROBLEMS 115 3 3–79. The joint of a space frame is subjected to four member forces. Member OA lies in the plane and member OB lies in the plane. Determine the forces acting in each of the members required for equilibrium of the joint. y–z x–y z y x 20 F3 200 lb P (1 ft, 7 ft, 4 ft) F4 300 lb F1 360 lb F2 120 lb Prob. 3–75 40 B C A l 2 ft 200 lb u Prob. 3–76 z P F3 F1 F2 y x 3 800 lb 200 lb 4 5 60 60 135 Prob. 3–77 D y x C A B 6 ft 8 ft 2 ft 2 ft 6 ft z Prob. 3–78 3–75. Determine the magnitude of P and the coordinate direction angles of required for equilibrium of the particle. Note that acts in the octant shown. F3 F3 *3–76. The ring of negligible size is subjected to a vertical force of 200 lb. Determine the longest length l of cord AC such that the tension acting in AC is 160 lb.Also, what is the force acting in cord AB? Hint: Use the equilibrium condition to determine the required angle for attachment, then determine l using trigonometry applied to . ¢ABC u •3–77. Determine the magnitudes of , , and for equilibrium of the particle. F3 F2 F1 x 45 A B 200 lb F1 z y 40 F2 F3 O Prob. 3–79 3–78. Determine the force in each cable needed to support the 500-lb load.
- 133. Application of forces to the handles of these wrenches will produce a tendency to rotate each wrench about its end. It is important to know how to calculate this effect and, in some cases, to be able to simplify this system to its resultants.
- 134. Force System Resultants CHAPTER OBJECTIVES • To discuss the concept of the moment of a force and show how to calculate it in two and three dimensions. • To provide a method for finding the moment of a force about a specified axis. • To define the moment of a couple. • To present methods for determining the resultants of nonconcurrent force systems. • To indicate how to reduce a simple distributed loading to a resultant force having a specified location. 4.1 Moment of a Force— Scalar Formulation When a force is applied to a body it will produce a tendency for the body to rotate about a point that is not on the line of action of the force. This tendency to rotate is sometimes called a torque, but most often it is called the moment of a force or simply the moment. For example, consider a wrench used to unscrew the bolt in Fig. 4–1a. If a force is applied to the handle of the wrench it will tend to turn the bolt about point O (or the z axis). The magnitude of the moment is directly proportional to the magnitude of F and the perpendicular distance or moment arm d. The larger the force or the longer the moment arm, the greater the moment or turning effect. Note that if the force F is applied at an angle , Fig. 4–1b, then it will be more difficult to turn the bolt since the moment arm will be smaller than d. If F is applied along the wrench, Fig. 4–1c, its moment arm will be zero since the line of action of F will intersect point O (the z axis).As a result, the moment of F about O is also zero and no turning can occur. d¿ = d sinu u Z 90° 4 z O d F (a) z O F d¿ d sin u (b) u d z O (c) F Fig. 4–1 117
- 135. 118 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 We can generalize the above discussion and consider the force F and point O which lie in the shaded plane as shown in Fig. 4–2a.The moment about point O, or about an axis passing through O and perpendicular to the plane, is a vector quantity since it has a specified magnitude and direction. Magnitude. The magnitude of is (4–1) where d is the moment arm or perpendicular distance from the axis at point O to the line of action of the force. Units of moment magnitude consist of force times distance, e.g., or Direction. The direction of is defined by its moment axis, which is perpendicular to the plane that contains the force F and its moment arm d. The right-hand rule is used to establish the sense of direction of . According to this rule, the natural curl of the fingers of the right hand, as they are drawn towards the palm, represent the tendency for rotation caused by the moment. As this action is performed, the thumb of the right hand will give the directional sense of , Fig. 4–2a. Notice that the moment vector is represented three-dimensionally by a curl around an arrow. In two dimensions this vector is represented only by the curl as in Fig. 4–2b. Since in this case the moment will tend to cause a counterclockwise rotation, the moment vector is actually directed out of the page. Resultant Moment. For two-dimensional problems, where all the forces lie within the x–y plane, Fig. 4–3, the resultant moment about point O (the z axis) can be determined by finding the algebraic sum of the moments caused by all the forces in the system. As a convention, we will generally consider positive moments as counterclockwise since they are directed along the positive z axis (out of the page). Clockwise moments will be negative. Doing this, the directional sense of each moment can be represented by a plus or minus sign. Using this sign convention, the resultant moment in Fig. 4–3 is therefore a If the numerical result of this sum is a positive scalar, will be a counterclockwise moment (out of the page); and if the result is negative, will be a clockwise moment (into the page). (MR)O (MR)O + (MR)O = ©Fd; (MR)O = F1d1 - F2d2 + F3d3 (MR)O MO MO MO lb # ft. N # m MO = Fd MO MO Sense of rotation O Moment axis d F MO MO F d O (a) (b) Fig. 4–2 y x O F3 F2 F1 M3 M2 M1 d3 d2 d1 Fig. 4–3
- 136. 4.1 MOMENT OF A FORCE—SCALAR FORMULATION 119 4 EXAMPLE 4.1 For each case illustrated in Fig. 4–4, determine the moment of the force about point O. SOLUTION (SCALAR ANALYSIS) The line of action of each force is extended as a dashed line in order to establish the moment arm d.Also illustrated is the tendency of rotation of the member as caused by the force. Furthermore, the orbit of the force about O is shown as a colored curl.Thus, Fig. 4–4a b Ans. Fig. 4–4b b Ans. Fig. 4–4c b Ans. Fig. 4–4d d Ans. Fig. 4–4e d Ans. MO = 17 kN214 m - 1 m2 = 21.0 kN # m MO = 160 lb211 sin 45° ft2 = 42.4 lb # ft MO = 140 lb214 ft + 2 cos 30° ft2 = 229 lb # ft MO = 150 N210.75 m2 = 37.5 N # m MO = 1100 N212 m2 = 200 N # m 2 m O (a) 100 N Fig. 4–4 2 m O (b) 50 N 0.75 m (d) O 1 sin 45 ft 60 lb 3 ft 45 1 ft 2 m O (e) 4 m 1 m 7 kN 2 ft (c) O 4 ft 2 cos 30 ft 40 lb 30
- 137. 120 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 EXAMPLE 4.2 50 N 40 N 20 N 3 m 2 m 2 m O x y 60 N 30 Fig. 4–5 As illustrated by the example problems, the moment of a force does not always cause a rotation.For example,the force F tends to rotate the beam clockwise about its support at A with a moment The actual rotation would occur if the support at B were removed. MA = FdA. O FN FH The ability to remove the nail will require the moment of about point O to be larger than the moment of the force about O that is needed to pull the nail out. FN FH MA FdA dA F A B Determine the resultant moment of the four forces acting on the rod shown in Fig. 4–5 about point O. SOLUTION Assuming that positive moments act in the direction, i.e., counterclockwise, we have a b Ans. For this calculation, note how the moment-arm distances for the 20-N and 40-N forces are established from the extended (dashed) lines of action of each of these forces. MRO = -334 N # m = 334 N # m -40 N14 m + 3 cos 30° m2 MRO = -50 N12 m2 + 60 N102 + 20 N13 sin 30° m2 +MRO = ©Fd; +k
- 138. 4.2 CROSS PRODUCT 121 4 4.2 Cross Product The moment of a force will be formulated using Cartesian vectors in the next section. Before doing this, however, it is first necessary to expand our knowledge of vector algebra and introduce the cross-product method of vector multiplication. The cross product of two vectors A and B yields the vector C, which is written (4–2) and is read “C equals A cross B.” Magnitude. The magnitude of C is defined as the product of the magnitudes of A and B and the sine of the angle between their tails Thus, Direction. Vector C has a direction that is perpendicular to the plane containing A and B such that C is specified by the right-hand rule; i.e., curling the fingers of the right hand from vector A (cross) to vector B, the thumb points in the direction of C, as shown in Fig. 4–6. Knowing both the magnitude and direction of C, we can write (4–3) where the scalar defines the magnitude of C and the unit vector defines the direction of C. The terms of Eq. 4–3 are illustrated graphically in Fig. 4–6. uC AB sin u C = A * B = 1AB sin u2uC C = AB sin u. 10° … u … 180°2. u C = A * B C A B A B u uC Fig. 4–6
- 139. B A C B A C A B B A Fig. 4–7 122 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 Laws of Operation. ● The commutative law is not valid; i.e., . Rather, This is shown in Fig. 4–7 by using the right-hand rule. The cross product yields a vector that has the same magnitude but acts in the opposite direction to C; i.e., ● If the cross product is multiplied by a scalar a, it obeys the assoc- iative law; This property is easily shown since the magnitude of the resultant vector and its direction are the same in each case. ● The vector cross product also obeys the distributive law of addition, ● The proof of this identity is left as an exercise (see Prob. 4–1). It is important to note that proper order of the cross products must be maintained, since they are not commutative. Cartesian Vector Formulation. Equation 4–3 may be used to find the cross product of any pair of Cartesian unit vectors. For example, to find the magnitude of the resultant vector is and its direction is determined using the right-hand rule. As shown in Fig. 4–8, the resultant vector points in the direction.Thus, In a similar manner, These results should not be memorized; rather, it should be clearly understood how each is obtained by using the right-hand rule and the definition of the cross product. A simple scheme shown in Fig. 4–9 is helpful for obtaining the same results when the need arises. If the circle is constructed as shown, then “crossing” two unit vectors in a counterclockwise fashion around the circle yields the positive third unit vector; e.g., “Crossing” clockwise, a negative unit vector is obtained; e.g., i * k = -j. k * i = j. k * i = j k * j = -i k * k = 0 j * k = i j * i = -k j * j = 0 i * j = k i * k = -j i * i = 0 i * j = 112k. +k 1i21j21sin 90°2 = 112112112 = 1, i * j, A * 1B + D2 = 1A * B2 + 1A * D2 1ƒaƒAB sin u2 a1A * B2 = 1aA2 * B = A * 1aB2 = 1A * B2a B * A = -C. B * A A * B = -B * A A * B Z B * A y x z k i j j i Fig. 4–8 i j k Fig. 4–9
- 140. 4.2 CROSS PRODUCT 123 4 Let us now consider the cross product of two general vectors A and B which are expressed in Cartesian vector form.We have Carrying out the cross-product operations and combining terms yields (4–4) This equation may also be written in a more compact determinant form as (4–5) Thus, to find the cross product of any two Cartesian vectors A and B, it is necessary to expand a determinant whose first row of elements consists of the unit vectors i, j, and k and whose second and third rows represent the x, y, z components of the two vectors A and B, respectively.* A * B = 3 i j k Ax Ay Az Bx By Bz 3 A * B = 1AyBz - AzBy2i - 1AxBz - AzBx2j + 1AxBy - AyBx2k + AzBx1k * i2 + AzBy1k * j2 + AzBz1k * k2 + AyBx1j * i2 + AyBy1j * j2 + AyBz1j * k2 = AxBx1i * i2 + AxBy1i * j2 + AxBz1i * k2 A * B = 1Axi + Ayj + Azk2 * 1Bxi + Byj + Bzk2 *A determinant having three rows and three columns can be expanded using three minors, each of which is multiplied by one of the three terms in the first row. There are four elements in each minor, for example, By definition, this determinant notation represents the terms which is simply the product of the two elements intersected by the arrow slanting downward to the right minus the product of the two elements intersected by the arrow slanting downward to the left For a determinant, such as Eq. 4–5, the three minors can be generated in accordance with the following scheme: 3 * 3 1A12A212. 1A11A222 1A11A22 - A12A212, Adding the results and noting that the j element must include the minus sign yields the expanded form of given by Eq. 4–4. A * B A11 A12 A21 A22 For element k: For element j: For element i: Ax Bx Ay By Az Bz i j k Ax Bx Ay By Az Bz i j k i j k Ax Bx Ay By Az Bz Remember the negative sign
- 141. r1 r3 r2 O F MO r1 F r2 F r3 F Line of action Fig. 4–11 124 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 4.3 Moment of a Force—Vector Formulation The moment of a force F about point O,or actually about the moment axis passing through O and perpendicular to the plane containing O and F, Fig. 4–10a, can be expressed using the vector cross product, namely, (4–6) Here r represents a position vector directed from O to any point on the line of action of F. We will now show that indeed the moment when determined by this cross product, has the proper magnitude and direction. Magnitude. The magnitude of the cross product is defined from Eq. 4–3 as where the angle is measured between the tails of r and F. To establish this angle, r must be treated as a sliding vector so that can be constructed properly, Fig. 4–10b. Since the moment arm then which agrees with Eq. 4–1. Direction. The direction and sense of in Eq. 4–6 are determined by the right-hand rule as it applies to the cross product. Thus, sliding r to the dashed position and curling the right-hand fingers from r toward F,“r cross F,” the thumb is directed upward or perpendicular to the plane containing r and F and this is in the same direction as the moment of the force about point O, Fig. 4–10b. Note that the “curl” of the fingers, like the curl around the moment vector, indicates the sense of rotation caused by the force. Since the cross product does not obey the commutative law, the order of must be maintained to produce the correct sense of direction for . Principle of Transmissibility. The cross product operation is often used in three dimensions since the perpendicular distance or moment arm from point O to the line of action of the force is not needed. In other words, we can use any position vector r measured from point O to any point on the line of action of the force F, Fig. 4–11.Thus, Since F can be applied at any point along its line of action and still create this same moment about point O, then F can be considered a sliding vector.This property is called the principle of transmissibility of a force. MO = r1 * F = r2 * F = r3 * F MO r * F MO, MO MO = rF sin u = F1r sin u2 = Fd d = r sin u, u u MO = rF sin u, MO, MO = r * F O Moment axis MO r A F (a) Fig. 4–10 O Moment axis d MO r A r F (b) u u
- 142. 4.3 MOMENT OF A FORCE—VECTOR FORMULATION 125 4 Cartesian Vector Formulation. If we establish x, y, z coordinate axes, then the position vector r and force F can be expressed as Cartesian vectors, Fig. 4–12a.Applying Eq. 4–5 we have (4–7) where represent the x, y, z components of the position vector drawn from point O to any point on the line of action of the force represent the x, y, z components of the force vector If the determinant is expanded, then like Eq. 4–4 we have (4–8) The physical meaning of these three moment components becomes evident by studying Fig. 4–12b. For example, the i component of can be determined from the moments of and about the x axis. The component does not create a moment or tendency to cause turning about the x axis since this force is parallel to the x axis.The line of action of passes through point B, and so the magnitude of the moment of about point A on the x axis is . By the right-hand rule this component acts in the negative i direction. Likewise, passes through point C and so it contributes a moment component of about the axis. Thus, as shown in Eq. 4–8. As an exercise, establish the j and k components of in this manner and show that indeed the expanded form of the determinant, Eq. 4–8, represents the moment of F about point O. Once is determined, realize that it will always be perpendicular to the shaded plane containing vectors r and F, Fig. 4–12a. Resultant Moment of a System of Forces. If a body is acted upon by a system of forces, Fig. 4–13, the resultant moment of the forces about point O can be determined by vector addition of the moment of each force.This resultant can be written symbolically as (4–9) MRO = ©1r * F2 MO MO 1MO2x = 1ryFz - rzFy2 ryFzi Fz rzFy Fy Fy Fx Fz Fy, Fx, MO MO = 1ryFz - rzFy2i - 1rxFz - rzFx2j + 1rxFy - ryFx2k Fx, Fy, Fz rx, ry, rz MO = r * F = 3 i j k rx ry rz Fx Fy Fz 3 z C y Fy Fx rz r ry rx x A B O F (b) Fz Fig. 4–12 z MO Moment axis x y O F (a) r z x y O r2 r1 r3 F3 F1 F2 MRO Fig. 4–13
- 143. 126 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 EXAMPLE 4.3 Determine the moment produced by the force F in Fig. 4–14a about point O. Express the result as a Cartesian vector. SOLUTION As shown in Fig. 4–14a, either or can be used to determine the moment about point O.These position vectors are Force F expressed as a Cartesian vector is Thus Ans. or Ans. NOTE: As shown in Fig. 4–14b, acts perpendicular to the plane that contains . Had this problem been worked using notice the difficulty that would arise in obtaining the moment arm d. MO = Fd, F, rA, and rB MO = 5-16.5i + 5.51j6 kN–m + [4(1.376) - 12(0.4588)]k = [12(-1.376) - 0(1.376)]i - [4(-1.376) - 0(0.4588)]j MO = rB * F = 3 i j k 4 12 0 0.4588 1.376 -1.376 3 = 5-16.5i + 5.51j6 kN–m + [0(1.376) - 0(0.4588)]k = [0(-1.376) - 12(1.376)]i - [0(-1.376) - 12(0.4588)] j MO = rA * F = 3 i j k 0 0 12 0.4588 1.376 -1.376 3 = 50.4588i + 1.376j - 1.376k6 kN F = FuAB = 2 kNc 54i + 12j - 12k6 m 214 m22 + 112 m22 + 1-12 m22 d rA = 512k6 m and rB = 54i + 12j6 m rB rA 12 m 4 m 12 m A B O x y z (a) F 2 kN uAB Fig. 4–14 (b) A B O x y z F rB rA MO
- 144. 4.3 MOMENT OF A FORCE—VECTOR FORMULATION 127 4 EXAMPLE 4.4 x z O 5 ft 4 ft 2 ft A B F2 {80i 40j 30k} lb F1 {60i 40j 20k} lb (a) y Fig. 4–15 x y z O A B (b) rA rB F1 F2 x y z O 39.8 67.4 121 MRO {30i 40j 60k} lb · ft (c) a g b SOLUTION Position vectors are directed from point O to each force as shown in Fig. 4–15b.These vectors are The resultant moment about O is therefore rB = 54i + 5j - 2k6 ft rA = 55j6 ft NOTE: This result is shown in Fig. 4–15c. The coordinate direction angles were determined from the unit vector for Realize that the two forces tend to cause the rod to rotate about the moment axis in the manner shown by the curl indicated on the moment vector. MRO . Ans. = 530i - 40j + 60k6 lb # ft + [51-302 - 1-221402]i - [41-302 - (-2)1802]j + [41402 - 51802]k = [51202 - 01402]i - [0]j + [01402 - (5)1-602]k = 3 i j k 0 5 0 -60 40 20 3 + 3 i j k 4 5 -2 80 40 -30 3 = rA * F1 + rB * F3 MRO = ©1r * F2 Two forces act on the rod shown in Fig. 4–15a. Determine the resultant moment they create about the flange at O. Express the result as a Cartesian vector.
- 145. 128 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 Important Points ● The moment of a force creates the tendency of a body to turn about an axis passing through a specific point O. ● Using the right-hand rule, the sense of rotation is indicated by the curl of the fingers, and the thumb is directed along the moment axis, or line of action of the moment. ● The magnitude of the moment is determined from where d is called the moment arm, which represents the perpendicular or shortest distance from point O to the line of action of the force. ● In three dimensions the vector cross product is used to determine the moment, i.e., Remember that r is directed from point O to any point on the line of action of F. ● The principle of moments states that the moment of a force about a point is equal to the sum of the moments of the force’s components about the point.This is a very convenient method to use in two dimensions. MO = r * F. MO = Fd, F2 O r F1 F Fig 4–16 MO Fx F Fy O d x y Fig. 4–17 F Fy Fy Fx Fx F d O The moment of the applied force F about point O is easy to determine if we use the principle of moments. It is simply . MO = Fxd 4.4 Principle of Moments A concept often used in mechanics is the principle of moments, which is sometimes referred to as Varignon’s theorem since it was originally developed by the French mathematician Varignon (1654–1722). It states that the moment of a force about a point is equal to the sum of the moments of the components of the force about the point.This theorem can be proven easily using the vector cross product since the cross product obeys the distributive law. For example, consider the moments of the force and two of its components about point O. Fig. 4–16. Since we have For two-dimensional problems, Fig. 4–17, we can use the principle of moments by resolving the force into its rectangular components and then determine the moment using a scalar analysis.Thus, This method is generally easier than finding the same moment using . MO = Fd MO = F xy - F yx MO = r * F = r * 1F1 + F22 = r * F1 + r * F2 F = F1 + F2 F
- 146. 4.4 PRINCIPLE OF MOMENTS 129 4 EXAMPLE 4.5 Determine the moment of the force in Fig. 4–18a about point O. Fig. 4–18 x y (c) 45 30 30 3 m O Fx (5 kN) sin 75 Fy (5 kN) sin 75 SOLUTION I The moment arm d in Fig. 4–18a can be found from trigonometry. Thus, b Ans. Since the force tends to rotate or orbit clockwise about point O, the moment is directed into the page. SOLUTION II The x and y components of the force are indicated in Fig. 4–18b. Considering counterclockwise moments as positive, and applying the principle of moments, we have MO = Fd = (5kN)(2.898 m2 = 14.5 kN # m d = (3 m) sin 75° = 2.898 m a b Ans. = -14.5 kN # m = 14.5 kN # m = -15 cos 45° kN213 sin 30° m2 - 15 sin 45° kN213 cos 30° m2 + MO = - Fxdy - Fydx SOLUTION III The x and y axes can be set parallel and perpendicular to the rod’s axis as shown in Fig. 4-18c. Here produces no moment about point O since its line of action passes through this point.Therefore, a b Ans. = -14.5 kN # m = 14.5 kN # m = -(5 sin 75° kN)(3 m) + MO = -Fy dx Fx 30 (a) 45 F 5 kN 3 m O d 75 y x (b) 30 45 O dy 3 sin 30 m dx 3 cos 30 m Fx (5 kN) cos 45 Fy (5 kN) sin 45
- 147. 130 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 EXAMPLE 4.6 Force F acts at the end of the angle bracket shown in Fig. 4–19a. Determine the moment of the force about point O. SOLUTION I (SCALAR ANALYSIS) The force is resolved into its x and y components as shown in Fig. 4–19b, then a b or Ans. SOLUTION II (VECTOR ANALYSIS) Using a Cartesian vector approach, the force and position vectors shown in Fig. 4–19c are The moment is therefore Ans. NOTE: It is seen that the scalar analysis (Solution I) provides a more convenient method for analysis than Solution II since the direction of the moment and the moment arm for each component force are easy to establish. Hence, this method is generally recommended for solving problems displayed in two dimensions, whereas a Cartesian vector analysis is generally recommended only for solving three-dimensional problems. = 5-98.6k6 N # m = 0i - 0j + [0.41-346.42 - 1-0.221200.02]k MO = r * F = 3 i j k 0.4 -0.2 0 200.0 -346.4 0 3 = 5200.0i - 346.4j6 N F = 5400 sin 30°i - 400 cos 30°j6 N r = 50.4i - 0.2j6 m MO = 5-98.6k6 N # m = -98.6 N # m = 98.6 N # m +MO = 400 sin 30° N10.2 m2 - 400 cos 30° N10.4 m2 0.4 m 0.2 m 30 O F = 400 N (a) Fig. 4–19 0.4 m 0.2 m (b) x 400 cos 30 N 400 sin 30 N O y y x 0.4 m 0.2 m 30 O F (c) r
- 148. 4.4 PRINCIPLE OF MOMENTS 131 4 FUNDAMENTAL PROBLEMS F4–1. Determine the moment of the force about point O. 5 ft 0.5 ft 600 lb 20 30 O F4–1 5 m 2 m 100 N 3 4 5 O F4–2 30 45 F 300 N 0.4 m 0.3 m O F4–3 4 ft 3 ft 1 ft 600 lb O 45 F4–4 50 N 60 45 100 mm 100 mm 200 mm O F4–5 500 N 3 m O 45 F4–6 F4–4. Determine the moment of the force about point O. F4–5. Determine the moment of the force about point O. Neglect the thickness of the member. F4–6. Determine the moment of the force about point O. F4–3. Determine the moment of the force about point O. F4–2. Determine the moment of the force about point O.
- 149. 132 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 F4–10. Determine the moment of force F about point O. Express the result as a Cartesian vector. F4–9. Determine the resultant moment produced by the forces about point O. F 4–11. Determine the moment of force F about point O. Express the result as a Cartesian vector. F4–12. If and , determine the resultant moment produced by these forces about point O. Express the result as a Cartesian vector. + 250j + 100k6 lb F2 = 5-200i F1 = 5100i - 120j + 75k6 lb O 2 m 2.5 m 45 1 m 600 N 300 N 500 N F4–7 F1 500 N F2 600 N A 0.25 m 0.3 m 0.125 m 60 4 3 5 O F4–8 O 30 30 6 ft 6 ft F2 200 lb F1 300 lb F4–9 x z y O A B 4 m 3 m F 500 N F 4–10 x z y O A B C 2 ft 1 ft 4 ft 4 ft F 120 lb F4–11 z O A x y 4 ft 3 ft 5 ft F1 F2 F4–12 F4–7. Determine the resultant moment produced by the forces about point O. F4–8. Determine the resultant moment produced by the forces about point O.
- 150. 4.4 PRINCIPLE OF MOMENTS 133 4 PROBLEMS A P F B C 6 ft 45 12 ft 3 4 5 Probs. 4–4/5 3 m 0.45 m 4 kN A u Probs. 4–6/7 F B A 18 in. 5 in. 30 Probs. 4–8/9 4 kN 800 N 800 N 4 kN Case 1 Case 2 0.4 m 0.05 m 0.05 m 0.4 m O O Prob. 4–10 4–6. If , determine the moment produced by the 4-kN force about point A. 4–7. If the moment produced by the 4-kN force about point A is clockwise, determine the angle , where . 0° … u … 90° u 10 kN # m u = 45° •4–1. If A, B, and D are given vectors, prove the distributive law for the vector cross product, i.e., . 4–2. Prove the triple scalar product identity . 4–3. Given the three nonzero vectors A, B, and C, show that if , the three vectors must lie in the same plane. *4–4. Two men exert forces of and on the ropes. Determine the moment of each force about A. Which way will the pole rotate,clockwise or counterclockwise? •4–5. If the man at B exerts a force of on his rope, determine the magnitude of the force F the man at C must exert to prevent the pole from rotating, i.e., so the resultant moment about A of both forces is zero. P = 30 lb P = 50 lb F = 80 lb A # (B : C) = 0 A # B : C = A : B # C A : (B + D) = (A : B) + (A : D) *4–8. The handle of the hammer is subjected to the force of Determine the moment of this force about the point A. •4–9. In order to pull out the nail at B, the force F exerted on the handle of the hammer must produce a clockwise moment of about point A. Determine the required magnitude of force F. 500 lb # in. F = 20 lb. 4–10. The hub of the wheel can be attached to the axle either with negative offset (left) or with positive offset (right). If the tire is subjected to both a normal and radial load as shown, determine the resultant moment of these loads about point O on the axle for both cases.
- 151. 134 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 4–11. The member is subjected to a force of . If , determine the moment produced by F about point A. *4–12. Determine the angle of the force F so that it produces a maximum moment and a minimum moment about point A. Also, what are the magnitudes of these maximum and minimum moments? •4–13. Determine the moment produced by the force F about point A in terms of the angle . Plot the graph of versus , where . 0° … u … 180° u MA u u (0° … u … 180°) u = 45° F = 6 kN A 6 m 1.5 m u F 6 kN Probs. 4–11/12/13 2 in. 4 in. 6 in. 30 60 P 50 lb F A Probs. 4–14 100 mm 65 mm 200 mm A Nf 400 N Ft 5 Probs. 4–15/16 60 6 ft C B A 3 ft 3 4 5 FB FA Probs. 4–17/18 4–14. Serious neck injuries can occur when a football player is struck in the face guard of his helmet in the manner shown, giving rise to a guillotine mechanism. Determine the moment of the knee force about point A. What would be the magnitude of the neck force F so that it gives the counterbalancing moment about A? P = 50 lb 4–15. The Achilles tendon force of is mobilized when the man tries to stand on his toes.As this is done, each of his feet is subjected to a reactive force of Determine the resultant moment of and about the ankle joint A. *4–16. The Achilles tendon force is mobilized when the man tries to stand on his toes.As this is done, each of his feet is subjected to a reactive force of If the resultant moment produced by forces and about the ankle joint A is required to be zero, determine the magnitude of . Ft Nt Ft Nt = 400 N. Ft Nf Ft Nf = 400 N. Ft = 650 N •4–17. The two boys push on the gate with forces of and as shown. Determine the moment of each force about C. Which way will the gate rotate, clockwise or counterclockwise? Neglect the thickness of the gate. 4–18. Two boys push on the gate as shown. If the boy at B exerts a force of , determine the magnitude of the force the boy at A must exert in order to prevent the gate from turning. Neglect the thickness of the gate. FA FB = 30 lb FA = 30 lb
- 152. 4.4 PRINCIPLE OF MOMENTS 135 4 4–26. The foot segment is subjected to the pull of the two plantarflexor muscles. Determine the moment of each force about the point of contact A on the ground. •4–21. Determine the direction for of the force F so that it produces the maximum moment about point A. Calculate this moment. 4–22. Determine the moment of the force F about point A as a function of . Plot the results of M (ordinate) versus (abscissa) for . 4–23. Determine the minimum moment produced by the force F about point A. Specify the angle . u … 180°) u (0° … 0° … u … 180° u u 0° … u … 180° u *4–24. In order to raise the lamp post from the position shown, force F is applied to the cable. If determine the moment produced by F about point A. •4–25. In order to raise the lamp post from the position shown,the force F on the cable must create a counterclockwise moment of about point A. Determine the magnitude of F that must be applied to the cable. 1500 lb # ft F = 200 lb, 43 in. 6 in. F P MP u Probs. 4–19/20 F 400 N 3 m 2 m A u Probs. 4–21/22/23 F 75 C A B 10 ft 20 ft Probs. 4–24/25 60 30 4 in. A 1 in. 3.5 in. 70 F2 30 lb F1 20 lb Prob. 4–26 4–19. The tongs are used to grip the ends of the drilling pipe P. Determine the torque (moment) that the applied force exerts on the pipe about point P as a function of . Plot this moment versus for . *4–20. The tongs are used to grip the ends of the drilling pipe P. If a torque (moment) of is needed at P to turn the pipe, determine the cable force F that must be applied to the tongs. Set . u = 30° MP = 800 lb # ft 0 … u … 90° u MP u F = 150 lb MP
- 153. 136 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 *4–32. The towline exerts a force of at the end of the 20-m-long crane boom. If determine the placement x of the hook at A so that this force creates a maximum moment about point O.What is this moment? •4–33. The towline exerts a force of at the end of the 20-m-long crane boom. If determine the position of the boom so that this force creates a maximum moment about point O.What is this moment? u x = 25 m, P = 4 kN u = 30°, P = 4 kN •4–29. Determine the moment of each force about the bolt located at A.Take 4–30. If and determine the resultant moment about the bolt located at A. FC = 45 lb, FB = 30 lb FB = 40 lb, FC = 50 lb. 4–31. The rod on the power control mechanism for a business jet is subjected to a force of 80 N. Determine the moment of this force about the bearing at A. 4–27. The 70-N force acts on the end of the pipe at B. Determine (a) the moment of this force about point A, and (b) the magnitude and direction of a horizontal force,applied at C, which produces the same moment.Take *4–28. The 70-N force acts on the end of the pipe at B. Determine the angles of the force that will produce maximum and minimum moments about point A.What are the magnitudes of these moments? u 10° … u … 180°2 u = 60°. A C 0.3 m 0.7 m 0.9 m B 70 N u Probs. 4–27/28 Probs. 4–29/30 20 60 A 80 N 150 mm Prob. 4–31 1.5 m O 20 m A B P 4 kN x u Probs. 4–32/33
- 154. 4.4 PRINCIPLE OF MOMENTS 137 4 4–43. Determine the moment produced by each force about point O located on the drill bit. Express the results as Cartesian vectors. *4–40. Determine the moment produced by force about point O. Express the result as a Cartesian vector. •4–41. Determine the moment produced by force about point O. Express the result as a Cartesian vector. 4–42. Determine the resultant moment produced by forces and about point O. Express the result as a Cartesian vector. FC FB FC FB B 0.65 m 0.5 m 1.2 m 30 0.3 m F G A Prob. 4–34/35/36 y x z 1 ft 2 ft 2 ft A O 3 ft F2 {10i 30j 50k} lb F1 {20i 10j 30k} lb Probs. 4–37/38/39 y x z C O B A 6 m 3 m 2 m 2.5 m FC 420 N FB 780 N Probs. 4–40/41/42 x z A B O y 150 mm 600 mm 300 mm 150 mm FA {40i 100j 60k} N FB {50i 120j 60k} N Prob. 4–43 4–34. In order to hold the wheelbarrow in the position shown, force F must produce a counterclockwise moment of about the axle at A. Determine the required magnitude of force F. 4–35. The wheelbarrow and its contents have a mass of 50 kg and a center of mass at G. If the resultant moment produced by force F and the weight about point A is to be zero, determine the required magnitude of force F. *4–36. The wheelbarrow and its contents have a center of mass at G. If and the resultant moment produced by force F and the weight about the axle at A is zero, determine the mass of the wheelbarrow and its contents. F = 100 N 200 N # m •4–37. Determine the moment produced by about point O. Express the result as a Cartesian vector. 4–38. Determine the moment produced by about point O. Express the result as a Cartesian vector. 4–39. Determine the resultant moment produced by the two forces about point O. Express the result as a Cartesian vector. F2 F1 *4–44. A force of produces a moment of about the origin of coordinates, point O. If the force acts at a point having an x coordinate of determine the y and z coordinates. x = 1 m, MO = 54i + 5j - 14k6 kN # m F = 56i - 2j + 1k6 kN
- 155. 138 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 4–50. A 20-N horizontal force is applied perpendicular to the handle of the socket wrench. Determine the magnitude and the coordinate direction angles of the moment created by this force about point O. *4–48. Force F acts perpendicular to the inclined plane. Determine the moment produced by F about point A. Express the result as a Cartesian vector. •4–49. Force F acts perpendicular to the inclined plane. Determine the moment produced by F about point B. Express the result as a Cartesian vector. 4–47. The force creates a moment about point O of . If the force passes through a point having an x coordinate of 1 m, determine the y and z coordinates of the point. Also, realizing that , determine the perpendicular distance d from point O to the line of action of F. MO = Fd MO = 5-14i + 8j + 2k6 N # m F = 56i + 8j + 10k6 N 400 mm y 300 mm 200 mm 250 mm x z 30 40 F 80 N B C A Probs. 4–45/46 d z x y O y 1 m z P F MO Prob. 4–47 z x y 3 m 3 m 4 m A B C F 400 N Probs. 4–48/49 15 200 mm 75 mm 20 N A O x y z Prob. 4–50 •4–45. The pipe assembly is subjected to the 80-N force. Determine the moment of this force about point A. 4–46. The pipe assembly is subjected to the 80-N force. Determine the moment of this force about point B.
- 156. 4.5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS 139 4.5 Moment of a Force about a Specified Axis Sometimes, the moment produced by a force about a specified axis must be determined. For example, suppose the lug nut at O on the car tire in Fig. 4–20a needs to be loosened. The force applied to the wrench will create a tendency for the wrench and the nut to rotate about the moment axis passing through O; however, the nut can only rotate about the y axis. Therefore, to determine the turning effect, only the y component of the moment is needed, and the total moment produced is not important. To determine this component, we can use either a scalar or vector analysis. Scalar Analysis. To use a scalar analysis in the case of the lug nut in Fig.4–20a,the moment arm perpendicular distance from the axis to the line of action of the force is Thus, the moment of F about the y dy = d cos u. F A B F x y d (a) z O dy MO My Moment Axis u Fig. 4–20 4 If large enough,the cable force F on the boom of this crane can cause the crane to topple over. To investigate this, the moment of the force must be calculated about an axis passing through the base of the legs at A and B. axis is According to the right-hand rule, is directed along the positive y axis as shown in the figure. In general, for any axis a, the moment is (4–10) Ma = Fda My My = F dy = F(d cos u).
- 157. 140 CHAPTER 4 FORCE SYSTEM RESULTANTS Vector Analysis. To find the moment of force F in Fig. 4–20b about the y axis using a vector analysis, we must first determine the moment of the force about any point O on the y axis by applying Eq. 4–7, .The component along the y axis is the projection of onto the y axis. It can be found using the dot product discussed in Chapter 2, so that where j is the unit vector for the y axis. We can generalize this approach by letting be the unit vector that specifies the direction of the axis shown in Fig. 4–21.Then the moment of F about the axis is . This combination is referred to as the scalar triple product. If the vectors are written in Cartesian form, we have This result can also be written in the form of a determinant, making it easier to memorize.* (4–11) where represent the x, y, z components of the unit vector defining the direction of the axis represent the x, y, z components of the position vector extended from any point O on the axis to any point A on the line of action of the force represent the x, y, z components of the force vector. When is evaluated from Eq. 4–11, it will yield a positive or negative scalar.The sign of this scalar indicates the sense of direction of along the axis. If it is positive, then will have the same sense as whereas if it is negative, then will act opposite to Once is determined, we can then express as a Cartesian vector, namely, (4–12) The examples which follow illustrate numerical applications of the above concepts. Ma = Maua Ma Ma ua. Ma ua, Ma a Ma Ma Fz Fy, Fx, a rz ry, rx, a uax , uay , uaz Ma = ua # 1r * F2 = 3 uax uay uaz rx ry rz Fx Fy Fz 3 = uax (ryFz - rzFy) - uay (rxFz - rzFx) + uaz (rxFy - ryFx) Ma = [uax i + uay j + uaz k] # 3 i j k rx ry rz Fx Fy Fz 3 Ma = ua # (r * F) a ua My = j # MO = j # (r * F), MO My MO = r * F *Take a moment to expand this determinant, to show that it will yield the above result. r O MO r F Ma ua a Axis of projection F A Fig. 4–21 x y r j (b) z O M0 r F F u u My 4 Fig. 4–20
- 158. 4.5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS 141 4 Important Points ● The moment of a force about a specified axis can be determined provided the perpendicular distance from the force line of action to the axis can be determined. ● If vector analysis is used, where defines the direction of the axis and r is extended from any point on the axis to any point on the line of action of the force. ● If is calculated as a negative scalar, then the sense of direction of is opposite to ● The moment expressed as a Cartesian vector is determined from Ma = Maua. Ma ua. Ma Ma ua Ma = ua # 1r * F2, Ma = Fda. da EXAMPLE 4.7 Determine the resultant moment of the three forces in Fig. 4–22 about the x axis, the y axis, and the z axis. SOLUTION A force that is parallel to a coordinate axis or has a line of action that passes through the axis does not produce any moment or tendency for turning about that axis.Therefore, defining the positive direction of the moment of a force according to the right-hand rule, as shown in the figure, we have Ans. Ans. Ans. The negative signs indicate that and act in the and directions, respectively. -z -y Mz My Mz = 0 + 0 - (40 lb)(2 ft) = -80 lb # ft My = 0 - (50 lb)(3 ft) - (40 lb)(2 ft) = -230 lb # ft Mx = (60 lb)(2 ft) + (50 lb)(2 ft) + 0 = 220 lb # ft 2 ft 2 ft 2 ft 3 ft x y z B C A O F3 40 lb F2 50 lb F1 60 lb Fig. 4–22
- 159. 142 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 EXAMPLE 4.8 Determine the moment produced by the force F in Fig. 4–23a, which tends to rotate the rod about the AB axis. SOLUTION A vector analysis using will be considered for the solution rather than trying to find the moment arm or perpendicular distance from the line of action of F to the AB axis. Each of the terms in the equation will now be identified. Unit vector defines the direction of the AB axis of the rod, Fig. 4–23b, where Vector r is directed from any point on the AB axis to any point on the line of action of the force. For example, position vectors and are suitable, Fig. 4–23b. (Although not shown, or can also be used.) For simplicity, we choose where The force is Substituting these vectors into the determinant form and expanding, we have This positive result indicates that the sense of is in the same direction as Expressing as a Cartesian vector yields Ans. The result is shown in Fig. 4–23b. NOTE: If axis AB is defined using a unit vector directed from B toward A,then in the above formulation would have to be used.This would lead to Consequently, and the same result would be obtained. MAB = MAB1-uB2, MAB = -80.50 N # m. -uB = 572.0i + 36.0j6 N # m MAB = MABuB = 180.50 N # m210.8944i + 0.4472j2 MAB uB. MAB = 80.50 N # m + 0[0.6102 - 0102] = 0.8944[01-3002 - 0102] - 0.4472[0.61-3002 - 0102] MAB = uB # 1rD * F2 = 3 0.8944 0.4472 0 0.6 0 0 0 0 -300 3 F = 5-300k6 N rD = 50.6i6 m rD, rBD rBC rD rC uB = rB rB = {0.4i + 0.2j} m 210.4 m2 2 + 10.2 m2 2 = 0.8944i + 0.4472j uB MAB = uB # 1r * F2 MAB 0.4 m (a) 0.3 m 0.6 m 0.2 m C F = 300 N B x y z A Fig. 4–23 (b) F C B x z MAB uB rC rD A D y
- 160. 4.5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS 143 4 EXAMPLE 4.9 Determine the magnitude of the moment of force F about segment OA of the pipe assembly in Fig. 4–24a. SOLUTION The moment of F about the OA axis is determined from where r is a position vector extending from any point on the OA axis to any point on the line of action of F. As indicated in Fig. 4–24b, either can be used; however, will be considered since it will simplify the calculation. The unit vector , which specifies the direction of the OA axis, is and the position vector is rOD = 50.5i + 0.5k6 m rOD uOA = rOA rOA = 50.3i + 0.4j6 m 210.3 m22 + 10.4 m22 = 0.6i + 0.8j uOA rOD rOD, rOC, rAD, or rAC MOA = uOA # (r * F), The force F expressed as a Cartesian vector is = {200i - 200j + 100k} N = (300 N)B 50.4i - 0.4j + 0.2k6 m 2(0.4 m)2 + (-0.4 m)2 + (0.2 m)2 R F = Fa rCD rCD b Therefore, Ans. = 100 N#m = 0.6[011002 - (0.5)1-2002] - 0.8[0.511002 - (0.5)12002] + 0 = 3 0.6 0.8 0 0.5 0 0.5 200 -200 100 3 MOA = uOA # 1rOD * F2 0.1 m 0.3 m 0.2 m 0.4 m 0.5 m 0.5 m (a) x y C A O D z F 300 N B Fig. 4–24 x y z F (b) D A C O rOD rAD rAC rOC uOA
- 161. 144 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 FUNDAMENTAL PROBLEMS F4–18. Determine the moment of force F about the x, the y, and the z axes. Use a scalar analysis. F4–15. Determine the magnitude of the moment of the 200-N force about the x axis. F4–13. Determine the magnitude of the moment of the force about the x axis. Express the result as a Cartesian vector. F4–14. Determine the magnitude of the moment of the force about the OA axis. Express the result as a Cartesian vector. F = 5300i - 200j + 150k6 N F = 5300i - 200j + 150k6 N z O A B F x y 0.4 m 0.2 m 0.3 m F4–13/14 x O A 45 120 60 F 200 N z y 0.25 m 0.3 m F4–15 2 m F {30i 20j 50k} N 4 m z x y A 3 m F4–16 2 ft 4 ft 3 ft x y z B C A F F4–17 z A O y x F 500 N 3 m 2 m 2 m 3 3 4 4 5 5 F4–18 F4–16. Determine the magnitude of the moment of the force about the y axis. F4–17. Determine the moment of the force about the AB axis. Express the result as a Cartesian vector. F = 550i - 40j + 20k6 lb
- 162. 4.5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS 145 4 3 m 1.5 m 3 m x C A B G F y z O D F {6i 3j 10k} N Probs. 4–51/52 x y 0.4 m F {60i 20j 15k} N 30 z 0.25 m Prob. 4–53 4 ft 3 ft 2 ft y z C A B F {4i 12j 3k} lb x Probs. 4–54/55 y x z 4 m 4 m 3 m A B C F {20i 10j 15k} N Prob. 4–56 •4–53. The tool is used to shut off gas valves that are difficult to access. If the force F is applied to the handle, determine the component of the moment created about the z axis of the valve. 4–51. Determine the moment produced by force F about the diagonal AF of the rectangular block. Express the result as a Cartesian vector. *4–52. Determine the moment produced by force F about the diagonal OD of the rectangular block. Express the result as a Cartesian vector. *4–56. Determine the moment produced by force F about segment AB of the pipe assembly. Express the result as a Cartesian vector. PROBLEMS 4–54. Determine the magnitude of the moments of the force F about the x, y, and z axes. Solve the problem (a) using a Cartesian vector approach and (b) using a scalar approach. 4–55. Determine the moment of the force F about an axis extending between A and C. Express the result as a Cartesian vector.
- 163. 146 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 4–58. If , determine the magnitude of the moment produced by this force about the x axis. 4–59. The friction at sleeve A can provide a maximum resisting moment of about the x axis. Determine the largest magnitude of force F that can be applied to the bracket so that the bracket will not turn. 125 N # m F = 450 N •4–57. Determine the magnitude of the moment that the force F exerts about the y axis of the shaft. Solve the problem using a Cartesian vector approach and using a scalar approach. *4–60. Determine the magnitude of the moment produced by the force of about the hinged axis (the x axis) of the door. F = 200 N 200 mm 250 mm 45 B x y z A O 30 50 mm 16 N F Prob. 4–57 300 mm x y z A B 60 60 45 F 100 mm 150 mm Probs. 4–58/59 y x z 15 A B 2.5 m 2 m F 200 N 0.5 m 1 m Prob. 4–60 6 ft 4 ft 4 ft 6 ft y z A C F D B 6 ft x Probs. 4–61/62 •4–61. If the tension in the cable is , determine the magnitude of the moment produced by this force about the hinged axis, CD, of the panel. 4–62. Determine the magnitude of force F in cable AB in order to produce a moment of about the hinged axis CD, which is needed to hold the panel in the position shown. 500 lb # ft F = 140 lb
- 164. 60 A 10 in. 0.75 in. P Probs. 4–66/67 4.5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS 147 4 4–66. The flex-headed ratchet wrench is subjected to a force of applied perpendicular to the handle as shown. Determine the moment or torque this imparts along the vertical axis of the bolt at A. 4–67. If a torque or moment of is required to loosen the bolt at A, determine the force P that must be applied perpendicular to the handle of the flex-headed ratchet wrench. 80 lb # in. P = 16 lb, *4–68. The pipe assembly is secured on the wall by the two brackets. If the flower pot has a weight of 50 lb, determine the magnitude of the moment produced by the weight about the OA axis. •4–69. The pipe assembly is secured on the wall by the two brackets. If the frictional force of both brackets can resist a maximum moment of , determine the largest weight of the flower pot that can be supported by the assembly without causing it to rotate about the OA axis. 150 lb # ft A O z x y 4 ft 3 ft 3 ft 4 ft 60 30 B 30 15 6 ft y y¿ x¿ C A B F x z 3 ft 3 ft Probs. 4–68/69 45 z y A C B 500 mm 200 mm 150 mm F x Probs. 4–70/71 4–63. The A-frame is being hoisted into an upright position by the vertical force of . Determine the moment of this force about the axis passing through points A and B when the frame is in the position shown. *4–64. The A-frame is being hoisted into an upright position by the vertical force of . Determine the moment of this force about the x axis when the frame is in the position shown. •4–65. The A-frame is being hoisted into an upright position by the vertical force of . Determine the moment of this force about the y axis when the frame is in the position shown. F = 80 lb F = 80 lb y¿ F = 80 lb 4–70. A vertical force of is applied to the handle of the pipe wrench. Determine the moment that this force exerts along the axis AB (x axis) of the pipe assembly. Both the wrench and pipe assembly ABC lie in the plane. Suggestion: Use a scalar analysis. 4–71. Determine the magnitude of the vertical force F acting on the handle of the wrench so that this force produces a component of moment along the AB axis (x axis) of the pipe assembly of . Both the pipe assembly ABC and the wrench lie in the plane. Suggestion: Use a scalar analysis. x-y (MA)x = 5-5i6 N # m x-y F = 60 N Probs. 4–63/64/65
- 165. 148 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 4.6 Moment of a Couple A couple is defined as two parallel forces that have the same magnitude, but opposite directions, and are separated by a perpendicular distance d, Fig. 4–25. Since the resultant force is zero, the only effect of a couple is to produce a rotation or tendency of rotation in a specified direction. For example,imagine that you are driving a car with both hands on the steering wheel and you are making a turn. One hand will push up on the wheel while the other hand pulls down,which causes the steering wheel to rotate. The moment produced by a couple is called a couple moment. We can determine its value by finding the sum of the moments of both couple forces about any arbitrary point. For example, in Fig. 4–26, position vectors and are directed from point O to points A and B lying on the line of action of and F. The couple moment determined about O is therefore However or , so that (4–13) This result indicates that a couple moment is a free vector, i.e., it can act at any point since M depends only upon the position vector r directed between the forces and not the position vectors and directed from the arbitrary point O to the forces.This concept is unlike the moment of a force, which requires a definite point (or axis) about which moments are determined. Scalar Formulation. The moment of a couple, M, Fig. 4–27, is defined as having a magnitude of (4–14) where F is the magnitude of one of the forces and d is the perpendicular distance or moment arm between the forces. The direction and sense of the couple moment are determined by the right-hand rule, where the thumb indicates this direction when the fingers are curled with the sense of rotation caused by the couple forces. In all cases, M will act perpendicular to the plane containing these forces. Vector Formulation. The moment of a couple can also be expressed by the vector cross product using Eq. 4–13, i.e., (4–15) Application of this equation is easily remembered if one thinks of taking the moments of both forces about a point lying on the line of action of one of the forces. For example, if moments are taken about point A in Fig. 4–26, the moment of is zero about this point, and the moment of F is defined from Eq. 4–15.Therefore, in the formulation r is crossed with the force F to which it is directed. -F M = r * F M = Fd rB, rA M = r * F r = rB - rA rB = rA + r M = rB * F + rA * -F = (rB - rA) * F -F rB rA F ⴚF d Fig. 4–25 O B A F ⴚF rA rB r Fig. 4–26 F ⴚF d M Fig. 4–27
- 166. M2 M1 (a) 4.6 MOMENT OF A COUPLE 149 4 Equivalent Couples. If two couples produce a moment with the same magnitude and direction,then these two couples are equivalent.For example, the two couples shown in Fig. 4–28 are equivalent because each couple moment has a magnitude of , and each is directed into the plane of the page. Notice that larger forces are required in the second case to create the same turning effect because the hands are placed closer together. Also, if the wheel was connected to the shaft at a point other than at its center, then the wheel would still turn when each couple is applied since the couple is a free vector. Resultant Couple Moment. Since couple moments are vectors, their resultant can be determined by vector addition. For example, consider the couple moments acting on the pipe in Fig. 4–29a. Since each couple moment is a free vector, we can join their tails at any arbitrary point and find the resultant couple moment, as shown in Fig. 4–29b. If more than two couple moments act on the body, we may generalize this concept and write the vector resultant as (4–16) These concepts are illustrated numerically in the examples that follow. In general, problems projected in two dimensions should be solved using a scalar analysis since the moment arms and force components are easy to determine. MR = ©1r * F2 MR = M1 + M2 M1 and M2 12 N # m M = 30 N(0.4 m) = 40 N(0.3 m) = 12 N # m 0.3 m 0.4 m 30 N 40 N 40 N 30 N Fig. 4–28 MR (b) M2 M1 Fig. 4–29
- 167. 150 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 Steering wheels on vehicles have been made smaller than on older vehicles because power steering does not require the driver to apply a large couple moment to the rim of the wheel. F F Important Points ● A couple moment is produced by two noncollinear forces that are equal in magnitude but opposite in direction. Its effect is to produce pure rotation, or tendency for rotation in a specified direction. ● A couple moment is a free vector, and as a result it causes the same rotational effect on a body regardless of where the couple moment is applied to the body. ● The moment of the two couple forces can be determined about any point. For convenience, this point is often chosen on the line of action of one of the forces in order to eliminate the moment of this force about the point. ● In three dimensions the couple moment is often determined using the vector formulation, where r is directed from any point on the line of action of one of the forces to any point on the line of action of the other force F. ● A resultant couple moment is simply the vector sum of all the couple moments of the system. M = r * F, EXAMPLE 4.10 Determine the resultant couple moment of the three couples acting on the plate in Fig. 4–30. SOLUTION As shown the perpendicular distances between each pair of couple forces are and . Considering counterclockwise couple moments as positive, we have a b Ans. The negative sign indicates that has a clockwise rotational sense. MR = -950 lb # ft = 950 lb # ft = (-200 lb)(4 ft) + (450 lb)(3 ft) - (300 lb)(5 ft) +MR = ©M; MR = -F1d1 + F2d2 - F3d3 d3 = 5 ft d2 = 3 ft, d1 = 4 ft, F2 450 lb F1 200 lb F3 300 lb F3 300 lb F2 450 lb d3 5 ft F1 200 lb A B d2 3 ft d1 4 ft Fig. 4–30
- 168. 4.6 MOMENT OF A COUPLE 151 4 EXAMPLE 4.11 Determine the magnitude and direction of the couple moment acting on the gear in Fig. 4–31a. SOLUTION The easiest solution requires resolving each force into its components as shown in Fig. 4–31b. The couple moment can be determined by summing the moments of these force components about any point, for example, the center O of the gear or point A. If we consider counterclockwise moments as positive, we have a d Ans. or a d Ans. = 43.9 N#m +M = ©MA; M = (600 cos 30° N)(0.2 m) - (600 sin 30° N)(0.2 m) = 43.9 N#m +M = ©MO; M = (600 cos 30° N)(0.2 m) - (600 sin 30° N)(0.2 m) This positive result indicates that M has a counterclockwise rotational sense, so it is directed outward, perpendicular to the page. NOTE: The same result can also be obtained using , where d is the perpendicular distance between the lines of action of the couple forces, Fig. 4–31c. However, the computation for d is more involved. Realize that the couple moment is a free vector and can act at any point on the gear and produce the same turning effect about point O. M = Fd (b) 30 F 600 N 600 sin 30 N 600 cos 30 N 30 F 600 N 600 sin 30 N 600 cos 30 N 0.2 m O A 30 30 (c) F 600 N F 600 N O d Fig. 4–31 30 30 (a) F 600 N F 600 N 0.2 m O
- 169. 152 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 EXAMPLE 4.12 Determine the couple moment acting on the pipe shown in Fig. 4–32a. Segment AB is directed 30° below the x–y plane. SOLUTION I (VECTOR ANALYSIS) The moment of the two couple forces can be found about any point. If point O is considered, Fig. 4–32b, we have Ans. It is easier to take moments of the couple forces about a point lying on the line of action of one of the forces, e.g., point A, Fig. 4–32c. In this case the moment of the force at A is zero, so that Ans. SOLUTION II (SCALAR ANALYSIS) Although this problem is shown in three dimensions, the geometry is simple enough to use the scalar equation The perpendicular distance between the lines of action of the couple forces is Fig. 4–32d. Hence, taking moments of the forces about either point A or point B yields Applying the right-hand rule, M acts in the direction.Thus, Ans. M = 5-130j6 lb # in. -j M = Fd = 25 lb 15.196 in.2 = 129.9 lb # in. d = 6 cos 30° = 5.196 in., M = Fd. = 5-130j6 lb # in. = 16 cos 30°i - 6 sin 30°k2 * 125k2 M = rAB * 125k2 = 5-130j6 lb # in. = -200i - 129.9j + 200i = 18j2 * 1-25k2 + 16 cos 30°i + 8j - 6 sin 30°k2 * 125k2 M = rA * 1-25k2 + rB * 125k2 O z 30 x y 25 lb A 25 lb B 8 in. 6 in. (a) Fig. 4–32 z x 25 lb A 25 lb B (b) y rB rA O z x y 25 lb A 25 lb B (c) rAB O 6 in. z x y 25 lb A 25 lb B (d) 30 d O
- 170. 4.6 MOMENT OF A COUPLE 153 4 EXAMPLE 4.13 Replace the two couples acting on the pipe column in Fig. 4–33a by a resultant couple moment. SOLUTION (VECTOR ANALYSIS) The couple moment developed by the forces at A and B, can easily be determined from a scalar formulation. By the right-hand rule, acts in the direction, Fig. 4–33b. Hence, Vector analysis will be used to determine caused by forces at C and D. If moments are computed about point D, Fig. 4–33a, then Since and are free vectors, they may be moved to some arbitrary point and added vectorially, Fig. 4–33c. The resultant couple moment becomes Ans. MR = M1 + M2 = 560i + 22.5j + 30k6 N # m M2 M1 = 522.5j + 30k6 N # m = 10.3i2 * [100j - 75k] = 301i * j2 - 22.51i * k2 M2 = rDC * FC = 10.3i2 * C125A4 5 Bj - 125A3 5 BkD M2 = rDC * FC, M2, M1 = 560i6 N # m +i M1 M1 = Fd = 150 N10.4 m2 = 60 N # m M1, 0.3 m 150 N 125 N 125 N 3 4 5 D z y 5 3 4 C 0.4 m 150 N A B x (a) Fig. 4–33 M2 37.5 N m M1 60 N m 3 4 5 (b) (c) M1 M2 MR
- 171. 154 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 FUNDAMENTAL PROBLEMS F4–23. Determine the resultant couple moment acting on the pipe assembly. F4–24. Determine the couple moment acting on the pipe assembly and express the result as a Cartesian vector. F4–19. Determine the resultant couple moment acting on the beam. F4–22. Determine the couple moment acting on the beam. 0.2 m 200 N 200 N A 300 N 300 N 400 N 400 N 3 m 2 m F4–19 4 ft 4 ft 4 ft 300 lb 200 lb 200 lb 300 lb 150 lb 150 lb F4–20 2 kN 2 kN 0.3 m A F F B 0.9 m F4–21 A B 4 m 1 m 1 m 10 kN 10 kN 4 3 5 4 3 5 F4–22 y z (Mc)3 300 lbft (Mc)1 450 lbft (Mc)2 250 lbft 2 ft 2 ft 2 ft 1.5 ft 3.5 ft x F4–23 B A 0.4 m z y x FA 450 N FB 450 N 3 3 4 4 5 5 C O 0.3 m F4–24 F4–21. Determine the magnitude of F so that the resultant couple moment acting on the beam is clockwise. 1.5 kN # m F4–20. Determine the resultant couple moment acting on the triangular plate.
- 172. 4.6 MOMENT OF A COUPLE 155 4 PROBLEMS 4–74. The caster wheel is subjected to the two couples. Determine the forces F that the bearings exert on the shaft so that the resultant couple moment on the caster is zero. *4–72. The frictional effects of the air on the blades of the standing fan creates a couple moment of on the blades. Determine the magnitude of the couple forces at the base of the fan so that the resultant couple moment on the fan is zero. MO = 6 N # m 0.15 m 0.15 m F F MO Prob. 4–72 M3 M2 45 M1 300 Nm Prob. 4–73 40 mm 45 mm 100 mm 500 N 500 N 50 mm F F A B Prob. 4–74 A B F F 2 ft 2 ft 2 ft 2 ft 150 lb 150 lb 3 3 4 4 5 5 2 ft 30 30 Probs. 4–75/76 •4–73. Determine the required magnitude of the couple moments and so that the resultant couple moment is zero. M3 M2 4–75. If , determine the resultant couple moment. *4–76. Determine the required magnitude of force F if the resultant couple moment on the frame is , clockwise. 200 lb # ft F = 200 lb
- 173. 156 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 *4–80. Two couples act on the beam. Determine the magnitude of F so that the resultant couple moment is counterclockwise. Where on the beam does the resultant couple moment act? 450 lb # ft, •4–77. The floor causes a couple moment of and on the brushes of the polishing machine. Determine the magnitude of the couple forces that must be developed by the operator on the handles so that the resultant couple moment on the polisher is zero. What is the magnitude of these forces if the brush at B suddenly stops so that MB = 0? MB = 30 N # m MA = 40 N # m •4–81. The cord passing over the two small pegs A and B of the square board is subjected to a tension of 100 N. Determine the required tension P acting on the cord that passes over pegs C and D so that the resultant couple produced by the two couples is acting clockwise. Take . 4–82. The cord passing over the two small pegs A and B of the board is subjected to a tension of 100 N. Determine the minimum tension P and the orientation of the cord passing over pegs C and D, so that the resultant couple moment produced by the two cords is , clockwise. 20 N # m u u = 15° 15 N # m 4–78. If , determine the magnitude of force F so that the resultant couple moment is , clockwise. 4–79. If , determine the required angle so that the resultant couple moment is zero. u F = 200 N 100 N # m u = 30° 0.3 m MB MA F ⴚF A B Prob. 4–77 30 15 15 F F 300 N 300 N 300 mm 30 u u Probs. 4–78/79 200 lb 200 lb 2 ft 1.5 ft 1.25 ft 30 30 ⴚF F Prob. 4–80 100 N 100 N P P C B 30 300 mm 300 mm 30 A D 45 u u Probs. 4–81/82
- 174. 4.6 MOMENT OF A COUPLE 157 4 •4–85. Determine the resultant couple moment acting on the beam. Solve the problem two ways: (a) sum moments about point O; and (b) sum moments about point A. 4–83. A device called a rolamite is used in various ways to replace slipping motion with rolling motion. If the belt, which wraps between the rollers, is subjected to a tension of 15 N, determine the reactive forces N of the top and bottom plates on the rollers so that the resultant couple acting on the rollers is equal to zero. 4–86. Two couples act on the cantilever beam. If , determine the resultant couple moment. 4–87. Determine the required magnitude of force F, if the resultant couple moment on the beam is to be zero. F = 6 kN *4–84. Two couples act on the beam as shown. Determine the magnitude of F so that the resultant couple moment is counterclockwise. Where on the beam does the resultant couple act? 300 lb # ft N N 30 25 mm A B 25 mm T 15 N T 15 N Prob. 4–83 200 lb 200 lb 1.5 ft Prob. 4–84 1.5 m 1.8 m 45 45 30 30 A 2 kN 2 kN 8 kN B 0.3 m 8 kN O Prob. 4–85 F F 5 kN 5 kN 0.5 m 0.5 m 30 30 4 4 3 3 5 5 3 m A B 3 m Probs. 4–86/87
- 175. 158 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 3 ft 60 lb 40 lb 40 lb 30 d y x A B 1 ft 30 3 4 5 4 ft 2 ft 3 4 5 60 lb Probs. 4–88/89/90 30 x z y M1 M2 M3 Probs. 4–91/92 x z y 300 mm 200 mm 200 mm 300 mm 300 mm F F Probs. 4–93/94 y y x¿ x ¿ 25 My Mx Prob. 4–95 •4–93. If , determine the magnitude and coordinate direction angles of the couple moment.The pipe assembly lies in the x–y plane. 4–94. If the magnitude of the couple moment acting on the pipe assembly is , determine the magnitude of the couple forces applied to each wrench. The pipe assembly lies in the x–y plane. 50 N # m F = 80 N 4–95. From load calculations it is determined that the wing is subjected to couple moments and . Determine the resultant couple moments created about the and axes.The axes all lie in the same horizontal plane. y¿ x¿ My = 25 kip # ft Mx = 17 kip # ft 4–91. If , ,and , determine the magnitude and coordinate direction angles of the resultant couple moment. *4–92. Determine the required magnitude of couple moments so that the resultant couple moment is . MR = 5-300i + 450j - 600k6 N # m M1, M2, and M3 M3 = 450 N#m M2 = 600 N#m M1 = 500 N#m *4–88. Two couples act on the frame. If the resultant couple moment is to be zero, determine the distance d between the 40-lb couple forces. •4–89. Two couples act on the frame. If , determine the resultant couple moment.Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 4–13) and (b) summing the moments of all the force components about point A. 4–90. Two couples act on the frame. If , determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 4–13) and (b) summing the moments of all the force components about point B. d = 4 ft d = 4 ft
- 176. 4.6 MOMENT OF A COUPLE 159 4 F x y z O 1.5 m 3 m 30 F Probs. 4–96/97 x 30 y z 350 mm 250 mm {35k} N {35k} N {50i} N {50i} N A B d C Probs. 4–98/99 *4–100. If , ,and , determine the magnitude and coordinate direction angles of the resultant couple moment. •4–101. Determine the magnitudes of couple moments so that the resultant couple moment is zero. M1, M2, and M3 M3 = 120 lb#ft M2 = 90 lb#ft M1 = 180 lb#ft 4–98. Determine the resultant couple moment of the two couples that act on the pipe assembly.The distance from A to B is . Express the result as a Cartesian vector. 4–99. Determine the distance d between A and B so that the resultant couple moment has a magnitude of . MR = 20 N # m d = 400 mm *4–96. Express the moment of the couple acting on the frame in Cartesian vector form. The forces are applied perpendicular to the frame. What is the magnitude of the couple moment? Take . •4–97. In order to turn over the frame, a couple moment is applied as shown. If the component of this couple moment along the x axis is , determine the magnitude F of the couple forces. Mx = 5-20i6 N # m F = 50 N x z y 2 ft 2 ft 2 ft 3 ft 150 lbft 1 ft 45 45 M1 M2 M3 Probs. 4–100/101 2 ft 3 ft 4 ft z y x F2 F1 F2 250 lb 250 lb F1 Probs. 4–102/103 4–102. If , determine the magnitude and coordinate direction angles of the resultant couple moment. 4–103. Determine the magnitude of couple forces and so that the resultant couple moment acting on the block is zero. F2 F1 F1 = 100 lb and F2 = 200 lb
- 177. 160 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 4.7 Simplification of a Force and Couple System Sometimes it is convenient to reduce a system of forces and couple moments acting on a body to a simpler form by replacing it with an equivalent system, consisting of a single resultant force acting at a specific point and a resultant couple moment.A system is equivalent if the external effects it produces on a body are the same as those caused by the original force and couple moment system.In this context,the external effects of a system refer to the translating and rotating motion of the body if the body is free to move, or it refers to the reactive forces at the supports if the body is held fixed. For example, consider holding the stick in Fig. 4–34a, which is subjected to the force F at point A. If we attach a pair of equal but opposite forces F and –F at point B, which is on the line of action of F, Fig. 4–34b, we observe that –F at B and F at A will cancel each other, leaving only F at B, Fig. 4–34c. Force F has now been moved from A to B without modifying its external effects on the stick; i.e., the reaction at the grip remains the same.This demonstrates the principle of transmissibility, which states that a force acting on a body (stick) is a sliding vector since it can be applied at any point along its line of action. We can also use the above procedure to move a force to a point that is not on the line of action of the force.If F is applied perpendicular to the stick,as in Fig.4–35a,then we can attach a pair of equal but opposite forces F and –F to B,Fig.4–35b.Force F is now applied at B,and the other two forces,F at A and –F at B, form a couple that produces the couple moment , Fig. 4–35c. Therefore, the force F can be moved from A to B provided a couple moment M is added to maintain an equivalent system. This couple moment is determined by taking the moment of F about B. Since M is actually a free vector, it can act at any point on the stick. In both cases the systems are equivalent which causes a downward force F and clockwise couple moment M = Fd to be felt at the grip. M = Fd F F F F A B (a) A B F (b) (c) Fig. 4–34 F F F A d (a) F F m Fd (b) (c) Fig. 4–35
- 178. 4.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 161 4 System of Forces and Couple Moments. Using the above method, a system of several forces and couple moments acting on a body can be reduced to an equivalent single resultant force acting at a point O and a resultant couple moment. For example, in Fig. 4–36a, O is not on the line of action of , and so this force can be moved to point O provided a couple moment is added to the body. Similarly, the couple moment should be added to the body when we move to point O. Finally, since the couple moment M is a free vector, it can just be moved to point O. By doing this, we obtain the equivalent system shown in Fig. 4–36b, which produces the same external effects (support reactions) on the body as that of the force and couple system shown in Fig. 4–36a. If we sum the forces and couple moments, we obtain the resultant force and the resultant couple moment Fig. 4–36c. Notice that is independent of the location of point O;however, depends upon this location since the moments and are determined using the position vectors and Also note that is a free vector and can act at any point on the body, although point O is generally chosen as its point of application. We can generalize the above method of reducing a force and couple system to an equivalent resultant force acting at point O and a resultant couple moment by using the following two equations. (4–17) The first equation states that the resultant force of the system is equivalent to the sum of all the forces; and the second equation states that the resultant couple moment of the system is equivalent to the sum of all the couple moments plus the moments of all the forces about point O. If the force system lies in the x–y plane and any couple moments are perpendicular to this plane, then the above equations reduce to the following three scalar equations. ©MO ©M FR = ©F (MR)O = ©MO + ©M (MR)O FR (MR)O r2. r1 M2 M1 (MR)O FR (MR)O = M + M1 + M2, FR = F1 + F2 F2 M2 = r2 * F2 M1 = r1 * F F1 O F1 (a) F2 r2 r1 M (b) O (c) ⴝ O F1 F2 M M2 r2 F2 M1 r1 F1 FR MRO ⴝ u Fig. 4–36 (4–18) Here the resultant force is determined from the vector sum of its two components and (FR)y. (FR)x (FR)x = ©F x (FR)y = ©F y (MR)O = ©MO + ©M
- 179. 162 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 W1 W2 d1 d2 O WR (MR)O O Procedure for Analysis The following points should be kept in mind when simplifying a force and couple moment system to an equivalent resultant force and couple system. • Establish the coordinate axes with the origin located at point O and the axes having a selected orientation. Force Summation. • If the force system is coplanar, resolve each force into its x and y components. If a component is directed along the positive x or y axis, it represents a positive scalar; whereas if it is directed along the negative x or y axis, it is a negative scalar. • In three dimensions, represent each force as a Cartesian vector before summing the forces. Moment Summation. • When determining the moments of a coplanar force system about point O, it is generally advantageous to use the principle of moments, i.e., determine the moments of the components of each force, rather than the moment of the force itself. • In three dimensions use the vector cross product to determine the moment of each force about point O. Here the position vectors extend from O to any point on the line of action of each force. The weights of these traffic lights can be replaced by their equivalent resultant force and a couple moment at the support, O. In both cases the support must provide the same resistance to translation and rotation in order to keep the member in the horizontal position. (MR)O = W1d1 + W2 d2 WR = W1 + W2
- 180. EXAMPLE 4.14 4.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 163 4 0.2 m 0.3 m 4 kN 5 kN 3 kN O (a) 5 4 3 30 0.1 m 0.1 m Fig. 4–37 (c) (FR)y 6.50 kN (MR)O 2.46 kNm (FR)x 5.598 kN FR u O Replace the force and couple system shown in Fig. 4–37a by an equivalent resultant force and couple moment acting at point O. Using the Pythagorean theorem, Fig. 4–37c, the magnitude of is FR Its direction is Ans. Moment Summation. The moments of 3 kN and 5 kN about point O will be determined using their x and y components. Referring to Fig. 4–37b, we have u = tan-1 a (FR)y (FR)x b = tan-1 a 6.50 kN 5.598 kN b = 49.3° u This clockwise moment is shown in Fig. 4–37c. NOTE: Realize that the resultant force and couple moment in Fig. 4–37c will produce the same external effects or reactions at the supports as those produced by the force system, Fig 4–37a. a b Ans. = -2.46 kN # m = 2.46 kN # m - A4 5 B (5 kN) (0.5 m) - (4 kN)(0.2 m) (MR)O = (3 kN)sin 30°(0.2 m) - (3 kN)cos 30°(0.1 m) + A3 5 B (5 kN) (0.1 m) + (MR)O = ©MO; SOLUTION Force Summation. The 3 kN and 5 kN forces are resolved into their x and y components as shown in Fig. 4–37b.We have = -6.50 kN = 6.50 kNT (FR)y = (3 kN)sin 30° - A4 5 B (5 kN) - 4 kN + c(FR)y = ©Fy; = 5.598 kN : (FR)x = (3 kN)cos 30° + A3 5 B (5 kN) : + (FR)x = ©Fx; Ans. F R = 21F R2x 2 + 1F R2y 2 = 215.598 kN22 + 16.50 kN22 = 8.58 kN (3 kN)cos 30 (3 kN)sin 30 y x 0.2 m 0.3 m 4 kN (5 kN) O (b) 4 5 3 5 (5 kN) 0.1 m 0.1 m
- 181. 164 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 Replace the force and couple system acting on the member in Fig. 4–38a by an equivalent resultant force and couple moment acting at point O. SOLUTION Force Summation. Since the couple forces of 200 N are equal but opposite, they produce a zero resultant force, and so it is not necessary to consider them in the force summation. The 500-N force is resolved into its x and y components, thus, From Fig. 4–15b, the magnitude of is Ans. And the angle is Ans. Moment Summation. Since the couple moment is a free vector, it can act at any point on the member. Referring to Fig. 4–38a, we have a b Ans. This clockwise moment is shown in Fig. 4–38b. = -37.5 N#m = 37.5 N#m - (750 N)(1.25 m) + 200 N#m (MR)O = (500 N)A4 5 B(2.5 m) - (500 N)A3 5 B(1 m) + (MR)O = ©MO + ©Mc; u = tan-1 a (FR)y (FR)x b = tan-1 a 350 N 300 N b = 49.4° u = 2(300 N)2 + (350 N)2 = 461 N F R = 2(F R2x 2 + (F R2y 2 FR (F R)y = (500 N)A4 5 B - 750 N = -350 N = 350 NT + c(FR)y = ©Fy; (FR)x = A3 5 B (500 N) = 300 N : : + (FR)x = ©Fx; EXAMPLE 4.15 (b) O y x (FR)x 300 N (FR)y 350 N (MR)O 37.5 Nm FR u O 4 3 5 1 m 1 m 1.25 m 1.25 m (a) 200 N 200 N 500 N 750 N Fig. 4–38
- 182. 4.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 165 4 The structural member is subjected to a couple moment M and forces and in Fig. 4–39a. Replace this system by an equivalent resultant force and couple moment acting at its base, point O. SOLUTION (VECTOR ANALYSIS) The three-dimensional aspects of the problem can be simplified by using a Cartesian vector analysis. Expressing the forces and couple moment as Cartesian vectors, we have Force Summation. Ans. Moment Summation. Ans. The results are shown in Fig. 4–39b. = 5-166i - 650j + 300k6 N # m = 1-400j + 300k2 + 102 + 1-166.4i - 249.6j2 MRO = 1- 400j + 300k2 + 11k2 * 1- 800k2+ 3 i j k -0.15 0.1 1 -249.6 166.4 0 3 MRO = M + rC * F1 + rB * F2 MRO = ©M + ©MO = 5-250i + 166j - 800k6 N FR = F1 + F2 = -800k - 249.6i + 166.4j FR = ©F; M = -500A4 5 Bj + 500A3 5 Bk = 5-400j + 300k6 N # m = 300 Nc {-0.15i + 0.1j} m 21-0.15 m22 + 10.1 m22 d = 5-249.6i + 166.4j6 N = 1300 N2a rCB rCB b F2 = 1300 N2uCB F1 = 5-800k6 N F2 F1 EXAMPLE 4.16 F1 800 N 0.1 m F2 300 N 0.15 m rB 1 m y C 5 3 4 M 500 N m O x (a) z rC B Fig. 4–39 y x z MRO FR (b) O
- 183. 166 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 F4–28. Replace the loading system by an equivalent resultant force and couple moment acting at point A. F4–29. Replace the loading system by an equivalent resultant force and couple moment acting at point O. F4–30. Replace the loading system by an equivalent resultant force and couple moment acting at point O. F4–26. Replace the loading system by an equivalent resultant force and couple moment acting at point A. F4–25. Replace the loading system by an equivalent resultant force and couple moment acting at point A. F4–27. Replace the loading system by an equivalent resultant force and couple moment acting at point A. A 3 ft 3 ft 4 ft 150 lb 200 lb 100 lb 3 4 5 50 N 200 N m 30 N 40 N A B 3 m 3 m 900 N 30 300 Nm 0.75 m 0.75 m 0.75 m 0.75 m A 300 N 50 lb 100 lb 4 3 5 A 4 3 5 150 lb 3 ft 3 ft 1 ft x z y O A B 2 m 1 m 1.5 m F1 {300i 150j 200k} N F2 {450k} N 0.5 m 0.4 m z y x F2 200 N F1 100 N 0.3 m Mc 75 Nm O FUNDAMENTAL PROBLEMS F4–25 F4–28 F4–29 F4–30 F4–26 F4–27
- 184. 4.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 167 4 4–107. Replace the two forces by an equivalent resultant force and couple moment at point O. Set . *4–108. Replace the two forces by an equivalent resultant force and couple moment at point O. Set . F = 15 lb F = 20 lb •4–105. Replace the force system acting on the beam by an equivalent force and couple moment at point A. 4–106. Replace the force system acting on the beam by an equivalent force and couple moment at point B. *4–104. Replace the force system acting on the truss by a resultant force and couple moment at point C. •4–109. Replace the force system acting on the post by a resultant force and couple moment at point A. PROBLEMS B A C 2 ft 6 ft 2 ft 200 lb 150 lb 100 lb 2 ft 2 ft 500 lb 3 4 5 Prob. 4–104 2.5 kN 1.5 kN 3 kN A B 4 m 3 4 5 2 m 2 m 30 Probs. 4–105/106 6 in. 30 4 3 5 1.5 in. F 20 lb 2 in. x y O 40 Probs. 4–107/108 250 N 500 N 0.2 m 0.5 m 3 4 5 300 N 1 m 1 m 1 m A B 30 Prob. 4–109
- 185. 168 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 *4–112. Replace the two forces acting on the grinder by a resultant force and couple moment at point O. Express the results in Cartesian vector form. 4–111. Replace the force system by a resultant force and couple moment at point O. 4–110. Replace the force and couple moment system acting on the overhang beam by a resultant force and couple moment at point A. •4–113. Replace the two forces acting on the post by a resultant force and couple moment at point O. Express the results in Cartesian vector form. B A 5 12 13 30 kN 45 kNm 26 kN 0.3 m 0.3 m 2 m 2 m 1 m 1 m 30 Prob. 4–110 200 N 200 N 500 N 4 3 5 O 750 N 1.25 m 1.25 m 1 m Prob. 4–111 z A D B C O x y 8 m 6 m 6 m 3 m 2 m FB 5 kN FD 7 kN Prob. 4–113 250 mm y x z 25 mm 40 mm 150 mm 100 mm O A B F2 {15i 20j 30k} N F1 {10i 15j 40k} N Prob. 4–112
- 186. 4.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 169 4 *4–116. Replace the force system acting on the pipe assembly by a resultant force and couple moment at point O. Express the results in Cartesian vector form. 4–115. Handle forces and are applied to the electric drill. Replace this force system by an equivalent resultant force and couple moment acting at point O. Express the results in Cartesian vector form. F2 F1 4–114. The three forces act on the pipe assembly. If and replace this force system by an equivalent resultant force and couple moment acting at O. Express the results in Cartesian vector form. F2 = 80 N, F1 = 50 N •4–117. The slab is to be hoisted using the three slings shown. Replace the system of forces acting on slings by an equivalent force and couple moment at point O. The force is vertical. F1 y O z x 1.25 m 180 N 0.75 m 0.5 m F2 F1 Prob. 4–114 x y z 0.25 m 0.3 m O F1 {6i 3j 10k} N F2 {2j 4k} N 0.15 m Prob. 4–115 x z 2 ft 1.5 ft 2 ft 2 ft O x F1 {20i 10j 25k}lb F2 {10i 25j 20k} lb Prob. 4–116 y x z 45 60 60 45 30 6 m 2 m 2 m F2 5 kN F3 4 kN O F1 6 kN Prob. 4–117
- 187. 170 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 4.8 Further Simplification of a Force and Couple System In the preceding section, we developed a way to reduce a force and couple moment system acting on a rigid body into an equivalent resultant force acting at a specific point O and a resultant couple moment .The force system can be further reduced to an equivalent single resultant force provided the lines of action of and are perpendicular to each other. Because of this condition, only concurrent, coplanar, and parallel force systems can be further simplified. Concurrent Force System. Since a concurrent force system is one in which the lines of action of all the forces intersect at a common point O, Fig. 4–40a, then the force system produces no moment about this point. As a result, the equivalent system can be represented by a single resultant force acting at O, Fig. 4–40b. FR = ©F (MR)O FR (MR)O FR Coplanar Force System. In the case of a coplanar force system, the lines of action of all the forces lie in the same plane, Fig. 4–41a, and so the resultant force of this system also lies in this plane. Furthermore, the moment of each of the forces about any point O is directed perpendicular to this plane. Thus, the resultant moment and resultant force will be mutually perpendicular, Fig. 4–41b. The resultant moment can be replaced by moving the resultant force a perpendicular or moment arm distance d away from point O such that produces the same moment about point O, Fig. 4–41c. This distance d can be determined from the scalar equation . (MR)O = FRd = ©MO or d = (MR)OFR (MR)O FR FR FR (MR)O FR = ©F F2 FR F2 F4 F3 O O (a) (b) Fig. 4–40
- 188. 4.8 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 171 4 Parallel Force System. The parallel force system shown in Fig.4–42a consists of forces that are all parallel to the z axis. Thus, the resultant force at point O must also be parallel to this axis, Fig. 4–42b. The moment produced by each force lies in the plane of the plate, and so the resultant couple moment, , will also lie in this plane, along the moment axis a since and are mutually perpendicular. As a result, the force system can be further reduced to an equivalent single resultant force , acting through point P located on the perpendicular b axis, Fig. 4–42c. The distance d along this axis from point O requires . (MR)O = FRd = ©MO or d = ©MOFR FR (MR)O FR (MR)O FR = ©F z F1 F2 F3 O (a) z a O b b (b) FR F FR F z O d (c) a P (MR)O Fig. 4–42 (a) (b) (c) O (MR)O FR O FR O d F3 F4 F1 F2 Fig. 4–41
- 189. 172 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 Procedure for Analysis The technique used to reduce a coplanar or parallel force system to a single resultant force follows a similar procedure outlined in the previous section. ● Establish the x, y, z, axes and locate the resultant force an arbitrary distance away from the origin of the coordinates. Force Summation. ● The resultant force is equal to the sum of all the forces in the system. ● For a coplanar force system, resolve each force into its x and y components. Positive components are directed along the positive x and y axes, and negative components are directed along the negative x and y axes. Moment Summation. ● The moment of the resultant force about point O is equal to the sum of all the couple moments in the system plus the moments of all the forces in the system about O. ● This moment condition is used to find the location of the resultant force from point O. FR O FR The four cable forces are all concurrent at point O on this bridge tower. Consequently they produce no resultant moment there, only a resultant force . Note that the designers have positioned the cables so that is directed along the bridge tower directly to the support, so that it does not cause any bending of the tower. FR FR
- 190. 4.8 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 173 4 Reduction to a Wrench In general, a three-dimensional force and couple moment system will have an equivalent resultant force acting at point O and a resultant couple moment that are not perpendicular to one another, as shown in Fig. 4–43a. Although a force system such as this cannot be further reduced to an equivalent single resultant force, the resultant couple moment can be resolved into components parallel and perpendicular to the line of action of , Fig. 4–43a. The perpendicular component can be replaced if we move to point P, a distance d from point O along the b axis, Fig. 4–43b. As seen, this axis is perpendicular to both the a axis and the line of action of . The location of P can be determined from . Finally, because is a free vector, it can be moved to point P, Fig. 4–43c.This combination of a resultant force and collinear couple moment will tend to translate and rotate the body about its axis and is referred to as a wrench or screw. A wrench is the simplest system that can represent any general force and couple moment system acting on a body. M|| FR M|| d = M⬜FR FR FR M⬜ FR (MR)O (MR)O FR (a) b a M M FR (b) P d O O FR (c) b P O FR (MR)O z z z M M b a a Fig. 4–43 W1 W2 d1 d2 O O WR d Here the weights of the traffic lights are replaced by their resultant force which acts at a distance from O. Both systems are equivalent. d = (W1d1 + W2 d2) WR WR = W1 + W2
- 191. 174 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 Replace the force and couple moment system acting on the beam in Fig. 4–44a by an equivalent resultant force, and find where its line of action intersects the beam, measured from point O. EXAMPLE 4.17 SOLUTION Force Summation. Summing the force components, From Fig. 4–44b, the magnitude of is Ans. The angle is Ans. Moment Summation. We must equate the moment of about point O in Fig. 4–44b to the sum of the moments of the force and couple moment system about point O in Fig. 4–44a. Since the line of action of acts through point O, only produces a moment about this point.Thus, a Ans. d = 2.25 m -[8 kNA3 5 B] (0.5 m) + [8 kNA4 5 B](4.5 m) 2.40 kN(d) = -(4 kN)(1.5 m) - 15 kN#m +(MR)O = ©MO; (FR)y (FR)x FR u = tan-1 a 2.40 kN 4.80 kN b = 26.6° u FR = 214.80 kN22 + 12.40 kN22 = 5.37 kN FR (FR)y = -4 kN + 8 kNA4 5 B = 2.40 kNc + c(FR)y = ©Fy; (FR)x = 8 kNA3 5 B = 4.80 kN : : + (FR)x = ©Fx; (a) O 4 kN 15 kNm 8 kN 3 4 5 1.5 m 1.5 m 1.5 m 1.5 m 0.5 m y x Fig. 4–44 (b) d O FR (FR)x 4.80 kN (FR)y 2.40 kN u
- 192. 4.8 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 175 4 EXAMPLE 4.18 The jib crane shown in Fig. 4–45a is subjected to three coplanar forces. Replace this loading by an equivalent resultant force and specify where the resultant’s line of action intersects the column AB and boom BC. SOLUTION Force Summation. Resolving the 250-lb force into x and y components and summing the force components yields As shown by the vector addition in Fig. 4–45b, Ans. Ans. Moment Summation. Moments will be summed about point A. Assuming the line of action of intersects AB at a distance y from A, Fig. 4–45b, we have a Ans. By the principle of transmissibility, can be placed at a distance x where it intersects BC, Fig. 4–45b. In this case we have a Ans. x = 10.9 ft + 250 lbA3 5 B111 ft2 - 250 lbA4 5 B18 ft2 = 175 lb 15 ft2 - 60 lb 13 ft2 325 lb 111 ft2 - 260 lb 1x2 +MRA = ©MA; FR y = 2.29 ft = 175 lb 15 ft2 - 60 lb 13 ft2 + 250 lbA3 5 B111 ft2 - 250 lbA4 5 B18 ft2 325 lb 1y2 + 260 lb 102 +MRA = ©MA; FR u = tan-1 a 260 lb 325 lb b = 38.7° u FR = 2(325 lb)2 + (260 lb)2 = 416 lb FRy = -250 lbA4 5 B - 60 lb = -260 lb = 260 lbT + cFRy = ©Fy; FRx = -250 lbA3 5 B - 175 lb = -325 lb = 325 lb ; : + FRx = ©Fx; 6 ft y x 5 ft 175 lb 60 lb (a) 250 lb 5 4 3 3 ft 5 ft 3 ft B C A Fig. 4–45 y (b) x x FR FR y C A 260 lb 325 lb 260 lb 325 lb B u
- 193. 176 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 The slab in Fig. 4–46a is subjected to four parallel forces. Determine the magnitude and direction of a resultant force equivalent to the given force system and locate its point of application on the slab. EXAMPLE 4.19 SOLUTION (SCALAR ANALYSIS) Force Summation. From Fig. 4–46a, the resultant force is Ans. Moment Summation. We require the moment about the x axis of the resultant force, Fig. 4–46b, to be equal to the sum of the moments about the x axis of all the forces in the system, Fig. 4–46a.The moment arms are determined from the y coordinates since these coordinates represent the perpendicular distances from the x axis to the lines of action of the forces. Using the right-hand rule, we have Ans. In a similar manner, a moment equation can be written about the y axis using moment arms defined by the x coordinates of each force. Ans. NOTE: A force of placed at point P(3.00 m, 2.50 m) on the slab, Fig. 4–46b, is therefore equivalent to the parallel force system acting on the slab in Fig. 4–46a. FR = 1400 N x = 3 m 1400x = 4200 11400 N2x = 600 N18 m2 - 100 N16 m2 + 400 N102 + 500 N102 (MR)y = ©My; -1400y = -3500 y = 2.50 m -11400 N2y = 600 N102 + 100 N15 m2 - 400 N110 m2 + 500 N102 (MR)x = ©Mx; = -1400 N = 1400 NT - FR = -600 N + 100 N - 400 N - 500 N + cFR = ©F; y x B 2 m O 600 N 500 N z 100 N 5 m 5 m 400 N C 8 m (a) Fig. 4–46 y x O FR z (b) x P(x, y) y
- 194. 4.8 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 177 4 EXAMPLE 4.20 Replace the force system in Fig. 4–47a by an equivalent resultant force and specify its point of application on the pedestal. SOLUTION Force Summation. Here we will demonstrate a vector analysis. Summing forces, Ans. Location. Moments will be summed about point O. The resultant force is assumed to act through point P (x, y, 0), Fig. 4–47b.Thus Equating the i and j components, (1) Ans. (2) Ans. The negative sign indicates that the x coordinate of point P is negative. NOTE: It is also possible to establish Eq. 1 and 2 directly by summing moments about the x and y axes. Using the right-hand rule, we have 700x = 300 lb14 in.2 - 500 lb14 in.) (MR)y = ©My; -700y = -100 lb(4 in.) - 500 lb(2 in.) (MR)x = ©Mx; x = -1.14 in. 700x = -800 y = 2 in. -700y = -1400 700xj - 700yi = 1200j - 2000j - 1000i - 400i - 1000(j * k) - 400(j * k) -700x(i * k) - 700y(j * k) = -1200(i * k) + 2000(i * k) + [1-4i + 2j2 * 1-500k2] + [(-4j) * (100k)] 1xi + yj2 * 1-700k2 = [14i2 * 1-300k2] rP * FR = (rA * FA) + (rB * FB) + (rC * FC) (MR)O = ©MO; FR = 5-700k6 lb = 5-300k6 lb + 5-500k6 lb + 5100k6 lb FR = FA + FB + FC FR = ©F; x y z (a) FB 500 lb FA 300 lb FC 100 lb 2 in. 4 in. 4 in. 4 in. B O A C rB rA rC Fig. 4–47 x y z (b) FR {700k} lb rP O P y x
- 195. 178 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 FUNDAMENTAL PROBLEMS F4–34. Replace the loading system by an equivalent resultant force and specify where the resultant’s line of action intersects the member AB measured from A. F4–32. Replace the loading system by an equivalent resultant force and specify where the resultant’s line of action intersects the member measured from A. F4–31. Replace the loading system by an equivalent resultant force and specify where the resultant’s line of action intersects the beam measured from O. F4–35. Replace the loading shown by an equivalent single resultant force and specify the x and y coordinates of its line of action. F4–33. Replace the loading system by an equivalent resultant force and specify where the resultant’s line of action intersects the member measured from A. F4–36. Replace the loading shown by an equivalent single resultant force and specify the x and y coordinates of its line of action. 500 lb 500 lb 250 lb O x y 3 ft 3 ft 3 ft 3 ft 30 200 lb 50 lb 100 lb 3 ft 3 ft 3 ft 4 3 5 A 4495 4 4 9 5 2 m 2 m 2 m 2 m A B 20 kN 15 kN 4 3 5 A 5 kN 6 kN 8 kN 4 3 5 1.5 m 3 m 0.5 m 0.5 m 0.5 m B y x z x y 100 N 400 N 500 N 4 m 4 m 3 m 3 m 2 m 3 m 3 m 1 m 1 m z y x 2m 200 N 200 N 100 N 100 N F4–31 F4–32 F4–33 F4–36 F4–35 F4–34
- 196. 4.8 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 179 4 PROBLEMS •4–121. The system of four forces acts on the roof truss. Determine the equivalent resultant force and specify its location along AB, measured from point A. *4–120. The system of parallel forces acts on the top of the Warren truss. Determine the equivalent resultant force of the system and specify its location measured from point A. 4–118. The weights of the various components of the truck are shown. Replace this system of forces by an equivalent resultant force and specify its location measured from B. 4–119. The weights of the various components of the truck are shown. Replace this system of forces by an equivalent resultant force and specify its location measured from point A. 4–122. Replace the force and couple system acting on the frame by an equivalent resultant force and specify where the resultant’s line of action intersects member AB, measured from A. 4–123. Replace the force and couple system acting on the frame by an equivalent resultant force and specify where the resultant’s line of action intersects member BC, measured from B. A 500 N 500 N 500 N 1 kN 2 kN 1 m 1 m 1 m 1 m Prob. 4–120 4 ft 150 lb B A 300 lb 30 30 275 lb 200 lb 4 ft 4 ft Prob. 4–121 3 ft 30 4 ft 3 5 4 2 ft 150 lb 50 lb 500 lb ft C B A Probs. 4–122/123 14 ft 6 ft 2 ft 3 ft A B 3500 lb 5500 lb 1750 lb Probs. 4–118/119
- 197. 180 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 4–127. Replace the force system acting on the post by a resultant force, and specify where its line of action intersects the post AB measured from point A. *4–128. Replace the force system acting on the post by a resultant force, and specify where its line of action intersects the post AB measured from point B. •4–125. Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant’s line of action intersects member AB, measured from point A. 4–126. Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant’s line of action intersects member BC, measured from point B. *4–124. Replace the force and couple moment system acting on the overhang beam by a resultant force, and specify its location along AB measured from point A. •4–129. The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x, y) on the slab. Take 4–130. The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x, y) on the slab.Take F2 = 50 kN. F1 = 20 kN, F2 = 40 kN. F1 = 30 kN, B A 5 12 13 30 kN 45 kNm 26 kN 0.3 m 0.3 m 2 m 2 m 1 m 1 m 30 Prob. 4–124 2 ft 4 ft 3 ft 25 lb 2 ft 20 lb A B C 30 35 lb Probs. 4–125/126 250 N 500 N 0.2 m 0.5 m 3 4 5 300 N 1 m 30 1 m 1 m A B Probs. 4–127/128 y x 20 kN 3 m 2 m 8 m 6 m 4 m 50 kN F1 F2 z Probs. 4–129/130
- 198. 4.8 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 181 4 4–134. If , determine the magnitude of the resultant force and specify the location of its point of application (x, y) on the slab. 4–135. If the resultant force is required to act at the center of the slab, determine the magnitude of the column loadings and and the magnitude of the resultant force. FB FA FA = 40 kN and FB = 35 kN *4–132. Three parallel bolting forces act on the circular plate. Determine the resultant force, and specify its location (x, z) on the plate. , , and . •4–133. The three parallel bolting forces act on the circular plate. If the force at A has a magnitude of , determine the magnitudes of and so that the resultant force of the system has a line of action that coincides with the y axis. Hint: This requires and . ©Mz = 0 ©Mx = 0 FR FC FB FA = 200 lb FC = 400 lb FB = 100 lb F A = 200 lb 4–131. The tube supports the four parallel forces. Determine the magnitudes of forces and acting at C and D so that the equivalent resultant force of the force system acts through the midpoint O of the tube. FD FC *4–136. Replace the parallel force system acting on the plate by a resultant force and specify its location on the x–z plane. x z A D C y z B O 400 mm 400 mm 500 N 200 mm 200 mm 600 N FC FD Prob. 4–131 45 30 1.5 ft z x y A B C FB FA FC Probs. 4–132/133 2.5 m 2.5 m 0.75 m 0.75 m 0.75 m 3 m 3 m 0.75 m 90 kN 30 kN 20 kN x y z FA FB Probs. 4–134/135 1 m 1 m 1 m 0.5 m 0.5 m 5 kN 3 kN x y z 2 kN Prob. 4–136
- 199. 182 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 *4–140. Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(y, z) where its line of action intersects the plate. 4–139. Replace the force and couple moment system acting on the rectangular block by a wrench. Specify the magnitude of the force and couple moment of the wrench and where its line of action intersects the x–y plane. •4–137. If , represent the force system acting on the corbels by a resultant force, and specify its location on the x–y plane. 4–138. Determine the magnitudes of and so that the resultant force passes through point O of the column. FB FA FA = 7 kN and FB = 5 kN •4–141. Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(x, y) where its line of action intersects the plate. 750 mm z x y 650 mm 100 mm 150 mm 600 mm 700 mm 100 mm 150 mm 8kN 6 kN FA FB O Probs. 4–137/138 y x z 300 lb 450 lb 600 lb 2 ft 4 ft 3 ft 600 lbft Prob. 4–139 y y x z P A C B z FB {60j} lb FC {40i} lb FA {80k}lb 12 ft 12 ft Prob. 4–140 4 m 6 m y y x x P A C B z FA {500i} N FC {300j} N FB {800k} N Prob. 4–141
- 200. 4.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING 183 4 4.9 Reduction of a Simple Distributed Loading Sometimes, a body may be subjected to a loading that is distributed over its surface. For example, the pressure of the wind on the face of a sign, the pressure of water within a tank, or the weight of sand on the floor of a storage container,are all distributed loadings.The pressure exerted at each point on the surface indicates the intensity of the loading. It is measured using pascals Pa (or ) in SI units or in the U.S. Customary system. Uniform Loading Along a Single Axis. The most common type of distributed loading encountered in engineering practice is generally uniform along a single axis.* For example, consider the beam (or plate) in Fig. 4–48a that has a constant width and is subjected to a pressure loading that varies only along the x axis. This loading can be described by the function . It contains only one variable x, and for this reason, we can also represent it as a coplanar distributed load. To do so, we multiply the loading function by the width b m of the beam, so that , Fig. 4-48b. Using the methods of Sec. 4.8, we can replace this coplanar parallel force system with a single equivalent resultant force acting at a specific location on the beam, Fig. 4–48c. Magnitude of Resultant Force. From Eq. 4–17 the magnitude of is equivalent to the sum of all the forces in the system. In this case integration must be used since there is an infinite number of parallel forces dF acting on the beam, Fig. 4–48b. Since dF is acting on an element of length dx, and w(x) is a force per unit length, then In other words, the magnitude of dF is determined from the colored differential area dA under the loading curve. For the entire length L, (4–19) Therefore, the magnitude of the resultant force is equal to the total area A under the loading diagram, Fig. 4–48c. FR = L L w1x2 dx = L A dA = A + TFR = ©F; dF = w1x2 dx = dA. FR 1FR = ©F2, FR w(x) = p(x)b N/m p = p(x) N/m2 lb/ft2 N/m2 *The more general case of a nonuniform surface loading acting on a body is considered in Sec. 9.5. p L p p(x) x (a) C x FR b Fig. 4–48 x w O L x dx dF dA w w(x) (b) x w O C A L x FR (c)
- 201. 184 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 Location of Resultant Force. Applying Eq.4–17 the location of the line of action of can be determined by equating the moments of the force resultant and the parallel force distribution about point O (the y axis). Since dF produces a moment of about O, Fig. 4–48b, then for the entire length, Fig. 4–48c, a Solving for using Eq. 4–19, we have (4–20) This coordinate locates the geometric center or centroid of the area under the distributed loading. In other words, the resultant force has a line of action which passes through the centroid C (geometric center) of the area under the loading diagram, Fig. 4–48c. Detailed treatment of the integration techniques for finding the location of the centroid for areas is given in Chapter 9. In many cases, however, the distributed-loading diagram is in the shape of a rectangle, triangle, or some other simple geometric form.The centroid location for such common shapes does not have to be determined from the above equation but can be obtained directly from the tabulation given on the inside back cover. Once is determined, by symmetry passes through point on the surface of the beam, Fig. 4–48a. Therefore, in this case the resultant force has a magnitude equal to the volume under the loading curve and a line of action which passes through the centroid (geometric center) of this volume. p = p1x2 1x, 02 FR x x, x = L L xw1x2 dx L L w1x2 dx = L A x dA L A dA x, - xFR = - L L xw1x2 dx + (MR)O = ©MO; x dF = xw1x2 dx FR x 1MRO = ©MO2, Important Points ● Coplanar distributed loadings are defined by using a loading function that indicates the intensity of the loading along the length of a member. This intensity is measured in or ● The external effects caused by a coplanar distributed load acting on a body can be represented by a single resultant force. ● This resultant force is equivalent to the area under the loading diagram, and has a line of action that passes through the centroid or geometric center of this area. lbft. Nm w = w1x2 The beam supporting this stack of lumber is subjected to a uniform loading of . The resultant force is therefore equal to the area under the loading diagram It acts trough the centroid or geometric center of this area, from the support. b2 FR = w0 b. w0 w0 b b 2 a FR p L p p(x) x (a) C x FR b x w O L x dx dF dA w w(x) (b) x w O C A L x FR (c)
- 202. 4.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING 185 4 EXAMPLE 4.21 Determine the magnitude and location of the equivalent resultant force acting on the shaft in Fig. 4–49a. w (60 x2 )N/m (a) dA w dx 2 m x dx O x 240 N/m w Fig. 4–49 (b) O x w C x 1.5 m FR 160 N SOLUTION Since is given, this problem will be solved by integration. The differential element has an area Applying Eq. 4–19, Ans. The location of measured from O, Fig. 4–49b, is determined from Eq. 4–20. Ans. NOTE: These results can be checked by using the table on the inside back cover, where it is shown that for an exparabolic area of length a, height b, and shape shown in Fig. 4–49a, we have = 3 4 12 m2 = 1.5 m A = ab 3 = 2 m1240 Nm2 3 = 160 N and x = 3 4 a = 1.5 m x = L A x dA L A dA = L 2 m 0 x160x2 2 dx 160 N = 60¢ x4 4 ≤ ` 0 2 m 160 N = 60¢ 24 4 - 04 4 ≤ 160 N FR x = 160 N FR = L A dA = L 2 m 0 60x2 dx = 60¢ x3 3 ≤ ` 0 2 m = 60¢ 23 3 - 03 3 ≤ + T FR = ©F; dA = w dx = 60x2 dx. w = w1x2
- 203. 186 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 A distributed loading of Pa acts over the top surface of the beam shown in Fig. 4–50a. Determine the magnitude and location of the equivalent resultant force. p = (800x) EXAMPLE 4.22 (a) p 7200 Pa x 9 m 0.2 m y p = 800x Pa x Fig. 4–50 w 160x N/m (b) 9 m x w 1440 N/m x C FR 6.48 kN 3 m x 6 m (c) SOLUTION Since the loading intensity is uniform along the width of the beam (the y axis), the loading can be viewed in two dimensions as shown in Fig. 4–50b. Here At note that Although we may again apply Eqs. 4–19 and 4–20 as in the previous example, it is simpler to use the table on the inside back cover. The magnitude of the resultant force is equivalent to the area of the triangle. Ans. The line of action of passes through the centroid C of this triangle. Hence, Ans. The results are shown in Fig. 4–50c. NOTE: We may also view the resultant as acting through the centroid of the volume of the loading diagram in Fig. 4–50a. Hence intersects the x–y plane at the point (6 m, 0). Furthermore, the magnitude of is equal to the volume under the loading diagram; i.e., Ans. FR = V = 1 217200 Nm2 219 m210.2 m2 = 6.48 kN FR FR p = p1x2 FR x = 9 m - 1 319 m2 = 6 m FR FR = 1 219 m211440 Nm2 = 6480 N = 6.48 kN w = 1440 Nm. x = 9 m, = 1160x2 Nm w = 1800x Nm2 210.2 m2
- 204. 4.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING 187 4 EXAMPLE 4.23 The granular material exerts the distributed loading on the beam as shown in Fig. 4–51a. Determine the magnitude and location of the equivalent resultant of this load. SOLUTION The area of the loading diagram is a trapezoid, and therefore the solution can be obtained directly from the area and centroid formulas for a trapezoid listed on the inside back cover. Since these formulas are not easily remembered, instead we will solve this problem by using “composite” areas. Here we will divide the trapezoidal loading into a rectangular and triangular loading as shown in Fig. 4–51b. The magnitude of the force represented by each of these loadings is equal to its associated area, The lines of action of these parallel forces act through the centroid of their associated areas and therefore intersect the beam at The two parallel forces and can be reduced to a single resultant The magnitude of is Ans. We can find the location of with reference to point A, Fig. 4–51b and 4–51c.We require c Ans. NOTE: The trapezoidal area in Fig. 4–51a can also be divided into two triangular areas as shown in Fig. 4–51d. In this case and NOTE: Using these results, show that again and x = 4 ft. FR = 675 lb x4 = 9 ft - 1 319 ft2 = 6 ft x3 = 1 319 ft2 = 3 ft F4 = 1 219 ft2150 lbft2 = 225 lb F3 = 1 219 ft21100 lbft2 = 450 lb x = 4 ft x16752 = 312252 + 4.514502 + MRA = ©MA; FR FR = 225 + 450 = 675 lb + TFR = ©F; FR FR. F2 F1 x2 = 1 219 ft2 = 4.5 ft x1 = 1 319 ft2 = 3 ft F2 = 19 ft2150 lbft2 = 450 lb F1 = 1 219 ft2150 lbft2 = 225 lb 9 ft B A (b) 50 lb/ft 50 lb/ft F1 F2 x1 x2 B A (c) FR x F3 F4 50 lb/ft x3 9 ft x4 (d) 100 lb/ft A 100 lb/ft 50 lb/ft 9 ft B A (a) Fig. 4–51
- 205. F4–37 F4–40 F4–41 F4–38 F4–39 F4–42 188 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 6 kN/m 9 kN/m 3 kN/m 3 m 1.5 m 1.5 m A B A B 6 ft 8 ft 150 lb/ft 6 kN/m 6 m 3 m A B B A 6 ft 3 ft 3 ft 500 lb 200 lb/ft 150 lb/ft 6 kN/m 3 kN/m 1.5 m 4.5 m A B 4 m w 2.5x3 160 N/m w A x F4–40. Determine the resultant force and specify where it acts on the beam measured from A. F4–38. Determine the resultant force and specify where it acts on the beam measured from A. F4–39. Determine the resultant force and specify where it acts on the beam measured from A. F4–37. Determine the resultant force and specify where it acts on the beam measured from A. F4–41. Determine the resultant force and specify where it acts on the beam measured from A. F4–42. Determine the resultant force and specify where it acts on the beam measured from A. FUNDAMENTAL PROBLEMS
- 206. 4.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING 189 4 A B 3 m 3 m 15 kN/m 10 kN/m 3 m Prob. 4–142 B A 8 kN/m 4 kN/m 3 m 3 m Prob. 4–143 3 m 2 m A B 800 N/m 200 N/m Prob. 4–144 A B L –– 2 L –– 2 w0 w0 Prob. 4–145 •4–145. Replace the distributed loading with an equivalent resultant force, and specify its location on the beam measured from point A. 4–143. Replace the distributed loading with an equivalent resultant force, and specify its location on the beam measured from point A. *4–144. Replace the distributed loading by an equivalent resultant force and specify its location, measured from point A. 4–142. Replace the distributed loading with an equivalent resultant force, and specify its location on the beam measured from point A. 4–146. The distribution of soil loading on the bottom of a building slab is shown. Replace this loading by an equivalent resultant force and specify its location, measured from point O. 4–147. Determine the intensities and of the distributed loading acting on the bottom of the slab so that this loading has an equivalent resultant force that is equal but opposite to the resultant of the distributed loading acting on the top of the plate. w2 w1 PROBLEMS 12 ft 9 ft 100 lb/ft 50 lb/ft 300 lb/ft O Prob. 4–146 300 lb/ft A B 3 ft 6 ft 1.5 ft w2 w1 Prob. 4–147
- 207. 190 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 1.2 m 1 m O 1.2 m 0.1 m 150 Pa y x z Prob. 4–149 4–150. The beam is subjected to the distributed loading. Determine the length b of the uniform load and its position a on the beam such that the resultant force and couple moment acting on the beam are zero. *4–148. The bricks on top of the beam and the supports at the bottom create the distributed loading shown in the second figure. Determine the required intensity w and dimension d of the right support so that the resultant force and couple moment about point A of the system are both zero. A w B x w 12(1 2x2 ) lb/ft 0.5 ft 12 lb/ft 18 lb/ft Prob. 4–151 3 m 0.5 m d 3 m 75 N/m A 200 N/m 0.5 m d w Prob. 4–148 6 ft 10 ft b a 60 lb/ft 40 lb/ft Prob. 4–150 •4–149. The wind pressure acting on a triangular sign is uniform. Replace this loading by an equivalent resultant force and couple moment at point O. 4–151. Currently eighty-five percent of all neck injuries are caused by rear-end car collisions. To alleviate this problem, an automobile seat restraint has been developed that provides additional pressure contact with the cranium. During dynamic tests the distribution of load on the cranium has been plotted and shown to be parabolic. Determine the equivalent resultant force and its location, measured from point A.
- 208. 4.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING 191 4 4–154. Replace the distributed loading with an equivalent resultant force, and specify its location on the beam measured from point A. •4–153. Wet concrete exerts a pressure distribution along the wall of the form. Determine the resultant force of this distribution and specify the height h where the bracing strut should be placed so that it lies through the line of action of the resultant force.The wall has a width of 5 m. *4–152. Wind has blown sand over a platform such that the intensity of the load can be approximated by the function Simplify this distributed loading to an equivalent resultant force and specify its magnitude and location measured from A. w = 10.5x3 2 Nm. 4–155. Replace the loading by an equivalent resultant force and couple moment at point A. *4–156. Replace the loading by an equivalent resultant force and couple moment acting at point B. x w A 10 m 500 N/m w (0.5x3 ) N/m Prob. 4–152 4 m h (4 ) kPa p 1/2 z 8 kPa z p Prob. 4–153 w x A B 4 m 8 kN/m w (4 x)2 1 –– 2 Prob. 4–154 60 6 ft 50 lb/ft 50 lb/ft 100 lb/ft 4 ft A B Probs. 4–155/156
- 209. 192 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 *4–160. The distributed load acts on the beam as shown. Determine the magnitude of the equivalent resultant force and specify its location, measured from point A. 4–158. The distributed load acts on the beam as shown. Determine the magnitude of the equivalent resultant force and specify where it acts, measured from point A. 4–159. The distributed load acts on the beam as shown. Determine the maximum intensity . What is the magnitude of the equivalent resultant force? Specify where it acts, measured from point B. wmax •4–157. The lifting force along the wing of a jet aircraft consists of a uniform distribution along AB, and a semiparabolic distribution along BC with origin at B. Replace this loading by a single resultant force and specify its location measured from point A. •4–161. If the distribution of the ground reaction on the pipe per foot of length can be approximated as shown, determine the magnitude of the resultant force due to this loading. x w 24 ft 12 ft w (2880 5x2) lb/ft 2880 lb/ft A B C Prob. 4–157 w (2x2 4x 16) lb/ft x B A w 4 ft Probs. 4–158/159 w ( x2 x 4) lb/ft x B A w 10 ft 2 lb/ft 4 lb/ft 2 15 17 15 Prob. 4–160 2.5 ft 50 lb/ft 25 lb/ft w 25 (1 cos u) lb/ft u Prob. 4–161
- 210. CHAPTER REVIEW 193 4 CHAPTER REVIEW Moment of Force—Scalar Definition A force produces a turning effect or moment about a point O that does not lie on its line of action. In scalar form, the moment magnitude is the product of the force and the moment arm or perpendicular distance from point O to the line of action of the force. The direction of the moment is defined using the right-hand rule. always acts along an axis perpendicular to the plane containing F and d, and passes through the point O. MO Rather than finding d, it is normally easier to resolve the force into its x and y components, determine the moment of each component about the point, and then sum the results. This is called the principle of moments. MO = Fd = Fxy - Fyx Moment of a Force—Vector Definition Since three-dimensional geometry is generally more difficult to visualize, the vector cross product should be used to determine the moment. Here , where r is a position vector that extends from point O to any point A, B, or C on the line of action of F. MO = r * F MO = Fd MO = rA * F = rB * F = rC * F If the position vector r and force F are expressed as Cartesian vectors, then the cross product results from the expansion of a determinant. MO = r * F = 3 i j k rx ry rz Fx Fy Fz 3 O Moment axis d F MO F Fy y y O d x x Fx z x y F O A B C rA rB MO rC
- 211. 194 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 a da Ma r F r Ma ua a a¿ Axis of projection F F F d B A F ⴚF r Moment about an Axis If the moment of a force F is to be determined about an arbitrary axis a, then the projection of the moment onto the axis must be obtained. Provided the distance that is perpendicular to both the line of action of the force and the axis can be found, then the moment of the force about the axis can be determined from a scalar equation. Note that when the line of action of F intersects the axis then the moment of F about the axis is zero.Also, when the line of action of F is parallel to the axis, the moment of F about the axis is zero. da Ma = Fda In three dimensions, the scalar triple product should be used. Here is the unit vector that specifies the direction of the axis, and r is a position vector that is directed from any point on the axis to any point on the line of action of the force. If is calculated as a negative scalar, then the sense of direction of is opposite to ua. Ma Ma ua Ma = ua # 1r * F2 = 3 uax uay uz rx ry rz Fx Fy Fz 3 Couple Moment A couple consists of two equal but opposite forces that act a perpendicular distance d apart. Couples tend to produce a rotation without translation. The magnitude of the couple moment is , and its direction is established using the right-hand rule. If the vector cross product is used to determine the moment of a couple, then r extends from any point on the line of action of one of the forces to any point on the line of action of the other force F that is used in the cross product. M = Fd M = Fd M = r * F
- 212. CHAPTER REVIEW 195 4 Simplification of a Force and Couple System Any system of forces and couples can be reduced to a single resultant force and resultant couple moment acting at a point. The resultant force is the sum of all the forces in the system, and the resultant couple moment is equal to the sum of all the moments of the forces about the point and couple moments. . MRO = ©MO + ©M FR = ©F, Further simplification to a single resultant force is possible provided the force system is concurrent, coplanar, or parallel. To find the location of the resultant force from a point, it is necessary to equate the moment of the resultant force about the point to the moment of the forces and couples in the system about the same point. If the resultant force and couple moment at a point are not perpendicular to one another, then this system can be reduced to a wrench, which consists of the resultant force and collinear couple moment. Coplanar Distributed Loading A simple distributed loading can be represented by its resultant force, which is equivalent to the area under the loading curve.This resultant has a line of action that passes through the centroid or geometric center of the area or volume under the loading diagram. O r2 r1 O ⴝ FR MRO u F1 F2 M O FR a b a b MRO a b a b FR d MRO FR P ⴝ O O ⴝ FR MRO u M兩兩 O a b a b FR P d x L w w w(x) O x O FR C L A
- 213. 196 CHAPTER 4 FORCE SYSTEM RESULTANTS 4 REVIEW PROBLEMS *4–164. Determine the coordinate direction angles , , of F, which is applied to the end of the pipe assembly, so that the moment of F about O is zero. •4–165. Determine the moment of the force F about point O. The force has coordinate direction angles of , , . Express the result as a Cartesian vector. g = 45° b = 120° a = 60° g b a 4–163. Two couples act on the frame. If the resultant couple moment is to be zero, determine the distance d between the 100-lb couple forces. 4–162. The beam is subjected to the parabolic loading. Determine an equivalent force and couple system at point A. 4–166. The snorkel boom lift is extended into the position shown. If the worker weighs 160 lb, determine the moment of this force about the connection at A. w (25 x2)lb/ft 4 ft 400 lb/ft x w A O Prob. 4–162 25 ft 50 A 2 ft Prob. 4–166 d 3 ft 4 ft A B 3 ft 30° 100 lb 150 lb 150 lb 100 lb 3 4 5 3 4 5 30° Prob. 4–163 x 10 in. F 20 lb 6 in. 6 in. 8 in. z O y Probs. 4–164/165
- 214. REVIEW PROBLEMS 197 4 4–171. Replace the force at A by an equivalent resultant force and couple moment at point P. Express the results in Cartesian vector form. •4–169. Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve the problem (a) using Eq. 4–13 and (b) summing the moment of each force about point O.Take . 4–170. If the couple moment acting on the pipe has a magnitude of , determine the magnitude F of the vertical force applied to each wrench. 400 N # m F = 525k6 N 4–167. Determine the moment of the force about the door hinge at A. Express the result as a Cartesian vector. *4–168. Determine the magnitude of the moment of the force about the hinged axis aa of the door. FC FC *4–172. The horizontal 30-N force acts on the handle of the wrench. Determine the moment of this force about point O. Specify the coordinate direction angles , , of the moment axis. •4–173. The horizontal 30-N force acts on the handle of the wrench. What is the magnitude of the moment of this force about the z axis? g b a 0.5 m 1 m 30 2.5 m 1.5 m z C A B a a x y FC 250 N Probs. 4–167/168 z y x O B F 200 mm A –F 300 mm 400 mm 150 mm 200 mm Probs. 4–169/170 z A F 120 lb y x P 4 ft 10 ft 8 ft 8 ft 6 ft 6 ft Prob. 4–171 O z x B y 50 mm 200 mm 10 mm 30 N 45 45 Probs. 4–172/173
- 215. The crane is subjected to its weight and the load it supports. In order to calculate the support reactions on the crane, it is necessary to apply the principles of equilibrium.
- 216. Equilibrium of a Rigid Body CHAPTER OBJECTIVES • To develop the equations of equilibrium for a rigid body. • To introduce the concept of the free-body diagram for a rigid body. • To show how to solve rigid-body equilibrium problems using the equations of equilibrium. 5.1 Conditions for Rigid-Body Equilibrium In this section,we will develop both the necessary and sufficient conditions for the equilibrium of the rigid body in Fig. 5–1a. As shown, this body is subjected to an external force and couple moment system that is the result of the effects of gravitational,electrical,magnetic,or contact forces caused by adjacent bodies. The internal forces caused by interactions between particles within the body are not shown in this figure because these forces occur in equal but opposite collinear pairs and hence will cancel out, a consequence of Newton’s third law. 5 F1 M2 M1 F2 F3 F4 O (a) Fig. 5–1
- 217. 200 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a) Fig. 5–1 R W 2T G Fig. 5–2 FR 0 (MR)O 0 O (b) FR 0 (MR)O 0 O A r (c) Using the methods of the previous chapter, the force and couple moment system acting on a body can be reduced to an equivalent resultant force and resultant couple moment at any arbitrary point O on or off the body, Fig. 5–1b. If this resultant force and couple moment are both equal to zero, then the body is said to be in equilibrium. Mathematically, the equilibrium of a body is expressed as (5–1) The first of these equations states that the sum of the forces acting on the body is equal to zero.The second equation states that the sum of the moments of all the forces in the system about point O, added to all the couple moments, is equal to zero. These two equations are not only necessary for equilibrium, they are also sufficient.To show this, consider summing moments about some other point, such as point A in Fig. 5–1c. We require Since , this equation is satisfied only if Eqs. 5–1 are satisfied, namely and . When applying the equations of equilibrium, we will assume that the body remains rigid. In reality, however, all bodies deform when subjected to loads.Although this is the case, most engineering materials such as steel and concrete are very rigid and so their deformation is usually very small. Therefore, when applying the equations of equilibrium, we can generally assume that the body will remain rigid and not deform under the applied load without introducing any significant error. This way the direction of the applied forces and their moment arms with respect to a fixed reference remain unchanged before and after the body is loaded. EQUILIBRIUM IN TWO DIMENSIONS In the first part of the chapter, we will consider the case where the force system acting on a rigid body lies in or may be projected onto a single plane and, furthermore, any couple moments acting on the body are directed perpendicular to this plane.This type of force and couple system is often referred to as a two-dimensional or coplanar force system. For example, the airplane in Fig. 5–2 has a plane of symmetry through its center axis, and so the loads acting on the airplane are symmetrical with respect to this plane. Thus, each of the two wing tires will support the same load T, which is represented on the side (two-dimensional) view of the plane as 2T. (MR)O = 0 FR = 0 r Z 0 ©MA = r * FR + (MR)O = 0 (MR)O = ©MO = 0 FR = ©F = 0
- 218. 5.2 FREE-BODY DIAGRAMS 201 5.2 Free-Body Diagrams Successful application of the equations of equilibrium requires a complete specification of all the known and unknown external forces that act on the body.The best way to account for these forces is to draw a free-body diagram.This diagram is a sketch of the outlined shape of the body, which represents it as being isolated or “free” from its surroundings, i.e., a “free body.” On this sketch it is necessary to show all the forces and couple moments that the surroundings exert on the body so that these effects can be accounted for when the equations of equilibrium are applied. A thorough understanding of how to draw a free-body diagram is of primary importance for solving problems in mechanics. Support Reactions. Before presenting a formal procedure as to how to draw a free-body diagram, we will first consider the various types of reactions that occur at supports and points of contact between bodies subjected to coplanar force systems.As a general rule, • If a support prevents the translation of a body in a given direction, then a force is developed on the body in that direction. • If rotation is prevented, a couple moment is exerted on the body. For example, let us consider three ways in which a horizontal member, such as a beam, is supported at its end. One method consists of a roller or cylinder, Fig. 5–3a. Since this support only prevents the beam from translating in the vertical direction, the roller will only exert a force on the beam in this direction, Fig. 5–3b. The beam can be supported in a more restrictive manner by using a pin, Fig. 5–3c.The pin passes through a hole in the beam and two leaves which are fixed to the ground. Here the pin can prevent translation of the beam in any direction Fig. 5–3d, and so the pin must exert a force F on the beam in this direction. For purposes of analysis, it is generally easier to represent this resultant force F by its two rectangular components and Fig. 5–3e. If and are known, then F and can be calculated. The most restrictive way to support the beam would be to use a fixed support as shown in Fig. 5–3f. This support will prevent both translation and rotation of the beam. To do this a force and couple moment must be developed on the beam at its point of connection, Fig. 5–3g. As in the case of the pin, the force is usually represented by its rectangular components and Table 5–1 lists other common types of supports for bodies subjected to coplanar force systems. (In all cases the angle is assumed to be known.) Carefully study each of the symbols used to represent these supports and the types of reactions they exert on their contacting members. u Fy. Fx f Fy Fx Fy, Fx f, 5 (a) roller (b) F (c) pin or Fy Fx F (e) (d) f (f) fixed support Fy Fx M (g) Fig. 5–3
- 219. 202 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 (3) Types of Connection Reaction Number of Unknowns One unknown. The reaction is a tension force which acts away from the member in the direction of the cable. One unknown. The reaction is a force which acts along the axis of the link. One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. One unknown. The reaction is a force which acts perpendicular to the slot. One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. One unknown. The reaction is a force which acts perpendicular to the rod. continued (1) cable F (2) weightless link F roller F or (4) roller or pin in confined smooth slot (5) rocker (6) smooth contacting surface F F F (7) or or F F F TABLE 5–1 Supports for Rigid Bodies Subjected to Two-Dimensional Force Systems member pin connected to collar on smooth rod u u u u u u u u u u u u u u u u u
- 220. 5.2 FREE-BODY DIAGRAMS 203 Typical examples of actual supports are shown in the following sequence of photos. The numbers refer to the connection types in Table 5–1. 5 The cable exerts a force on the bracket in the direction of the cable. (1) The rocker support for this bridge girder allows horizontal movement so the bridge is free to expand and contract due to a change in temperature. (5) This concrete girder rests on the ledge that is assumed to act as a smooth contacting surface. (6) This utility building is pin supported at the top of the column. (8) The floor beams of this building are welded together and thus form fixed connections. (10) Types of Connection Reaction Number of Unknowns Two unknowns. The reactions are two components of force, or the magnitude and direction of the resultant force. Note that and are not necessarily equal [usually not, unless the rod shown is a link as in (2)]. Three unknowns. The reactions are the couple moment and the two force components, or the couple moment and the magnitude and direction of the resultant force. Two unknowns. The reactions are the couple moment and the force which acts perpendicular to the rod. F Fy M or Fx F fixed support Fy Fx F or M M f f f u TABLE 5–1 Continued member fixed connected to collar on smooth rod smooth pin or hinge (8) (9) (10) u f f
- 221. 204 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 Internal Forces. As stated in Sec. 5.1, the internal forces that act between adjacent particles in a body always occur in collinear pairs such that they have the same magnitude and act in opposite directions (Newton’s third law). Since these forces cancel each other, they will not create an external effect on the body.It is for this reason that the internal forces should not be included on the free-body diagram if the entire body is to be considered. For example, the engine shown in Fig. 5–4a has a free-body diagram shown in Fig. 5–4b. The internal forces between all its connected parts such as the screws and bolts, will cancel out because they form equal and opposite collinear pairs. Only the external forces and , exerted by the chains and the engine weight W, are shown on the free-body diagram. T2 T1 (a) (b) W T2 T1 G Fig. 5–4 Weight and the Center of Gravity. When a body is within a gravitational field, then each of its particles has a specified weight. It was shown in Sec. 4.8 that such a system of forces can be reduced to a single resultant force acting through a specified point. We refer to this force resultant as the weight W of the body and to the location of its point of application as the center of gravity. The methods used for its determination will be developed in Chapter 9. In the examples and problems that follow, if the weight of the body is important for the analysis, this force will be reported in the problem statement. Also, when the body is uniform or made from the same material, the center of gravity will be located at the body’s geometric center or centroid; however, if the body consists of a nonuniform distribution of material, or has an unusual shape, then the location of its center of gravity G will be given. Idealized Models. When an engineer performs a force analysis of any object, he or she considers a corresponding analytical or idealized model that gives results that approximate as closely as possible the actual situation. To do this, careful choices have to be made so that selection of the type of supports, the material behavior, and the object’s dimensions can be justified. This way one can feel confident that any design or analysis will yield results which can be trusted. In complex
- 222. 5.2 FREE-BODY DIAGRAMS 205 5 cases this process may require developing several different models of the object that must be analyzed. In any case, this selection process requires both skill and experience. The following two cases illustrate what is required to develop a proper model. In Fig. 5–5a, the steel beam is to be used to support the three roof joists of a building. For a force analysis it is reasonable to assume the material (steel) is rigid since only very small deflections will occur when the beam is loaded. A bolted connection at A will allow for any slight rotation that occurs here when the load is applied, and so a pin can be considered for this support. At B a roller can be considered since this support offers no resistance to horizontal movement. Building code is used to specify the roof loading A so that the joist loads F can be calculated. These forces will be larger than any actual loading on the beam since they account for extreme loading cases and for dynamic or vibrational effects. Finally, the weight of the beam is generally neglected when it is small compared to the load the beam supports. The idealized model of the beam is therefore shown with average dimensions a, b, c, and d in Fig. 5–5b. As a second case, consider the lift boom in Fig. 5–6a. By inspection, it is supported by a pin at A and by the hydraulic cylinder BC, which can be approximated as a weightless link. The material can be assumed rigid, and with its density known, the weight of the boom and the location of its center of gravity G are determined.When a design loading P is specified, the idealized model shown in Fig. 5–6b can be used for a force analysis. Average dimensions (not shown) are used to specify the location of the loads and the supports. Idealized models of specific objects will be given in some of the examples throughout the text. It should be realized, however, that each case represents the reduction of a practical situation using simplifying assumptions like the ones illustrated here. (a) B A F F F A B (b) a b c d Fig. 5–5 (a) A C B (b) B C G A P Fig. 5–6
- 223. 206 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 Procedure for Analysis To construct a free-body diagram for a rigid body or any group of bodies considered as a single system, the following steps should be performed: Draw Outlined Shape. Imagine the body to be isolated or cut “free” from its constraints and connections and draw (sketch) its outlined shape. Show All Forces and Couple Moments. Identify all the known and unknown external forces and couple moments that act on the body.Those generally encountered are due to (1) applied loadings,(2) reactions occurring at the supports or at points of contact with other bodies (see Table 5–1), and (3) the weight of the body. To account for all these effects, it may help to trace over the boundary, carefully noting each force or couple moment acting on it. Identify Each Loading and Give Dimensions. The forces and couple moments that are known should be labeled with their proper magnitudes and directions. Letters are used to represent the magnitudes and direction angles of forces and couple moments that are unknown. Establish an x, y coordinate system so that these unknowns, etc., can be identified. Finally, indicate the dimensions of the body necessary for calculating the moments of forces. Ay, Ax, Important Points • No equilibrium problem should be solved without first drawing the free-body diagram, so as to account for all the forces and couple moments that act on the body. • If a support prevents translation of a body in a particular direction, then the support exerts a force on the body in that direction. • If rotation is prevented, then the support exerts a couple moment on the body. • Study Table 5–1. • Internal forces are never shown on the free-body diagram since they occur in equal but opposite collinear pairs and therefore cancel out. • The weight of a body is an external force, and its effect is represented by a single resultant force acting through the body’s center of gravity G. • Couple moments can be placed anywhere on the free-body diagram since they are free vectors. Forces can act at any point along their lines of action since they are sliding vectors.
- 224. 5.2 FREE-BODY DIAGRAMS 207 5 EXAMPLE 5.1 Draw the free-body diagram of the uniform beam shown in Fig. 5–7a. The beam has a mass of 100 kg. SOLUTION The free-body diagram of the beam is shown in Fig. 5–7b. Since the support at A is fixed, the wall exerts three reactions on the beam, denoted as and . The magnitudes of these reactions are unknown, and their sense has been assumed. The weight of the beam, acts through the beam’s center of gravity G, which is 3 m from A since the beam is uniform. W = 10019.812 N = 981 N, MA Ay, Ax, (a) 2 m 1200 N 6 m A Ay Ax 2 m 1200 N 3 m A 981 N MA G Effect of applied force acting on beam Effect of gravity (weight) acting on beam Effect of fixed support acting on beam (b) y x Fig. 5–7
- 225. 208 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 Draw the free-body diagram of the foot lever shown in Fig. 5–8a. The operator applies a vertical force to the pedal so that the spring is stretched 1.5 in. and the force in the short link at B is 20 lb. EXAMPLE 5.2 A B (a) F 5 in. 1.5 in. 1 in. A B k 20 lb/in. (b) F 30 lb 5 in. 1.5 in. 1 in. A B 20 lb Ay Ax (c) Fig. 5–8 SOLUTION By inspection of the photo the lever is loosely bolted to the frame at A. The rod at B is pinned at its ends and acts as a “short link.” After making the proper measurements, the idealized model of the lever is shown in Fig. 5–8b. From this, the free-body diagram is shown in Fig. 5–8c. The pin support at A exerts force components and on the lever. The link at B exerts a force of 20 lb, acting in the direction of the link. In addition the spring also exerts a horizontal force on the lever. If the stiffness is measured and found to be then since the stretch using Eq. 3–2, Finally, the operator’s shoe applies a vertical force of F on the pedal. The dimensions of the lever are also shown on the free-body diagram, since this information will be useful when computing the moments of the forces. As usual, the senses of the unknown forces at A have been assumed. The correct senses will become apparent after solving the equilibrium equations. Fs = ks = 20 lbin. 11.5 in.2 = 30 lb. s = 1.5 in., k = 20 lbin., Ay Ax
- 226. 5.2 FREE-BODY DIAGRAMS 209 5 EXAMPLE 5.3 Two smooth pipes, each having a mass of 300 kg, are supported by the forked tines of the tractor in Fig. 5–9a. Draw the free-body diagrams for each pipe and both pipes together. (a) Fig. 5–9 (b) 30 A B 0.35 m 0.35 m 30 B 30 P R 2943 N (d) 30 A 30 30 Effect of gravity (weight) acting on A Effect of sloped fork acting on A Effect of B acting on A Effect of sloped blade acting on A T F R 2943 N (c) 30 A 30 T F 2943 N (e) 30 B P 2943 N SOLUTION The idealized model from which we must draw the free-body diagrams is shown in Fig. 5–9b. Here the pipes are identified, the dimensions have been added, and the physical situation reduced to its simplest form. The free-body diagram for pipe A is shown in Fig. 5–9c. Its weight is Assuming all contacting surfaces are smooth, the reactive forces T, F, R act in a direction normal to the tangent at their surfaces of contact. The free-body diagram of pipe B is shown in Fig. 5–9d. Can you identify each of the three forces acting on this pipe? In particular, note that R, representing the force of A on B, Fig. 5–9d, is equal and opposite to R representing the force of B on A, Fig. 5–9c. This is a consequence of Newton’s third law of motion. The free-body diagram of both pipes combined (“system”) is shown in Fig. 5–9e. Here the contact force R, which acts between A and B, is considered as an internal force and hence is not shown on the free- body diagram. That is, it represents a pair of equal but opposite collinear forces which cancel each other. W = 30019.812 N = 2943 N.
- 227. 210 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 Draw the free-body diagram of the unloaded platform that is suspended off the edge of the oil rig shown in Fig. 5–10a.The platform has a mass of 200 kg. EXAMPLE 5.4 (a) 1.40 m 1 m 70 0.8 m (b) A G B 1.40 m 1 m 70 0.8 m 1962 N (c) Ax Ay G A T Fig. 5–10 SOLUTION The idealized model of the platform will be considered in two dimensions because by observation the loading and the dimensions are all symmetrical about a vertical plane passing through its center, Fig. 5–10b.The connection at A is considered to be a pin, and the cable supports the platform at B. The direction of the cable and average dimensions of the platform are listed, and the center of gravity G has been determined. It is from this model that we have drawn the free-body diagram shown in Fig. 5–10c. The platform’s weight is The force components and along with the cable force T represent the reactions that both pins and both cables exert on the platform, Fig. 5–10a. Consequently, after the solution for these reactions, half their magnitude is developed at A and half is developed at B. Ay Ax 20019.812 = 1962 N.
- 228. 5.2 FREE-BODY DIAGRAMS 211 5 PROBLEMS *5–4. Draw the free-body diagram of the beam which supports the 80-kg load and is supported by the pin at A and a cable which wraps around the pulley at D. Explain the significance of each force on the diagram. (See Fig. 5–7b.) 5–2. Draw the free-body diagram of member AB, which is supported by a roller at A and a pin at B. Explain the significance of each force on the diagram. (See Fig. 5–7b.) •5–1. Draw the free-body diagram of the 50-kg paper roll which has a center of mass at G and rests on the smooth blade of the paper hauler. Explain the significance of each force acting on the diagram. (See Fig. 5–7b.) •5–5. Draw the free-body diagram of the truss that is supported by the cable AB and pin C.Explain the significance of each force acting on the diagram. (See Fig. 5–7b.) 5–3. Draw the free-body diagram of the dumpster D of the truck, which has a weight of 5000 lb and a center of gravity at G. It is supported by a pin at A and a pin-connected hydraulic cylinder BC (short link). Explain the significance of each force on the diagram. (See Fig. 5–7b.) B 30 35 mm A G Prob. 5–1 A B 8 ft 30 4 ft 3 ft 13 12 5 800 lb ft 390 lb Prob. 5–2 1.5 m 3 m 1 m 20 30 B A D G C Prob. 5–3 2 m 2 m 4 3 5 1.5 m B A C E D Prob. 5–4 A B C 2 m 2 m 2 m 2 m 30 3 kN 4 kN Prob. 5–5
- 229. 212 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 •5–9. Draw the free-body diagram of the bar, which has a negligible thickness and smooth points of contact at A, B, and C. Explain the significance of each force on the diagram. (See Fig. 5–7b.) 5–7. Draw the free-body diagram of the “spanner wrench” subjected to the 20-lb force. The support at A can be considered a pin, and the surface of contact at B is smooth. Explain the significance of each force on the diagram. (See Fig. 5–7b.) 5–6. Draw the free-body diagram of the crane boom AB which has a weight of 650 lb and center of gravity at G.The boom is supported by a pin at A and cable BC. The load of 1250 lb is suspended from a cable attached at B. Explain the significance of each force acting on the diagram. (See Fig. 5–7b.) 5–10. Draw the free-body diagram of the winch, which consists of a drum of radius 4 in. It is pin-connected at its center C, and at its outer rim is a ratchet gear having a mean radius of 6 in. The pawl AB serves as a two-force member (short link) and prevents the drum from rotating. Explain the significance of each force on the diagram. (See Fig. 5–7b.) *5–8. Draw the free-body diagram of member ABC which is supported by a smooth collar at A, roller at B, and short link CD. Explain the significance of each force acting on the diagram. (See Fig. 5–7b.) 12 13 5 G C A B 30 18 ft 12 ft Prob. 5–6 A B 6 in. 20 lb 1 in. Prob. 5–7 6 m 2.5 kN 60 3 m 4 kN m 4 m 45 A B C D Prob. 5–8 3 in. 5 in. 8 in. A 30 10 lb 30 B C Prob. 5–9 3 in. 2 in. 6 in. B A 500 lb C 4 in. Prob. 5–10
- 230. 5.2 FREE-BODY DIAGRAMS 213 5 CONCEPTUAL PROBLEMS P5–3. Draw the free-body diagram of the wing on the passenger plane. The weights of the engine and wing are significant.The tires at B are free to roll. P5–2. Draw the free-body diagram of the outrigger ABC used to support a backhoe. The top pin B is connected to the hydraulic cylinder, which can be considered to be a short link (two-force member), the bearing shoe at A is smooth, and the outrigger is pinned to the frame at C. P5–1. Draw the free-body diagram of the uniform trash bucket which has a significant weight. It is pinned at A and rests against the smooth horizontal member at B. Show your result in side view. Label any necessary dimensions. *P5–4. Draw the free-body diagram of the wheel and member ABC used as part of the landing gear on a jet plane. The hydraulic cylinder AD acts as a two-force member, and there is a pin connection at B. A B P5–1 B C A P5–2 A B P5–3 C D A B P5–4
- 231. 214 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 5.3 Equations of Equilibrium In Sec. 5.1 we developed the two equations which are both necessary and sufficient for the equilibrium of a rigid body, namely, and When the body is subjected to a system of forces, which all lie in the x–y plane, then the forces can be resolved into their x and y components. Consequently, the conditions for equilibrium in two dimensions are (5–2) Here and represent, respectively, the algebraic sums of the x and y components of all the forces acting on the body, and represents the algebraic sum of the couple moments and the moments of all the force components about the z axis, which is perpendicular to the x–y plane and passes through the arbitrary point O. Alternative Sets of Equilibrium Equations. Although Eqs. 5–2 are most often used for solving coplanar equilibrium problems, two alternative sets of three independent equilibrium equations may also be used. One such set is (5–3) When using these equations it is required that a line passing through points A and B is not parallel to the y axis.To prove that Eqs. 5–3 provide the conditions for equilibrium, consider the free-body diagram of the plate shown in Fig. 5–11a. Using the methods of Sec. 4.8, all the forces on the free-body diagram may be replaced by an equivalent resultant force acting at point A, and a resultant couple moment Fig. 5–11b. If is satisfied, it is necessary that Furthermore, in order that satisfy it must have no component along the x axis, and therefore must be parallel to the y axis, Fig. 5–11c. Finally, if it is required that where B does not lie on the line of action of then Since Eqs. 5–3 show that both of these resultants are zero, indeed the body in Fig. 5–11a must be in equilibrium. FR = 0. FR, ©MB = 0, FR ©Fx = 0, FR MRA = 0. ©MA = 0 MRA = ©MA, FR = ©F, ©Fx = 0 ©MA = 0 ©MB = 0 ©MO ©Fy ©Fx ©Fx = 0 ©Fy = 0 ©MO = 0 ©MO = 0. ©F = 0 B A C (a) F4 F3 F1 F2 x y A MRA FR (b) B C x y (c) A FR B C x y Fig. 5–11
- 232. 5.3 EQUATIONS OF EQUILIBRIUM 215 5 A second alternative set of equilibrium equations is (5–4) Here it is necessary that points A, B, and C do not lie on the same line.To prove that these equations, when satisfied, ensure equilibrium, consider again the free-body diagram in Fig. 5–11b. If is to be satisfied, then is satisfied if the line of action of passes through point C as shown in Fig. 5–11c. Finally, if we require it is necessary that and so the plate in Fig. 5–11a must then be in equilibrium. FR = 0, ©MB = 0, FR ©MC = 0 MRA = 0. ©MA = 0 ©MA = 0 ©MB = 0 ©MC = 0 Procedure for Analysis Coplanar force equilibrium problems for a rigid body can be solved using the following procedure. Free-Body Diagram. • Establish the x, y coordinate axes in any suitable orientation. • Draw an outlined shape of the body. • Show all the forces and couple moments acting on the body. • Label all the loadings and specify their directions relative to the x or y axis. The sense of a force or couple moment having an unknown magnitude but known line of action can be assumed. • Indicate the dimensions of the body necessary for computing the moments of forces. Equations of Equilibrium. • Apply the moment equation of equilibrium, about a point (O) that lies at the intersection of the lines of action of two unknown forces. In this way, the moments of these unknowns are zero about O, and a direct solution for the third unknown can be determined. • When applying the force equilibrium equations, and orient the x and y axes along lines that will provide the simplest resolution of the forces into their x and y components. • If the solution of the equilibrium equations yields a negative scalar for a force or couple moment magnitude, this indicates that the sense is opposite to that which was assumed on the free-body diagram. ©Fy = 0, ©Fx = 0 ©MO = 0,
- 233. 216 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 (a) 600 N D 100 N A B 200 N 2 m 3 m 2 m 0.2 m By 2 m 600 sin 45 N 3 m 2 m A B 200 N 600 cos 45 N Ay Bx x y (b) 100 N 0.2 m D Fig. 5–12 Determine the horizontal and vertical components of reaction on the beam caused by the pin at B and the rocker at as shown in Fig. 5–12a. Neglect the weight of the beam. A EXAMPLE 5.5 SOLUTION Free-Body Diagram. Identify each of the forces shown on the free- body diagram of the beam, Fig. 5–12b. (See Example 5.1.) For simplicity, the 600-N force is represented by its x and y components as shown in Fig. 5–12b. Equations of Equilibrium. Summing forces in the x direction yields Ans. A direct solution for can be obtained by applying the moment equation about point B. a Ans. Summing forces in the y direction, using this result, gives Ans. NOTE: We can check this result by summing moments about point A. a Ans. By = 405 N -1100 N215 m2 - 1200 N217 m2 + By17 m2 = 0 -1600 sin 45° N212 m2 - 1600 cos 45° N210.2 m2 +©MA = 0; By = 405 N 319 N - 600 sin 45° N - 100 N - 200 N + By = 0 + c ©Fy = 0; Ay = 319 N - 1600 cos 45° N210.2 m2 - Ay17 m2 = 0 100 N12 m2 + 1600 sin 45° N215 m2 +©MB = 0; ©MB = 0 Ay Bx = 424 N 600 cos 45° N - Bx = 0 : + ©Fx = 0;
- 234. 5.3 EQUATIONS OF EQUILIBRIUM 217 5 EXAMPLE 5.6 The cord shown in Fig. 5–13a supports a force of 100 lb and wraps over the frictionless pulley. Determine the tension in the cord at C and the horizontal and vertical components of reaction at pin A. 100 lb 0.5 ft 30 C (a) A u T 100 lb 30 p Ax Ay A (b) p Ax Ay A T 100 lb 0.5 ft 30 (c) x y u Fig. 5–13 SOLUTION Free-Body Diagrams. The free-body diagrams of the cord and pulley are shown in Fig. 5–13b. Note that the principle of action, equal but opposite reaction must be carefully observed when drawing each of these diagrams: the cord exerts an unknown load distribution p on the pulley at the contact surface, whereas the pulley exerts an equal but opposite effect on the cord. For the solution, however, it is simpler to combine the free-body diagrams of the pulley and this portion of the cord, so that the distributed load becomes internal to this “system” and is therefore eliminated from the analysis, Fig. 5–13c. Equations of Equilibrium. Summing moments about point A to eliminate and Fig. 5–13c, we have a Ans. Using the result, Ans. Ans. NOTE: It is seen that the tension remains constant as the cord passes over the pulley. (This of course is true for any angle at which the cord is directed and for any radius r of the pulley.) u Ay = 187 lb Ay - 100 lb - 100 cos 30° lb = 0 + c©Fy = 0; Ax = 50.0 lb -Ax + 100 sin 30° lb = 0 : + ©Fx = 0; T = 100 lb 100 lb 10.5 ft2 - T10.5 ft2 = 0 +©MA = 0; Ay, Ax
- 235. 218 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 EXAMPLE 5.7 0.75 m 30 1 m 0.5 m 60 N 90 N m A B (a) NB 30 0.75 m 1 m 60 N A Ax Ay 30 (b) x y 90 N m Fig. 5–14 SOLUTION Free-Body Diagram. As shown in Fig. 5–14b, the reaction is perpendicular to the member at B. Also, horizontal and vertical components of reaction are represented at A. Equations of Equilibrium. Summing moments about A, we obtain a direct solution for a Using this result, Ans. Ans. Ay = 233 N Ay - 200 cos 30° N - 60 N = 0 + c©Fy = 0; Ax = 100 N Ax - 200 sin 30° N = 0 : + ©Fx = 0; NB = 200 N -90 N # m - 60 N11 m2 + NB10.75 m2 = 0 +©MA = 0; NB, NB The member shown in Fig. 5–14a is pin-connected at A and rests against a smooth support at B. Determine the horizontal and vertical components of reaction at the pin A.
- 236. 5.3 EQUATIONS OF EQUILIBRIUM 219 5 EXAMPLE 5.8 The box wrench in Fig. 5–15a is used to tighten the bolt at A. If the wrench does not turn when the load is applied to the handle, determine the torque or moment applied to the bolt and the force of the wrench on the bolt. SOLUTION Free-Body Diagram. The free-body diagram for the wrench is shown in Fig. 5–15b. Since the bolt acts as a “fixed support,” it exerts force components and and a moment on the wrench at A. Equations of Equilibrium. Ans. Ans. a Ans. Note that must be included in this moment summation. This couple moment is a free vector and represents the twisting resistance of the bolt on the wrench. By Newton’s third law, the wrench exerts an equal but opposite moment or torque on the bolt. Furthermore, the resultant force on the wrench is Ans. NOTE: Although only three independent equilibrium equations can be written for a rigid body, it is a good practice to check the calculations using a fourth equilibrium equation. For example, the above computations may be verified in part by summing moments about point C: a 19.2 N # m + 32.6 N # m - 51.8 N # m = 0 C52A12 13 B ND 10.4 m2 + 32.6 N # m - 74.0 N10.7 m2 = 0 +©MC = 0; FA = 215.0022 + 174.022 = 74.1 N MA MA = 32.6 N # m MA - C52A12 13 B ND 10.3 m2 - 130 sin 60° N210.7 m2 = 0 +©M A = 0; Ay = 74.0 N Ay - 52A12 13 B N - 30 sin 60° N = 0 + c©Fy = 0; Ax = 5.00 N Ax - 52A 5 13 B N + 30 cos 60° N = 0 : + ©Fx = 0; MA Ay Ax 300 mm 400 mm 13 12 5 B C 60 52 N 30 N (a) A C 0.3 m 0.4 m 13 12 5 60 52 N 30 N (b) Ay MA Ax y x Fig. 5–15
- 237. 220 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 Determine the horizontal and vertical components of reaction on the member at the pin A, and the normal reaction at the roller B in Fig. 5–16a. SOLUTION Free-Body Diagram. The free-body diagram is shown in Fig. 5–16b. The pin at A exerts two components of reaction on the member, and . Ay Ax EXAMPLE 5.9 3 ft A B 3 ft 2 ft (a) 30 750 lb A B 2 ft 3 ft 3 ft 750 lb Ax Ay NB 30 y x (b) Fig. 5–16 Equations of Equilibrium. The reaction NB can be obtained directly by summing moments about point A since and produce no moment about A. a Ans. Using this result, Ans. Ans. Ay = 286 lb Ay + (536.2 lb) cos 30° - 750 lb = 0 + c©Fy = 0; Ax = 268 lb Ax - (536.2 lb ) sin 30° = 0 : + © Fx = 0; NB = 536.2 lb = 536 lb [NB cos 30°](6 ft) - [NB sin 30°](2 ft) - 750 lb(3 ft) = 0 +©MA = 0; Ay Ax
- 238. 5.3 EQUATIONS OF EQUILIBRIUM 221 5 EXAMPLE 5.10 The uniform smooth rod shown in Fig. 5–17a is subjected to a force and couple moment. If the rod is supported at A by a smooth wall and at B and C either at the top or bottom by rollers, determine the reactions at these supports. Neglect the weight of the rod. (a) A 2 m 300 N 4000 N m 4 m 2 m C B 30 2 m (b) 2 m 300 N 4000 N m 4 m 2 m 30 30 Cy¿ By¿ 30 30 Ax y y¿ x x¿ 30 SOLUTION Free-Body Diagram. As shown in Fig. 5–17b, all the support reactions act normal to the surfaces of contact since these surfaces are smooth. The reactions at B and C are shown acting in the positive direction. This assumes that only the rollers located on the bottom of the rod are used for support. Equations of Equilibrium. Using the x, y coordinate system in Fig. 5–17b, we have (1) (2) a (3) When writing the moment equation, it should be noted that the line of action of the force component 300 sin 30° N passes through point A, and therefore this force is not included in the moment equation. Solving Eqs. 2 and 3 simultaneously, we obtain Ans. Ans. Since is a negative scalar, the sense of is opposite to that shown on the free-body diagram in Fig. 5–17b. Therefore, the top roller at B serves as the support rather than the bottom one.Retaining the negative sign for (Why?) and substituting the results into Eq. 1, we obtain Ans. Ax = 173 N 1346.4 sin 30° N + (-1000.0 sin 30° N) - Ax = 0 By¿ By¿ By¿ Cy¿ = 1346.4 N = 1.35 kN By¿ = -1000.0 N = -1 kN + 1300 cos 30° N218 m2 = 0 -By¿12 m2 + 4000 N # m - Cy¿16 m2 +©MA = 0; -300 N + Cy¿ cos 30° + By¿ cos 30° = 0 + c ©Fy = 0; Cy¿ sin 30° + By¿ sin 30° - Ax = 0 : + ©Fx = 0; y¿ Fig. 5–17
- 239. 222 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 The uniform truck ramp shown in Fig. 5–18a has a weight of 400 lb and is pinned to the body of the truck at each side and held in the position shown by the two side cables. Determine the tension in the cables. SOLUTION The idealized model of the ramp, which indicates all necessary dimensions and supports, is shown in Fig. 5–18b. Here the center of gravity is located at the midpoint since the ramp is considered to be uniform. Free-Body Diagram. Working from the idealized model, the ramp’s free-body diagram is shown in Fig. 5–18c. Equations of Equilibrium. Summing moments about point A will yield a direct solution for the cable tension. Using the principle of moments, there are several ways of determining the moment of T about A. If we use x and y components, with T applied at B, we have a The simplest way to determine the moment of T about A is to resolve it into components along and perpendicular to the ramp at B.Then the moment of the component along the ramp will be zero about A, so that a Since there are two cables supporting the ramp, Ans. NOTE: As an exercise, show that and Ay = 887.4 lb. Ax = 1339 lb T¿ = T 2 = 712 lb T = 1425 lb -T sin 10°17 ft2 + 400 lb 15 cos 30° ft2 = 0 +©MA = 0; T = 1425 lb + 400 lb 15 cos 30° ft2 = 0 -T cos 20°17 sin 30° ft2 + T sin 20°17 cos 30° ft2 +©MA = 0; EXAMPLE 5.11 (a) (b) G B A 30 20 2 ft 5 ft (c) G B A Ay Ax T 30 2 ft 10 20 5 ft 400 lb x y Fig. 5–18
- 240. 5.3 EQUATIONS OF EQUILIBRIUM 223 5 EXAMPLE 5.12 Determine the support reactions on the member in Fig. 5–19a. The collar at A is fixed to the member and can slide vertically along the vertical shaft. A B (a) 1.5 m 1.5 m 1 m 45 900 N 500 N m A B Ax MA 900 N NB 45 500 N m 1 m 1.5 m 1.5 m y x (b) Fig. 5–19 SOLUTION Free-Body Diagram. The free-body diagram of the member is shown in Fig. 5–19b.The collar exerts a horizontal force and moment on the member. The reaction of the roller on the member is vertical. Equations of Equilibrium. The forces and can be determined directly from the force equations of equilibrium. Ans. Ans. The moment can be determined by summing moments either about point A or point B. a b Ans. or a b Ans. The negative sign indicates that has the opposite sense of rotation to that shown on the free-body diagram. MA MA = -1486 N # m = 1.49 kN # m MA + 900 N [1.5 m + (1 m) cos 45°] - 500 N # m = 0 +©MB = 0; MA = -1486 N # m = 1.49 kN # m MA - 900 N(1.5 m) - 500 N # m + 900 N [3 m + (1 m) cos 45°] = 0 + ©MA = 0; MA NB - 900 N NB - 900 N = 0 +c©Fy = 0; Ax = 0 : + © Fx = 0; NB Ax NB MA Ax
- 241. A B 224 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 5.4 Two- and Three-Force Members The solutions to some equilibrium problems can be simplified by recognizing members that are subjected to only two or three forces. Two-Force Members As the name implies, a two-force member has forces applied at only two points on the member. An example of a two- force member is shown in Fig. 5–20a. To satisfy force equilibrium, and must be equal in magnitude, , but opposite in direction , Fig. 5–20b. Furthermore, moment equilibrium requires that and share the same line of action, which can only happen if they are directed along the line joining points A and B ( or ), Fig. 5–20c. Therefore, for any two-force member to be in equilibrium, the two forces acting on the member must have the same magnitude, act in opposite directions, and have the same line of action, directed along the line joining the two points where these forces act. ©MB = 0 ©MA = 0 FB FA (©F = 0) FA = FB = F FB FA Three-Force Members If a member is subjected to only three forces, it is called a three-force member. Moment equilibrium can be satisfied only if the three forces form a concurrent or parallel force system. To illustrate, consider the member subjected to the three forces , , and , shown in Fig. 5–21a. If the lines of action of and intersect at point O, then the line of action of must also pass through point O so that the forces satisfy .As a special case, if the three forces are all parallel, Fig. 5–21b, the location of the point of intersection, O, will approach infinity. ©MO = 0 F3 F2 F1 F3 F2 F1 B FB (a) A FA (b) Two-force member A FA F FB F A FB F (c) B FA F Fig. 5–20 F3 F1 O F1 F3 Three-force member F2 F2 (b) (a) Fig. 5–21 The bucket link AB on the back-hoe is a typical example of a two-force member since it is pin connected at its ends and, provided its weight is neglected,no other force acts on this member. The link used for this railroad car brake is a three-force member. Since the force in the tie rod at B and from the link at C are parallel, then for equilibrium the resultant force at the pin A must also be parallel with these two forces. FA FC FB FB FA FC B A C The boom on this lift is a three-force member, provided its weight is neglected. Here the lines of action of the weight of the worker, W, and the force of the two-force member (hydraulic cylinder) at B, , intersect at O.For moment equilibrium,the resultant force at the pin A, , must also be directed towards O. FA FB FA B W O A FB
- 242. 5.4 TWO- AND THREE-FORCE MEMBERS 225 5 EXAMPLE 5.13 The lever ABC is pin supported at A and connected to a short link BD as shown in Fig. 5–22a. If the weight of the members is negligible, determine the force of the pin on the lever at A. SOLUTION Free-Body Diagrams. As shown in Fig. 5–22b, the short link BD is a two-force member, so the resultant forces at pins D and B must be equal, opposite, and collinear. Although the magnitude of the force is unknown, the line of action is known since it passes through B and D. Lever ABC is a three-force member, and therefore, in order to satisfy moment equilibrium, the three nonparallel forces acting on it must be concurrent at O, Fig. 5–22c. In particular, note that the force F on the lever at B is equal but opposite to the force F acting at B on the link. Why? The distance CO must be 0.5 m since the lines of action of F and the 400-N force are known. Equations of Equilibrium. By requiring the force system to be concurrent at O, since the angle which defines the line of action of can be determined from trigonometry, Using the x, y axes and applying the force equilibrium equations, Solving, we get Ans. NOTE: We can also solve this problem by representing the force at A by its two components and and applying to the lever. Once and are determined, we can get and . u FA Ay Ax ©Fy = 0 ©Fx = 0, ©MA = 0, Ay Ax F = 1.32 kN F A = 1.07 kN F A sin 60.3° - F sin 45° = 0 + c©F y = 0; F A cos 60.3° - F cos 45° + 400 N = 0 : + ©F x = 0; u = tan-1 a 0.7 0.4 b = 60.3° FA u ©MO = 0, 0.5 m 0.2 m B A D C 0.1 m 0.2 m (a) 400 N 45 F F B D (b) 0.2 m B A C 0.5 m 0.5 m F 45 45 45 O 0.1 m (c) 0.4 m FA 400 N u Fig. 5–22
- 243. 226 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 B A 5 ft 5 ft 5 ft 500 lb 600 lb ft 4 3 5 F5–1 FUNDAMENTAL PROBLEMS F5–4. Determine the components of reaction at the fixed support A. Neglect the thickness of the beam. F5–2. Determine the horizontal and vertical components of reaction at the pin A and the reaction on the beam at C. All problem solutions must include an FBD. F5–1. Determine the horizontal and vertical components of reaction at the supports. Neglect the thickness of the beam. F5–5. The 25-kg bar has a center of mass at G. If it is supported by a smooth peg at C, a roller at A, and cord AB, determine the reactions at these supports. F5–3. The truss is supported by a pin at A and a roller at B. Determine the support reactions. F5–6. Determine the reactions at the smooth contact points A, B, and C on the bar. 1.5 m C B A 1.5 m 1.5 m D 4 kN F5–22 A B 2 m 5 kN 10 kN 2 m 4 m 4 m 45 F5–3 60 30 3 m 1 m 1 m 1 m 400 N 200 N 200 N 200 N A F5–4 A B G C D 30 15 0.5 m 0.2 m 0.3 m F5–5 0.4 m 250 N 0.2 m 0.15 m 30 A B C 30 F5–6
- 244. 5.4 TWO- AND THREE-FORCE MEMBERS 227 5 PROBLEMS *5–20. The train car has a weight of 24 000 lb and a center of gravity at G. It is suspended from its front and rear on the track by six tires located at A, B, and C. Determine the normal reactions on these tires if the track is assumed to be a smooth surface and an equal portion of the load is supported at both the front and rear tires. All problem solutions must include an FBD. 5–11. Determine the normal reactions at A and B in Prob. 5–1. *5–12. Determine the tension in the cord and the horizontal and vertical components of reaction at support A of the beam in Prob. 5–4. •5–13. Determine the horizontal and vertical components of reaction at C and the tension in the cable AB for the truss in Prob. 5–5. 5–14. Determine the horizontal and vertical components of reaction at A and the tension in cable BC on the boom in Prob. 5–6. 5–15. Determine the horizontal and vertical components of reaction at A and the normal reaction at B on the spanner wrench in Prob. 5–7. *5–16. Determine the normal reactions at A and B and the force in link CD acting on the member in Prob. 5–8. •5–17. Determine the normal reactions at the points of contact at A, B, and C of the bar in Prob. 5–9. 5–18. Determine the horizontal and vertical components of reaction at pin C and the force in the pawl of the winch in Prob. 5–10. 5–19. Compare the force exerted on the toe and heel of a 120-lb woman when she is wearing regular shoes and stiletto heels. Assume all her weight is placed on one foot and the reactions occur at points A and B as shown. •5–21. Determine the horizontal and vertical components of reaction at the pin A and the tension developed in cable BC used to support the steel frame. A A B B 5.75 in. 3.75 in. 0.75 in. 1.25 in. 120 lb 120 lb Prob. 5–19 5 ft A C B G 4 ft 6 ft Prob. 5–20 A B C 30 kN m 60 kN 1 m 3 m 1 m 1 m 5 4 3 Prob. 5–21
- 245. 228 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 •5–25. The 300-lb electrical transformer with center of gravity at G is supported by a pin at A and a smooth pad at B. Determine the horizontal and vertical components of reaction at the pin A and the reaction of the pad B on the transformer. 5–23. The airstroke actuator at D is used to apply a force of F = 200 N on the member at B. Determine the horizontal and vertical components of reaction at the pin A and the force of the smooth shaft at C on the member. *5–24. The airstroke actuator at D is used to apply a force of F on the member at B. The normal reaction of the smooth shaft at C on the member is 300 N. Determine the magnitude of F and the horizontal and vertical components of reaction at pin A. 5–22. The articulated crane boom has a weight of 125 lb and center of gravity at G.If it supports a load of 600 lb,determine the force acting at the pin A and the force in the hydraulic cylinder BC when the boom is in the position shown. 5–26. A skeletal diagram of a hand holding a load is shown in the upper figure. If the load and the forearm have masses of 2 kg and 1.2 kg, respectively, and their centers of mass are located at and , determine the force developed in the biceps CD and the horizontal and vertical components of reaction at the elbow joint B. The forearm supporting system can be modeled as the structural system shown in the lower figure. G2 G1 C 40 B G A 1 ft 4 ft 1 ft 8 ft Prob. 5–22 A C B D 60 600 mm 600 mm 15 200 mm F Probs. 5–23/24 B A 1.5 ft 3 ft G Prob. 5–25 B B C C D D G2 G2 G1 G1 A A 135 mm 65 mm 75 100 mm Prob. 5–26
- 246. 5.4 TWO- AND THREE-FORCE MEMBERS 229 5 •5–29. The mass of 700 kg is suspended from a trolley which moves along the crane rail from to . Determine the force along the pin-connected knee strut BC (short link) and the magnitude of force at pin A as a function of position d. Plot these results of and (vertical axis) versus d (horizontal axis). F A F BC d = 3.5 m d = 1.7 m *5–28. The 1.4-Mg drainpipe is held in the tines of the fork lift. Determine the normal forces at A and B as functions of the blade angle and plot the results of force (vertical axis) versus (horizontal axis) for 0 … u … 90°. u u 5–27. As an airplane’s brakes are applied, the nose wheel exerts two forces on the end of the landing gear as shown. Determine the horizontal and vertical components of reaction at the pin C and the force in strut AB. 5–30. If the force of F = 100 lb is applied to the handle of the bar bender, determine the horizontal and vertical components of reaction at pin A and the reaction of the roller B on the smooth bar. 5–31. If the force of the smooth roller at B on the bar bender is required to be 1.5 kip, determine the horizontal and vertical components of reaction at pin A and the required magnitude of force F applied to the handle. 20 30 2 kN 6 kN B A 600 mm 400 mm C Prob. 5–27 0.4 m A B u Prob. 5–28 A B C 2 m 1.5 m d Prob. 5–29 60 F 40 in. 5 in. B A C Probs. 5–30/31
- 247. 230 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 G D 4 m 0.2 m 3.2 m B C A x Probs. 5–32/33 5–35. The framework is supported by the member AB which rests on the smooth floor. When loaded, the pressure distribution on AB is linear as shown.Determine the length d of member AB and the intensity w for this case. 5–34. Determine the horizontal and vertical components of reaction at the pin A and the normal force at the smooth peg B on the member. *5–32. The jib crane is supported by a pin at C and rod AB. If the load has a mass of 2 Mg with its center of mass located at G, determine the horizontal and vertical components of reaction at the pin C and the force developed in rod AB on the crane when x = 5 m. •5–33. The jib crane is supported by a pin at C and rod AB. The rod can withstand a maximum tension of 40 kN. If the load has a mass of 2 Mg, with its center of mass located at G, determine its maximum allowable distance x and the corresponding horizontal and vertical components of reaction at C. *5–36. Outriggers A and B are used to stabilize the crane from overturning when lifting large loads. If the load to be lifted is 3 Mg, determine the maximum boom angle so that the crane does not overturn. The crane has a mass of 5 Mg and center of mass at , whereas the boom has a mass of 0.6 Mg and center of mass at . GB GC u A C F 600 N B 30 0.4 m 0.4 m 30 Prob. 5–34 4 ft 800 lb d w 7 ft A B Prob. 5–35 2.8 m 4.5 m A B 5 m 0.7 m 2.3 m GB GC u Prob. 5–36
- 248. 5.4 TWO- AND THREE-FORCE MEMBERS 231 5 *5–40. The platform assembly has a weight of 250 lb and center of gravity at If it is intended to support a maximum load of 400 lb placed at point determine the smallest counterweight W that should be placed at B in order to prevent the platform from tipping over. G2, G1. 5–38. Spring CD remains in the horizontal position at all times due to the roller at D. If the spring is unstretched when and the bracket achieves its equilibrium position when , determine the stiffness k of the spring and the horizontal and vertical components of reaction at pin A. 5–39. Spring CD remains in the horizontal position at all times due to the roller at D. If the spring is unstretched when and the stiffness is , determine the smallest angle for equilibrium and the horizontal and vertical components of reaction at pin A. u k = 1.5 kNm u = 0° u = 30° u = 0° •5–37. The wooden plank resting between the buildings deflects slightly when it supports the 50-kg boy. This deflection causes a triangular distribution of load at its ends, having maximum intensities of and . Determine and , each measured in , when the boy is standing 3 m from one end as shown. Neglect the mass of the plank. Nm wB wA wB wA •5–41. Determine the horizontal and vertical components of reaction at the pin A and the reaction of the smooth collar B on the rod. 3 m 0.45 m 0.3 m 6 m A B wA wB Prob. 5–37 0.45 m 0.6 m k D C B A F 300 N u Probs. 5–38/39 6 ft 8 ft 1 ft 1 ft C B G1 D 2 ft 6 ft G2 Prob. 5–40 A B D C 2 ft 300 lb 4 ft 1 ft 1 ft 30 450 lb Prob. 5–41
- 249. 232 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 *5–44. Determine the horizontal and vertical components of force at the pin A and the reaction at the rocker B of the curved beam. 5–43. The uniform rod AB has a weight of 15 lb. Determine the force in the cable when the rod is in the position shown. 5–42. Determine the support reactions of roller A and the smooth collar B on the rod. The collar is fixed to the rod AB, but is allowed to slide along rod CD. •5–45. The floor crane and the driver have a total weight of 2500 lb with a center of gravity at G. If the crane is required to lift the 500-lb drum, determine the normal reaction on both the wheels at A and both the wheels at B when the boom is in the position shown. 5–46. The floor crane and the driver have a total weight of 2500 lb with a center of gravity at G. Determine the largest weight of the drum that can be lifted without causing the crane to overturn when its boom is in the position shown. A 1 m 2 m 600 N m 1 m B D C 900 N 45 45 Prob. 5–42 A 10 30 5 ft C B T Prob. 5–43 A B 500 N 200 N 10 15 2 m Prob. 5–44 12 ft 30 3 ft 6 ft 8.4 ft 2.2 ft 1.4 ft A B D E F C G Probs. 5–45/46
- 250. 5.4 TWO- AND THREE-FORCE MEMBERS 233 5 5–50. The winch cable on a tow truck is subjected to a force of when the cable is directed at . Determine the magnitudes of the total brake frictional force F for the rear set of wheels B and the total normal forces at both front wheels A and both rear wheels B for equilibrium. The truck has a total mass of 4 Mg and mass center at G. 5–51. Determine the minimum cable force T and critical angle which will cause the tow truck to start tipping, i.e., for the normal reaction at A to be zero.Assume that the truck is braked and will not slip at B. The truck has a total mass of 4 Mg and mass center at G.x u u = 60° T = 6 kN *5–48. Determine the force P needed to pull the 50-kg roller over the smooth step.Take •5–49. Determine the magnitude and direction of the minimum force P needed to pull the 50-kg roller over the smooth step. u u = 60°. 5–47. The motor has a weight of 850 lb. Determine the force that each of the chains exerts on the supporting hooks at A, B, and C. Neglect the size of the hooks and the thickness of the beam. *5–52. Three uniform books, each having a weight W and length a, are stacked as shown. Determine the maximum distance d that the top book can extend out from the bottom one so the stack does not topple over. 1.5 ft 1 ft 0.5 ft C A B 30 10 10 850 lb Prob. 5–47 20 A B P 0.6 m 0.1 m u Probs. 5–48/49 1.25 m 3 m A G B F T 1.5 m 2 m 2.5 m u Probs. 5–50/51 a d Prob. 5–52
- 251. 234 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 5–55. The horizontal beam is supported by springs at its ends. Each spring has a stiffness of and is originally unstretched so that the beam is in the horizontal position. Determine the angle of tilt of the beam if a load of 800 N is applied at point C as shown. *5–56. The horizontal beam is supported by springs at its ends. If the stiffness of the spring at A is , determine the required stiffness of the spring at B so that if the beam is loaded with the 800 N it remains in the horizontal position. The springs are originally constructed so that the beam is in the horizontal position when it is unloaded. kA = 5 kNm k = 5 kNm 5–54. The uniform rod AB has a weight of 15 lb and the spring is unstretched when . If , determine the stiffness k of the spring. u = 30° u = 0° •5–53. Determine the angle at which the link ABC is held in equilibrium if member BD moves 2 in. to the right. The springs are originally unstretched when . Each spring has the stiffness shown. The springs remain horizontal since they are attached to roller guides. u = 0° u •5–57. The smooth disks D and E have a weight of 200 lb and 100 lb, respectively. If a horizontal force of is applied to the center of disk E, determine the normal reactions at the points of contact with the ground at A, B, and C. 5–58. The smooth disks D and E have a weight of 200 lb and 100 lb, respectively. Determine the largest horizontal force P that can be applied to the center of disk E without causing the disk D to move up the incline. P = 200 lb kCF 100 lb/ft kAE 500 lb/ft E F C A B D F 6 in. 6 in. u Prob. 5–53 6 ft u B A 3 ft k Prob. 5–54 800 N B C A 3 m 1 m Probs. 5–55/56 P 1.5 ft A B D E C 3 5 4 1 ft Probs. 5–57/58
- 252. 5.4 TWO- AND THREE-FORCE MEMBERS 235 5 •5–61. If spring BC is unstretched with and the bell crank achieves its equilibrium position when , determine the force F applied perpendicular to segment AD and the horizontal and vertical components of reaction at pin A. Spring BC remains in the horizontal postion at all times due to the roller at C. u = 15° u = 0° *5–60. The uniform rod has a length l and weight W. It is supported at one end A by a smooth wall and the other end by a cord of length s which is attached to the wall as shown. Show that for equilibrium it is required that . h = [(s2 - l2 )3]12 5–59. A man stands out at the end of the diving board, which is supported by two springs A and B, each having a stiffness of . In the position shown the board is horizontal. If the man has a mass of 40 kg, determine the angle of tilt which the board makes with the horizontal after he jumps off. Neglect the weight of the board and assume it is rigid. k = 15 kNm 5–62. The thin rod of length l is supported by the smooth tube. Determine the distance a needed for equilibrium if the applied load is P. B A 1 m 3 m Prob. 5–59 Prob. 5–60 300 mm 400 mm B k 2 kN/m D C A 150 F u Prob. 5–61 P B A 2r a l Prob. 5–62 h s C B A l
- 253. 236 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 CONCEPTUAL PROBLEMS P5–7. Like all aircraft, this jet plane rests on three wheels. Why not use an additional wheel at the tail for better support? (Can you think of any other reason for not including this wheel?) If there was a fourth tail wheel, draw a free-body diagram of the plane from a side (2 D) view, and show why one would not be able to determine all the wheel reactions using the equations of equilibrium. P5–6. The man attempts to pull the four wheeler up the incline and onto the truck bed. From the position shown, is it more effective to keep the rope attached at A, or would it be better to attach it to the axle of the front wheels at B? Draw a free-body diagram and do an equilibrium analysis to explain your answer. P5–5. The tie rod is used to support this overhang at the entrance of a building. If it is pin connected to the building wall at A and to the center of the overhang B, determine if the force in the rod will increase, decrease, or remain the same if (a) the support at A is moved to a lower position D, and (b) the support at B is moved to the outer position C. Explain your answer with an equilibrium analysis, using dimensions and loads. Assume the overhang is pin supported from the building wall. *P5–8. Where is the best place to arrange most of the logs in the wheelbarrow so that it minimizes the amount of force on the backbone of the person transporting the load? Do an equilibrium analysis to explain your answer. C B D A P5–5 A B P5–6 P5–7 P5–8
- 254. 5.5 FREE-BODY DIAGRAMS 237 5 EQUILIBRIUM IN THREE DIMENSIONS 5.5 Free-Body Diagrams The first step in solving three-dimensional equilibrium problems, as in the case of two dimensions, is to draw a free-body diagram. Before we can do this, however, it is first necessary to discuss the types of reactions that can occur at the supports. Support Reactions. The reactive forces and couple moments acting at various types of supports and connections, when the members are viewed in three dimensions, are listed in Table 5–2. It is important to recognize the symbols used to represent each of these supports and to understand clearly how the forces and couple moments are developed. As in the two-dimensional case: • A force is developed by a support that restricts the translation of its attached member. • A couple moment is developed when rotation of the attached member is prevented. For example, in Table 5–2, item (4), the ball-and-socket joint prevents any translation of the connecting member; therefore, a force must act on the member at the point of connection.This force has three components having unknown magnitudes, Fx, Fy, Fz. Provided these components are known, one can obtain the magnitude of force, , and the force’s orientation defined by its coordinate direction angles Eqs. 2–7.* Since the connecting member is allowed to rotate freely about any axis, no couple moment is resisted by a ball-and-socket joint. It should be noted that the single bearing supports in items (5) and (7), the single pin (8), and the single hinge (9) are shown to resist both force and couple-moment components. If, however, these supports are used in conjunction with other bearings, pins, or hinges to hold a rigid body in equilibrium and the supports are properly aligned when connected to the body, then the force reactions at these supports alone are adequate for supporting the body. In other words, the couple moments become redundant and are not shown on the free-body diagram. The reason for this should become clear after studying the examples which follow. g, b, a, F = 2Fx 2 + Fy 2 + Fz 2 * The three unknowns may also be represented as an unknown force magnitude F and two unknown coordinate direction angles. The third direction angle is obtained using the identity Eq. 2–8. cos2 a + cos2 b + cos2 g = 1,
- 255. 238 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 Types of Connection Reaction Number of Unknowns continued One unknown. The reaction is a force which acts away from the member in the known direction of the cable. One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. Three unknowns. The reactions are three rectangular force components. Four unknowns. The reactions are two force and two couple-moment components which act perpendicular to the shaft. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples. F F F Fz Fy Fx single journal bearing Fz Fx Mz Mx (1) cable (2) (3) roller ball and socket (4) (5) smooth surface support TABLE 5–2 Supports for Rigid Bodies Subjected to Three-Dimensional Force Systems
- 256. 5.5 FREE-BODY DIAGRAMS 239 5 Reaction Number of Unknowns Five unknowns. The reactions are two force and three couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples. Five unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples. Five unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples. Five unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples. Six unknowns. The reactions are three force and three couple-moment components. Fz Fx Mz Mx Fy Fz Fx Mz Mx My Fz Mz Fx Fy My Mz Fx Fy Mx Fz Mz Fx My Mx Fy Fz Types of Connection TABLE 5–2 Continued single hinge fixed support single thrust bearing single journal bearing with square shaft single smooth pin (7) (6) (8) (10) (9)
- 257. 240 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 Typical examples of actual supports that are referenced to Table 5–2 are shown in the following sequence of photos. This ball-and-socket joint provides a connection for the housing of an earth grader to its frame. (4) This journal bearing supports the end of the shaft. (5) This thrust bearing is used to support the drive shaft on a machine. (7) Free-Body Diagrams. The general procedure for establishing the free-body diagram of a rigid body has been outlined in Sec. 5.2. Essentially it requires first “isolating” the body by drawing its outlined shape. This is followed by a careful labeling of all the forces and couple moments with reference to an established x, y, z coordinate system. It is suggested to show the unknown components of reaction as acting on the free-body diagram in the positive sense. In this way, if any negative values are obtained, they will indicate that the components act in the negative coordinate directions. This pin is used to support the end of the strut used on a tractor. (8)
- 258. 5.5 FREE-BODY DIAGRAMS 241 5 EXAMPLE 5.14 Consider the two rods and plate, along with their associated free-body diagrams shown in Fig. 5–23. The x, y, z axes are established on the diagram and the unknown reaction components are indicated in the positive sense.The weight is neglected. SOLUTION 45 N m 500 N Properly aligned journal bearings at A, B, C. A B C 45 N m 500 N The force reactions developed by the bearings are sufficient for equilibrium since they prevent the shaft from rotating about each of the coordinate axes. Bz Bx Cx Cy x y Ay Az z 400 lb A B C Properly aligned journal bearing at A and hinge at C. Roller at B. Ax 400 lb Bz z y x Az Cx Cz Cy Only force reactions are developed by the bearing and hinge on the plate to prevent rotation about each coordinate axis. No moments at the hinge are developed. C 200 lb ft Pin at A and cable BC. A B 300 lb 200 lb ft Moment components are developed by the pin on the rod to prevent rotation about the x and z axes. x B 300 lb y Az z MAz MAx Ax Ay T Fig. 5–23
- 259. 242 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 5.6 Equations of Equilibrium As stated in Sec. 5.1, the conditions for equilibrium of a rigid body subjected to a three-dimensional force system require that both the resultant force and resultant couple moment acting on the body be equal to zero. Vector Equations of Equilibrium. The two conditions for equilibrium of a rigid body may be expressed mathematically in vector form as (5–5) where is the vector sum of all the external forces acting on the body and is the sum of the couple moments and the moments of all the forces about any point O located either on or off the body. Scalar Equations of Equilibrium. If all the external forces and couple moments are expressed in Cartesian vector form and substituted into Eqs. 5–5, we have Since the i, j, and k components are independent from one another, the above equations are satisfied provided (5–6a) and (5–6b) These six scalar equilibrium equations may be used to solve for at most six unknowns shown on the free-body diagram. Equations 5–6a require the sum of the external force components acting in the x, y, and z directions to be zero, and Eqs. 5–6b require the sum of the moment components about the x, y, and z axes to be zero. ©Mx = 0 ©My = 0 ©Mz = 0 ©Fx = 0 ©Fy = 0 ©Fz = 0 ©MO = ©Mxi + ©Myj + ©Mzk = 0 ©F = ©Fxi + ©Fy j + ©Fzk = 0 ©MO ©F ©F = 0 ©MO = 0
- 260. 5.7 CONSTRAINTS AND STATICAL DETERMINACY 243 5 5.7 Constraints and Statical Determinacy To ensure the equilibrium of a rigid body, it is not only necessary to satisfy the equations of equilibrium, but the body must also be properly held or constrained by its supports.Some bodies may have more supports than are necessary for equilibrium, whereas others may not have enough or the supports may be arranged in a particular manner that could cause the body to move. Each of these cases will now be discussed. Redundant Constraints. When a body has redundant supports, that is, more supports than are necessary to hold it in equilibrium, it becomes statically indeterminate. Statically indeterminate means that there will be more unknown loadings on the body than equations of equilibrium available for their solution. For example, the beam in Fig. 5–24a and the pipe assembly in Fig. 5–24b, shown together with their free-body diagrams, are both statically indeterminate because of additional (or redundant) support reactions. For the beam there are five unknowns, and for which only three equilibrium equations can be written ( and Eqs. 5–2). The pipe assembly has eight unknowns, for which only six equilibrium equations can be written, Eqs. 5–6. The additional equations needed to solve statically indeterminate problems of the type shown in Fig. 5–24 are generally obtained from the deformation conditions at the points of support.These equations involve the physical properties of the body which are studied in subjects dealing with the mechanics of deformation, such as “mechanics of materials.”* ©MO = 0, ©Fy = 0, ©Fx = 0, Cy, By, Ay, Ax, MA, 500 N B C A 2 kN m 500 N 2 kN m Ax Ay MA By Cy (a) x y B A 400 N 200 N 400 N 200 N Ay Az By Bx Mx My Bz Mz (b) y z x Fig. 5–24 * See R. C. Hibbeler, Mechanics of Materials, 7th edition, Pearson Education/Prentice Hall, Inc.
- 261. 244 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 Improper Constraints. Having the same number of unknown reactive forces as available equations of equilibrium does not always guarantee that a body will be stable when subjected to a particular loading. For example, the pin support at A and the roller support at B for the beam in Fig. 5–25a are placed in such a way that the lines of action of the reactive forces are concurrent at point A. Consequently, the applied loading P will cause the beam to rotate slightly about A, and so the beam is improperly constrained, . In three dimensions, a body will be improperly constrained if the lines of action of all the reactive forces intersect a common axis. For example, the reactive forces at the ball-and-socket supports at A and B in Fig. 5–25b all intersect the axis passing through A and B. Since the moments of these forces about A and B are all zero, then the loading P will rotate the member about the AB axis, . ©MAB Z 0 ©MA Z 0 Fig. 5–25 A B FB Ay Ax A P P (a) A Az Bz Ax Bx Ay By z x B y A z x B y P P (b)
- 262. 5.7 CONSTRAINTS AND STATICAL DETERMINACY 245 5 Another way in which improper constraining leads to instability occurs when the reactive forces are all parallel. Two- and three- dimensional examples of this are shown in Fig. 5–26. In both cases, the summation of forces along the x axis will not equal zero. In some cases, a body may have fewer reactive forces than equations of equilibrium that must be satisfied.The body then becomes only partially constrained. For example, consider member AB in Fig. 5–27a with its corresponding free-body diagram in Fig. 5–27b. Here will not be satisfied for the loading conditions and therefore equilibrium will not be maintained. To summarize these points, a body is considered improperly constrained if all the reactive forces intersect at a common point or pass through a common axis, or if all the reactive forces are parallel. In engineering practice, these situations should be avoided at all times since they will cause an unstable condition. ©Fy = 0 A FA A B FB P P (a) y x FB 100 N A B C 100 N FA FC x (b) z y Fig. 5–26 A B B FB (a) (b) FA 100 N 100 N Fig. 5–27
- 263. 246 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 Important Points • Always draw the free-body diagram first when solving any equilibrium problem. • If a support prevents translation of a body in a specific direction, then the support exerts a force on the body in that direction. • If a support prevents rotation about an axis, then the support exerts a couple moment on the body about the axis. • If a body is subjected to more unknown reactions than available equations of equilibrium,then the problem is statically indeterminate. • A stable body requires that the lines of action of the reactive forces do not intersect a common axis and are not parallel to one another. Procedure for Analysis Three-dimensional equilibrium problems for a rigid body can be solved using the following procedure. Free-Body Diagram. • Draw an outlined shape of the body. • Show all the forces and couple moments acting on the body. • Establish the origin of the x, y, z axes at a convenient point and orient the axes so that they are parallel to as many of the external forces and moments as possible. • Label all the loadings and specify their directions. In general, show all the unknown components having a positive sense along the x, y, z axes. • Indicate the dimensions of the body necessary for computing the moments of forces. Equations of Equilibrium. • If the x, y, z force and moment components seem easy to determine, then apply the six scalar equations of equilibrium; otherwise use the vector equations. • It is not necessary that the set of axes chosen for force summation coincide with the set of axes chosen for moment summation. Actually, an axis in any arbitrary direction may be chosen for summing forces and moments. • Choose the direction of an axis for moment summation such that it intersects the lines of action of as many unknown forces as possible. Realize that the moments of forces passing through points on this axis and the moments of forces which are parallel to the axis will then be zero. • If the solution of the equilibrium equations yields a negative scalar for a force or couple moment magnitude, it indicates that the sense is opposite to that assumed on the free-body diagram.
- 264. 5.7 CONSTRAINTS AND STATICAL DETERMINACY 247 5 EXAMPLE 5.15 The homogeneous plate shown in Fig. 5–28a has a mass of 100 kg and is subjected to a force and couple moment along its edges. If it is supported in the horizontal plane by a roller at A, a ball-and-socket joint at B, and a cord at C, determine the components of reaction at these supports. SOLUTION (SCALAR ANALYSIS) Free-Body Diagram. There are five unknown reactions acting on the plate, as shown in Fig. 5–28b. Each of these reactions is assumed to act in a positive coordinate direction. Equations of Equilibrium. Since the three-dimensional geometry is rather simple, a scalar analysis provides a direct solution to this problem.A force summation along each axis yields Ans. Ans. (1) Recall that the moment of a force about an axis is equal to the product of the force magnitude and the perpendicular distance (moment arm) from the line of action of the force to the axis. Also, forces that are parallel to an axis or pass through it create no moment about the axis. Hence, summing moments about the positive x and y axes, we have (2) (3) The components of the force at B can be eliminated if moments are summed about the and axes.We obtain (4) (5) Solving Eqs. 1 through 3 or the more convenient Eqs. 1, 4, and 5 yields Ans. The negative sign indicates that acts downward. NOTE: The solution of this problem does not require a summation of moments about the z axis. The plate is partially constrained since the supports cannot prevent it from turning about the z axis if a force is applied to it in the x–y plane. Bz Az = 790 N Bz = -217 N TC = 707 N -300 N11.5m2 - 981 N11.5m2 - 200N#m + TC13m2 = 0 ©My¿ = 0; 981 N11 m2 + 300 N12 m2 - Az12 m2 = 0 ©Mx¿ = 0; y¿ x¿ 300 N11.5m2 + 981 N11.5m2 - Bz13m2 - Az 13m2 - 200N#m = 0 ©My = 0; TC12 m2 - 981 N11 m2 + Bz12 m2 = 0 ©Mx = 0; Az + Bz + TC - 300 N - 981 N = 0 ©Fz = 0; By = 0 ©Fy = 0; Bx = 0 ©Fx = 0; A B C 200 N m 1.5 m 2 m 3 m (a) 300 N 200 N m 1.5 m 1.5 m y y¿ x¿ 1 m 1 m Az Bz Bx By z z¿ 981 N TC (b) 300 N x Fig. 5–28
- 265. 248 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 Determine the components of reaction that the ball-and-socket joint at A, the smooth journal bearing at B, and the roller support at C exert on the rod assembly in Fig. 5–29a. EXAMPLE 5.16 x y z A B C D 0.4 m 0.4 m (a) 0.6 m 900 N 0.4 m 0.4 m A x y z 0.4 m 0.4 m (b) 0.6 m 0.4 m 0.4 m FC Bz Az Bx Ax Ay 900 N Fig. 5–29 SOLUTION Free-Body Diagram. As shown on the free-body diagram, Fig. 5–29b, the reactive forces of the supports will prevent the assembly from rotating about each coordinate axis, and so the journal bearing at B only exerts reactive forces on the member. Equations of Equilibrium. A direct solution for can be obtained by summing forces along the y axis. Ans. The force can be determined directly by summing moments about the y axis. Ans. Using this result, can be determined by summing moments about the x axis. Ans. The negative sign indicates that acts downward. The force can be found by summing moments about the z axis. Ans. Thus, Ans. Finally, using the results of and . Ans. Az = 750 N Az + (-450 N) + 600 N - 900 N = 0 ©Fz = 0; F C Bz Ax + 0 = 0 Ax = 0 ©Fx = 0; -Bx(0.8 m) = 0 Bx = 0 ©Mz = 0; Bx Bz Bz = -450 N Bz(0.8 m) + 600 N(1.2 m) - 900 N(0.4 m) = 0 ©Mx = 0; Bz FC = 600 N FC(0.6 m) - 900 N(0.4 m) = 0 ©My = 0; FC Ay = 0 ©Fy = 0; Ay
- 266. 5.7 CONSTRAINTS AND STATICAL DETERMINACY 249 5 EXAMPLE 5.17 The boom is used to support the 75-lb flowerpot in Fig. 5–30a. Determine the tension developed in wires AB and AC. SOLUTION Free-Body Diagram. The free-body diagram of the boom is shown in Fig. 5–30b. Equations of Equilibrium. We will use a vector analysis. We can eliminate the force reaction at O by writing the moment equation of equilibrium about point O. rA * (FAB + FAC + W) = 0 ©MO = 0; = - 2 7 F ACi - 6 7 F ACj + 3 7 F ACk FAC = FACa rAC rAC b = FACa 5-2i-6j + 3k6 ft 2 (-2 ft)2 + (-6 ft)2 + (3 ft)2 b = 2 7 FABi - 6 7 F ABj + 3 7 F ABk FAB = FABa rAB rAB b = FABa 52i - 6j + 3k6 ft 2 (2 ft)2 + (-6 ft)2 + (3 ft)2 b (6j) * c a 2 7 F ABi - 6 7 F AB j + 3 7 F ABkb + a- 2 7 F ACi - 6 7 F ACj + 3 7 F ACkb + (-75k)d = 0 (1) (2) Solving Eqs. (1) and (2) simultaneously, Ans. FAB = FAC = 87.5 lb - 12 7 FAB + 12 7 FAC = 0 ©Mz = 0; 0 = 0 ©My = 0; 18 7 F AB + 18 7 F AC - 450 = 0 ©Mx = 0; a 18 7 FAB + 18 7 FAC - 450bi + a- 12 7 FAB + 12 7 FACbk = 0 x y O A z 6 ft (a) 3 ft 2 ft 2 ft B C B A (b) 6 ft x y O z 3 ft 2 ft 2 ft W 75 lb Oz Oy Ox C rA FAB FAC Fig. 5–30
- 267. 250 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 EXAMPLE 5.18 1.5 m 2 m 200 N 1.5 m 2 m E A B D C (a) 1 m 200 N y B C x z rB rC TD TE Az A Ay Ax (b) Fig. 5–31 Rod AB shown in Fig. 5–31a is subjected to the 200-N force. Determine the reactions at the ball-and-socket joint A and the tension in the cables BD and BE. SOLUTION (VECTOR ANALYSIS) Free-Body Diagram. Fig. 5–31b. Equations of Equilibrium. Representing each force on the free-body diagram in Cartesian vector form, we have Applying the force equation of equilibrium. (1) (2) (3) Summing moments about point A yields Since then Expanding and rearranging terms gives (4) (5) (6) Solving Eqs. 1 through 5, we get Ans. Ans. Ans. Ans. Ans. NOTE: The negative sign indicates that and have a sense which is opposite to that shown on the free-body diagram, Fig. 5–31b. Ay Ax Az = 200 N Ay = -100 N Ax = -50 N TE = 50 N TD = 100 N TD - 2TE = 0 ©Mz = 0; -2TE + 100 = 0 ©My = 0; 2TD - 200 = 0 ©Mx = 0; 12TD - 2002i + 1-2TE + 1002j + 1TD - 2TE2k = 0 10.5i + 1j - 1k2 * 1-200k2 + 11i + 2j - 2k2 * 1T Ei + T Dj2 = 0 rC = 1 2 rB, rC * F + rB * 1TE + TD2 = 0 ©MA = 0; Az - 200 = 0 ©Fz = 0; Ay + TD = 0 ©Fy = 0; Ax + TE = 0 ©Fx = 0; 1Ax + TE2i + 1Ay + TD2j + 1Az - 2002k = 0 FA + TE + TD + F = 0 ©F = 0; F = 5-200k6 N TD = TDj TE = TEi FA = Axi + Ayj + Azk
- 268. 5.7 CONSTRAINTS AND STATICAL DETERMINACY 251 5 EXAMPLE 5.19 The bent rod in Fig. 5–32a is supported at A by a journal bearing, at D by a ball-and-socket joint, and at B by means of cable BC. Using only one equilibrium equation, obtain a direct solution for the tension in cable BC. The bearing at A is capable of exerting force components only in the z and y directions since it is properly aligned on the shaft. SOLUTION (VECTOR ANALYSIS) Free-Body Diagram. As shown in Fig. 5–32b, there are six unknowns. Equations of Equilibrium. The cable tension may be obtained directly by summing moments about an axis that passes through points D and A. Why? The direction of this axis is defined by the unit vector u, where Hence, the sum of the moments about this axis is zero provided Here r represents a position vector drawn from any point on the axis DA to any point on the line of action of force F (see Eq. 4–11). With reference to Fig. 5–32b, we can therefore write Ans. Since the moment arms from the axis to and W are easy to obtain, we can also determine this result using a scalar analysis. As shown in Fig. 5–32b, Ans. T B = 490.5 N ©MDA = 0; TB(1 m sin 45°) - 981 N(0.5 m sin 45°) = 0 TB T B = 490.5 N - 0.70711-TB + 490.52 + 0 + 0 = 0 1-0.7071i - 0.7071j2 # [1-TB + 490.52i] = 0 + 1-0.5j2 * 1-981k2D = 0 1-0.7071i - 0.7071j2 # C1-1j2 * 1TBk2 u # 1rB * TB + rE * W2 = 0 ©MDA = u # ©1r * F2 = 0 = -0.7071i - 0.7071j u = rDA rDA = - 1 22 i - 1 22 j TB 0.5 m 0.5 m x z y E B A D 100 kg C (a) 1 m TB x z y B A D Az Ay Dy Dz Dx rE rB W 981 N u (b) 45 0.5 m 0.5 m Fig. 5–32
- 269. 252 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 FUNDAMENTAL PROBLEMS F5–10. Determine the support reactions at the smooth journal bearings A, B, and C of the pipe assembly. F5–8. Determine the reactions at the roller support A, the ball-and-socket joint D, and the tension in cable BC for the plate. All problem solutions must include an FBD. F5–7. The uniform plate has a weight of 500 lb. Determine the tension in each of the supporting cables. F5–11. Determine the force developed in cords BD, CE, and CF and the reactions of the ball-and-socket joint A on the block. z A B C y x 200 lb 3 ft 2 ft 2 ft F5–7 z x y 0.6 m 0.6 m 0.6 m 450 N 0.4 m A B C F5–10 F5–9. The rod is supported by smooth journal bearings at A, B and C and is subjected to the two forces. Determine the reactions at these supports. F5–12. Determine the components of reaction that the thrust bearing A and cable BC exert on the bar. x y D B C A z 0.4 m 0.5 m 600 N 900 N 0.3 m 0.4 m 0.1 m 0.2 m F5–8 z x y A B D C 600N 400N 0.6 m 0.6 m 0.6 m 0.4 m F5–9 x 3 m 9 kN 6 kN 1.5 m 4 m C A B E y z D F F5–11 F 80 lb x y z B D C A 1.5 ft 1.5 ft 6 ft F5–12
- 270. 5.7 CONSTRAINTS AND STATICAL DETERMINACY 253 5 PROBLEMS •5–65. If and , determine the tension developed in cables AB, CD, and EF. Neglect the weight of the plate. 5–66. Determine the location x and y of the point of application of force P so that the tension developed in cables AB, CD, and EF is the same. Neglect the weight of the plate. y = 1 m P = 6 kN, x = 0.75 m *5–64. The pole for a power line is subjected to the two cable forces of 60 lb, each force lying in a plane parallel to the plane. If the tension in the guy wire AB is 80 lb, determine the x, y, z components of reaction at the fixed base of the pole, O. x-y All problem solutions must include an FBD. 5–63. The cart supports the uniform crate having a mass of 85 kg. Determine the vertical reactions on the three casters at A, B, and C. The caster at B is not shown. Neglect the mass of the cart. 5–67. Due to an unequal distribution of fuel in the wing tanks, the centers of gravity for the airplane fuselage A and wings B and C are located as shown. If these components have weights and determine the normal reactions of the wheels D, E, and F on the ground. W C = 6000 lb, W B = 8000 lb, W A = 45 000 lb, B A C 0.2 m 0.5 m 0.6 m 0.35 m 0.1 m 0.4 m 0.2 m 0.35 m Prob. 5–63 8 ft 20 ft A B D E F 8 ft 6 ft 6 ft 4 ft 3 ft z x y C Prob. 5–67 z 45 60 lb 60 lb 80 lb 1 ft 10 ft 4 ft 45 3 ft y B A O x Prob. 5–64 z F B D A y x x y E C P 2 m 2 m Probs. 5–65/66
- 271. 254 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 5–70. Determine the tension in cables BD and CD and the x, y, z components of reaction at the ball-and-socket joint at A. •5–69. The shaft is supported by three smooth journal bearings at A, B, and C. Determine the components of reaction at these bearings. *5–68. Determine the magnitude of force F that must be exerted on the handle at C to hold the 75-kg crate in the position shown.Also, determine the components of reaction at the thrust bearing A and smooth journal bearing B. 5–71. The rod assembly is used to support the 250-lb cylinder. Determine the components of reaction at the ball-and- socket joint A, the smooth journal bearing E, and the force developed along rod CD. The connections at C and D are ball-and-socket joints. x y z D A C E F 1 ft 1 ft 1 ft 1.5 ft 1 ft Prob. 5–71 F 0.1 m 0.2 m 0.5 m 0.6 m 0.1 m z x y A B C Prob. 5–68 0.6 m x B C A z 0.9 m 0.6 m 0.9 m y 0.9 m 0.9 m 0.9 m 900 N 500 N 450 N 600 N Prob. 5–69 z y x C B A 3 m 300 N D 1 m 0.5 m 1.5 m Prob. 5–70
- 272. 5.7 CONSTRAINTS AND STATICAL DETERMINACY 255 5 5–74. If the load has a weight of 200 lb, determine the x, y, z components of reaction at the ball-and-socket joint A and the tension in each of the wires. •5–73. Determine the force components acting on the ball- and-socket at A, the reaction at the roller B and the tension on the cord CD needed for equilibrium of the quarter circular plate. *5–72. Determine the components of reaction acting at the smooth journal bearings A, B, and C. 5–75. If the cable can be subjected to a maximum tension of 300 lb, determine the maximum force F which may be applied to the plate. Compute the x, y, z components of reaction at the hinge A for this loading. 0.6 m 45 x y C z B A 0.4 m 0.8 m 0.4 m 450 N 300 N m Prob. 5–72 y x z C A D E F G B 2 ft 2 ft 2 ft 2 ft 3 ft 2 ft 4 ft Prob. 5–74 z x 350 N 1 m 2 m 60 3 m 200 N 200 N y B A C D Prob. 5–73 9 ft F 3 ft z x y A B 2 ft 3 ft 1 ft C Prob. 5–75
- 273. 256 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 5–79. The boom is supported by a ball-and-socket joint at A and a guy wire at B. If the 5-kN loads lie in a plane which is parallel to the x–y plane, determine the x, y, z components of reaction at A and the tension in the cable at B. •5–77. The plate has a weight of W with center of gravity at G. Determine the distance d along line GH where the vertical force P = 0.75W will cause the tension in wire CD to become zero. 5–78. The plate has a weight of W with center of gravity at G. Determine the tension developed in wires AB, CD, and EF if the force P = 0.75W is applied at d = L/2. *5–76. The member is supported by a pin at A and a cable BC. If the load at D is 300 lb, determine the x, y, z components of reaction at the pin A and the tension in cable B C. *5–80. The circular door has a weight of 55 lb and a center of gravity at G. Determine the x, y, z components of reaction at the hinge A and the force acting along strut CB needed to hold the door in equilibrium. Set . •5–81. The circular door has a weight of 55 lb and a center of gravity at G. Determine the x, y, z components of reaction at the hinge A and the force acting along strut CB needed to hold the door in equilibrium. Set . u = 90° u = 45° C 1 ft z A B D x 6 ft 2 ft 2 ft 2 ft 2 ft y Prob. 5–76 z F B D A H y x G d E C P L –– 2 L –– 2 L –– 2 L –– 2 Probs. 5–77/78 z 5 kN 5 kN y x 3 m 2 m 1.5 m 30 30 B A Prob. 5–79 C z x y B G A 3 ft 3 ft u Probs. 5–80/81
- 274. 5.7 CONSTRAINTS AND STATICAL DETERMINACY 257 5 •5–85. The circular plate has a weight W and center of gravity at its center. If it is supported by three vertical cords tied to its edge, determine the largest distance d from the center to where any vertical force P can be applied so as not to cause the force in any one of the cables to become zero. 5–86. Solve Prob. 5–85 if the plate’s weight W is neglected. 5–82. Member AB is supported at B by a cable and at A by a smooth fixed square rod which fits loosely through the square hole of the collar. If , determine the x, y, z components of reaction at A and the tension in the cable. 5–83. Member AB is supported at B by a cable and at A by a smooth fixed square rod which fits loosely through the square hole of the collar. Determine the tension in cable BC if the force . F = 5-45k6 lb F = 520i - 40j - 75k6 lb 8 ft C z 6 ft 12 ft 4 ft F B x A y Probs. 5–82/83 a/2 a/2 a Prob. 5–87 x z C G B A y 3 ft 1.5ft 10 ft 4 ft 2 ft 2.5 ft 2.5 ft 1 ft 30 Prob. 5–84 A d 120 120 120 C r P B Probs. 5–85/86 *5–84. Determine the largest weight of the oil drum that the floor crane can support without overturning.Also, what are the vertical reactions at the smooth wheels A, B, and C for this case. The floor crane has a weight of 300 lb, with its center of gravity located at G. 5–87. A uniform square table having a weight W and sides a is supported by three vertical legs. Determine the smallest vertical force P that can be applied to its top that will cause it to tip over.
- 275. 258 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 CHAPTER REVIEW roller u F u Equilibrium A body in equilibrium does not rotate but can translate with constant velocity, or it does not move at all. ©M = 0 ©F = 0 F3 y x z F4 F1 F2 O Two Dimensions Before analyzing the equilibrium of a body,it is first necessary to draw its free-body diagram. This is an outlined shape of the body, which shows all the forces and couple moments that act on it. Couple moments can be placed anywhere on a free-body diagram since they are free vectors. Forces can act at any point along their line of action since they are sliding vectors. Angles used to resolve forces, and dimensions used to take moments of the forces, should also be shown on the free-body diagram. Some common types of supports and their reactions are shown below in two dimensions. Remember that a support will exert a force on the body in a particular direction if it prevents translation of the body in that direction, and it will exert a couple moment on the body if it prevents rotation. The three scalar equations of equilibrium can be applied when solving problems in two dimensions, since the geometry is easy to visualize. A B C 500 N m 30 Ax T Ay 500 N m 30 y x 1 m 1 m 2 m 2 m smooth pin or hinge u Fy Fx fixed support Fy Fx M ©MO = 0 ©Fy = 0 ©Fx = 0
- 276. CHAPTER REVIEW 259 5 For the most direct solution, try to sum forces along an axis that will eliminate as many unknown forces as possible. Sum moments about a point A that passes through the line of action of as many unknown forces as possible. By = P 1d1 - P 2d2 dB P2d2 + BydB - P1d1 = 0 ©MA = 0; Ax - P2 = 0 Ax = P2 ©Fx = 0; Three Dimensions Some common types of supports and their reactions are shown here in three dimensions. In three dimensions, it is often advantageous to use a Cartesian vector analysis when applying the equations of equilibrium. To do this, first express each known and unknown force and couple moment shown on the free-body diagram as a Cartesian vector. Then set the force summation equal to zero. Take moments about a point O that lies on the line of action of as many unknown force components as possible. From point O direct position vectors to each force, and then use the cross product to determine the moment of each force. The six scalar equations of equilibrium are established by setting the respective i, j, and k components of these force and moment summations equal to zero. ©Mz = 0 ©My = 0 ©Mx = 0 ©Fz = 0 ©Fy = 0 ©Fx = 0 ©MO = 0 ©F = 0 Determinacy and Stability If a body is supported by a minimum number of constraints to ensure equilibrium, then it is statically determinate. If it has more constraints than required, then it is statically indeterminate. To properly constrain the body, the reactions must not all be parallel to one another or concurrent. By d1 P1 P2 d2 Ay Ax A dB roller F ball and socket Fz Fy Fx fixed support Fz Mz Fx My Mx Fy 500 N Statically indeterminate, five reactions, three equilibrium equations 2 kN m 600 N 100 N Proper constraint, statically determinate 200 N 45
- 277. 260 CHAPTER 5 EQUILIBRIUM OF A RIGID BODY 5 B C A 4.5 m 4 m 100 N 3 m 200 N/m 30 Prob. 5–88 A B 2 m 2 m 2 m 45 2 m F F F Probs. 5–89/90 0.4 m 60 0.8 m 10 kN 0.6 m 0.6 m 6 kN A B Prob. 5–91 REVIEW PROBLEMS 5–91. Determine the normal reaction at the roller A and horizontal and vertical components at pin B for equilibrium of the member. •5–89. Determine the horizontal and vertical components of reaction at the pin A and the reaction at the roller B required to support the truss. Set . 5–90. If the roller at B can sustain a maximum load of 3 kN, determine the largest magnitude of each of the three forces F that can be supported by the truss. F = 600 N *5–88. Determine the horizontal and vertical components of reaction at the pin A and the force in the cable BC. Neglect the thickness of the members. *5–92. The shaft assembly is supported by two smooth journal bearings A and B and a short link DC. If a couple moment is applied to the shaft as shown, determine the components of force reaction at the journal bearings and the force in the link. The link lies in a plane parallel to the y–z plane and the bearings are properly aligned on the shaft. 250 mm 300 mm 400 mm 250 N m y A x 20 120 mm 30 D B z C Prob. 5–92
- 278. REVIEW PROBLEMS 261 5 5–95. A vertical force of 80 lb acts on the crankshaft. Determine the horizontal equilibrium force P that must be applied to the handle and the x, y, z components of force at the smooth journal bearing A and the thrust bearing B.The bearings are properly aligned and exert only force reactions on the shaft. 5–94. A skeletal diagram of the lower leg is shown in the lower figure. Here it can be noted that this portion of the leg is lifted by the quadriceps muscle attached to the hip at A and to the patella bone at B. This bone slides freely over cartilage at the knee joint. The quadriceps is further extended and attached to the tibia at C. Using the mechanical system shown in the upper figure to model the lower leg, determine the tension in the quadriceps at C and the magnitude of the resultant force at the femur (pin), D, in order to hold the lower leg in the position shown. The lower leg has a mass of 3.2 kg and a mass center at ; the foot has a mass of 1.6 kg and a mass center at . G2 G1 •5–93. Determine the reactions at the supports A and B of the frame. *5–96. The symmetrical shelf is subjected to a uniform load of 4 kPa. Support is provided by a bolt (or pin) located at each end A and and by the symmetrical brace arms, which bear against the smooth wall on both sides at B and . Determine the force resisted by each bolt at the wall and the normal force at B for equilibrium. B¿ A¿ A P B z x y 80 lb 14 in. 10 in. 14 in. 6 in. 4 in. 8 in. Prob. 5–95 A B 8 ft 6 ft 8 ft 6 ft 0.5 kip 2 kip 10 kip 7 kip 5 kip 6 ft Prob. 5–93 A A B C D D C B 350 mm 300 mm 75 mm 75 25 mm G1 G2 Prob. 5–94 0.2 m 0.15 m 4 kPa 1.5 m A A¿ B B¿ Prob. 5–96
- 279. The forces within the members of each truss bridge must be determined if the members are to be properly designed.
- 280. Structural Analysis CHAPTER OBJECTIVES • To show how to determine the forces in the members of a truss using the method of joints and the method of sections. • To analyze the forces acting on the members of frames and machines composed of pin-connected members. 6.1 Simple Trusses A truss is a structure composed of slender members joined together at their end points.The members commonly used in construction consist of wooden struts or metal bars. In particular, planar trusses lie in a single plane and are often used to support roofs and bridges.The truss shown in Fig. 6–1a is an example of a typical roof-supporting truss. In this figure, the roof load is transmitted to the truss at the joints by means of a series of purlins. Since this loading acts in the same plane as the truss, Fig. 6–1b, the analysis of the forces developed in the truss members will be two-dimensional. 6 (a) A Purlin Fig. 6–1 (b) Roof truss
- 281. 264 CHAPTER 6 STRUCTURAL ANALYSIS 6 In the case of a bridge, such as shown in Fig. 6–2a, the load on the deck is first transmitted to stringers, then to floor beams, and finally to the joints of the two supporting side trusses. Like the roof truss, the bridge truss loading is also coplanar, Fig. 6–2b. When bridge or roof trusses extend over large distances, a rocker or roller is commonly used for supporting one end, for example, joint A in Figs. 6–1a and 6–2a.This type of support allows freedom for expansion or contraction of the members due to a change in temperature or application of loads. Assumptions for Design. To design both the members and the connections of a truss, it is necessary first to determine the force developed in each member when the truss is subjected to a given loading.To do this we will make two important assumptions: • All loadings are applied at the joints. In most situations, such as for bridge and roof trusses, this assumption is true. Frequently the weight of the members is neglected because the force supported by each member is usually much larger than its weight. However, if the weight is to be included in the analysis, it is generally satisfactory to apply it as a vertical force, with half of its magnitude applied at each end of the member. • The members are joined together by smooth pins. The joint connections are usually formed by bolting or welding the ends of the members to a common plate, called a gusset plate, as shown in Fig. 6–3a, or by simply passing a large bolt or pin through each of the members, Fig. 6–3b.We can assume these connections act as pins provided the center lines of the joining members are concurrent, as in Fig. 6–3. (a) Floor beam Stringer Deck A Fig. 6–2 (b) Bridge truss (a) Fig. 6–3 (b)
- 282. 6.1 SIMPLE TRUSSES 265 6 Because of these two assumptions, each truss member will act as a two- force member, and therefore the force acting at each end of the member will be directed along the axis of the member. If the force tends to elongate the member, it is a tensile force (T), Fig. 6–4a; whereas if it tends to shorten the member, it is a compressive force (C), Fig. 6–4b. In the actual design of a truss it is important to state whether the nature of the force is tensile or compressive. Often, compression members must be made thicker than tension members because of the buckling or column effect that occurs when a member is in compression. Simple Truss. If three members are pin connected at their ends they form a triangular truss that will be rigid, Fig. 6–5. Attaching two more members and connecting these members to a new joint D forms a larger truss, Fig. 6–6. This procedure can be repeated as many times as desired to form an even larger truss. If a truss can be constructed by expanding the basic triangular truss in this way, it is called a simple truss. T C T C Compression Tension (b) (a) Fig. 6–4 A B C P Fig. 6–5 A C D B P Fig. 6–6 The use of metal gusset plates in the construction of these Warren trusses is clearly evident.
- 283. 266 CHAPTER 6 STRUCTURAL ANALYSIS 6 6.2 The Method of Joints In order to analyze or design a truss, it is necessary to determine the force in each of its members. One way to do this is to use the method of joints. This method is based on the fact that if the entire truss is in equilibrium, then each of its joints is also in equilibrium. Therefore, if the free-body diagram of each joint is drawn, the force equilibrium equations can then be used to obtain the member forces acting on each joint. Since the members of a plane truss are straight two-force members lying in a single plane, each joint is subjected to a force system that is coplanar and concurrent.As a result,only and need to be satisfied for equilibrium. For example, consider the pin at joint B of the truss in Fig. 6–7a. Three forces act on the pin, namely, the 500-N force and the forces exerted by members BA and BC. The free-body diagram of the pin is shown in Fig. 6–7b. Here, is “pulling” on the pin, which means that member BA is in tension; whereas is “pushing” on the pin, and consequently member BC is in compression. These effects are clearly demonstrated by isolating the joint with small segments of the member connected to the pin, Fig. 6–7c. The pushing or pulling on these small segments indicates the effect of the member being either in compression or tension. When using the method of joints, always start at a joint having at least one known force and at most two unknown forces, as in Fig. 6–7b. In this way, application of and yields two algebraic equations which can be solved for the two unknowns. When applying these equations, the correct sense of an unknown member force can be determined using one of two possible methods. ©Fy = 0 ©Fx = 0 FBC FBA ©Fy = 0 ©Fx = 0 B 2 m 500 N A C 45 2 m (a) Fig. 6–7 B 45 500 N FBC (compression) FBA(tension) (b) FBA(tension) B 45 500 N FBC (compression) (c)
- 284. 6.2 THE METHOD OF JOINTS 267 6 • The correct sense of direction of an unknown member force can, in many cases, be determined “by inspection.” For example, in Fig. 6–7b must push on the pin (compression) since its horizontal component, must balance the 500-N force Likewise, is a tensile force since it balances the vertical component, In more complicated cases, the sense of an unknown member force can be assumed; then, after applying the equilibrium equations, the assumed sense can be verified from the numerical results.A positive answer indicates that the sense is correct, whereas a negative answer indicates that the sense shown on the free-body diagram must be reversed. • Always assume the unknown member forces acting on the joint’s free-body diagram to be in tension; i.e., the forces “pull” on the pin. If this is done, then numerical solution of the equilibrium equations will yield positive scalars for members in tension and negative scalars for members in compression. Once an unknown member force is found, use its correct magnitude and sense (T or C) on subsequent joint free-body diagrams. FBC cos 45° 1©Fy = 02. FBA 1©Fx = 02. F BC sin 45°, FBC Procedure for Analysis The following procedure provides a means for analyzing a truss using the method of joints. • Draw the free-body diagram of a joint having at least one known force and at most two unknown forces. (If this joint is at one of the supports, then it may be necessary first to calculate the external reactions at the support.) • Use one of the two methods described above for establishing the sense of an unknown force. • Orient the x and y axes such that the forces on the free-body diagram can be easily resolved into their x and y components and then apply the two force equilibrium equations and Solve for the two unknown member forces and verify their correct sense. • Using the calculated results, continue to analyze each of the other joints. Remember that a member in compression “pushes” on the joint and a member in tension “pulls” on the joint.Also, be sure to choose a joint having at most two unknowns and at least one known force. ©Fy = 0. ©Fx = 0 The forces in the members of this simple roof truss can be determined using the method of joints.
- 285. 268 CHAPTER 6 STRUCTURAL ANALYSIS 6 B 2 m 2 m 500 N A C (a) 45 Fig. 6–8 (b) B 45 500 N FBC FBA (c) 45 707.1 N FCA C Cy (d) A FBA 500 N FCA 500 N Ay Ax (e) B 45 500 N A 45 500 N 500 N 500 N 500 N 500 N C 707.1 N 707.1 N 500 N 500 N Tension C o m p r e s s i o n Tension Determine the force in each member of the truss shown in Fig. 6–8a and indicate whether the members are in tension or compression. SOLUTION Since we should have no more than two unknown forces at the joint and at least one known force acting there, we will begin our analysis at joint B. Joint B. The free-body diagram of the joint at B is shown in Fig.6–8b. Applying the equations of equilibrium, we have Ans. Ans. Since the force in member BC has been calculated, we can proceed to analyze joint C to determine the force in member CA and the support reaction at the rocker. Joint C. From the free-body diagram of joint C, Fig. 6–8c, we have Ans. Ans. Joint A. Although it is not necessary, we can determine the components of the support reactions at joint A using the results of and . From the free-body diagram, Fig. 6–8d, we have NOTE: The results of the analysis are summarized in Fig. 6–8e. Note that the free-body diagram of each joint (or pin) shows the effects of all the connected members and external forces applied to the joint, whereas the free-body diagram of each member shows only the effects of the end joints on the member. 500 N - Ay = 0 Ay = 500 N + c©Fy = 0; 500 N - Ax = 0 Ax = 500 N : + ©Fx = 0; FBA FCA Cy - 707.1 sin 45° N = 0 Cy = 500 N + c ©Fy = 0; -FCA + 707.1 cos 45° N = 0 FCA = 500 N 1T2 : + ©Fx = 0; FBA = 500 N 1T2 FBC cos 45° - FBA = 0 + c©Fy = 0; FBC = 707.1 N 1C2 500 N - FBC sin 45° = 0 : + ©Fx = 0; EXAMPLE 6.1
- 286. 6.2 THE METHOD OF JOINTS 269 6 EXAMPLE 6.2 Determine the force in each member of the truss in Fig. 6–9a and indicate if the members are in tension or compression. SOLUTION Since joint has one known and only two unknown forces acting on it, it is possible to start at this joint, then analyze joint , and finally joint .This way the support reactions will not have to be determined prior to starting the analysis. Joint C. By inspection of the force equilibrium, Fig. 6–9b, it can be seen that both members and must be in compression. Ans. Ans. Joint D. Using the result , the force in members and AD can be found by analyzing the equilibrium of joint .We will assume and are both tensile forces, Fig. 6–9c. The , coordinate system will be established so that the axis is directed along .This way, we will eliminate the need to solve two equations simultaneously. Now can be obtained directly by applying . ; Ans. The negative sign indicates that is a compressive force. Using this result, ; Ans. Joint A. The force in member AB can be found by analyzing the equilibrium of joint A, Fig. 6–9d.We have Ans. F AB = 546.41 N (C) = 546 N (C) (772.74 N) cos 45° - F AB = 0 : + ©Fx = 0; FBD = 1092.82 N = 1.09 kN (T) F BD + (-772.74 cos 15°) - 400 cos 30° = 0 + R©F x¿ = 0 FAD F AD = -772.74 N = 773 N (C) - F AD sin 15° - 400 sin 30° = 0 +Q©F y¿ = 0 ©F y¿ = 0 FAD FBD x¿ y¿ x¿ FBD FAD D BD FCD = 400 N (C) F CD = 400 N (C) F CD - (565.69 N) cos 45° = 0 : + ©Fx = 0; F BC = 565.69 N = 566 N (C) F BC sin 45° - 400 N = 0 + c ©F y = 0; CD BC A D C 2 m 2 m (a) 400 N 2 m D B A C 30 45 45 2 m 2 m Fig. 6–9 (b) 400 N C y x 45 FCD FBC (c) D y¿ x¿ 15 30 FCD 400 N FAD FBD (d) A y x 45 FAB Ay FAD 772.74 N
- 287. 270 CHAPTER 6 STRUCTURAL ANALYSIS 6 Determine the force in each member of the truss shown in Fig. 6–10a. Indicate whether the members are in tension or compression. EXAMPLE 6.3 4 m (a) 3 m 400 N B C D A 3 m 600 N Fig. 6–10 4 m (b) 400 N C A 6 m 600 N 3 m Ay Cy Cx SOLUTION Support Reactions. No joint can be analyzed until the support reactions are determined, because each joint has more than three unknown forces acting on it.A free-body diagram of the entire truss is given in Fig. 6–10b.Applying the equations of equilibrium, we have a The analysis can now start at either joint A or C. The choice is arbitrary since there are one known and two unknown member forces acting on the pin at each of these joints. Joint A. (Fig. 6–10c). As shown on the free-body diagram, is assumed to be compressive and is tensile.Applying the equations of equilibrium, we have Ans. Ans. FAD - 3 51750 N2 = 0 FAD = 450 N 1T2 : + ©Fx = 0; 600 N - 4 5 FAB = 0 FAB = 750 N 1C2 + c ©Fy = 0; FAD FAB Cy = 200 N 600 N - 400 N - Cy = 0 + c©F y = 0; Ay = 600 N -Ay16 m2 + 400 N13 m2 + 600 N14 m2 = 0 +©MC = 0; Cx = 600 N 600 N - Cx = 0 : + ©F x = 0; 3 4 5 x y FAB FAD 600 N (c) A
- 288. 6.2 THE METHOD OF JOINTS 271 6 Joint D. (Fig. 6–10d). Using the result for and summing forces in the horizontal direction, Fig. 6–10d, we have The negative sign indicates that acts in the opposite sense to that shown in Fig. 6–10d.* Hence, Ans. To determine we can either correct the sense of on the free- body diagram, and then apply or apply this equation and retain the negative sign for i.e., Ans. Joint C. (Fig. 6–10e). Ans. NOTE: The analysis is summarized in Fig. 6–10f, which shows the free-body diagram for each joint and member. 200 N - 200 N K 0 1check2 + c ©Fy = 0; FCB = 600 N 1C2 FCB - 600 N = 0 : + ©Fx = 0; FDC = 200 N 1C2 -FDC - 4 51-250 N2 = 0 + c ©Fy = 0; FDB, ©Fy = 0, FDB FDC, FDB = 250 N 1T2 FDB -450 N + 3 5 FDB + 600 N = 0 FDB = -250 N : + ©Fx = 0; FAD 3 4 5 x y FDB 600 N (d) FDC D 450 N (f) 750 N 250 N 600 N 400 N Compression 600 N 200 N 600 N 200 N T e n s i o n Compression C o m p r e s s i o n 750 N 450 N 600 N A Tension 450 N 250 N 200 N 600 N D C B x y 200 N (e) C 600 N 200 N FCB * The proper sense could have been determined by inspection, prior to applying ©Fx = 0.
- 289. 272 CHAPTER 6 STRUCTURAL ANALYSIS 6 6.3 Zero-Force Members Truss analysis using the method of joints is greatly simplified if we can first identify those members which support no loading. These zero-force members are used to increase the stability of the truss during construction and to provide added support if the loading is changed. The zero-force members of a truss can generally be found by inspection of each of the joints. For example, consider the truss shown in Fig. 6–11a. If a free-body diagram of the pin at joint A is drawn, Fig. 6–11b, it is seen that members AB and AF are zero-force members. (We could not have come to this conclusion if we had considered the free-body diagrams of joints F or B simply because there are five unknowns at each of these joints.) In a similar manner, consider the free- body diagram of joint D, Fig. 6–11c. Here again it is seen that DC and DE are zero-force members. From these observations, we can conclude that if only two members form a truss joint and no external load or support reaction is applied to the joint, the two members must be zero- force members. The load on the truss in Fig. 6–11a is therefore supported by only five members as shown in Fig. 6–11d. (a) D C E F B A P u Fig. 6–11 FAB y x FAF A (b) Fx 0; FAB 0 Fy 0; FAF 0 FDC y x FDE D (c) Fy 0; FDC sin u = 0; FDC 0 since sin u 0 Fx 0; FDE 0 0; FDE 0 u (d) B C E F P
- 290. 6.3 ZERO-FORCE MEMBERS 273 6 Now consider the truss shown in Fig. 6–12a. The free-body diagram of the pin at joint D is shown in Fig. 6–12b. By orienting the y axis along members DC and DE and the x axis along member DA, it is seen that DA is a zero-force member. Note that this is also the case for member CA, Fig. 6–12c. In general then, if three members form a truss joint for which two of the members are collinear, the third member is a zero-force member provided no external force or support reaction is applied to the joint. The truss shown in Fig. 6–12d is therefore suitable for supporting the load . P (a) E A D C B P u Fig. 6–12 D FDE (b) Fx 0; Fy 0; FDA FDC y x FDA 0 FDC FDE FCD C FCB FCA y x u (c) Fx 0; FCA sin u = 0; FCA 0 since sin u 0; Fy 0; FCB FCD (d) E P B A
- 291. 274 CHAPTER 6 STRUCTURAL ANALYSIS 6 C A E 5 kN 2 kN D F G H B (a) Fig. 6–13 Using the method of joints, determine all the zero-force members of the Fink roof truss shown in Fig. 6–13a. Assume all joints are pin connected. EXAMPLE 6.4 SOLUTION Look for joint geometries that have three members for which two are collinear.We have Joint G. (Fig. 6–13b). Ans. Realize that we could not conclude that GC is a zero-force member by considering joint C, where there are five unknowns. The fact that GC is a zero-force member means that the 5-kN load at C must be supported by members CB, CH, CF, and CD. Joint D. (Fig. 6–13c). Ans. Joint F. (Fig. 6–13d). Ans. NOTE: If joint B is analyzed, Fig. 6–13e, Also, must satisfy Fig. 6–13f, and therefore HC is not a zero-force member. ©Fy = 0, F HC 2 kN - FBH = 0 FBH = 2 kN 1C2 +R©Fx = 0; F FC = 0 F FC cos u = 0 Since u Z 90°, + c©Fy = 0; FDF = 0 +b©Fx = 0; FGC = 0 + c©Fy = 0; (b) y x G FGC FGF FGH (c) D FDC FDF FDE y x (d) y x F FFE FFG 0 FFC u (e) B FBH FBC FBA 2 kN x y (f) y x H FHG FHA 2 kN FHC
- 292. 6.3 ZERO-FORCE MEMBERS 275 6 FUNDAMENTAL PROBLEMS F6–4. Determine the greatest load P that can be applied to the truss so that none of the members are subjected to a force exceeding either 2 kN in tension or 1.5 kN in compression. F6–2. Determine the force in each member of the truss. State if the members are in tension or compression. F6–1. Determine the force in each member of the truss. State if the members are in tension or compression. F6–5. Identify the zero-force members in the truss. F6–3. Determine the force in members AE and DC. State if the members are in tension or compression. F6–6. Determine the force in each member of the truss. State if the members are in tension or compression. 4 ft 4 ft 4 ft A B C D 450 lb F6–1 D A C B 2 ft 2 ft 300 lb 3 ft F6–2 A C B F E D 4 ft 4 ft 3 ft 800 lb F6–3 A B P C 3 m 60 60 F6–4 A B C D E 1.5 m 2 m 2 m 3 kN F6–5 B D C E 600 lb 450 lb 3 ft 3 ft 30 A F6–6
- 293. 276 CHAPTER 6 STRUCTURAL ANALYSIS 6 PROBLEMS *6–4. Determine the force in each member of the truss and state if the members are in tension or compression. Assume each joint as a pin. Set P = 4 kN. •6–5. Assume that each member of the truss is made of steel having a mass per length of 4 kg/m. Set , determine the force in each member, and indicate if the members are in tension or compression.Neglect the weight of the gusset plates and assume each joint is a pin. Solve the problem by assuming the weight of each member can be represented as a vertical force, half of which is applied at the end of each member. P = 0 6–2. The truss, used to support a balcony, is subjected to the loading shown. Approximate each joint as a pin and determine the force in each member. State whether the members are in tension or compression. Set 6–3. The truss, used to support a balcony, is subjected to the loading shown. Approximate each joint as a pin and determine the force in each member. State whether the members are in tension or compression. Set P2 = 0. P1 = 800 lb, P2 = 400 lb. P1 = 600 lb, •6–1. Determine the force in each member of the truss, and state if the members are in tension or compression. 6–6. Determine the force in each member of the truss and state if the members are in tension or compression. Set and . 6–7. Determine the force in each member of the truss and state if the members are in tension or compression. Set . P1 = P2 = 4 kN P2 = 1.5 kN P1 = 2 kN 600 N 900 N 2 m 2 m 2 m A C E D B Prob. 6–1 45 4 ft 4 ft 45 D E C B P2 A 4 ft P1 Probs. 6–2/3Prob. 6–1 A E D C B P P 2P 4 m 4 m 4 m Probs. 6–4/5 A E D 30 30 B C 3 m 3 m P2 P1 Probs. 6–6/7
- 294. 6.3 ZERO-FORCE MEMBERS 277 6 *6–12. Determine the force in each member of the truss and state if the members are in tension or compression. Set , . •6–13. Determine the largest load that can be applied to the truss so that the force in any member does not exceed 500 lb (T) or 350 lb (C).Take . P1 = 0 P2 P2 = 100 lb P1 = 240 lb 6–10. Determine the force in each member of the truss and state if the members are in tension or compression. Set , . 6–11. Determine the force in each member of the truss and state if the members are in tension or compression. Set , . P2 = 400 lb P1 = 600 lb P2 = 0 P1 = 800 lb *6–8. Determine the force in each member of the truss, and state if the members are in tension or compression. Set . •6–9. Remove the 500-lb force and then determine the greatest force P that can be applied to the truss so that none of the members are subjected to a force exceeding either 800 lb in tension or in compression. 600 lb P = 800 lb 6–14. Determine the force in each member of the truss, and state if the members are in tension or compression. Set . 6–15. Remove the 1200-lb forces and determine the greatest force P that can be applied to the truss so that none of the members are subjected to a force exceeding either 2000 lb in tension or 1500 lb in compression. P = 2500 lb 3 ft 3 ft 3 ft P 3 ft 500 lb A C B D F E Probs. 6–8/9 6 ft A G B C F D E P1 P2 4 ft 4 ft 4 ft 4 ft Probs. 6–10/11 B C D A 12 ft 5 ft P1 P2 Probs. 6–12/13 4 ft 4 ft 1200 lb 1200 lb P 4 ft 4 ft 4 ft A B F E D C G 30 30 Probs. 6–14/15
- 295. 278 CHAPTER 6 STRUCTURAL ANALYSIS 6 *6–20. Determine the force in each member of the truss and state if the members are in tension or compression.The load has a mass of 40 kg. •6–21. Determine the largest mass m of the suspended block so that the force in any member does not exceed 30 kN (T) or 25 kN (C). 6–18. Determine the force in each member of the truss, and state if the members are in tension or compression. 6–19. The truss is fabricated using members having a weight of . Remove the external forces from the truss, and determine the force in each member due to the weight of the members. State whether the members are in tension or compression. Assume that the total force acting on a joint is the sum of half of the weight of every member connected to the joint. 10 lbft *6–16. Determine the force in each member of the truss, and state if the members are in tension or compression. Set . •6–17. Determine the greatest force P that can be applied to the truss so that none of the members are subjected to a force exceeding either in tension or in compression. 2 kN 2.5 kN P = 5 kN 6–22. Determine the force in each member of the truss, and state if the members are in tension or compression. 6–23. The truss is fabricated using uniform members having a mass of . Remove the external forces from the truss, and determine the force in each member due to the weight of the truss. State whether the members are in tension or compression. Assume that the total force acting on a joint is the sum of half of the weight of every member connected to the joint. 5 kgm A C B D E P 1.5 m 1.5 m 2 m 2 m 1.5 m Probs. 6–16/17 3 ft 4 ft 900 lb 600 lb 4 ft 4 ft A B C D E F Probs. 6–18/19 G A B F C E D 0.1 m 6 m 2.5 m 3.5 m Probs. 6–20/21 A E D B C 2 m 400 N 45 45 45 45 2 m 600 N Probs. 6–22/23
- 296. B C A 4 ft 3 ft 300 lb u Prob. 6–30 6–30. The two-member truss is subjected to the force of 300 lb.Determine the range of for application of the load so that the force in either member does not exceed 400 lb (T) or 200 lb (C). u A B C D F E P d d d d/2 d/2 d Probs. 6–28/29 6.3 ZERO-FORCE MEMBERS 279 6 6–27. Determine the force in each member of the double scissors truss in terms of the load P and state if the members are in tension or compression. 6–26. A sign is subjected to a wind loading that exerts horizontal forces of 300 lb on joints B and C of one of the side supporting trusses. Determine the force in each member of the truss and state if the members are in tension or compression. *6–24. Determine the force in each member of the truss, and state if the members are in tension or compression. Set . •6–25. Determine the greatest force P that can be applied to the truss so that none of the members are subjected to a force exceeding either in tension or in compression. 1 kN 1.5 kN P = 4 kN *6–28. Determine the force in each member of the truss in terms of the load P, and indicate whether the members are in tension or compression. •6–29. If the maximum force that any member can support is 4 kN in tension and 3 kN in compression, determine the maximum force P that can be applied at joint B. Take . d = 1 m P 3 m A C B E D F P 3 m 3 m 3 m Probs. 6–24/25 A C B D E 13 ft 13 ft 12 ft 5 ft 300 lb 300 lb 12 ft 45 Prob. 6–26 A D F E P P B C L/3 L/3 L/3 L/3 Prob. 6–27
- 297. B 2 m 1000 N (a) 2 m 2 m C D G F E A 2 m a a 280 CHAPTER 6 STRUCTURAL ANALYSIS 6 6.4 The Method of Sections When we need to find the force in only a few members of a truss, we can analyze the truss using the method of sections. It is based on the principle that if the truss is in equilibrium then any segment of the truss is also in equilibrium. For example, consider the two truss members shown on the left in Fig.6–14.If the forces within the members are to be determined,then an imaginary section, indicated by the blue line, can be used to cut each member into two parts and thereby “expose” each internal force as “external” to the free-body diagrams shown on the right. Clearly, it can be seen that equilibrium requires that the member in tension (T) be subjected to a“pull,”whereas the member in compression (C) is subjected to a“push.” The method of sections can also be used to“cut”or section the members of an entire truss. If the section passes through the truss and the free-body diagram of either of its two parts is drawn,we can then apply the equations of equilibrium to that part to determine the member forces at the “cut section.” Since only three independent equilibrium equations ( ) can be applied to the free-body diagram of any segment, then we should try to select a section that, in general, passes through not more than three members in which the forces are unknown. For example, consider the truss in Fig. 6–15a. If the forces in members BC, GC, and GF are to be determined, then section aa would be appropriate. The free-body diagrams of the two segments are shown in Figs. 6–15b and 6–15c. Note that the line of action of each member force is specified from the geometry of the truss,since the force in a member is along its axis.Also, the member forces acting on one part of the truss are equal but opposite to those acting on the other part—Newton’s third law. Members BC and GC are assumed to be in tension since they are subjected to a “pull,” whereas GF in compression since it is subjected to a “push.” The three unknown member forces and can be obtained by applying the three equilibrium equations to the free-body diagram in Fig. 6–15b. If, however, the free-body diagram in Fig. 6–15c is considered, the three support reactions and will have to be known, because only three equations of equilibrium are available. (This, of course, is done in the usual manner by considering a free-body diagram of the entire truss.) Ex Dy Dx, FGF FGC, FBC, ©MO = 0 ©Fy = 0, ©Fx = 0, C Compression C Internal compressive forces C C C C Fig. 6–14 Tension T T T Internal tensile forces T T T
- 298. Fig. 6–15 6.4 THE METHOD OF SECTIONS 281 6 When applying the equilibrium equations, we should carefully consider ways of writing the equations so as to yield a direct solution for each of the unknowns, rather than having to solve simultaneous equations. For example, using the truss segment in Fig. 6–15b and summing moments about C would yield a direct solution for since and create zero moment about C. Likewise, can be directly obtained by summing moments about G. Finally, can be found directly from a force summation in the vertical direction since and have no vertical components. This ability to determine directly the force in a particular truss member is one of the main advantages of using the method of sections.* As in the method of joints, there are two ways in which we can determine the correct sense of an unknown member force: • The correct sense of an unknown member force can in many cases be determined “by inspection.” For example, is a tensile force as represented in Fig. 6–15b since moment equilibrium about G requires that create a moment opposite to that of the 1000-N force. Also, is tensile since its vertical component must balance the 1000-N force which acts downward. In more complicated cases, the sense of an unknown member force may be assumed. If the solution yields a negative scalar, it indicates that the force’s sense is opposite to that shown on the free-body diagram. • Always assume that the unknown member forces at the cut section are tensile forces, i.e., “pulling” on the member. By doing this, the numerical solution of the equilibrium equations will yield positive scalars for members in tension and negative scalars for members in compression. FGC FBC FBC FBC FGF FGC FBC FGC FBC FGF 2 m 1000 N 2 m 2 m C FBC 45 FGC G (b) FGF (c) 2 m 2 m 45 C Dy Dx Ex FGC FBC FGF G The forces in selected members of this Pratt truss can readily be determined using the method of sections. *Notice that if the method of joints were used to determine, say, the force in member GC, it would be necessary to analyze joints A, B, and G in sequence.
- 299. 282 CHAPTER 6 STRUCTURAL ANALYSIS 6 Simple trusses are often used in the construction of large cranes in order to reduce the weight of the boom and tower. Procedure for Analysis The forces in the members of a truss may be determined by the method of sections using the following procedure. Free-Body Diagram. • Make a decision on how to “cut” or section the truss through the members where forces are to be determined. • Before isolating the appropriate section, it may first be necessary to determine the truss’s support reactions. If this is done then the three equilibrium equations will be available to solve for member forces at the section. • Draw the free-body diagram of that segment of the sectioned truss which has the least number of forces acting on it. • Use one of the two methods described above for establishing the sense of the unknown member forces. Equations of Equilibrium. • Moments should be summed about a point that lies at the intersection of the lines of action of two unknown forces, so that the third unknown force can be determined directly from the moment equation. • If two of the unknown forces are parallel, forces may be summed perpendicular to the direction of these unknowns to determine directly the third unknown force.
- 300. 6.4 THE METHOD OF SECTIONS 283 6 EXAMPLE 6.5 Determine the force in members GE, GC, and BC of the truss shown in Fig. 6–16a. Indicate whether the members are in tension or compression. SOLUTION Section aa in Fig. 6–16a has been chosen since it cuts through the three members whose forces are to be determined. In order to use the method of sections, however, it is first necessary to determine the external reactions at A or D.Why? A free-body diagram of the entire truss is shown in Fig. 6–16b. Applying the equations of equilibrium, we have a Free-Body Diagram. For the analysis the free-body diagram of the left portion of the sectioned truss will be used, since it involves the least number of forces, Fig. 6–16c. Equations of Equilibrium. Summing moments about point G eliminates and and yields a direct solution for . a Ans. In the same manner, by summing moments about point C we obtain a direct solution for a Ans. Since and have no vertical components, summing forces in the y direction directly yields i.e., Ans. NOTE: Here it is possible to tell, by inspection, the proper direction for each unknown member force. For example, requires to be compressive because it must balance the moment of the 300-N force about C. FGE ©MC = 0 FGC = 500 N 1T2 300 N - 3 5 FGC = 0 + c©Fy = 0; FGC, FGE FBC FGE = 800 N 1C2 -300 N18 m2 + FGE13 m2 = 0 +©MC = 0; FGE. FBC = 800 N 1T2 -300 N14 m2 - 400 N13 m2 + FBC13 m2 = 0 +©MG = 0; FBC FGC FGE Ay = 300 N Ay - 1200 N + 900 N = 0 + c ©Fy = 0; Dy = 900 N -1200 N18 m2 - 400 N13 m2 + Dy112 m2 = 0 +©MA = 0; Ax = 400 N 400 N - Ax = 0 : + ©Fx = 0; 3 m 4 m 400 N G 4 m E B C D A a a 1200 N (a) 4 m Fig. 6–16 3 m 8 m 400 N D A 1200 N (b) Ax Ay Dy 4 m 3 m 4 m 400 N A (c) FGE FGC FBC 3 4 5 G 300 N C 4 m
- 301. 284 CHAPTER 6 STRUCTURAL ANALYSIS 6 Determine the force in member CF of the truss shown in Fig. 6–17a. Indicate whether the member is in tension or compression. Assume each member is pin connected. EXAMPLE 6.6 SOLUTION Free-Body Diagram. Section aa in Fig. 6–17a will be used since this section will “expose” the internal force in member CF as “external” on the free-body diagram of either the right or left portion of the truss. It is first necessary, however, to determine the support reactions on either the left or right side. Verify the results shown on the free- body diagram in Fig. 6–17b. The free-body diagram of the right portion of the truss, which is the easiest to analyze, is shown in Fig. 6–17c. There are three unknowns, and Equations of Equilibrium. We will apply the moment equation about point O in order to eliminate the two unknowns and . The location of point O measured from E can be determined from proportional triangles, i.e., Or, stated in another manner, the slope of member GF has a drop of 2 m to a horizontal distance of 4 m. Since FD is 4 m, Fig. 6–17c, then from D to O the distance must be 8 m. An easy way to determine the moment of about point O is to use the principle of transmissibility and slide to point C, and then resolve into its two rectangular components.We have a Ans. FCF = 0.589 kN 1C2 -FCF sin 45°112 m2 + 13 kN218 m2 - 14.75 kN214 m2 = 0 +©MO = 0; FCF FCF FCF x = 4 m. 414 + x2 = 618 + x2, FCD FFG FCD. FCF, F FG, G H F E A B C D 3 kN 5 kN 4 m 2 m (a) a a 4 m 4 m 4 m 4 m Fig. 6–17 4 m 5 kN 3 kN (b) 8 m 3.25 kN 4.75 kN 4 m 4 m 4 m 4 m 2 m 3 kN (c) 4.75 kN D E F x 6 m 45 C FCF cos 45 FCF sin 45 FCF FFG FCD O G
- 302. 6.4 THE METHOD OF SECTIONS 285 6 EXAMPLE 6.7 Determine the force in member EB of the roof truss shown in Fig. 6–18a. Indicate whether the member is in tension or compression. SOLUTION Free-Body Diagrams. By the method of sections, any imaginary section that cuts through EB, Fig. 6–18a, will also have to cut through three other members for which the forces are unknown. For example, section aa cuts through ED, EB, FB, and AB. If a free-body diagram of the left side of this section is considered, Fig. 6–18b, it is possible to obtain by summing moments about B to eliminate the other three unknowns; however, cannot be determined from the remaining two equilibrium equations. One possible way of obtaining is first to determine from section aa, then use this result on section bb, Fig. 6–18a, which is shown in Fig. 6–18c. Here the force system is concurrent and our sectioned free-body diagram is the same as the free-body diagram for the joint at E. FED FEB FEB FED Equations of Equilibrium. In order to determine the moment of about point B, Fig. 6–18b, we will use the principle of transmissibility and slide the force to point C and then resolve it into its rectangular components as shown. Therefore, a Considering now the free-body diagram of section bb,Fig.6–18c,we have Ans. FEB = 2000 N 1T2 213000 sin 30° N2 - 1000 N - FEB = 0 + c©Fy = 0; FEF = 3000 N 1C2 FEF cos 30° - 3000 cos 30° N = 0 : + ©Fx = 0; FED = 3000 N 1C2 + FED sin 30°14 m2 = 0 1000 N14 m2 + 3000 N12 m2 - 4000 N14 m2 +©MB = 0; FED 1000 N 1000 N 1000 N 3000 N A B C D E F a a b b (a) 4000 N 2000 N 30 2 m 2 m 2 m 2 m Fig. 6–18 1000 N 1000 N 3000 N B C E 4000 N FED sin 30 30 2 m 2 m 4 m A FED cos 30 FAB FEB FED 30 (b) FFB 1000 N E 30 y x FEB FEF FED 3000 N (c) 30
- 303. 286 CHAPTER 6 STRUCTURAL ANALYSIS 6 FUNDAMENTAL PROBLEMS F6–10. Determine the force in members EF, CF, and BC of the truss. State if the members are in tension or compression. F6–8. Determine the force in members LK, KC, and CD of the Pratt truss. State if the members are in tension or compression. F6–9. Determine the force in members KJ, KD, and CD of the Pratt truss. State if the members are in tension or compression. F6–7. Determine the force in members BC, CF, and FE. State if the members are in tension or compression. F6–11. Determine the force in members GF, GD, and CD of the truss. State if the members are in tension or compression. F6–12. Determine the force in members DC, HI, and JI of the truss. State if the members are in tension or compression. A D C B G F E 4 ft 4 ft 4 ft 4 ft 600 lb 600 lb 800 lb F6–7 B C D A E F G H I J K L 2 m 3 m 2 m 20 kN 30 kN 40 kN 2 m 2 m 2 m 2 m f F6–8/9 A B C D E F G 30 30 6 ft 6 ft 6 ft 300 lb 300 lb f F6–10 A B C D E F G H 2 m 2 m 1 m 2 m 2 m 2 m 10 kN 25 kN 15 kN f F6–11 B t s s t C A D I K H E F G 1600 lb 1200 lb 9 ft 6 ft 6 ft 6 ft 12 ft 9 ft 6 ft 6 ft 6 ft J F6–12
- 304. A E B C D 6 kN 8 kN G F 3 m 3 m 3 m 3 m 3 m 3 m 3 m Probs. 6–36/37 6.4 THE METHOD OF SECTIONS 287 6 PROBLEMS 6–34. Determine the force in members JK, CJ, and CD of the truss, and state if the members are in tension or compression. 6–35. Determine the force in members HI, FI, and EF of the truss, and state if the members are in tension or compression. *6–32. The Howe bridge truss is subjected to the loading shown. Determine the force in members HD, CD, and GD, and state if the members are in tension or compression. •6–33. The Howe bridge truss is subjected to the loading shown. Determine the force in members HI, HB, and BC, and state if the members are in tension or compression. 6–31. The internal drag truss for the wing of a light airplane is subjected to the forces shown. Determine the force in members BC, BH, and HC, and state if the members are in tension or compression. *6–36. Determine the force in members BC, CG, and GF of the Warren truss. Indicate if the members are in tension or compression. •6–37. Determine the force in members CD, CF, and FG of the Warren truss. Indicate if the members are in tension or compression. 2 ft A B C D J I H G E F 2 ft 2 ft 2 ft 1.5 ft 80 lb 80 lb 60 lb 40 lb Prob. 6–31 A E B C D I J 30 kN 20 kN 20 kN 40 kN H G F 4 m 16 m, 4@4m Probs. 6–32/33 A B C D F E G H I J L K 6 kN 8 kN 5 kN 4 kN 3 m 2 m 2 m 2 m 2 m 2 m 2 m Probs. 6–34/35
- 305. 288 CHAPTER 6 STRUCTURAL ANALYSIS 6 6–42. Determine the force in members IC and CG of the truss and state if these members are in tension or compression.Also, indicate all zero-force members. 6–43. Determine the force in members JE and GF of the truss and state if these members are in tension or compression.Also, indicate all zero-force members. *6–40. Determine the force in members GF, GD, and CD of the truss and state if the members are in tension or compression. •6–41. Determine the force in members BG, BC, and HG of the truss and state if the members are in tension or compression. 6–38. Determine the force in members DC, HC, and HI of the truss, and state if the members are in tension or compression. 6–39. Determine the force in members ED, EH, and GH of the truss, and state if the members are in tension or compression. *6–44. Determine the force in members JI, EF, EI, and JE of the truss, and state if the members are in tension or compression. •6–45. Determine the force in members CD, LD, and KL of the truss, and state if the members are in tension or compression. A C G E D H F I B 2 m 2 m 2 m 1.5 m 50 kN 40 kN 40 kN 30 kN 1.5 m 1.5 m Probs. 6–38/39 260 lb 4 ft 4 ft 4 ft 3 ft 3 ft 4 ft 4 ft 5 12 13 A H G F B C D E Probs. 6–40/41 1.5 m 1.5 m 1.5 m A H B C D J I G F E 1.5 m 2 m 2 m 6 kN 6 kN Probs. 6–42/43 8 ft 8 ft 8 ft 900 lb 1500 lb 1000 lb 1000 lb A G N B H F M C D E I J L K 8 ft 8 ft 8 ft 8 ft 8 ft 8 ft Probs. 6–44/45
- 306. 6.4 THE METHOD OF SECTIONS 289 6 6–50. Determine the force in each member of the truss and state if the members are in tension or compression. Set , . 6–51. Determine the force in each member of the truss and state if the members are in tension or compression. Set , . P2 = 20 kN P1 = 40 kN P2 = 10 kN P1 = 20 kN *6–48. Determine the force in members IJ, EJ, and CD of the Howe truss, and state if the members are in tension or compression. •6–49. Determine the force in members KJ, KC, and BC of the Howe truss, and state if the members are in tension or compression. 6–46. Determine the force developed in members BC and CH of the roof truss and state if the members are in tension or compression. 6–47. Determine the force in members CD and GF of the truss and state if the members are in tension or compression.Also indicate all zero-force members. *6–52. Determine the force in members KJ, NJ, ND, and CD of the K truss. Indicate if the members are in tension or compression. Hint: Use sections aa and bb. •6–53. Determine the force in members JI and DE of the K truss. Indicate if the members are in tension or compression. 2 kN 3 kN 4 kN 5 kN 4 kN 6 kN 5 kN B A C D E F G H I J K L 2 m 4 m 2 m 2 m 2 m 2 m 2 m Probs. 6–48/49 A G F E D C B P2 P1 1.5 m 1.5 m 1.5 m 1.5 m 2 m Probs. 6–50/51 1800 lb 15 ft 15 ft 20 ft 20 ft 20 ft 20 ft 20 ft A B I H 20 ft L M N O P G F E D C 1500 lb 1200 lb a b J K a b Probs. 6–52/53 1.5 m 2 m 2 m 1 m 1 m 0.8 m 2 kN 1.5 kN A H B D G C F E Probs. 6–46/47
- 307. 290 CHAPTER 6 STRUCTURAL ANALYSIS 6 Procedure for Analysis Either the method of joints or the method of sections can be used to determine the forces developed in the members of a simple space truss. Method of Joints. If the forces in all the members of the truss are to be determined, then the method of joints is most suitable for the analysis. Here it is necessary to apply the three equilibrium equations to the forces acting at each joint. Remember that the solution of many simultaneous equations can be avoided if the force analysis begins at a joint having at least one known force and at most three unknown forces.Also,if the three-dimensional geometry of the force system at the joint is hard to visualize,it is recommended that a Cartesian vector analysis be used for the solution. Method of Sections. If only a few member forces are to be determined, the method of sections can be used. When an imaginary section is passed through a truss and the truss is separated into two parts, the force system acting on one of the segments must satisfy the six equilibrium equations: (Eqs.5–6).By proper choice of the section and axes for summing forces and moments, many of the unknown member forces in a space truss can be computed directly, using a single equilibrium equation. ©Mz = 0 ©My = 0, ©Mx = 0, ©Fz = 0, ©Fy = 0, ©Fx = 0, ©Fz = 0 ©Fy = 0, ©Fx = 0, P For economic reasons, large electrical transmission towers are often constructed using space trusses. Fig. 6–19 *6.5 Space Trusses A space truss consists of members joined together at their ends to form a stable three-dimensional structure.The simplest form of a space truss is a tetrahedron, formed by connecting six members together, as shown in Fig. 6–19. Any additional members added to this basic element would be redundant in supporting the force P. A simple space truss can be built from this basic tetrahedral element by adding three additional members and a joint, and continuing in this manner to form a system of multiconnected tetrahedrons. Assumptions for Design The members of a space truss may be treated as two-force members provided the external loading is applied at the joints and the joints consist of ball-and-socket connections. These assumptions are justified if the welded or bolted connections of the joined members intersect at a common point and the weight of the members can be neglected. In cases where the weight of a member is to be included in the analysis, it is generally satisfactory to apply it as a vertical force, half of its magnitude applied at each end of the member. Typical roof-supporting space truss. Notice the use of ball-and- socket joints for the connections
- 308. 6.5 SPACE TRUSSES 291 6 EXAMPLE 6.8 Determine the forces acting in the members of the space truss shown in Fig. 6–20a. Indicate whether the members are in tension or compression. SOLUTION Since there are one known force and three unknown forces acting at joint A, the force analysis of the truss will begin at this joint. Joint A. (Fig. 6–20b). Expressing each force acting on the free-body diagram of joint A as a Cartesian vector, we have For equilibrium, Ans. Ans. Since is known, joint B can be analyzed next. Joint B. (Fig. 6–20c). Ans. The scalar equations of equilibrium may also be applied directly to the forces acting on the free-body diagrams of joints D and C since the force components are easily determined. Show that Ans. FDE = FDC = FCE = 0 FBD = 2 kN 1C2 RB = FBE = 5.66 kN 1T2, 2 + FBD - 0.707FBE = 0 ©Fz = 0; -4 + RB sin 45° = 0 ©Fy = 0; -RB cos 45° + 0.707FBE = 0 ©Fx = 0; F AB FAB = 4 kN 1T2 FAC = FAE = 0 -FAC - 0.577FAE = 0 ©Fz = 0; -4 + FAB + 0.577FAE = 0 ©Fy = 0; 0.577FAE = 0 ©Fx = 0; -4j + F ABj - F ACk + 0.577F AEi + 0.577F AEj - 0.577F AEk = 0 P + FAB + FAC + FAE = 0 ©F = 0; FAE = FAEa rAE rAE b = FAE10.577i + 0.577j - 0.577k2 FAC = -F ACk, FAB = F ABj, P = 5-4j6 kN, (a) 2 m 2 m 2 m P 4 kN 2 kN 45 z y x A B C D E Fig. 6–20 x y z P 4 kN FAC FAE FAB A (b) x y z FAB 4 kN 45 1 1 FBE FBD RB B 2 kN (c)
- 309. 292 CHAPTER 6 STRUCTURAL ANALYSIS 6 PROBLEMS 6–58. Determine the force in members BE, DF, and BC of the space truss and state if the members are in tension or compression. 6–59. Determine the force in members AB, CD, ED, and CF of the space truss and state if the members are in tension or compression. *6–56. Determine the force in each member of the space truss and state if the members are in tension or compression. The truss is supported by ball-and-socket joints at A, B, and E. Set . Hint: The support reaction at E acts along member EC.Why? •6–57. Determine the force in each member of the space truss and state if the members are in tension or compression. The truss is supported by ball-and-socket joints at A, B, and E. Set . Hint: The support reaction at E acts along member EC.Why? F = 5-200i + 400j6 N F = 5800j6 N 6–54. The space truss supports a force . Determine the force in each member, and state if the members are in tension or compression. 6–55. The space truss supports a force . Determine the force in each member, and state if the members are in tension or compression. F = 5600i + 450j - 750k6 lb F = 5-500i + 600j + 400k6 lb *6–60. Determine the force in the members AB, AE, BC, BF, BD, and BE of the space truss, and state if the members are in tension or compression. A B C D x y z F 8 ft 6 ft 6 ft 6 ft 6 ft Probs. 6–54/55 F D A z 2 m x y B C E 5 m 1 m 2 m 1.5 m Probs. 6–56/57 2 m 2 m 2 m E A 3 m F D C B {2k} kN {2k} kN 2 m Probs. 6–58/59 F E D x z y C B A 4 ft 4 ft 2 ft 2 ft 300 lb 600 lb 400 lb 4 ft Prob. 6–60
- 310. 6.5 SPACE TRUSSES 293 6 *6–64. Determine the force developed in each member of the space truss and state if the members are in tension or compression.The crate has a weight of 150 lb. 6–62. If the truss supports a force of , determine the force in each member and state if the members are in tension or compression. 6–63. If each member of the space truss can support a maximum force of 600 N in compression and 800 N in tension, determine the greatest force F the truss can support. F = 200 N •6–61. Determine the force in the members EF, DF, CF, and CD of the space truss, and state if the members are in tension or compression. •6–65. Determine the force in members FE and ED of the space truss and state if the members are in tension or compression. The truss is supported by a ball-and-socket joint at C and short links at A and B. 6–66. Determine the force in members GD, GE, and FD of the space truss and state if the members are in tension or compression. F E D x z y C B A 4 ft 4 ft 2 ft 2 ft 300 lb 600 lb 400 lb 4 ft Prob. 6–61 y D E F x z C B A 200 mm 200 mm 200 mm 200 mm 500 mm 300 mm Probs. 6–62/63 x y z A B C D 6 ft 6 ft 6 ft 6 ft Prob. 6–64 z x y {500k} lb G {200j} lb 6 ft 6 ft F E D C 4 ft 2 ft 3 ft 3 ft A B Probs. 6–65/66
- 311. 294 CHAPTER 6 STRUCTURAL ANALYSIS 6 6.6 Frames and Machines Frames and machines are two types of structures which are often composed of pin-connected multiforce members, i.e., members that are subjected to more than two forces. Frames are used to support loads, whereas machines contain moving parts and are designed to transmit and alter the effect of forces. Provided a frame or machine contains no more supports or members than are necessary to prevent its collapse, the forces acting at the joints and supports can be determined by applying the equations of equilibrium to each of its members. Once these forces are obtained, it is then possible to design the size of the members, connections, and supports using the theory of mechanics of materials and an appropriate engineering design code. Free-Body Diagrams. In order to determine the forces acting at the joints and supports of a frame or machine, the structure must be disassembled and the free-body diagrams of its parts must be drawn.The following important points must be observed: • Isolate each part by drawing its outlined shape Then show all the forces and/or couple moments that act on the part. Make sure to label or identify each known and unknown force and couple moment with reference to an established x, y coordinate system. Also, indicate any dimensions used for taking moments. Most often the equations of equilibrium are easier to apply if the forces are represented by their rectangular components.As usual, the sense of an unknown force or couple moment can be assumed. • Identify all the two-force members in the structure and represent their free-body diagrams as having two equal but opposite collinear forces acting at their points of application. (See Sec. 5.4.) By recognizing the two-force members, we can avoid solving an unnecessary number of equilibrium equations. • Forces common to any two contacting members act with equal magnitudes but opposite sense on the respective members. If the two members are treated as a “system” of connected members, then these forces are “internal” and are not shown on the free-body diagram of the system; however, if the free-body diagram of each member is drawn, the forces are “external” and must be shown on each of the free-body diagrams. The following examples graphically illustrate how to draw the free- body diagrams of a dismembered frame or machine. In all cases, the weight of the members is neglected. Common tools such as these pliers act as simple machines. Here the applied force on the handles creates a much larger force at the jaws. This large crane is a typical example of a framework.
- 312. 6.6 FRAMES AND MACHINES 295 6 EXAMPLE 6.9 For the frame shown in Fig. 6–21a, draw the free-body diagram of (a) each member, (b) the pin at B, and (c) the two members connected together. P B A C (a) M Fig. 6–21 P (b) M Bx By By Ax Ay Cy Cx Bx Effect of pin on member Bx By Bx By (c) Effect of member BC on the pin Effect of member AB on the pin B Equilibrium P M Ax Ay Cy Cx (d) SOLUTION Part (a). By inspection, members BA and BC are not two-force members. Instead, as shown on the free-body diagrams, Fig. 6–21b, BC is subjected to a force from the pins at B and C and the external force P. Likewise, AB is subjected to a force from the pins at A and B and the external couple moment M. The pin forces are represented by their x and y components. Part (b). The pin at B is subjected to only two forces, i.e., the force of member BC and the force of member AB. For equilibrium these forces or their respective components must be equal but opposite, Fig. 6–21c. Realize that Newton’s third law is applied between the pin and its connected members, i.e., the effect of the pin on the two members, Fig. 6–21b, and the equal but opposite effect of the two members on the pin, Fig. 6–21c. Part (c). The free-body diagram of both members connected together, yet removed from the supporting pins at A and C, is shown in Fig. 6–21d. The force components and are not shown on this diagram since they are internal forces (Fig. 6–21b) and therefore cancel out. Also, to be consistent when later applying the equilibrium equations, the unknown force components at A and C must act in the same sense as those shown in Fig. 6–21b. By Bx
- 313. 296 CHAPTER 6 STRUCTURAL ANALYSIS 6 A constant tension in the conveyor belt is maintained by using the device shown in Fig. 6–22a. Draw the free-body diagrams of the frame and the cylinder that the belt surrounds. The suspended block has a weight of W. EXAMPLE 6.10 SOLUTION The idealized model of the device is shown in Fig. 6–22b. Here the angle is assumed to be known. From this model, the free-body diagrams of the cylinder and frame are shown in Figs. 6–22c and 6–22d, respectively. Note that the force that the pin at B exerts on the cylinder can be represented by either its horizontal and vertical components and which can be determined by using the force equations of equilibrium applied to the cylinder, or by the two components T, which provide equal but opposite moments on the cylinder and thus keep it from turning. Also, realize that once the pin reactions at A have been determined, half of their values act on each side of the frame since pin connections occur on each side, Fig. 6–22a. By, Bx u (a) Fig. 6–22 T T B (b) A u T Bx By Bx Ax By Ay T (c) T T T T (d) or W u u u
- 314. 6.6 FRAMES AND MACHINES 297 6 EXAMPLE 6.11 For the frame shown in Fig. 6–23a, draw the free-body diagrams of (a) the entire frame including the pulleys and cords, (b) the frame without the pulleys and cords, and (c) each of the pulleys. SOLUTION Part (a). When the entire frame including the pulleys and cords is considered, the interactions at the points where the pulleys and cords are connected to the frame become pairs of internal forces which cancel each other and therefore are not shown on the free-body diagram, Fig. 6–23b. Part (b). When the cords and pulleys are removed, their effect on the frame must be shown, Fig. 6–23c. Part (c). The force components of the pins on the pulleys, Fig. 6–23d, are equal but opposite to the force components exerted by the pins on the frame, Fig. 6–23c.Why? Cy Cx, By, Bx, C A B 75 lb (a) D Fig. 6–23 75 lb (b) Ay Ax T 75 lb 75 lb By Bx Cx Cy T T (c) (d) Ax Ay 75 lb T Bx By Cy Cx
- 315. 298 CHAPTER 6 STRUCTURAL ANALYSIS 6 Draw the free-body diagrams of the bucket and the vertical boom of the backhoe shown in the photo, Fig. 6–24a. The bucket and its contents have a weight W. Neglect the weight of the members. SOLUTION The idealized model of the assembly is shown in Fig. 6–24b. By inspection, members AB, BC, BE, and HI are all two-force members since they are pin connected at their end points and no other forces act on them. The free-body diagrams of the bucket and the boom are shown in Fig. 6–24c. Note that pin C is subjected to only two forces, whereas the pin at B is subjected to three forces, Fig. 6–24d. These three forces are related by the two equations of force equilibrium applied to each pin. The free-body diagram of the entire assembly is shown in Fig. 6–24e. EXAMPLE 6.12 (a) Fig. 6–24 A B E C (b) D F H I G (c) Dy Dy FBA Fx Fy FBC FBE FHI Dx Dx W C FBC FBC B FBC FBE FBA (d) (e) Fx Fy FHI W
- 316. 6.6 FRAMES AND MACHINES 299 6 EXAMPLE 6.13 Draw the free-body diagram of each part of the smooth piston and link mechanism used to crush recycled cans, which is shown in Fig. 6–25a. SOLUTION By inspection, member AB is a two-force member. The free-body diagrams of the parts are shown in Fig. 6–25b. Since the pins at B and D connect only two parts together, the forces there are shown as equal but opposite on the separate free-body diagrams of their connected members. In particular, four components of force act on the piston: and represent the effect of the pin (or lever EBD), is the resultant force of the support, and P is the resultant compressive force caused by the can C. NOTE: A free-body diagram of the entire assembly is shown in Fig. 6–25c. Here the forces between the components are internal and are not shown on the free-body diagram. Nw Dy Dx Before proceeding, it is highly recommended that you cover the solutions to the previous examples and attempt to draw the requested free- body diagrams.When doing so, make sure the work is neat and that all the forces and couple moments are properly labeled.When finished, challenge yourself and solve the following four problems. C F 800 N A B D E 75 90 30 (a) Fig. 6–25 F 800 N E 75 D Dx Dy A B B FAB FAB FAB 30 Dx P D Nw Dy (b) F 800 N 75 30 P FAB Nw (c)
- 317. A B C E D F G H P6–3 300 CHAPTER 6 STRUCTURAL ANALYSIS 6 A B C E D F P6–1 F G J E C D B A I H P6–2 CONCEPTUAL PROBLEMS A C B D E P6–4 P6–3. Draw the free-body diagrams of the boom ABCDF and the stick FGH of the bucket lift. Neglect the weights of the member. The bucket weighs W. The two force members are BI, CE, DE and GE. Assume all indicated points of connection are pins. P6–2. Draw the free-body diagrams of the boom ABCD and the stick EDFGH of the backhoe. The weights of these two members are significant. Neglect the weights of all the other members, and assume all indicated points of connection are pins. P6–1. Draw the free-body diagrams of each of the crane boom segments AB, BC, and BD. Only the weights of AB and BC are significant.Assume A and B are pins. P6–4. To operate the can crusher one pushes down on the lever arm ABC which rotates about the fixed pin at B. This moves the side links CD downward, which causes the guide plate E to also move downward and thereby crush the can. Draw the free-body diagrams of the lever, side link, and guide plate. Make up some reasonable numbers and do an equilibrium analysis to shown how much an applied vertical force at the handle is magnified when it is transmitted to the can.Assume all points of connection are pins and the guides for the plate are smooth.
- 318. 6.6 FRAMES AND MACHINES 301 6 Procedure for Analysis The joint reactions on frames or machines (structures) composed of multiforce members can be determined using the following procedure. Free-Body Diagram. • Draw the free-body diagram of the entire frame or machine, a portion of it, or each of its members. The choice should be made so that it leads to the most direct solution of the problem. • When the free-body diagram of a group of members of a frame or machine is drawn, the forces between the connected parts of this group are internal forces and are not shown on the free-body diagram of the group. • Forces common to two members which are in contact act with equal magnitude but opposite sense on the respective free-body diagrams of the members. • Two-force members, regardless of their shape, have equal but opposite collinear forces acting at the ends of the member. • In many cases it is possible to tell by inspection the proper sense of the unknown forces acting on a member; however, if this seems difficult, the sense can be assumed. • Remember that a couple moment is a free vector and can act at any point on the free-body diagram. Also, a force is a sliding vector and can act at any point along its line of action. Equations of Equilibrium. • Count the number of unknowns and compare it to the total number of equilibrium equations that are available. In two dimensions, there are three equilibrium equations that can be written for each member. • Sum moments about a point that lies at the intersection of the lines of action of as many of the unknown forces as possible. • If the solution of a force or couple moment magnitude is found to be negative, it means the sense of the force is the reverse of that shown on the free-body diagram.
- 319. 302 CHAPTER 6 STRUCTURAL ANALYSIS 6 Determine the horizontal and vertical components of force which the pin at C exerts on member BC of the frame in Fig. 6–26a. SOLUTION I Free-Body Diagrams. By inspection it can be seen that AB is a two-force member.The free-body diagrams are shown in Fig. 6–26b. Equations of Equilibrium. The three unknowns can be determined by applying the three equations of equilibrium to member CB. a Ans. Ans. SOLUTION II Free-Body Diagrams. If one does not recognize that AB is a two- force member, then more work is involved in solving this problem. The free-body diagrams are shown in Fig. 6–26c. Equations of Equilibrium. The six unknowns are determined by applying the three equations of equilibrium to each member. Member AB a (1) (2) (3) Member BC a (4) (5) (6) The results for and can be determined by solving these equations in the following sequence: 4, 1, 5, then 6.The results are Ans. Ans. By comparison, Solution I is simpler since the requirement that in Fig. 6–26b be equal, opposite, and collinear at the ends of member AB automatically satisfies Eqs. 1, 2, and 3 above and therefore eliminates the need to write these equations. As a result, save yourself some time and effort by always identifying the two-force members before starting the analysis! F AB Cy = 1000 N Cx = 577 N Bx = 577 N By = 1000 N Cy Cx By - 2000 N + Cy = 0 + c©Fy = 0; Bx - Cx = 0 : + ©Fx = 0; 2000 N12 m2 - By14 m2 = 0 +©MC = 0; Ay - By = 0 + c ©Fy = 0; Ax - Bx = 0 : + ©Fx = 0; Bx13 sin 60° m2 - By13 cos 60° m2 = 0 +©MA = 0; 1154.7 sin 60° N - 2000 N + Cy = 0 Cy = 1000 N + c©Fy = 0; 1154.7 cos 60° N - Cx = 0 Cx = 577 N : + ©Fx = 0; F AB = 1154.7 N 2000 N12 m2- 1F AB sin 60°214 m2 = 0 +©MC = 0; EXAMPLE 6.14 A B C 2000 N 2 m 2 m 3 m 60 (a) Fig. 6–26 2 m 2 m 60 FAB Cy Cx FAB FAB 2000 N (b) B 2 m 2 m Cy Cx C By Bx 2000 N By Bx Ay A Ax (c) 3 m 60
- 320. 6.6 FRAMES AND MACHINES 303 6 EXAMPLE 6.15 The compound beam shown in Fig. 6–27a is pin connected at B. Determine the components of reaction at its supports. Neglect its weight and thickness. B C 4 kN/m 3 4 5 2 m 2 m 2 m (a) A 10 kN Fig. 6–27 2 m 4 m 3 4 5 10 kN B 2 m 1 m A Ay Ax MA By Bx Bx By Cy 8 kN (b) SOLUTION Free-Body Diagrams. By inspection, if we consider a free-body diagram of the entire beam ABC, there will be three unknown reactions at A and one at C. These four unknowns cannot all be obtained from the three available equations of equilibrium, and so for the solution it will become necessary to dismember the beam into its two segments, as shown in Fig. 6–27b. Equations of Equilibrium. The six unknowns are determined as follows: Segment BC a Segment AB a Solving each of these equations successively, using previously calculated results, we obtain Ans. Ans. Cy = 4 kN By = 4 kN Bx = 0 MA = 32 kN # m Ay = 12 kN Ax = 6 kN Ay - 110 kN2A4 5 B - By = 0 + c ©Fy = 0; MA - 110 kN2A4 5 B12 m2 - By14 m2 = 0 +©MA = 0; Ax - 110 kN2A3 5 B + Bx = 0 : + ©Fx = 0; By - 8 kN + Cy = 0 + c ©Fy = 0; -8 kN11 m2 + Cy12 m2 = 0 +©MB = 0; Bx = 0 ; + ©Fx = 0;
- 321. 304 CHAPTER 6 STRUCTURAL ANALYSIS 6 D E C B A F (a) Fig. 6–28 (b) T1 N1 N4 N2 N3 T1 T1 T2T2 500 (9.81) N C T1 T1 T2 A 500-kg elevator car in Fig. 6–28a is being hoisted by motor A using the pulley system shown. If the car is traveling with a constant speed, determine the force developed in the two cables. Neglect the mass of the cable and pulleys. EXAMPLE 6.16 SOLUTION Free-Body Diagram. We can solve this problem using the free- body diagrams of the elevator car and pulley C, Fig. 6–28b.The tensile forces developed in the cables are denoted as and . Equations of Equilibrium. For pulley C, ; or (1) For the elevator car, (2) Substituting Eq. (1) into Eq. (2) yields Ans. Substituting this result into Eq. (1), Ans. T2 = 2(700.71) N = 1401 N = 1.40 kN T1 = 700.71 N = 701 N 3T1 + 2(2T1) - 500(9.81) N = 0 3T1 + 2T2 - 500(9.81) N = 0 + c©Fy = 0; T2 = 2T1 T2 - 2T1 = 0 + c ©Fy = 0 T2 T1
- 322. 6.6 FRAMES AND MACHINES 305 6 EXAMPLE 6.17 The smooth disk shown in Fig. 6–29a is pinned at D and has a weight of 20 lb. Neglecting the weights of the other members, determine the horizontal and vertical components of reaction at pins B and D. SOLUTION Free-Body Diagrams. The free-body diagrams of the entire frame and each of its members are shown in Fig. 6–29b. Equations of Equilibrium. The eight unknowns can of course be obtained by applying the eight equilibrium equations to each member—three to member AB, three to member BCD, and two to the disk. (Moment equilibrium is automatically satisfied for the disk.) If this is done, however, all the results can be obtained only from a simultaneous solution of some of the equations. (Try it and find out.) To avoid this situation, it is best first to determine the three support reactions on the entire frame; then, using these results, the remaining five equilibrium equations can be applied to two other parts in order to solve successively for the other unknowns. Entire Frame a Member AB Ans. a Ans. Disk Ans. Ans. Dy = 20 lb 40 lb - 20 lb - Dy = 0 + c ©Fy = 0; Dx = 0 : + ©Fx = 0; By = 20 lb 20 lb - 40 lb + By = 0 + c ©Fy = 0; N D = 40 lb -20 lb 16 ft2 + ND13 ft2 = 0 +©MB = 0; Bx = 17.1 lb 17.1 lb - Bx = 0 : + ©Fx = 0; Ay = 20 lb Ay - 20 lb = 0 + c ©Fy = 0; Ax = 17.1 lb Ax - 17.1 lb = 0 : + ©Fx = 0; Cx = 17.1 lb -20 lb 13 ft2 + Cx13.5 ft2 = 0 +©MA = 0; 3.5 ft 3 ft D C A (a) B Fig. 6–29 3.5 ft 3 ft Ay Ax 20 lb Cx 3.5 ft 3 ft Cx Dx Dy By Bx 3 ft (b) 3 ft ND By Bx ND Dy Dx 20 lb 20 lb 17.1 lb
- 323. 306 CHAPTER 6 STRUCTURAL ANALYSIS 6 Determine the tension in the cables and also the force P required to support the 600-N force using the frictionless pulley system shown in Fig. 6–30a. EXAMPLE 6.18 SOLUTION Free-Body Diagram. A free-body diagram of each pulley including its pin and a portion of the contacting cable is shown in Fig. 6–30b. Since the cable is continuous, it has a constant tension P acting throughout its length.The link connection between pulleys B and C is a two-force member, and therefore it has an unknown tension T acting on it. Notice that the principle of action, equal but opposite reaction must be carefully observed for forces P and T when the separate free-body diagrams are drawn. Equations of Equilibrium. The three unknowns are obtained as follows: Pulley A Ans. Pulley B Ans. Pulley C Ans. R = 800 N R - 2P - T = 0 + c ©Fy = 0; T = 400 N T - 2P = 0 + c©Fy = 0; P = 200 N 3P - 600 N = 0 + c©Fy = 0; A P B C 600 N (a) Fig. 6–30 A B C R T P P P P T P P P (b) 600 N
- 324. 6.6 FRAMES AND MACHINES 307 6 EXAMPLE 6.19 The two planks in Fig. 6–31a are connected together by cable BC and a smooth spacer DE. Determine the reactions at the smooth supports A and F, and also find the force developed in the cable and spacer. SOLUTION Free-Body Diagrams. The free-body diagram of each plank is shown in Fig. 6–31b. It is important to apply Newton’s third law to the interaction forces as shown. Equations of Equilibrium. For plank AD, a ; For plank CF, a ; Solving simultaneously, Ans. Using these results, for plank AD, Ans. And for plank CF, Ans. N F = 180 lb NF + 160 lb - 140 lb - 200 lb = 0 + c©F y = 0; N A = 120 lb N A + 140 lb - 160 lb - 100 lb = 0 + c©F y = 0; F BC = 160 lb F DE = 140 lb F DE(4 ft) - FBC(6 ft) + 200 lb (2 ft) = 0 +©MF = 0 F DE(6 ft) - FBC(4 ft) - 100 lb (2 ft) = 0 +©MA = 0 F D E B C A 2 ft 2 ft 2 ft 100 lb 200 lb 2 ft 2 ft (a) D C F A 100 lb (b) 2 ft 2 ft 2 ft 2 ft 2 ft 2 ft 200 lb NA NF FDE FDE FBC FBC Fig. 6–31
- 325. 308 CHAPTER 6 STRUCTURAL ANALYSIS 6 The 75-kg man in Fig. 6–32a attempts to lift the 40-kg uniform beam off the roller support at B. Determine the tension developed in the cable attached to B and the normal reaction of the man on the beam when this is about to occur. SOLUTION Free-Body Diagrams. The tensile force in the cable will be denoted as .The free-body diagrams of the pulley E, the man, and the beam are shown in Fig. 6–32b. The beam has no contact with roller B, so .When drawing each of these diagrams, it is very important to apply Newton’s third law. Equations of Equilibrium. Using the free-body diagram of pulley E, or (1) Referring to the free-body diagram of the man using this result, ; (2) Summing moments about point A on the beam, a ; (3) Solving Eqs. 2 and 3 simultaneously for and , then using Eq. (1) for , we obtain Ans. SOLUTION II A direct solution for can be obtained by considering the beam, the man, and pulley as a single system. The free-body diagram is shown in Fig. 6–32c.Thus, a ; Ans. With this result Eqs. 1 and 2 can then be used to find and . T2 Nm T1 = 256 N -[40(9.81) N](1.5 m) + T1(3 m) = 0 2T1(0.8 m) - [75(9.81) N](0.8 m) +©MA = 0 E T1 T2 = 512 N Nm = 224 N T1 = 256 N T2 Nm T1 T1(3 m) - Nm(0.8 m) - [40(9.81) N](1.5 m) = 0 +©MA = 0 Nm + 2T1 - 75(9.81) N = 0 + c©Fy = 0 T2 = 2T1 2T1 - T2 = 0 + c©Fy = 0; NB = 0 T1 EXAMPLE 6.20 A B C D H E F 2.2 m (a) 0.8 m Fig. 6–32 G H E 1.5 m 75 (9.81) N 40 (9.81) N (b) 0.8 m 0.7 m Ay NB 0 Ax Nm T1 T1 T1 T2 2 T1 T2 Nm G 1.5 m 75 (9.81) N 40 (9.81) N (c) 0.8 m 0.7 m Ay NB 0 Ax T1 T1 T1
- 326. 6.6 FRAMES AND MACHINES 309 6 EXAMPLE 6.21 The frame in Fig. 6–33a supports the 50-kg cylinder. Determine the horizontal and vertical components of reaction at A and the force at C. SOLUTION Free-Body Diagrams. The free-body diagram of pulley D, along with the cylinder and a portion of the cord (a system), is shown in Fig. 6–33b. Member BC is a two-force member as indicated by its free-body diagram.The free-body diagram of member ABD is also shown. Equations of Equilibrium. We will begin by analyzing the equilibrium of the pulley. The moment equation of equilibrium is automatically satisfied with T = 50(9.81) N, and so Ans. Using these results, FBC can be determined by summing moments about point A on member ABD. Ans. Now Ax and Ay can be determined by summing forces. Ans. Ans. Ay = 490.5 N Ay - 490.5 N = 0 + c©Fy = 0; Ax - 245.25 N - 490.5 N = 0 Ax = 736 N : + ©Fx = 0; FBC = 245.25 N F BC (0.6 m) + 490.5 N(0.9 m) - 490.5 N(1.20 m) = 0 +©MA = 0; Dy - 50(9.81) N = 0 Dy = 490.5 N + c ©Fy = 0; Dx - 50(9.81) N = 0 Dx = 490.5 N : + ©Fx = 0; A B D C (a) 1.2 m 0.6 m 0.3 m 0.1 m Fig. 6–33 1.20 m 0.6 m 490.5 N 490.5 N (b) T 50 (9.81) N 50 (9.81) N Ax Dx FBC FBC FBC Dx Ay Dy Dy 0.9 m b
- 327. 310 CHAPTER 6 STRUCTURAL ANALYSIS 6 P F6–13 3 ft 3 ft 400 lb 500 lb 3 ft 3 ft 4 ft B A C F6–4 250 mm 50 mm 100 N 100 N 45 A B F6–15 A B C 400 N 800 N m 2 m 1 m 1 m 1 m 1 m F6–16 FUNDAMENTAL PROBLEMS F6–16. Determine the horizontal and vertical components of reaction at pin C. F6–15. If a 100-N force is applied to the handles of the pliers, determine the clamping force exerted on the smooth pipe B and the magnitude of the resultant force at pin A. F6–13. Determine the force P needed to hold the 60-lb weight in equilibrium. F6–17. Determine the normal force that the 100-lb plate A exerts on the 30-lb plate B. F6–14. Determine the horizontal and vertical components of reaction at pin C. F6–18. Determine the force P needed to lift the load. Also, determine the proper placement x of the hook for equilibrium. Neglect the weight of the beam. 4 ft B A 1 ft 1 ft F6–17 P B C A 0.9 m 100 mm 100 mm 100 mm 6 kN x F6–18
- 328. 6.6 FRAMES AND MACHINES 311 6 •6–69. Determine the force required to hold the 50-kg mass in equilibrium. P *6–68. Determine the force required to hold the 150-kg crate in equilibrium. P 6–67. Determine the force required to hold the 100-lb weight in equilibrium. P 6–70. Determine the force needed to hold the 20-lb block in equilibrium. P PROBLEMS P A B C D Prob. 6–67 P A B C Prob. 6–68 P A B C Prob. 6–69 C B A P Prob. 6–70
- 329. 312 CHAPTER 6 STRUCTURAL ANALYSIS 6 •6–73. If the peg at B is smooth, determine the components of reaction at the pin A and fixed support C. *6–72. The cable and pulleys are used to lift the 600-lb stone. Determine the force that must be exerted on the cable at A and the corresponding magnitude of the resultant force the pulley at C exerts on pin B when the cables are in the position shown. 6–71. Determine the force needed to support the 100-lb weight. Each pulley has a weight of 10 lb.Also, what are the cord reactions at A and B? P 6–74. Determine the horizontal and vertical components of reaction at pins A and C. P 2 in. 2 in. 2 in. C A B Prob. 6–71 P A C B D 30 Prob. 6–72 A B C 600 mm 800 mm 900 Nm 600 mm 500 N 45 Prob. 6–73 B A C 2 ft 3 ft 150 lb 100 lb 2 ft 45 Prob. 6–74
- 330. 6.6 FRAMES AND MACHINES 313 6 6–78. Determine the horizontal and vertical components of reaction at pins A and C of the two-member frame. *6–76. The compound beam is pin-supported at C and supported by rollers at A and B.There is a hinge (pin) at D. Determine the components of reaction at the supports. Neglect the thickness of the beam. •6–77. The compound beam is supported by a rocker at B and is fixed to the wall at A. If it is hinged (pinned) together at C, determine the components of reaction at the supports. Neglect the thickness of the beam. 6–75. The compound beam is fixed at A and supported by rockers at B and C. There are hinges (pins) at D and E. Determine the components of reaction at the supports. 6–79. If a force of acts on the rope, determine the cutting force on the smooth tree limb at D and the horizontal and vertical components of force acting on pin A. The rope passes through a small pulley at C and a smooth ring at E. F = 50 N 6 m 2 m 6 m 30 kN m 2 m 2 m 15 kN A D B E C Prob. 6–75 A D B C 8 ft 3 4 5 8 ft 12 kip 15 kip ft 4 kip 30 8 kip 8 ft 4 ft 2 ft 6 ft Prob. 6–76 4 ft 4 ft 500 lb 200 lb 4000 lb ft 4 ft 8 ft A C B 12 13 5 60 Prob. 6–77 3 m 3 m 200 N/m A B C Prob. 6–78 F 50 N B C E 30 mm 100 mm A D Prob. 6–79
- 331. 314 CHAPTER 6 STRUCTURAL ANALYSIS 6 6–82. If the 300-kg drum has a center of mass at point G, determine the horizontal and vertical components of force acting at pin A and the reactions on the smooth pads C and D. The grip at B on member DAB resists both horizontal and vertical components of force at the rim of the drum. •6–81. The bridge frame consists of three segments which can be considered pinned at A, D, and E, rocker supported at C and F, and roller supported at B. Determine the horizontal and vertical components of reaction at all these supports due to the loading shown. *6–80. Two beams are connected together by the short link BC. Determine the components of reaction at the fixed support A and at pin D. 6–83. Determine the horizontal and vertical components of reaction that pins A and C exert on the two-member arch. A B C D 10 kN 12 kN 3 m 1.5 m 1 m 1.5 m Prob. 6–80 15 ft 20 ft 5 ft 5 ft 15 ft 2 kip/ft 30 ft A B C F D E Prob. 6–81 P 390 mm 100 mm 60 mm 60 mm 600 mm 30 B A C D G E Prob. 6–82 1 m 1.5 m 2 kN 1.5 kN 0.5 m A B C Prob. 6–83
- 332. 6.6 FRAMES AND MACHINES 315 6 6–87. The hoist supports the 125-kg engine. Determine the force the load creates in member DB and in member FB, which contains the hydraulic cylinder H. •6–85. The platform scale consists of a combination of third and first class levers so that the load on one lever becomes the effort that moves the next lever. Through this arrangement, a small weight can balance a massive object. If , determine the required mass of the counterweight S required to balance a 90-kg load, L. 6–86. The platform scale consists of a combination of third and first class levers so that the load on one lever becomes the effort that moves the next lever. Through this arrangement, a small weight can balance a massive object. If and, the mass of the counterweight S is 2 kg, determine the mass of the load L required to maintain the balance. x = 450 mm x = 450 mm *6–84. The truck and the tanker have weights of 8000 lb and 20 000 lb respectively. Their respective centers of gravity are located at points and . If the truck is at rest, determine the reactions on both wheels at A, at B, and at C. The tanker is connected to the truck at the turntable D which acts as a pin. G2 G1 *6–88. The frame is used to support the 100-kg cylinder E. Determine the horizontal and vertical components of reaction at A and D. G1 15 ft 10 ft 9 ft 5 ft A B D C G2 Prob. 6–84 350 mm 150 mm 150 mm 100 mm 250 mm B A C D E F H G x L S Probs. 6–85/86 C D E F G H 2 m 1 m 1 m 2 m 1 m 2 m A B Prob. 6–87 A C D E 0.6 m 1.2 m r 0.1 m Prob. 6–88
- 333. 316 CHAPTER 6 STRUCTURAL ANALYSIS 6 *6–92. The wall crane supports a load of 700 lb. Determine the horizontal and vertical components of reaction at the pins A and D.Also, what is the force in the cable at the winch W? •6–93. The wall crane supports a load of 700 lb. Determine the horizontal and vertical components of reaction at the pins A and D. Also, what is the force in the cable at the winch W? The jib ABC has a weight of 100 lb and member BD has a weight of 40 lb. Each member is uniform and has a center of gravity at its center. 6–91. The clamping hooks are used to lift the uniform smooth 500-kg plate. Determine the resultant compressive force that the hook exerts on the plate at A and B, and the pin reaction at C. •6–89. Determine the horizontal and vertical components of reaction which the pins exert on member AB of the frame. 6–90. Determine the horizontal and vertical components of reaction which the pins exert on member EDC of the frame. 6–94. The lever-actuated scale consists of a series of compound levers. If a load of weight is placed on the platform, determine the required weight of the counterweight S to balance the load. Is it necessary to place the load symmetrically on the platform? Explain. W = 150 lb A E B C D 500 lb 300 lb 3 ft 3 ft 4 ft 60 Probs. 6–89/90 A B 80 mm P P P 150 mm C Prob. 6–91 4 ft D A B C E W 4 ft 700 lb 60 4 ft Probs. 6–92/93 B A C D E F G H I J K S M W L 1.5 in. 1.5 in. 7.5 in. 7.5 in. 4.5 in. 4 in. 1.25 in. Prob. 6–94
- 334. 6.6 FRAMES AND MACHINES 317 6 6–98. A 300-kg counterweight, with center of mass at G, is mounted on the pitman crank AB of the oil-pumping unit. If a force of is to be developed in the fixed cable attached to the end of the walking beam DEF, determine the torque M that must be supplied by the motor. 6–99. A 300-kg counterweight, with center of mass at G, is mounted on the pitman crank AB of the oil-pumping unit. If the motor supplies a torque of , determine the force F developed in the fixed cable attached to the end of the walking beam DEF. M = 2500 N # m F = 5 kN •6–97. The pipe cutter is clamped around the pipe P. If the wheel at A exerts a normal force of on the pipe, determine the normal forces of wheels B and C on the pipe. The three wheels each have a radius of 7 mm and the pipe has an outer radius of 10 mm. FA = 80 N 6–95. If , determine the force F that the toggle clamp exerts on the wooden block. *6–96. If the wooden block exerts a force of on the toggle clamp, determine the force P applied to the handle. F = 600 N P = 75 N *6–100. The two-member structure is connected at C by a pin, which is fixed to BDE and passes through the smooth slot in member AC. Determine the horizontal and vertical components of reaction at the supports. 85 mm 140 mm 50 mm 50 mm 20 mm 140 mm P P F A B C D E Probs. 6–95/96 10 mm 10 mm P C B A Prob. 6–97 A B M D E F F 0.5 m 30 30 1.75 m 2.50 m G 0.65 m Probs. 6–98/99 3 ft 3 ft 2 ft 4 ft A B C D E 600 lb ft 500 lb Prob. 6–100
- 335. 318 CHAPTER 6 STRUCTURAL ANALYSIS 6 *6–104. The compound arrangement of the pan scale is shown. If the mass on the pan is 4 kg, determine the horizontal and vertical components at pins A, B, and C and the distance x of the 25-g mass to keep the scale in balance. 6–103. Determine the reactions at the fixed support E and the smooth support A. The pin, attached to member BD, passes through a smooth slot at D. •6–105. Determine the horizontal and vertical components of reaction that the pins at A, B, and C exert on the frame. The cylinder has a mass of 80 kg. A B C D 1.2 m 0.8 m 0.8 m 100 mm 100 mm Probs. 6–101/102 B C D E 0.3 m 0.3 m 0.3 m 0.3 m 0.4 m 0.4 m 600 N A Prob. 6–103 50 mm G 100 mm 75 mm 300 mm 350 mm x F E D B A 4 kg C Prob. 6–104 A B C 1 m 0.7 m 0.5 m D 100 mm Prob. 6–105 •6–101. The frame is used to support the 50-kg cylinder. Determine the horizontal and vertical components of reaction at A and D. 6–102. The frame is used to support the 50-kg cylinder. Determine the force of the pin at C on member ABC and on member CD.
- 336. 6.6 FRAMES AND MACHINES 319 6 •6–109. If a clamping force of is required at A, determine the amount of force F that must be applied to the handle of the toggle clamp. 6–110. If a force of is applied to the handle of the toggle clamp, determine the resulting clamping force at A. F = 350 N 300 N 6–107. A man having a weight of 175 lb attempts to hold himself using one of the two methods shown. Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. Neglect the weight of the platform. *6–108. A man having a weight of 175 lb attempts to hold himself using one of the two methods shown. Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C.The platform has a weight of 30 lb. 6–106. The bucket of the backhoe and its contents have a weight of 1200 lb and a center of gravity at G. Determine the forces of the hydraulic cylinder AB and in links AC and AD in order to hold the load in the position shown. The bucket is pinned at E. 6–111. Two smooth tubes A and B, each having the same weight, W, are suspended from a common point O by means of equal-length cords. A third tube, C, is placed between A and B. Determine the greatest weight of C without upsetting equilibrium. 120 45 1.5 ft 1 ft B G E A D C 0.25 ft Prob. 6–106 C C A B A B (a) (b) Probs. 6–107/108 275 mm 30 30 235 mm 30 mm 30 mm 70 mm F C E B D A Probs. 6–109/110 r/2 r B C 3r 3r O r A Prob. 6–111
- 337. 320 CHAPTER 6 STRUCTURAL ANALYSIS 6 6–114. The tractor shovel carries a 500-kg load of soil, having a center of mass at G. Compute the forces developed in the hydraulic cylinders IJ and BC due to this loading. •6–113. Show that the weight of the counterweight at H required for equilibrium is , and so it is independent of the placement of the load W on the platform. W1 = (ba)W W1 *6–112. The handle of the sector press is fixed to gear G, which in turn is in mesh with the sector gear C. Note that AB is pinned at its ends to gear C and the underside of the table EF, which is allowed to move vertically due to the smooth guides at E and F. If the gears only exert tangential forces between them, determine the compressive force developed on the cylinder S when a vertical force of 40 N is applied to the handle of the press. 6–115. If a force of is applied to the handle of the toggle clamp, determine the horizontal clamping force NE that the clamp exerts on the smooth wooden block at E. *6–116. If the horizontal clamping force that the toggle clamp exerts on the smooth wooden block at E is , determine the force applied to the handle of the clamp. P NE = 200 N P = 100 N 1.2 m E F A B C G D S 0.5 m 0.2 m 0.35 m 0.65 m 40 N H Prob. 6–112 A B W C E G H D F c b 3b a c 4 Prob. 6–113 100 mm 300 mm 300 mm 30 A C E G D F H J B 30 50 mm 400 mm 200 mm 200 mm 200 mm I 350 mm Prob. 6–114 B C D 160 mm 50 mm 75 mm 60 mm 30 45 A E P Probs. 6–115/116
- 338. 6.6 FRAMES AND MACHINES 321 6 *6–120. Determine the couple moment M that must be applied to member DC for equilibrium of the quick-return mechanism. Express the result in terms of the angles and , dimension L, and the applied vertical force P. The block at C is confined to slide within the slot of member AB. •6–121. Determine the couple moment M that must be applied to member DC for equilibrium of the quick-return mechanism. Express the result in terms of the angles and , dimension L, and the applied force P, which should be changed in the figure and instead directed horizontally to the right. The block at C is confined to slide within the slot of member AB. u f u f 6–118. Determine the force that the smooth roller C exerts on member AB. Also, what are the horizontal and vertical components of reaction at pin A? Neglect the weight of the frame and roller. 6–119. Determine the horizontal and vertical components of reaction which the pins exert on member ABC. •6–117. The engine hoist is used to support the 200-kg engine. Determine the force acting in the hydraulic cylinder AB, the horizontal and vertical components of force at the pin C, and the reactions at the fixed support D. 6–122. The kinetic sculpture requires that each of the three pinned beams be in perfect balance at all times during its slow motion. If each member has a uniform weight of 2 and length of 3 ft, determine the necessary counterweights and which must be added to the ends of each member to keep the system in balance for any position. Neglect the size of the counterweights. W3 W1, W2, lbft C D A G 1250 mm 350 mm 850 mm 550 mm 10 B Prob. 6–117 C 0.5 ft 3 ft A 60 lb ft 4 ft B D Prob. 6–118 3 ft Prob. 6–119 C M D A B 4 L L P u f Probs. 6–120/121 W Prob. 6–122
- 339. 322 CHAPTER 6 STRUCTURAL ANALYSIS 6 •6–125. The three-member frame is connected at its ends using ball-and-socket joints. Determine the x, y, z components of reaction at B and the tension in member ED. The force acting at D is F = 5135i + 200j - 180k6 lb. *6–124. The structure is subjected to the loading shown. Member AD is supported by a cable AB and roller at C and fits through a smooth circular hole at D. Member ED is supported by a roller at D and a pole that fits in a smooth snug circular hole at E. Determine the x, y, z components of reaction at E and the tension in cable AB. 6–123. The four-member “A” frame is supported at A and E by smooth collars and at G by a pin. All the other joints are ball-and-sockets. If the pin at G will fail when the resultant force there is 800 N, determine the largest vertical force P that can be supported by the frame. Also, what are the x, y, z force components which member BD exerts on members EDC and ABC? The collars at A and E and the pin at G only exert force components on the frame. 6–126. The structure is subjected to the loadings shown. Member AB is supported by a ball-and-socket at A and smooth collar at B. Member CD is supported by a pin at C. Determine the x, y, z components of reaction at A and C. x y C D B F G E A P Pk z 300 mm 300 mm 600 mm 600 mm 600 mm Prob. 6–123 z C A D B E 0.3 m y 0.3 m 0.5 m 0.4 m F {2.5k} kN x 0.8 m Prob. 6–124 y 6 ft 2 ft 1ft 3 ft 6 ft 3 ft 4 ft x A D F B C E z Prob. 6–125 2 m 3 m y 4 m 1.5 m B 800 N m A 250 N D 45 60 60 z x C Prob. 6–126
- 340. CHAPTER REVIEW 323 6 CHAPTER REVIEW Simple Truss A simple truss consists of triangular elements connected together by pinned joints. The forces within its members can be determined by assuming the members are all two-force members, connected concurrently at each joint. The members are either in tension or compression, or carry no force. Method of Joints The method of joints states that if a truss is in equilibrium, then each of its joints is also in equilibrium. For a plane truss, the concurrent force system at each joint must satisfy force equilibrium. To obtain a numerical solution for the forces in the members, select a joint that has a free-body diagram with at most two unknown forces and one known force. (This may require first finding the reactions at the supports.) Once a member force is determined, use its value and apply it to an adjacent joint. Remember that forces that are found to pull on the joint are tensile forces, and those that push on the joint are compressive forces. To avoid a simultaneous solution of two equations, set one of the coordinate axes along the line of action of one of the unknown forces and sum forces perpendicular to this axis.This will allow a direct solution for the other unknown. The analysis can also be simplified by first identifying all the zero-force members. ©Fy = 0 ©Fx = 0 B 45 500 N FBC (compression) FBA (tension) B 500 N A C 45 45 Roof truss
- 341. 324 CHAPTER 6 STRUCTURAL ANALYSIS 6 Method of Sections The method of sections states that if a truss is in equilibrium, then each segment of the truss is also in equilibrium. Pass a section through the truss and the member whose force is to be determined.Then draw the free-body diagram of the sectioned part having the least number of forces on it. Sectioned members subjected to pulling are in tension, and those that are subjected to pushing are in compression. Three equations of equilibrium are available to determine the unknowns. If possible, sum forces in a direction that is perpendicular to two of the three unknown forces. This will yield a direct solution for the third force. Sum moments about the point where the lines of action of two of the three unknown forces intersect, so that the third unknown force can be determined directly. ©MO = 0 ©Fy = 0 ©Fx = 0 FGC = 1.41 kN 1T2 -1000 N + FGC sin 45° = 0 + c©Fy = 0 a FGF = 2 kN 1C2 1000 N14 m2 - FGF (2 m) = 0 +©MC = 0 B 2 m 1000 N 2 m 2 m C D G F E A 2 m a a 2 m 1000 N 2 m 2 m C FBC 45 FGC G FGF
- 342. CHAPTER REVIEW 325 6 Space Truss A space truss is a three-dimensional truss built from tetrahedral elements, and is analyzed using the same methods as for plane trusses. The joints are assumed to be ball and socket connections. Frames and Machines Frames and machines are structures that contain one or more multiforce members, that is, members with three or more forces or couples acting on them. Frames are designed to support loads, and machines transmit and alter the effect of forces. The forces acting at the joints of a frame or machine can be determined by drawing the free-body diagrams of each of its members or parts. The principle of action–reaction should be carefully observed when indicating these forces on the free-body diagram of each adjacent member or pin. For a coplanar force system, there are three equilibrium equations available for each member. To simplify the analysis, be sure to recognize all two-force members. They have equal but opposite collinear forces at their ends. P FAB Cy Cx FAB FAB 2000 N Action–reaction B A B C 2000 N Two-force member Multi-force member
- 343. 326 CHAPTER 6 STRUCTURAL ANALYSIS 6 REVIEW PROBLEMS •6–129. Determine the force in each member of the truss and state if the members are in tension or compression. *6–128. Determine the forces which the pins at A and B exert on the two-member frame which supports the 100-kg crate. 6–127. Determine the clamping force exerted on the smooth pipe at B if a force of 20 lb is applied to the handles of the pliers.The pliers are pinned together at A. 6–130. The space truss is supported by a ball-and-socket joint at D and short links at C and E. Determine the force in each member and state if the members are in tension or compression.Take and . 6–131. The space truss is supported by a ball-and-socket joint at D and short links at C and E. Determine the force in each member and state if the members are in tension or compression. Take and . F2 = 5400j6 lb F1 = 5200i + 300j - 500k6 lb F2 = 5400j6 lb F1 = 5-500k6 lb A C B D 0.6 m 0.8 m 0.6 m 0.4 m Prob. 6–128 D A E 3 m 3 m 3 m 8 kN B 0.1 m C Prob. 6–129 3 ft 4 ft 3 ft x y z C D E A B F F2 F1 Probs. 6–130/131 A 20 lb 20 lb 10 in. 40 1.5 in. 0.5 in. B Prob. 6–127
- 344. REVIEW PROBLEMS 327 6 6–135. Determine the horizontal and vertical components of reaction at the pin supports A and E of the compound beam assembly. 6–134. The two-bar mechanism consists of a lever arm AB and smooth link CD, which has a fixed smooth collar at its end C and a roller at the other end D. Determine the force P needed to hold the lever in the position . The spring has a stiffness k and unstretched length 2L. The roller contacts either the top or bottom portion of the horizontal guide. u *6–132. Determine the horizontal and vertical components of reaction that the pins A and B exert on the two-member frame. Set . •6–133. Determine the horizontal and vertical components of reaction that pins A and B exert on the two-member frame. Set . F = 500 N F = 0 *6–136. Determine the force in members AB, AD, and AC of the space truss and state if the members are in tension or compression. 1.5 m 400 N/m 60 1 m 1 m B C A F Probs. 6–132/133 2 L L k C A B D P u Prob. 6–134 1.5 ft 1.5 ft 2 ft F {600k} lb 8 ft x y z B A C D Prob. 6–136 2 ft 2 kip/ft 1 ft 3 ft 6 ft 2 ft 1 ft A C E D B Prob. 6–135
- 345. These reinforcing rods will be encased in concrete in order to create a building column. The internal loadings developed within the material resist the external loading that is to be placed upon the column.
- 346. Internal Forces 7 A B (a) P1 P2 a a Fig. 7–1 (b) VB VB MB MB MA NB NB Ax Ay B B P1 P2 CHAPTER OBJECTIVES • To show how to use the method of sections to determine the internal loadings in a member. • To generalize this procedure by formulating equations that can be plotted so that they describe the internal shear and moment throughout a member. • To analyze the forces and study the geometry of cables supporting a load. 7.1 Internal Forces Developed in Structural Members To design a structural or mechanical member it is necessary to know the loading acting within the member in order to be sure the material can resist this loading. Internal loadings can be determined by using the method of sections.To illustrate this method, consider the cantilever beam in Fig. 7–1a. If the internal loadings acting on the cross section at point B are to be determined,we must pass an imaginary section a–a perpendicular to the axis of the beam through point B and then separate the beam into two segments.The internal loadings acting at B will then be exposed and become external on the free-body diagram of each segment, Fig. 7–1b.
- 347. 330 CHAPTER 7 INTERNAL FORCES 7 In each case, the link on the backhoe is a two-force member. In the top photo it is subjected to both bending and an axial load at its center. By making the member straight,as in the bottom photo,then only an axial force acts within the member. (b) VB VB MB MB MA NB NB Ax Ay B B P1 P2 (a) V N M Shear force Normal force Bending moment C Fig. 7–2 y z Ny Normal force My Torsional moment Vx Vz Mx x C Mz Shear force components Bending moment components (b) Fig. 7–1 The force component that acts perpendicular to the cross section, is termed the normal force. The force component that is tangent to the cross section is called the shear force, and the couple moment is referred to as the bending moment. The force components prevent the relative translation between the two segments, and the couple moment prevents the relative rotation. According to Newton’s third law, these loadings must act in opposite directions on each segment, as shown in Fig. 7–1b. They can be determined by applying the equations of equilibrium to the free-body diagram of either segment. In this case, however, the right segment is the better choice since it does not involve the unknown support reactions at A.A direct solution for is obtained by applying , is obtained from , and can be obtained by applying , since the moments of and about B are zero. In two dimensions, we have shown that three internal loading resultants exist, Fig. 7–2a; however in three dimensions, a general internal force and couple moment resultant will act at the section.The x, y, z components of these loadings are shown in Fig. 7–2b. Here is the normal force, and and are shear force components. is a torsional or twisting moment, and and are bending moment components. For most applications, these resultant loadings will act at the geometric center or centroid (C) of the section’s cross-sectional area. Although the magnitude for each loading generally will be different at various points along the axis of the member, the method of sections can always be used to determine their values. Mz Mx My Vz Vx Ny VB NB ©MB = 0 MB ©Fy = 0 VB ©Fx = 0 NB MB VB NB
- 348. 7.1 INTERNAL FORCES DEVELOPED IN STRUCTURAL MEMBERS 331 7 Procedure for Analysis The method of sections can be used to determine the internal loadings on the cross section of a member using the following procedure. Support Reactions. • Before the member is sectioned, it may first be necessary to determine its support reactions, so that the equilibrium equations can be used to solve for the internal loadings only after the member is sectioned. Free-Body Diagram. • Keep all distributed loadings, couple moments, and forces acting on the member in their exact locations, then pass an imaginary section through the member, perpendicular to its axis at the point where the internal loadings are to be determined. • After the section is made, draw a free-body diagram of the segment that has the least number of loads on it, and indicate the components of the internal force and couple moment resultants at the cross section acting in their postive directions to the established sign convention. Equations of Equilibrium. • Moments should be summed at the section. This way the normal and shear forces at the section are elminated, and we can obtain a direct solution for the moment. • If the solution of the equilibrium equations yields a negative scalar, the sense of the quantity is opposite to that shown on the free-body diagram. The designer of this shop crane realized the need for additional reinforcement around the joint in order to prevent severe internal bending of the joint when a large load is suspended from the chain hoist. Positive shear Positive normal force Positive moment M M V V N N V V M M Fig. 7–3 Sign Convention. Engineers generally use a sign convention to report the three internal loadings N, V, and M. Although this sign convention can be arbitrarily assigned, the one that is widely accepted will be used here, Fig. 7–3. The normal force is said to be positive if it creates tension, a positive shear force will cause the beam segment on which it acts to rotate clockwise, and a positive bending moment will tend to bend the segment on which it acts in a concave upward manner. Loadings that are opposite to these are considered negative. If the member is subjected to a three-dimensional external loading, then the internal loadings are usually expressed as positive or negative, in accordance with an established x, y, z coordinate system such as shown in Fig. 7–2.
- 349. 332 CHAPTER 7 INTERNAL FORCES 7 EXAMPLE 7.1 (a) A C B D 3 m 6 m 9 kN m 6 kN Fig. 7–4 Ay A D (b) 3 m 6 m Dy 9 kN m 6 kN A (c) 3 m VB NB MB 5 kN B 5 kN A (d) 6 kN 3 m C NC MC VC Determine the normal force, shear force, and bending moment acting just to the left, point B, and just to the right, point C, of the 6-kN force on the beam in Fig. 7–4a. SOLUTION Support Reactions. The free-body diagram of the beam is shown in Fig. 7–4b. When determining the external reactions, realize that the couple moment is a free vector and therefore it can be placed anywhere on the free-body diagram of the entire beam. Here we will only determine since the left segments will be used for the analysis. a Free-Body Diagrams. The free-body diagrams of the left segments AB and AC of the beam are shown in Figs. 7–4c and 7–4d. In this case the couple moment is not included on these diagrams since it must be kept in its original position until after the section is made and the appropriate segment is isolated. Equations of Equilibrium. Segment AB Ans. Ans. a Ans. Segment AC Ans. Ans. a Ans. NOTE: The negative sign indicates that VC acts in the opposite sense to that shown on the free-body diagram.Also, the moment arm for the 5-kN force in both cases is approximately 3 m since B and C are “almost” coincident. MC = 15 kN # m -15 kN213 m2 + MC = 0 +©MC = 0; V C = -1 kN 5 kN - 6 kN - V C = 0 + c©Fy = 0; NC = 0 : + ©Fx = 0; MB = 15 kN # m -15 kN213 m2 + MB = 0 +©MB = 0; V B = 5 kN 5 kN - V B = 0 + c©Fy = 0; NB = 0 : + ©Fx = 0; 9-kN # m Ay = 5 kN 9 kN # m + 16 kN216 m2 - Ay19 m2 = 0 +©MD = 0; Ay, 9-kN # m
- 350. 7.1 INTERNAL FORCES DEVELOPED IN STRUCTURAL MEMBERS 333 7 EXAMPLE 7.2 B C A 1.5 m 1.5 m 1200 N/m (a) Fig. 7–5 1.5 m (b) 1200 N/m 3 m wC (c) VC MC NC C B 0.5 m 600 N/m (600 N/m)(1.5 m) 1 2 Determine the normal force, shear force, and bending moment at C of the beam in Fig. 7–5a. SOLUTION Free-Body Diagram. It is not necessary to find the support reactions at A since segment BC of the beam can be used to determine the internal loadings at C. The intensity of the triangular distributed load at C is determined using similar triangles from the geometry shown in Fig. 7–5b, i.e., The distributed load acting on segment BC can now be replaced by its resultant force, and its location is indicated on the free-body diagram, Fig. 7–5c. Equations of Equilibrium Ans. Ans. a Ans. The negative sign indicates that acts in the opposite sense to that shown on the free-body diagram. MC MC = -225 N -MC - 1 2(600 Nm)(1.5 m)(0.5 m) = 0 +©MC = 0; VC = 450 N VC - 1 2(600 Nm)(1.5 m) = 0 + c©Fy = 0; NC = 0 : + ©Fx = 0; wC = (1200 Nm) a 1.5 m 3 m b = 600 Nm
- 351. 334 CHAPTER 7 INTERNAL FORCES 7 EXAMPLE 7.3 200 lb 266.7 lb 2 ft 2 ft 200 lb 3 4 5 200 lb 2 ft 2 ft 333.3 lb C (c) VB NB MB VB NB MB B A B (a) A 4 ft 4 ft 6 ft D B C 50 lb/ft Free-Body Diagrams. Passing an imaginary section perpendicular to the axis of member AC through point B yields the free-body diagrams of segments AB and BC shown in Fig. 7–6c. When constructing these diagrams it is important to keep the distributed loading where it is until after the section is made. Only then can it be replaced by a single resultant force. Equations of Equilibrium. Applying the equations of equilibrium to segment AB, we have Ans. Ans. a Ans. NOTE: As an exercise,try to obtain these same results using segment BC. MB = 400 lb # ft MB - 200 lb (4 ft) + 200 lb (2 ft) = 0 +©MB = 0; V B = 0 200 lb - 200 lb - VB = 0 + c©Fy = 0; NB = 267 lb NB - 266.7 lb = 0 : + ©Fx = 0; (b) 4 ft A C 4 ft Ay Ax 3 4 5 FDC FDC FDC 400 lb Fig. 7–6 Determine the normal force, shear force, and bending moment acting at point B of the two-member frame shown in Fig. 7–6a. SOLUTION Support Reactions. A free-body diagram of each member is shown in Fig. 7–6b. Since CD is a two-force member, the equations of equilibrium need to be applied only to member AC. a Ay = 200 lb Ay - 400 lb + A3 5 B(333.3 lb) = 0 + c©Fy = 0; Ax = 266.7 lb -Ax + A4 5 B1333.3 lb2 = 0 : + ©Fx = 0; FDC = 333.3 lb -400 lb (4 ft) + A3 5 B FDC (8 ft) = 0 +©MA = 0;
- 352. 7.1 INTERNAL FORCES DEVELOPED IN STRUCTURAL MEMBERS 335 7 EXAMPLE 7.4 (a) 1 m 1 m 1 m A E D B C 600 N 0.5 m 0.5 m Fig. 7–7 D C P P 45 A R C R (b) P R C 45 600 N VE NE ME C 848.5 N 0.5 m E 45 (c) Determine the normal force, shear force, and bending moment acting at point E of the frame loaded as shown in Fig. 7–7a. SOLUTION Support Reactions. By inspection, members AC and CD are two- force members, Fig. 7–7b. In order to determine the internal loadings at E, we must first determine the force R acting at the end of member AC.To obtain it, we will analyze the equilibrium of the pin at C. Summing forces in the vertical direction on the pin, Fig. 7–7b, we have Free-Body Diagram. The free-body diagram of segment CE is shown in Fig. 7–7c. Equations of Equilibrium. Ans. Ans. a Ans. NOTE: These results indicate a poor design. Member AC should be straight (from A to C) so that bending within the member is eliminated. If AC were straight then the internal force would only create tension in the member. ME = 300 N #m 848.5 cos 45° N10.5 m2 -ME = 0 +©ME = 0; NE = 600 N -848.5 sin 45° N + NE = 0 + c©Fy = 0; VE = 600 N 848.5 cos 45° N - VE = 0 : + ©Fx = 0; R sin 45° - 600 N = 0 R = 848.5 N + c ©Fy = 0;
- 353. 336 CHAPTER 7 INTERNAL FORCES 7 EXAMPLE 7.5 (a) Fig. 7–8 A 6 m 2.5 m 4 m 4 m (b) 3 m (c) 5.25 m 6.376 kN 13.5 kN z G A y x FA MA r The uniform sign shown in Fig. 7–8a has a mass of 650 kg and is supported on the fixed column. Design codes indicate that the expected maximum uniform wind loading that will occur in the area where it is located is 900 Pa. Determine the internal loadings at A. SOLUTION The idealized model for the sign is shown in Fig. 7–8b. Here the necessary dimensions are indicated. We can consider the free-body diagram of a section above point A since it does not involve the support reactions. Free-Body Diagram. The sign has a weight of and the wind creates a resultant force of , which acts perpendicular to the face of the sign. These loadings are shown on the free-body diagram, Fig. 7–8c. Equations of Equilibrium. Since the problem is three dimensional, a vector analysis will be used. Ans. Ans. NOTE: Here represents the normal force, whereas is the shear force. Also, the torsional moment is and the bending moment is determined from its components and ; i.e., . (Mb)A = 2(MA)2 x + (MA)2 y = 73.4 kN # m MAy = 570.9j6 kN # m MAx = 519.1i6 kN # m MAz = 5-40.5k6 kN # m, FAx = 513.5i6 kN FAz = 56.38k6 kN MA = 519.1i + 70.9j - 40.5k6 kN # m MA + ` i j k 0 3 5.25 -13.5 0 -6.376 ` = 0 MA + r * 1Fw + W2 = 0 ©MA = 0; FA = 513.5i + 6.38k6 kN FA - 13.5i - 6.376k = 0 ©F = 0; 900 Nm2 16 m212.5 m2 = 13.5 kN Fw = 6.376 kN, W = 65019.812 N =
- 354. 7.1 INTERNAL FORCES DEVELOPED IN STRUCTURAL MEMBERS 337 7 FUNDAMENTAL PROBLEMS F7–4. Determine the normal force, shear force, and moment at point C. F7–2. Determine the normal force, shear force, and moment at point C. F7–5. Determine the normal force, shear force, and moment at point C. A B C 15 kN 10 kN 1.5 m 1.5 m 1.5 m 1.5 m A B C 30 kN m 10 kN 1.5 m 1.5 m 1.5 m 1.5 m A B C 4.5 ft 4.5 ft 6 ft 3 kip/ft F7–3 A B C 12 kN 9 kN/m 1.5 m 1.5 m 1.5 m 1.5 m F7–4 A B C 3 m 3 m 9 kN/m F7–5 F7–3. Determine the normal force, shear force, and moment at point C. A C B 3 m 3 m 6 kN/m F7–6 F7–6. Determine the normal force, shear force, and moment at point C.Assume A is pinned and B is a roller. F7–1. Determine the normal force, shear force, and moment at point C. F7–1 F7–2
- 355. 338 CHAPTER 7 INTERNAL FORCES 7 PROBLEMS *7–4. Determine the internal normal force, shear force, and moment at points E and F in the beam. 7–2. Determine the shear force and moment at points C and D. 7–3. Determine the internal normal force, shear force, and moment at point C in the simply supported beam. Point C is located just to the right of the 1500-lb ft couple moment. – •7–5. Determine the internal normal force, shear force, and moment at point C. D B A E F 1.5 m 300 N/m 45 1.5 m 1.5 m 1.5 m C Prob. 7–4 3 m 2 m 1.5 m 1 m 0.2 m 400 N A C B Prob. 7–5 40 kip ft 8 ft 8 ft 8 ft 8 kip A B C D Prob. 7–1 6 ft A C D E B 6 ft 2 ft 4 ft 4 ft 300 lb 200 lb 500 lb Prob. 7–2 B A C 500 lb/ft 1500 lb ft 6 ft 30 6 ft Prob. 7–3 •7–1. Determine the internal normal force and shear force, and the bending moment in the beam at points C and D.Assume the support at B is a roller. Point C is located just to the right of the 8-kip load.
- 356. 7.1 INTERNAL FORCES DEVELOPED IN STRUCTURAL MEMBERS 339 7 7–10. Determine the internal normal force, shear force, and moment at point C in the double-overhang beam. *7–8. Determine the internal normal force, shear force, and moment at points C and D in the simply supported beam. Point D is located just to the left of the 5-kN force. 7–7. Determine the internal normal force, shear force, and moment at point C in the cantilever beam. 7–11. Determine the internal normal force, shear force, and moment at points C and D in the simply supported beam. Point D is located just to the left of the 10-kN concentrated load. C B A 3 m 4 kN/m 3 m Prob. 7–6 A B C w0 L –– 2 L –– 2 Prob. 7–7 A C D B 3 kN/m 5 kN 3 m 1.5 m 1.5 m Prob. 7–8 A B C 90 6 in. Prob. 7–9 A C B 1.5 m 3 kN/m 1.5 m 1.5 m 1.5 m Prob. 7–10 •7–9. The bolt shank is subjected to a tension of 80 lb. Determine the internal normal force, shear force, and moment at point C. A C D B 1.5 m 6 kN/m 10 kN 1.5 m 1.5 m 1.5 m Prob. 7–11 7–6. Determine the internal normal force, shear force, and moment at point C in the simply supported beam.
- 357. 340 CHAPTER 7 INTERNAL FORCES 7 *7–16. Determine the internal normal force, shear force, and moment in the cantilever beam at point B. •7–13. Determine the internal normal force, shear force, and moment at point D of the two-member frame. 7–14. Determine the internal normal force, shear force, and moment at point E of the two-member frame. •7–17. Determine the ratio of for which the shear force will be zero at the midpoint C of the double-overhang beam. ab 6 ft 6 ft 6 ft 6 ft 5 kip 0.5 kip/ft A C D B Prob. 7–12 2 m 1.5 m 250 N/m 300 N/m 4 m A C D E B Probs. 7–13/14 200 lb/ft 200 lb/ft 300 lb/ft 4 ft A F E C B D 4 ft 4 ft 4 ft Prob. 7–15 A 6 kip/ft B 12 ft 3 ft Prob. 7–16 B C a b/2 b/2 w0 a A B C Prob. 7–17 7–15. Determine the internal normal force, shear force, and moment acting at point C and at point D, which is located just to the right of the roller support at B. 7–18. Determine the internal normal force, shear force, and moment at points D and E in the overhang beam. Point D is located just to the left of the roller support at B, where the couple moment acts. 2 kN/m 5 kN 3 m 1.5 m 3 4 5 A D B E C 6 kN m 1.5 m Prob. 7–18 *7–12. Determine the internal normal force, shear force, and moment in the beam at points C and D. Point D is just to the right of the 5-kip load.
- 358. 7.1 INTERNAL FORCES DEVELOPED IN STRUCTURAL MEMBERS 341 7 7–22. The stacker crane supports a 1.5-Mg boat with the center of mass at G. Determine the internal normal force, shear force, and moment at point D in the girder.The trolley is free to roll along the girder rail and is located at the position shown. Only vertical reactions occur at A and B. *7–20. Determine the internal normal force, shear force, and moment at points D and E in the compound beam. Point E is located just to the left of the 10-kN concentrated load.Assume the support at A is fixed and the connection at B is a pin. 7–23. Determine the internal normal force, shear force, and moment at points D and E in the two members. w0 w0 A B L a –– 2 a –– 2 Prob. 7–19 10 kN 2 kN/m D B E C A 1.5 m 1.5 m 1.5 m 1.5 m Prob. 7–20 A F G E B D C 2 ft 2 ft 2 ft 2 ft 1.5 ft 2 ft 500 lb 600 lb Prob. 7–21 3.5 m D G C B A 5 m 7.5 m 1 m 1 m 2 m 2 m Prob. 7–22 •7–21. Determine the internal normal force, shear force, and moment at points F and G in the compound beam. Point F is located just to the right of the 500-lb force, while point G is located just to the right of the 600-lb force. 2 m 1 m 0.75 m 0.75 m 60 N D E B C A 60 30 Prob. 7–23 7–19. Determine the distance a in terms of the beam’s length L between the symmetrically placed supports A and B so that the internal moment at the center of the beam is zero.
- 359. 342 CHAPTER 7 INTERNAL FORCES 7 7–26. The beam has a weight w per unit length. Determine the internal normal force, shear force, and moment at point C due to its weight. •7–25. Determine the internal normal force, shear force, and moment at points D and E of the frame which supports the 200-lb crate. Neglect the size of the smooth peg at C. *7–24. Determine the internal normal force, shear force, and moment at points F and E in the frame. The crate weighs 300 lb. 7–27. Determine the internal normal force, shear force, and moment acting at point C. The cooling unit has a total mass of 225 kg with a center of mass at G. C B E A D 4 ft 4.5 ft 2 ft 1.5 ft 1.5 ft Prob. 7–25 B A C L –– 2 L –– 2 u Prob. 7–26 3 m F 3 m 30 30 0.2 m G A B E D C Prob. 7–27 1.5 ft 1.5 ft 1.5 ft 1.5 ft 0.4 ft 4 ft A B F C E D Prob. 7–24
- 360. 7.1 INTERNAL FORCES DEVELOPED IN STRUCTURAL MEMBERS 343 7 *7–32. Determine the internal normal force, shear force, and moment acting at points B and C on the curved rod. 7–30. The jib crane supports a load of 750 lb from the trolley which rides on the top of the jib. Determine the internal normal force, shear force, and moment in the jib at point C when the trolley is at the position shown.The crane members are pinned together at B, E and F and supported by a short link BH. 7–31. The jib crane supports a load of 750 lb from the trolley which rides on the top of the jib. Determine the internal normal force, shear force, and moment in the column at point D when the trolley is at the position shown. The crane members are pinned together at B, E and F and supported by a short link BH. *7–28. The jack AB is used to straighten the bent beam DE using the arrangement shown. If the axial compressive force in the jack is 5000 lb, determine the internal moment developed at point C of the top beam. Neglect the weight of the beams. •7–29. Solve Prob. 7–28 assuming that each beam has a uniform weight of . 150 lbft •7–33. Determine the internal normal force, shear force, and moment at point D which is located just to the right of the 50-N force. 1 ft 1 ft 3 ft 5 ft 1 ft 3 ft 750 lb 2 ft 3 ft G F C B H D E A Probs. 7–30/31 10 ft 10 ft 2 ft 2 ft A B C D E Probs. 7–28/29 45 30 2 ft B C A 3 4 5 500 lb Prob. 7–32 50 N 50 N 50 N 50 N 600 mm D C B A 30 30 30 30 30 Prob. 7–33
- 361. 344 CHAPTER 7 INTERNAL FORCES 7 •7–37. The shaft is supported by a thrust bearing at A and a journal bearing at B. Determine the x, y, z components of internal loading at point C. 7–35. Determine the x, y, z components of internal loading at a section passing through point C in the pipe assembly. Neglect the weight of the pipe.Take and *7–36. Determine the x,y,z components of internal loading at a section passing through point C in the pipe assembly.Neglect the weight of the pipe. Take and F2 = 5250i - 150j - 200k6 lb. F1 = 5-80i + 200j - 300k6 lb F2 = 5150i - 300k6 lb. F1 = 5350j - 400k6 lb 7–38. Determine the x, y, z components of internal loading in the rod at point D. There are journal bearings at A, B, and C. Take 7–39. Determine the x, y, z components of internal loading in the rod at point E.Take F = 57i - 12j - 5k6 kN. F = 57i - 12j - 5k6 kN. F1 F2 2 ft x z y 3 ft C B A M 1.5 ft Prob. 7–34 x z y C 1.5 ft 2 ft F1 F2 3 ft Probs. 7–35/36 1 m 1 m 0.5 m 0.2 m 0.2 m 1 m 750 N 750 N 600 N z C y x 900 N A B Prob. 7–37 0.75 m 0.2 m 0.2 m 0.5 m 0.5 m A 3 kN m C z x B D E F y 0.6 m Probs. 7–38/39 7–34. Determine the x, y, z components of internal loading at point C in the pipe assembly. Neglect the weight of the pipe. The load is , , and . M = 5-30k6 lb # ft F2 = 5-80i6 lb F1 = 5-24i -10k6 lb
- 362. 7.2 SHEAR AND MOMENT EQUATIONS AND DIAGRAMS 345 7 *7.2 Shear and Moment Equations and Diagrams Beams are structural members designed to support loadings applied perpendicular to their axes. In general, they are long and straight and have a constant cross-sectional area.They are often classified as to how they are supported. For example, a simply supported beam is pinned at one end and roller supported at the other, as in Fig. 7–9a, whereas a cantilevered beam is fixed at one end and free at the other.The actual design of a beam requires a detailed knowledge of the variation of the internal shear force V and bending moment M acting at each point along the axis of the beam.* These variations of V and M along the beam’s axis can be obtained by using the method of sections discussed in Sec. 7.1. In this case, however, it is necessary to section the beam at an arbitrary distance x from one end and then apply the equations of equilibrium to the segment having the length x. Doing this we can then obtain V and M as functions of x. In general, the internal shear and bending-moment functions will be discontinuous, or their slopes will be discontinuous, at points where a distributed load changes or where concentrated forces or couple moments are applied. Because of this, these functions must be determined for each segment of the beam located between any two discontinuities of loading. For example, segments having lengths and will have to be used to describe the variation of V and M along the length of the beam in Fig. 7–9a. These functions will be valid only within regions from O to a for from a to b for and from b to L for If the resulting functions of x are plotted, the graphs are termed the shear diagram and bending-moment diagram, Fig. 7–9b and Fig. 7–9c, respectively. x3. x2, x1, x3 x2, x1, *The internal normal force is not considered for two reasons. In most cases, the loads applied to a beam act perpendicular to the beam’s axis and hence produce only an internal shear force and bending moment. And for design purposes, the beam’s resistance to shear, and particularly to bending, is more important than its ability to resist a normal force. To save on material and thereby produce an efficient design, these beams, also called girders, have been tapered, since the internal moment in the beam will be larger at the supports, or piers, than at the center of the span. O L P b a x3 x2 x1 w (a) Fig. 7–9 V x (b) a b L M x (c) b a L
- 363. 346 CHAPTER 7 INTERNAL FORCES 7 Positive shear Positive moment Beam sign convention M M V V V V M M Fig. 7–10 This extended towing arm must resist both bending and shear loadings throughout its length due to the weight of the vehicle.The variation of these loadings must be known if the arm is to be properly designed. Procedure for Analysis The shear and bending-moment diagrams for a beam can be constructed using the following procedure. Support Reactions. • Determine all the reactive forces and couple moments acting on the beam and resolve all the forces into components acting perpendicular and parallel to the beam’s axis. Shear and Moment Functions. • Specify separate coordinates x having an origin at the beam’s left end and extending to regions of the beam between concentrated forces and/or couple moments, or where the distributed loading is continuous. • Section the beam at each distance x and draw the free-body diagram of one of the segments. Be sure V and M are shown acting in their positive sense, in accordance with the sign convention given in Fig. 7–10. • The shear V is obtained by summing forces perpendicular to the beam’s axis. • The moment M is obtained by summing moments about the sectioned end of the segment. Shear and Moment Diagrams. • Plot the shear diagram (V versus x) and the moment diagram (M versus x). If computed values of the functions describing V and M are positive, the values are plotted above the x axis, whereas negative values are plotted below the x axis. • Generally, it is convenient to plot the shear and bending-moment diagrams directly below the free-body diagram of the beam.
- 364. 7.2 SHEAR AND MOMENT EQUATIONS AND DIAGRAMS 347 7 EXAMPLE 7.6 Draw the shear and moment diagrams for the shaft shown in Fig.7–11a. The support at A is a thrust bearing and the support at C is a journal bearing. SOLUTION Support Reactions. The support reactions are shown on the shaft’s free-body diagram, Fig. 7–11d. Shear and Moment Functions. The shaft is sectioned at an arbitrary distance x from point A, extending within the region AB, and the free-body diagram of the left segment is shown in Fig. 7–11b. The unknowns V and M are assumed to act in the positive sense on the right-hand face of the segment according to the established sign convention.Applying the equilibrium equations yields (1) a (2) A free-body diagram for a left segment of the shaft extending a distance x within the region BC is shown in Fig. 7–11c. As always, V and M are shown acting in the positive sense. Hence, (3) a (4) Shear and Moment Diagrams. When Eqs. 1 through 4 are plotted within the regions in which they are valid, the shear and moment diagrams shown in Fig. 7–11d are obtained.The shear diagram indicates that the internal shear force is always 2.5 kN (positive) within segment AB. Just to the right of point B, the shear force changes sign and remains at a constant value of for segment BC. The moment diagram starts at zero, increases linearly to point B at where and thereafter decreases back to zero. NOTE: It is seen in Fig. 7–11d that the graphs of the shear and moment diagrams are discontinuous where the concentrated force acts, i.e., at points A, B, and C. For this reason, as stated earlier, it is necessary to express both the shear and moment functions separately for regions between concentrated loads. It should be realized, however, that all loading discontinuities are mathematical, arising from the idealization of a concentrated force and couple moment. Physically, loads are always applied over a finite area, and if the actual load variation could be accounted for, the shear and moment diagrams would then be continuous over the shaft’s entire length. Mmax = 2.5 kN12 m2 = 5 kN # m, x = 2 m, -2.5 kN M = 110 - 2.5x2 kN # m M + 5 kN1x - 2 m2 - 2.5 kN1x2 = 0 +©M = 0; V = -2.5 kN 2.5 kN - 5 kN - V = 0 + c©Fy = 0; M = 2.5x kN # m +©M = 0; V = 2.5 kN + c©Fy = 0; 2 m 5 kN (a) B A C 2 m Fig. 7–11 x 2.5 kN (b) A M V 0 x 2 m 2.5 kN x 5 kN M V 2 m x 2 m A B (c) 2 m x 4 m M (10 2.5x) 2.5 kN 2.5 kN V (kN) V 2.5 V 2.5 x (m) 5 kN C A (d) B M 2.5x M (kN m) Mmax 5 x (m) 2 2 4 4
- 365. 348 CHAPTER 7 INTERNAL FORCES 7 Draw the shear and moment diagrams for the beam shown in Fig. 7–12a. SOLUTION Support Reactions. The support reactions are shown on the beam’s free-body diagram, Fig. 7–12c. Shear and Moment Functions. A free-body diagram for a left segment of the beam having a length x is shown in Fig. 7–12b. Due to proportional triangles, the distributed loading acting at the end of this segment has an intensity of or . It is replaced by a resultant force after the segment is isolated as a free-body diagram. The magnitude of the resultant force is equal to This force acts through the centroid of the distributed loading area, a distance from the right end. Applying the two equations of equilibrium yields (1) a (2) Shear and Moment Diagrams. The shear and moment diagrams shown in Fig. 7–12c are obtained by plotting Eqs. 1 and 2. The point of zero shear can be found using Eq. 1: NOTE: It will be shown in Sec. 7–3 that this value of x happens to represent the point on the beam where the maximum moment occurs. Using Eq. 2, we have = 31.2 kN # m Mmax = a915.202 - 15.2023 9 b kN # m x = 5.20 m V = 9 - x2 3 = 0 M = a9x - x3 9 b kN # m M + 1 3 x2 a x 3 b - 9x = 0 +©M = 0; V = a9 - x2 3 b kN 9 - 1 3 x2 - V = 0 + c©Fy = 0; 1 3 x 1 21x2A2 3 xB = 1 3 x2 . w = (23)x wx = 69 EXAMPLE 7.7 (a) 9 m 6 kN/m Fig. 7–12 (b) x 1 3 2 3 x 3 x2 kN x kN/m M V 9 kN 6 kN/m 9 kN 18 kN V (kN) 5.20 m x (m) V 9 M (kN m) M 9x Mmax 31.2 (c) 9 18 x2 3 x3 9 x (m) 9 9 5.20
- 366. 7.2 SHEAR AND MOMENT EQUATIONS AND DIAGRAMS 349 7 FUNDAMENTAL PROBLEMS F7–10. Determine the shear and moment as a function of x, and then draw the shear and moment diagrams. F7–8. Determine the shear and moment as a function of x, and then draw the shear and moment diagrams. F7–7. Determine the shear and moment as a function of x, and then draw the shear and moment diagrams. F7–11. Determine the shear and moment as a function of x, where , and then draw the shear and moment diagrams. 0 … x 6 3 m and 3 m 6 x … 6 m F7–9. Determine the shear and moment as a function of x, and then draw the shear and moment diagrams. F7–12. Determine the shear and moment as a function of x, where , and then draw the shear and moment diagrams. 0 … x 6 3 m and 3 m 6 x … 6 m 3 m x 6 kN A 9 ft 2 kip/ft 15 kip ft x A 3 m 6 kN/m A x x B A 6 m 12 kN m B A C x 3 m 3 m 30 kN m B A C 12 kN m 4 kN 3 m 3 m x F7–7 F7–8 F7–9 F7–12 F7–11 F7–10
- 367. 350 CHAPTER 7 INTERNAL FORCES 7 2 kN/m 6 kN m 2 m A Prob. 7–43 L/2 L/2 M0 A B Probs. 7–44/45 A B w0 L –– 2 L –– 2 Prob. 7–46 4 m 2 m 9 kN A B Prob. 7–41 PROBLEMS 7–43. Draw the shear and moment diagrams for the cantilever beam. •7–41. Draw the shear and moment diagrams for the simply supported beam. *7–40. Draw the shear and moment diagrams for the beam (a) in terms of the parameters shown; (b) set L = 12 ft. a = 5 ft, P = 800 lb, *7–44. Draw the shear and moment diagrams for the beam (a) in terms of the parameters shown; (b) set , . •7–45. If , the beam will fail when the maximum shear force is or the maximum bending moment is . Determine the largest couple moment the beam will support. M0 Mmax = 22 kN # m V max = 5 kN L = 9 m L = 8 m M0 = 500 N # m 7–42. Draw the shear and moment diagrams for the beam ABCDE.All pulleys have a radius of 1 ft. Neglect the weight of the beam and pulley arrangement.The load weighs 500 lb. 7–46. Draw the shear and moment diagrams for the simply supported beam. a a L P P Prob. 7–40 A B C D E 8 ft 2 ft 2 ft 2 ft 3 ft 2 ft 3 ft Prob. 7–42
- 368. 7.2 SHEAR AND MOMENT EQUATIONS AND DIAGRAMS 351 7 300 N/m 4 m 300 N m A B Prob. 7–47 A B C 4 m 2 m 8 kN/m Prob. 7–48 5 m 5 m 2 kN/m 50 kN m A B C Prob. 7–49 *7–48. Draw the shear and moment diagrams for the overhang beam. 7–47. Draw the shear and moment diagrams for the simply supported beam. •7–49. Draw the shear and moment diagrams for the beam. A B 3 m 1.5 kN/m Prob. 7–51 250 lb/ft 150 lb ft 150 lb ft A B 20 ft Prob. 7–50 A B 150 lb/ft 12 ft 300 lb ft Prob. 7–52 7–50. Draw the shear and moment diagrams for the beam. 7–51. Draw the shear and moment diagrams for the beam. *7–52. Draw the shear and moment diagrams for the simply supported beam.
- 369. 352 CHAPTER 7 INTERNAL FORCES 7 4 kN/m 3 m 3 m A B Prob. 7–57 w0 2w0 6 ft 6 ft A B Prob. 7–58 A B C 9 ft 4.5 ft 30 lb/ft 180 lb ft Prob. 7–53 12 ft A 12 ft 4 kip/ft Prob. 7–55 300 lb 200 lb/ft A 6 ft Prob. 7–56 •7–53. Draw the shear and moment diagrams for the beam. L w A B Prob. 7.54 7–54. If the beam will fail when the maximum shear force is or the maximum moment is Determine the largest intensity of the distributed loading it will support. w Mmax = 1200 lb # ft. Vmax = 800 lb, L = 18 ft, 7–55. Draw the shear and moment diagrams for the beam. *7–56. Draw the shear and moment diagrams for the cantilevered beam. •7–57. Draw the shear and moment diagrams for the overhang beam. 7–58. Determine the largest intensity of the distributed load that the beam can support if the beam can withstand a maximum shear force of and a maximum bending moment of . Mmax = 600 lb # ft Vmax = 1200 lb w0
- 370. 7.2 SHEAR AND MOMENT EQUATIONS AND DIAGRAMS 353 7 L a A B w0 Prob. 7–59 w0 4.5 m 1.5 m A B C Prob. 7–60 A B C 500 lb/ft 6 ft 3 ft Prob. 7–61 A L x 2 r0 r0 Prob. 7–62 y z x y 4 ft 2 ft 4 lb/ft Prob. 7–63 r w u Prob. 7–64 *7–60. Determine the placement a of the roller support B so that the maximum moment within the span AB is equivalent to the moment at the support B. •7–61. The compound beam is fix supported at A, pin connected at B and supported by a roller at C. Draw the shear and moment diagrams for the beam. 7–59. Determine the largest intensity of the distributed load that the beam can support if the beam can withstand a maximum bending moment of and a maximum shear force of . V max = 80 kN Mmax = 20 kN # m w0 7–62. The frustum of the cone is cantilevered from point A. If the cone is made from a material having a specific weight of , determine the internal shear force and moment in the cone as a function of x. g 7–63. Express the internal shear and moment components acting in the rod as a function of y, where 0 … y … 4 ft. *7–64. Determine the normal force, shear force, and moment in the curved rod as a function of u.
- 371. 354 CHAPTER 7 INTERNAL FORCES 7 Relation Between the Distributed Load and Shear. If we apply the force equation of equilibrium to the segment, then Dividing by , and letting , we get (7–1) slope of shear diagram = distributed load intensity dV dx = w(x) ¢x : 0 ¢x ¢V = w(x)¢x V + w(x)¢x - (V + ¢V) = 0 + c ©Fy = 0; In order to design the beam used to support these power lines, it is important to first draw the shear and moment diagrams for the beam. x F1 F2 w w w (x) x B M1 M2 C x D A (a) Fig. 7–13 M V M V x M F w(x) x w(x) V (b) k (x) O *7.3 Relations between Distributed Load, Shear, and Moment If a beam is subjected to several concentrated forces, couple moments, and distributed loads, the method of constructing the shear and bending- moment diagrams discussed in Sec. 7–2 may become quite tedious. In this section a simpler method for constructing these diagrams is discussed—a method based on differential relations that exist between the load, shear, and bending moment. Distributed Load. Consider the beam AD shown in Fig. 7–13a, which is subjected to an arbitrary load and a series of concentrated forces and couple moments. In the following discussion, the distributed load will be considered positive when the loading acts upward as shown.A free-body diagram for a small segment of the beam having a length is chosen at a point x along the beam which is not subjected to a concentrated force or couple moment, Fig. 7–13b. Hence any results obtained will not apply at these points of concentrated loading. The internal shear force and bending moment shown on the free-body diagram are assumed to act in the positive sense according to the established sign convention. Note that both the shear force and moment acting on the right-hand face must be increased by a small, finite amount in order to keep the segment in equilibrium. The distributed loading has been replaced by a resultant force that acts at a fractional distance from the right end, where [for example, if w(x) is uniform, . 1 2 k = 0 6 k 6 1 k1¢x2 ¢F = w1x2 ¢x ¢x w = w1x2
- 372. 7.3 RELATIONS BETWEEN DISTRIBUTED LOAD, SHEAR, AND MOMENT 355 7 If we rewrite the above equation in the form dV = w(x)dx and perform an integration between any two points B and C on the beam, we see that (7–2) Relation Between the Shear and Moment. If we apply the moment equation of equilibrium about point O on the free-body diagram in Fig. 7–13b, we get a Dividing both sides of this equation by , and letting 0, yields (7–3) In particular, notice that the absolute maximum bending moment occurs at the point where the slope , since this is where the shear is equal to zero. If Eq. 7–3 is rewritten in the form and integrated between any two points B and C on the beam, we have (7–4) As stated previously, the above equations do not apply at points where a concentrated force or couple moment acts. These two special cases create discontinuities in the shear and moment diagrams, and as a result, each deserves separate treatment. Force. A free-body diagram of a small segment of the beam in Fig. 7–13a, taken from under one of the forces, is shown in Fig. 7–14a. Here force equilibrium requires (7–5) Since the change in shear is positive, the shear diagram will “jump” upward when F acts upward on the beam. Likewise, the jump in shear is downward when F acts downward. 1¢V2 ¢V = F + c ©Fy = 0; Change in moment = Area under shear diagram ¢M = L V dx dM = 1 V dx dMdx = 0 |M|max Slope of moment diagram = Shear dM dx = V ¢x : ¢x ¢M = V¢x + k w(x)¢x2 (M + ¢M) - [w(x)¢x] k¢x - V¢x - M = 0 +©M0 = 0; Change in shear = Area under loading curve ¢V = L w1x2 dx V M V x V M M (a) F Fig. 7–14
- 373. 356 CHAPTER 7 INTERNAL FORCES 7 Couple Moment. If we remove a segment of the beam in Fig. 7–13a that is located at the couple moment , the free-body diagram shown in Fig. 7–14b results. In this case letting moment equilibrium requires a (7–6) Thus, the change in moment is positive, or the moment diagram will “jump” upward if is clockwise. Likewise, the jump is downward when is counterclockwise. The examples which follow illustrate application of the above equations when used to construct the shear and moment diagrams.After working through these examples, it is recommended that you solve Examples 7.6 and 7.7 using this method. M0 ¢M M0 ¢M = M0 +©M = 0; ¢x : 0, M0 M V M V x M V (b) M0 This concrete beam is used to support the deck. Its size and the placement of steel reinforcement within it can be determined once the shear and moment diagrams have been established. Important Points • The slope of the shear diagram at a point is equal to the intensity of the distributed loading, where positive distributed loading is upward, i.e., . • If a concentrated force acts upward on the beam, the shear will jump upward by the same amount. • The change in the shear between two points is equal to the area under the distributed-loading curve between the points. • The slope of the moment diagram at a point is equal to the shear, i.e., . • The change in the moment between two points is equal to the area under the shear diagram between the two points. • If a clockwise couple moment acts on the beam, the shear will not be affected; however, the moment diagram will jump upward by the amount of the moment. • Points of zero shear represent points of maximum or minimum moment since • Because two integrations of are involved to first determine the change in shear, , then to determine the change in moment, , then if the loading curve is a polynomial of degree n, will be a curve of degree and will be a curve of degree n + 2. M = M1x2 n + 1, V = V1x2 w = w1x2 ¢M = 1 V dx ¢V = 1 w (x) dx w = w(x) dMdx = 0. ¢M dMdx = V ¢V dVdx = w1x2 Fig. 7–14
- 374. 7.3 RELATIONS BETWEEN DISTRIBUTED LOAD, SHEAR, AND MOMENT 357 7 EXAMPLE 7.8 Draw the shear and moment diagrams for the cantilever beam in Fig. 7–15a. 2 kN 1.5 kN/m (a) A B 2 m 2 m Fig. 7–15 SOLUTION The support reactions at the fixed support B are shown in Fig. 7–15b. Shear Diagram. The shear at end A is –2 kN. This value is plotted at x = 0, Fig. 7–15c. Notice how the shear diagram is constructed by following the slopes defined by the loading w. The shear at x = 4 m is –5 kN, the reaction on the beam. This value can be verified by finding the area under the distributed loading; i.e., Moment Diagram. The moment of zero at x = 0 is plotted in Fig. 7–15d. Construction of the moment diagram is based on knowing its slope which is equal to the shear at each point. The change of moment from x = 0 to x = 2 m is determined from the area under the shear diagram. Hence, the moment at x = 2 m is This same value can be determined from the method of sections, Fig. 7–15e. Mƒx=2 m = Mƒx=0 + ¢M = 0 + [-2 kN(2 m)] = -4 kN # m V ƒx=4 m = Vƒx=2 m + ¢V = -2 kN - (1.5 kNm)(2m) = -5 kN (d) (c) (b) 2 kN 2 m 2 m 2 4 5 2 By 5 kN MB 11 kN m x (m) V (kN) 2 0 4 11 4 x (m) M (kN m) w 0 slope 0 w negative constant slope negative constant V negative constant slope negative constant V negative increasing slope negative increasing 1.5 kN/m (e) 2 m V 2 kN M 4 kN m 2 kN
- 375. 358 CHAPTER 7 INTERNAL FORCES 7 Draw the shear and moment diagrams for the overhang beam in Fig. 7–16a. EXAMPLE 7.9 4 kN/m 4 m 2 m (a) A B Fig. 7–16 SOLUTION The support reactions are shown in Fig. 7–16b. Shear Diagram. The shear of –2 kN at end A of the beam is plotted at x = 0, Fig. 7–16c. The slopes are determined from the loading and from this the shear diagram is constructed, as indicated in the figure. In particular, notice the positive jump of 10 kN at x = 4 m due to the force , as indicated in the figure. Moment Diagram. The moment of zero at x = 0 is plotted, Fig. 7–16d, then following the behavior of the slope found from the shear diagram, the moment diagram is constructed. The moment at is found from the area under the shear diagram. We can also obtain this value by using the method of sections, as shown in Fig. 7–16e. Mƒ x=4 m = Mƒx=0 + ¢M = 0 + [-2 kN(4 m)] = -8 kN # m x = 4 m By 4 m 2 m Ay 2 kN By 10 kN A 2 8 4 6 4 0 x (m) V (kN) 6 8 0 x (m) M (kN m) w 0 slope 0 V positive decreasing slope positive decreasing V negative constant slope negative constant w negative constant slope negative constant (d) (c) (b) 4 kN/m 4 m 2 kN A (e) V 2 kN M 8 kN m
- 376. 7.3 RELATIONS BETWEEN DISTRIBUTED LOAD, SHEAR, AND MOMENT 359 7 EXAMPLE 7.10 The shaft in Fig. 7–17a is supported by a thrust bearing at A and a journal bearing at B. Draw the shear and moment diagrams. B A 12 ft (a) 120 lb/ft Fig. 7–17 SOLUTION The support reactions are shown in Fig. 7–17b. Shear Diagram. As shown in Fig. 7–17c, the shear at x = 0 is +240. Following the slope defined by the loading, the shear diagram is constructed, where at B its value is –480 lb. Since the shear changes sign, the point where V = 0 must be located.To do this we will use the method of sections. The free-body diagram of the left segment of the shaft, sectioned at an arbitrary position x within the region 0 x 9 ft, is shown in Fig. 7–17e. Notice that the intensity of the distributed load at x is w = 10x, which has been found by proportional triangles, i.e., . Thus, for V = 0, Moment Diagram. The moment diagram starts at 0 since there is no moment at A, then it is constructed based on the slope as determined from the shear diagram.The maximum moment occurs at x = 6.93 ft, where the shear is equal to zero, since dM/dx = V = 0, Fig. 7–17e, x = 6.93 ft 240 lb -1 2(10x)x = 0 + c©Fy = 0; 12012 = wx … x (ft) B 120 lb/ft A 12 ft Ay 240 lb By 480 lb 12 6.93 6.93 12 240 480 V (lb) x (ft) 0 0 M (lb ft) w negative increasing slope negative increasing V negative increasing slope negative increasing (d) (c) (b) linear parabolic cubic 1109 positive decreasing A x (e) Ay 240 lb x 3 [ ] x 1 2 10 x 10 x V M Finally, notice how integration, first of the loading w which is linear, produces a shear diagram which is parabolic, and then a moment diagram which is cubic. Mmax = 1109 lb # ft a ; Mmax + 1 2 [(10)(6.93)] 6.93 A1 3 (6.93)B -240(6.93) = 0 +©M = 0
- 377. 360 CHAPTER 7 INTERNAL FORCES 7 FUNDAMENTAL PROBLEMS F7–16. Draw the shear and moment diagrams for the beam. F7–14. Draw the shear and moment diagrams for the beam. F7–13. Draw the shear and moment diagrams for the beam. F7–17. Draw the shear and moment diagrams for the beam. 1 m 1 m 1 m 8 kN 6 kN 4 kN A 6 kN 8 kN/m 1.5 m 1.5 m A B A 6 ft 6 ft 6 ft 6 kip 12 kip B A 6 kN/m 1.5 m 3 m 6 kN/m 1.5 m A B 3 m 6 kN/m 6 kN/m 3 m A B 3 m 9 kN/m 3 m F7–15. Draw the shear and moment diagrams for the beam. F7–13 F7–16 F7–14 F7–17 F7–14 F7–15 F7–18. Draw the shear and moment diagrams for the beam.
- 378. 7.3 RELATIONS BETWEEN DISTRIBUTED LOAD, SHEAR, AND MOMENT 361 7 300 lb 600 lb 400 lb B A 2 ft 2 ft 2 ft 2 ft Prob. 7–65 A B 5 kN 10 kN 5 kN 2 m 2 m 2 m 2 m Prob. 7–66 PROBLEMS A B M = 10 kN m 2 m 2 m 2 m 6 kN 18 kN Prob. 7–67 A B M 2 kN m 4 kN 2 m 2 m 2 m Prob. 7–68 A B 2 m 2 m 2 m 10 kN 10 kN 15 kN m Prob. 7–69 P L –– 3 L –– 3 L –– 3 A B P Prob. 7–70 •7–65. The shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B. Draw the shear and moment diagrams for the shaft. 7–66. Draw the shear and moment diagrams for the double overhang beam. 7–67. Draw the shear and moment diagrams for the overhang beam. *7–68. Draw the shear and moment diagrams for the simply supported beam. •7–69. Draw the shear and moment diagrams for the simply supported beam. 7–70. Draw the shear and moment diagrams for the beam. The support at A offers no resistance to vertical load.
- 379. 362 CHAPTER 7 INTERNAL FORCES 7 200 mm 100 mm 50 mm 50 mm 50 mm 50 mm 200 mm 40 N 80 N 60 N 100 N 50 N 40 N 50 N A B Prob. 7–71 6 m 10 kN 3 kN/m A B Prob. 7–72 8 kN 15 kN/m 20 kN m 8 kN 1 m 1 m 1 m 0.75 m 0.25 m A B C D Prob. 7–74 500 N B A 1.5 m 1.5 m 300 N/m Prob. 7–75 10 kN 2 kN/m 5 m 3 m 2 m A B Prob. 7–76 7–71. Draw the shear and moment diagrams for the lathe shaft if it is subjected to the loads shown.The bearing at A is a journal bearing, and B is a thrust bearing. *7–72. Draw the shear and moment diagrams for the beam. •7–73. Draw the shear and moment diagrams for the shaft. The support at A is a thrust bearing and at B it is a journal bearing. 7–74. Draw the shear and moment diagrams for the beam. 7–75. The shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B. Draw the shear and moment diagrams for the shaft. *7–76. Draw the shear and moment diagrams for the beam. A B 2 kN/m 4 kN 0.8 m 0.2 m Prob. 7–73
- 380. Prob. 7–81 7.3 RELATIONS BETWEEN DISTRIBUTED LOAD, SHEAR, AND MOMENT 363 7 300 lb 200 lb/ft A 6 ft Prob. 7–79 1 ft 4 ft 1 ft 100 lb/ft A 300 lb ft 200 lb B Prob. 7–77 8 ft 4 ft 6 ft 700 lb 150 lb/ft 800 lb ft A B C Prob. 7–78 10 kN 10 kN/m A B 3 m 3 m Prob. 7–80 w0 A B L L Prob. 7–82 •7–77. Draw the shear and moment diagrams for the shaft. The support at A is a journal bearing and at B it is a thrust bearing. 7–78. The beam consists of two segments pin connected at B. Draw the shear and moment diagrams for the beam. 7–79. Draw the shear and moment diagrams for the cantilever beam. *7–80. Draw the shear and moment diagrams for the simply supported beam. •7–81. Draw the shear and moment diagrams for the beam. 7–82. Draw the shear and moment diagrams for the beam. A B 2000 lb 500 lb/ft 9 ft 9 ft
- 381. 364 CHAPTER 7 INTERNAL FORCES 7 3 m 8 kN/m 8 kN/m 3 m A Prob. 7–83 40 kN/m 20 kN 150 kN m A B 8 m 3 m Prob. 7–84 w 6 ft 6 ft A B Prob. 7–85 5 kN 3 kN/m A B C D 3 m 3 m 1.5 m 1.5 m Prob. 7–86 A B 2 kN/m 300 mm 450 mm 600 mm Prob. 7–87 A 6 ft 10 ft 6 ft 5 kip/ft B 15 kip ft 15 kip ft Prob. 7–88 7–83. Draw the shear and moment diagrams for the beam. *7–84. Draw the shear and moment diagrams for the beam. •7–85. The beam will fail when the maximum moment is or the maximum shear is Determine the largest intensity w of the distributed load the beam will support. Vmax = 8 kip. Mmax = 30 kip # ft 7–86. Draw the shear and moment diagrams for the compound beam. 7–87. Draw the shear and moment diagrams for the shaft. The supports at A and B are journal bearings. *7–88. Draw the shear and moment diagrams for the beam.
- 382. 7.4 CABLES 365 7 *As will be shown in the following example, the eight equilibrium equations also can be written for the entire cable, or any part thereof. But no more than eight equations are available. yC h P1 B P2 A L1 L2 L3 yD D C Fig. 7–18 Each of the cable segments remains approximately straight as they support the weight of these traffic lights. *7.4 Cables Flexible cables and chains combine strength with lightness and often are used in structures for support and to transmit loads from one member to another. When used to support suspension bridges and trolley wheels, cables form the main load-carrying element of the structure. In the force analysis of such systems, the weight of the cable itself may be neglected because it is often small compared to the load it carries. On the other hand, when cables are used as transmission lines and guys for radio antennas and derricks, the cable weight may become important and must be included in the structural analysis. Three cases will be considered in the analysis that follows. In each case we will make the assumption that the cable is perfectly flexible and inextensible. Due to its flexibility, the cable offers no resistance to bending, and therefore, the tensile force acting in the cable is always tangent to the cable at points along its length. Being inextensible, the cable has a constant length both before and after the load is applied.As a result, once the load is applied, the geometry of the cable remains unchanged, and the cable or a segment of it can be treated as a rigid body. Cable Subjected to Concentrated Loads. When a cable of negligible weight supports several concentrated loads, the cable takes the form of several straight-line segments, each of which is subjected to a constant tensile force. Consider, for example, the cable shown in Fig. 7–18, where the distances h, and and the loads and are known. The problem here is to determine the nine unknowns consisting of the tension in each of the three segments, the four components of reaction at A and B, and the two sags and at points C and D. For the solution we can write two equations of force equilibrium at each of points A, B, C, and D.This results in a total of eight equations.* To complete the solution, we need to know something about the geometry of the cable in order to obtain the necessary ninth equation. For example, if the cable’s total length L is specified, then the Pythagorean theorem can be used to relate each of the three segmental lengths, written in terms of h, and to the total length L. Unfortunately, this type of problem cannot be solved easily by hand. Another possibility, however, is to specify one of the sags, either or instead of the cable length. By doing this, the equilibrium equations are then sufficient for obtaining the unknown forces and the remaining sag. Once the sag at each point of loading is obtained, the length of the cable can then be determined by trigonometry. The following example illustrates a procedure for performing the equilibrium analysis for a problem of this type. yD, yC L3, L2, L1, yD, yC, yD yC P2 P1 L3 L2, L1,
- 383. 366 CHAPTER 7 INTERNAL FORCES 7 Determine the tension in each segment of the cable shown in Fig. 7–19a. EXAMPLE 7.11 A 12 m C B yB D 15 kN 4 kN 3 kN E 3 m 2 m 5 m 8 m (a) yD Fig. 7–19 E 15 kN 4 kN 3 kN 3 m 2 m 5 m 8 m (b) Ex Ey Ax Ay 4 kN 3 m 5 m (c) Ax C 12 m TBC 12 kN uBC SOLUTION By inspection, there are four unknown external reactions ( and ) and four unknown cable tensions, one in each cable segment. These eight unknowns along with the two unknown sags and can be determined from ten available equilibrium equations. One method is to apply the force equations of equilibrium ( ) to each of the five points A through E. Here, however, we will take a more direct approach. Consider the free-body diagram for the entire cable,Fig.7–19b.Thus, a Since the sag is known, we will now consider the leftmost section, which cuts cable BC, Fig. 7–19c. a Thus, Ans. TBC = 10.2 kN uBC = 51.6° 12 kN - 4 kN - TBC sin uBC = 0 + c ©Fy = 0; TBC cos uBC - 6.33 kN = 0 : + ©Fx = 0; Ax = Ex = 6.33 kN Ax112 m2 - 12 kN 18 m2 + 4 kN 15 m2 = 0 +©MC = 0; yC = 12 m Ey = 10 kN 12 kN - 4 kN - 15 kN - 3 kN + Ey = 0 + c ©Fy = 0; Ay = 12 kN -Ay118 m2 + 4 kN 115 m2 + 15 kN 110 m2 + 3 kN 12 m2 = 0 +©ME = 0; -Ax + Ex = 0 : + ©Fx = 0; ©Fy = 0 ©Fx = 0, yD yB Ey Ex, Ay, Ax,
- 384. 7.4 CABLES 367 7 Proceeding now to analyze the equilibrium of points A, C, and E in sequence, we have Point A (Fig. 7–19d). Ans. Point C (Fig. 7–19e). Ans. Point E (Fig. 7–19f). Ans. NOTE: By comparison, the maximum cable tension is in segment AB since this segment has the greatest slope and it is required that for any cable segment the horizontal component (a constant).Also, since the slope angles that the cable segments make with the horizontal have now been determined, it is possible to determine the sags and Fig. 7–19a, using trigonometry. yD, yB T cos u = Ax = Ex 1u2 TED = 11.8 kN uED = 57.7° 10 kN - TED sin uED = 0 + c ©Fy = 0; 6.33 kN - TED cos uED = 0 : + ©Fx = 0; TCD = 9.44 kN uCD = 47.9° TCD sin uCD + 10.2 sin 51.6° kN - 15 kN = 0 + c©Fy = 0; TCD cos uCD - 10.2 cos 51.6° kN = 0 : + ©Fx = 0; TAB = 13.6 kN uAB = 62.2° -TAB sin uAB + 12 kN = 0 + c©Fy = 0; TAB cos uAB - 6.33 kN = 0 : + ©Fx = 0; uAB A 12 kN 6.33 kN TAB (d) TCD 51.6 10.2 kN 15 kN (e) C uCD 10 kN 6.33 kN TED E (f) uED
- 385. 368 CHAPTER 7 INTERNAL FORCES Cable Subjected to a Distributed Load. Let us now consider the weightless cable shown in Fig. 7–20a, which is subjected to a distributed loading that is measured in the x direction. The free-body diagram of a small segment of the cable having a length is shown in Fig. 7–20b. Since the tensile force changes in both magnitude and direction along the cable’s length, we will denote this change on the free-body diagram by Finally, the distributed load is represented by its resultant force which acts at a fractional distance from point O, where Applying the equations of equilibrium, we have a Dividing each of these equations by and taking the limit as and therefore and we obtain (7–7) (7–8) (7–9) dy dx = tan u d1T sin u2 dx - w1x2 = 0 d1T cos u2 dx = 0 ¢T : 0, ¢u : 0, ¢y : 0, ¢x : 0, ¢x w1x21¢x2k1¢x2 - T cos u ¢y + T sin u ¢x = 0 +©MO = 0; -T sin u - w1x21¢x2 + 1T + ¢T2 sin1u + ¢u2 = 0 + c©Fy = 0; -T cos u + 1T + ¢T2 cos1u + ¢u2 = 0 : + ©Fx = 0; 0 6 k 6 1. k1¢x2 w1x21¢x2, ¢T. ¢s w = w1x2 7 368 CHAPTER 7 INTERNAL FORCES A (a) B w w(x) x x y x Fig. 7–20 The cable and suspenders are used to support the uniform load of a gas pipe which crosses the river.
- 386. 7.4 CABLES 369 7 Integrating Eq. 7–7, we have (7–10) where represents the horizontal component of tensile force at any point along the cable. Integrating Eq. 7–8 gives (7–11) Dividing Eq. 7–11 by Eq. 7–10 eliminates T. Then, using Eq. 7–9, we can obtain the slope of the cable. Performing a second integration yields (7–12) This equation is used to determine the curve for the cable, The horizontal force component and the additional two constants, say and resulting from the integration are determined by applying the boundary conditions for the curve. C2, C1 FH y = f1x2. y = 1 FH L a L w1x2 dxb dx tan u = dy dx = 1 FH L w1x2 dx T sin u = L w1x2 dx FH T cos u = constant = FH (b) T T w(x)(x) (x) k O T u u u x s y The cables of the suspension bridge exert very large forces on the tower and the foundation block which have to be accounted for in their design.
- 387. 370 CHAPTER 7 INTERNAL FORCES 7 The cable of a suspension bridge supports half of the uniform road surface between the two towers at A and B, Fig. 7–21a. If this distributed loading is determine the maximum force developed in the cable and the cable’s required length.The span length L and sag h are known. w0, EXAMPLE 7.12 L y x O h B A w0 (a) Fig. 7–21 SOLUTION We can determine the unknowns in the problem by first finding the equation of the curve that defines the shape of the cable using Eq. 7–12. For reasons of symmetry, the origin of coordinates has been placed at the cable’s center. Noting that we have Performing the two integrations gives (1) The constants of integration may be determined using the boundary conditions at and at Substituting into Eq. 1 and its derivative yields The equation of the curve then becomes (2) y = w0 2FH x2 C1 = C2 = 0. x = 0. dydx = 0 x = 0 y = 0 y = 1 FH a w0x2 2 + C1x + C2b y = 1 FH L a L w0 dxb dx w1x2 = w0,
- 388. 7.4 CABLES 371 7 This is the equation of a parabola. The constant may be obtained using the boundary condition at Thus, (3) Therefore, Eq. 2 becomes (4) Since is known, the tension in the cable may now be determined using Eq. 7–10, written as For the maximum tension will occur when is maximum, i.e., at point B, Fig. 7–21a. From Eq. 2, the slope at this point is or (5) Therefore, (6) Using the triangular relationship shown in Fig. 7–21b, which is based on Eq. 5, Eq. 6 may be written as Substituting Eq. 3 into the above equation yields Ans. For a differential segment of cable length ds, we can write Hence, the total length of the cable can be determined by integration. Using Eq. 4, we have (7) Integrating yields Ans. l = L 2 c A 1 + a 4h L b 2 + L 4h sinh-1 a 4h L b d l = L ds = 2 L L2 0 B 1 + a 8h L2 xb 2 dx ds = 21dx22 + 1dy22 = B 1 + a dy dx b 2 dx Tmax = w0L 2 B 1 + a L 4h b 2 Tmax = 24FH 2 + w0 2 L2 2 Tmax = FH cos1umax2 umax = tan-1 a w0L 2FH b dy dx ` x=L2 = tan umax = w0 FH x ` x=L2 u 0 … u 6 p2, T = FHcos u. FH y = 4h L2 x2 FH = w0L2 8h x = L2. y = h FH w0L 2FH 4FH 2 w0 2 L 2 (b) umax
- 389. 372 CHAPTER 7 INTERNAL FORCES 7 Cable Subjected to Its Own Weight. When the weight of a cable becomes important in the force analysis, the loading function along the cable will be a function of the arc length s rather than the projected length x. To analyze this problem, we will consider a generalized loading function acting along the cable as shown in Fig. 7–22a.The free- body diagram for a small segment of the cable is shown in Fig. 7–22b.Applying the equilibrium equations to the force system on this diagram, one obtains relationships identical to those given by Eqs.7–7 through 7–9,but with ds replacing dx.Therefore,we can show that (7–13) (7–14) To perform a direct integration of Eq. 7–14, it is necessary to replace by . Since then dy dx = A a ds dx b 2 - 1 ds = 2dx2 + dy2 dsdx dydx dy dx = 1 FH L w1s2 ds T sin u = L w1s2 ds T cos u = FH ¢s w = w1s2 s y x s (a) B w w(s) A Fig. 7–22
- 390. 7.4 CABLES 373 7 Therefore, Separating the variables and integrating we obtain (7–15) The two constants of integration, say and are found using the boundary conditions for the curve. C2, C1 x = L ds c1 + 1 FH 2 a L w1s2 dsb 2 d 1/2 ds dx = c1 + 1 FH 2 a L w1s2 dsb 2 d 1/2 (b) T T u u w(s)(x) k (x) O T y s x u Electrical transmission towers must be designed to support the weights of the suspended power lines.The weight and length of the cables can be determined since they each form a catenary curve.
- 391. 374 CHAPTER 7 INTERNAL FORCES 7 Determine the deflection curve, the length, and the maximum tension in the uniform cable shown in Fig. 7–23.The cable has a weight per unit length of SOLUTION For reasons of symmetry, the origin of coordinates is located at the center of the cable.The deflection curve is expressed as We can determine it by first applying Eq. 7–15, where Integrating the term under the integral sign in the denominator, we have Substituting so that a second integration yields or (1) To evaluate the constants note that, from Eq. 7–14, Since at then Thus, (2) The constant may be evaluated by using the condition at in Eq. 1, in which case To obtain the deflection curve, solve for s in Eq. 1, which yields (3) Now substitute into Eq. 2, in which case dy dx = sinha w0 FH xb s = FH w0 sinha w0 FH xb C2 = 0. x = 0 s = 0 C2 dy dx = w0s FH C1 = 0. s = 0, dydx = 0 dy dx = 1 FH L w0 ds or dy dx = 1 FH 1w0s + C12 x = FH w0 esinh-1 c 1 FH 1w0s + C12d + C2 f x = FH w0 1sinh-1 u + C22 du = 1w0FH2 ds, u = 11FH21w0s + C12 x = L ds [1 + 11FH 2 21w0s + C122 ]12 x = L ds c1 + 11FH 2 2a L w0 dsb 2 d 12 w1s2 = w0. y = f1x2. w0 = 5 Nm. EXAMPLE 7.13 y x s L 20 m h 6 m umax Fig. 7–23
- 392. 7.4 CABLES 375 7 Hence, If the boundary condition at is applied, the constant and therefore the deflection curve becomes (4) This equation defines the shape of a catenary curve. The constant is obtained by using the boundary condition that at in which case (5) Since and Eqs. 4 and 5 become (6) (7) Equation 7 can be solved for by using a trial-and-error procedure. The result is and therefore the deflection curve, Eq. 6, becomes Ans. Using Eq. 3, with the half-length of the cable is Hence, Ans. Since the maximum tension occurs when is maximum, i.e., at Using Eq. 2 yields And so, Ans. Tmax = FH cos umax = 45.9 N cos 52.8° = 75.9 N umax = 52.8° dy dx ` s=12.1 m = tan umax = 5 Nm112.1 m2 45.9 N = 1.32 s = l2 = 12.1 m. u T = FHcos u, l = 24.2 m l 2 = 45.9 N 5 Nm sinhc 5 Nm 45.9 N 110 m2d = 12.1 m x = 10 m, y = 9.19[cosh10.109x2 - 1] m FH = 45.9 N FH 6 m = FH 5 Nm ccosha 50 N FH b - 1d y = FH 5 Nm ccosha 5 Nm FH xb - 1d L = 20 m, h = 6 m, w0 = 5 Nm, h = FH w0 ccosha w0L 2FH b - 1d x = L2, y = h FH y = FH w0 ccosha w0 FH xb - 1d C3 = -FHw0, x = 0 y = 0 y = FH w0 cosha w0 FH xb + C3
- 393. 376 CHAPTER 7 INTERNAL FORCES 7 PROBLEMS Neglect the weight of the cable in the following problems, unless specified. •7–89. Determine the tension in each segment of the cable and the cable’s total length. Set . 7–90. If each cable segment can support a maximum tension of 75 lb, determine the largest load P that can be applied. P = 80 lb P A B C D 2 ft 3 ft 50 lb 5 ft 4 ft 3 ft Probs. 7–89/90 5 ft 2 ft 3 ft 60 lb D C B A xB 8 ft P Probs. 7–91/92 4 m 3 m 2 m 6 m 4 kN P 6 kN yB 3 m A B C D E Prob. 7–93 7–91. The cable segments support the loading shown. Determine the horizontal distance from the force at B to point A. Set . *7–92. The cable segments support the loading shown. Determine the magnitude of the horizontal force P so that . xB = 6 ft P = 40 lb xB 3 m 1 m 0.5 m yB 2 m A D B C E F Prob. 7–94 •7–93. Determine the force P needed to hold the cable in the position shown, i.e., so segment BC remains horizontal. Also, compute the sag and the maximum tension in the cable. yB 7–94. Cable ABCD supports the 10-kg lamp E and the 15-kg lamp F. Determine the maximum tension in the cable and the sag of point B. yB
- 394. 7.4 CABLES 377 7 4 ft 12 ft 20 ft 15 ft 12 ft A E B C D yB yD 14 ft P2 P2 P1 Probs. 7–95/96 5 ft 2 ft 3 ft 30 lb D C B A xB 5 4 3 8 ft P Probs. 7–97/98 60 m 7 m w0 Prob. 7–99 7–95. The cable supports the three loads shown. Determine the sags and of points B and D. Take *7–96. The cable supports the three loads shown. Determine the magnitude of if and Also find the sag yD. yB = 8 ft. P2 = 300 lb P1 P2 = 250 lb. P1 = 400 lb, yD yB •7–97. The cable supports the loading shown. Determine the horizontal distance the force at point B acts from A. Set 7–98. The cable supports the loading shown. Determine the magnitude of the horizontal force P so that xB = 6 ft. P = 40 lb. xB 7–99. Determine the maximum uniform distributed loading N/m that the cable can support if it is capable of sustaining a maximum tension of 60 kN. w0 *7–100. The cable supports the uniform distributed load of . Determine the tension in the cable at each support A and B. •7–101. Determine the maximum uniform distributed load the cable can support if the maximum tension the cable can sustain is 4000 lb. w0 w0 = 600 lbft A w0 B 25 ft 10 ft 15 ft Prob. 7–101
- 395. 378 CHAPTER 7 INTERNAL FORCES 7 12 ft h B D A C Prob. 7–103 A B 1000 m 150 m 75 m Probs. 7–104/105 10 ft 500 lb/ft 10 A B x y 40 ft Prob. 7–106 7–103. If cylinders C and D each weigh 900 lb, determine the maximum sag h, and the length of the cable between the smooth pulleys at A and B. The beam has a weight per unit length of . 100 lbft *7–104. The bridge deck has a weight per unit length of . It is supported on each side by a cable. Determine the tension in each cable at the piers A and B. •7–105. If each of the two side cables that support the bridge deck can sustain a maximum tension of 50 MN, determine the allowable uniform distributed load caused by the weight of the bridge deck. w0 80 kNm 7–106. If the slope of the cable at support A is 10°, determine the deflection curve y = f(x) of the cable and the maximum tension developed in the cable. 7–102. The cable is subjected to the triangular loading. If the slope of the cable at point O is zero, determine the equation of the curve which defines the cable shape OB, and the maximum tension developed in the cable. y = f1x2 Prob. 7–102 15 ft 15 ft 500 lb/ft 500 lb/ft 8 ft y x A O B
- 396. 7.4 CABLES 379 7 7–111. The cable has a mass per unit length of . Determine the shortest total length L of the cable that can be suspended in equilibrium. 10 kgm *7–112. The power transmission cable has a weight per unit length of . If the lowest point of the cable must be at least 90 ft above the ground, determine the maximum tension developed in the cable and the cable’s length between A and B. 15 lbft A B 50 m h 5 m Prob. 7–107 A B 150 ft 30 30 Prob. 7–108 A B 40 m Prob. 7–109 A B 8 m Prob. 7–111 7–110. Show that the deflection curve of the cable discussed in Example 7–13 reduces to Eq. 4 in Example 7–12 when the hyperbolic cosine function is expanded in terms of a series and only the first two terms are retained. (The answer indicates that the catenary may be replaced by a parabola in the analysis of problems in which the sag is small. In this case, the cable weight is assumed to be uniformly distributed along the horizontal.) •7–113. If the horizontal towing force is T = 20 kN and the chain has a mass per unit length of , determine the maximum sag h. Neglect the buoyancy effect of the water on the chain.The boats are stationary. 15 kgm A B 180 ft 90 ft 120 ft 300 ft Prob. 7–112 40 m h T T Prob. 7–113 7–107. If h = 5 m, determine the maximum tension developed in the chain and its length. The chain has a mass per unit length of . 8 kgm *7–108. A cable having a weight per unit length of is suspended between supports A and B. Determine the equation of the catenary curve of the cable and the cable’s length. 5 lbft •7–109. If the 45-m-long cable has a mass per unit length of , determine the equation of the catenary curve of the cable and the maximum tension developed in the cable. 5 kgm
- 397. 380 CHAPTER 7 INTERNAL FORCES 7 CHAPTER REVIEW Internal Loadings If a coplanar force system acts on a member, then in general a resultant internal normal force N, shear force V, and bending moment M will act at any cross section along the member. The positive directions of these loadings are shown in the figure. The resultant internal normal force, shear force, and bending moment are determined using the method of sections. To find them, the member is sectioned at the point C where the internal loadings are to be determined.A free-body diagram of one of the sectioned parts is then drawn and the internal loadings are shown in their positive directions. The resultant normal force is determined by summing forces normal to the cross section. The resultant shear force is found by summing forces tangent to the cross section, and the resultant bending moment is found by summing moments about the geometric center or centroid of the cross- sectional area. If the member is subjected to a three- dimensional loading, then, in general, a torsional moment will also act on the cross section. It can be determined by summing moments about an axis that is perpendicular to the cross section and passes through its centroid. ©MC = 0 ©Fy = 0 ©Fx = 0 B Ay Ax By A C F1 F2 A Ay Ax VC B By C NC MC VC C NC MC F1 F2 y z Ny Normal force My Torsional moment Vx Vz Mx x C Mz Shear force components Bending moment components (a) V N M Shear force Normal force Bending moment C
- 398. CHAPTER REVIEW 381 7 Positive shear Positive moment M M V V V V M M Shear and Moment Diagrams To construct the shear and moment diagrams for a member, it is necessary to section the member at an arbitrary point, located a distance x from the left end. If the external loading consists of changes in the distributed load, or a series of concentrated forces and couple moments act on the member, then different expressions for V and M must be determined within regions between any load discontinuities. The unknown shear and moment are indicated on the cross section in the positive direction according to the established sign convention, and then the internal shear and moment are determined as functions of x. Each of the functions of the shear and moment is then plotted to create the shear and moment diagrams. O L P b a x3 x2 x1 w
- 399. 382 CHAPTER 7 INTERNAL FORCES 7 Distributed load y = 1 FH L a L w1x2 dxb dx Cable weight x = L ds c1 + 1 FH 2 a L w1s2 dsb 2 d 12 P1 P2 Relations between Shear and Moment It is possible to plot the shear and moment diagrams quickly by using differential relationships that exist between the distributed loading and V and M. The slope of the shear diagram is equal to the distributed loading at any point. The slope is positive if the distributed load acts upward, and vice-versa. The slope of the moment diagram is equal to the shear at any point. The slope is positive if the shear is positive,or vice-versa. The change in shear between any two points is equal to the area under the distributed loading between the points. The change in the moment is equal to the area under the shear diagram between the points. w Cables When a flexible and inextensible cable is subjected to a series of concentrated forces, then the analysis of the cable can be performed by using the equations of equilibrium applied to free-body diagrams of either segments or points of application of the loading. If external distributed loads or the weight of the cable are to be considered, then the shape of the cable must be determined by first analyzing the forces on a differential segment of the cable and then integrating this result. The two constants, say and resulting from the integration are determined by applying the boundary conditions for the cable. C2, C1 dV dx = w ¢V = L w dx ¢M = L V dx dM dx = V
- 400. REVIEW PROBLEMS 383 7 4 kip · ft 10 kip A C B D 3 ft 3 ft 2 ft 2 ft 2 ft Prob. 7–115 5 m 5 m 3 m 2 kN/m 1 kN/m 7.5 kN 40 kN m 6 kN 1 m A B C Prob. 7–116 60 A D E C B 1 m 0.75 m 0.75 m 0.75 m 0.25 m 400 N/m Prob. 7–117 REVIEW PROBLEMS •7–117. Determine the internal normal force, shear force and moment at points D and E of the frame. *7–116. Determine the internal normal force, shear force, and moment at points B and C of the beam. 7–114. A 100-lb cable is attached between two points at a distance 50 ft apart having equal elevations. If the maximum tension developed in the cable is 75 lb, determine the length of the cable and the sag. 7–115. Draw the shear and moment diagrams for beam CD. 7–118. Determine the distance a between the supports in terms of the beam’s length L so that the moment in the symmetric beam is zero at the beam’s center. L a w Prob. 7–118
- 401. 384 CHAPTER 7 INTERNAL FORCES 384 CHAPTER 7 INTERNAL FORCES 7 x 2 m A C 5 m 8 kN B Prob. 7–122 2 ft 1 ft 150 lb 200 lb y A B C u Prob. 7–123 7–122. The traveling crane consists of a 5-m-long beam having a uniform mass per unit length of 20 kg/m.The chain hoist and its supported load exert a force of 8 kN on the beam when . Draw the shear and moment diagrams for the beam. The guide wheels at the ends A and B exert only vertical reactions on the beam. Neglect the size of the trolley at C. x = 2 m •7–121. Determine the internal shear and moment in member ABC as a function of x,where the origin for x is at A. 7–119. A chain is suspended between points at the same elevation and spaced a distance of 60 ft apart. If it has a weight per unit length of and the sag is 3 ft, determine the maximum tension in the chain. *7–120. Draw the shear and moment diagrams for the beam. 0.5 lbft *7–123. Determine the internal normal force, shear force, and the moment as a function of and for the member loaded as shown. 0 … y … 2 ft 0° … u … 180° 5 m 5 m 2 kN/m A 50 kN m B C Prob. 7–120 A C D B 3 m 1.5 m 1.5 m 1.5 m 6 kN 45 Prob. 7–121
- 402. REVIEW PROBLEMS 385 7 d A B s x y 60 d l Prob. 7–124 *7–124. The yacht is anchored with a chain that has a total length of 40 m and a mass per unit length of and the tension in the chain at A is 7 kN. Determine the length of chain which is lying at the bottom of the sea. What is the distance d? Assume that buoyancy effects of the water on the chain are negligible. Hint: Establish the origin of the coordinate system at B as shown in order to find the chain length BA. ld 18 kg/m, •7–125. Determine the internal normal force, shear force, and moment at points D and E of the frame. 7–126. The uniform beam weighs 500 lb and is held in the horizontal position by means of cable AB, which has a weight of 5 lb/ft. If the slope of the cable at A is 30°, determine the length of the cable. 7–127. The balloon is held in place using a 400-ft cord that weighs 0.8 lb/ft and makes a 60° angle with the horizontal. If the tension in the cord at point A is 150 lb, determine the length of the cord, l, that is lying on the ground and the height h. Hint: Establish the coordinate system at B as shown. 60 A l x y h s B Prob. 7–127 A B C 15 ft 30 Prob. 7–126 E 4 ft 1 ft 8 ft 3 ft D F C A 30 150 lb B Prob. 7–125
- 403. The effective design of a brake system, such as the one for this bicycle, requires an efficient capacity for the mechanism to resist frictional forces. In this chapter, we will study the nature of friction and show how these forces are considered in engineering analysis and design.
- 404. Friction CHAPTER OBJECTIVES • To introduce the concept of dry friction and show how to analyze the equilibrium of rigid bodies subjected to this force. • To present specific applications of frictional force analysis on wedges, screws, belts, and bearings. • To investigate the concept of rolling resistance. 8.1 Characteristics of Dry Friction Friction is a force that resists the movement of two contacting surfaces that slide relative to one another. This force always acts tangent to the surface at the points of contact and is directed so as to oppose the possible or existing motion between the surfaces. In this chapter, we will study the effects of dry friction, which is sometimes called Coulomb friction since its characteristics were studied extensively by C. A. Coulomb in 1781. Dry friction occurs between the contacting surfaces of bodies when there is no lubricating fluid.* 8 The heat generated by the abrasive action of friction can be noticed when using this grinder to sharpen a metal blade. *Another type of friction, called fluid friction, is studied in fluid mechanics.
- 405. 388 CHAPTER 8 FRICTION 8 Theory of Dry Friction. The theory of dry friction can be explained by considering the effects caused by pulling horizontally on a block of uniform weight W which is resting on a rough horizontal surface that is nonrigid or deformable, Fig. 8–1a.The upper portion of the block, however, can be considered rigid.As shown on the free-body diagram of the block, Fig. 8–1b, the floor exerts an uneven distribution of both normal force and frictional force along the contacting surface. For equilibrium, the normal forces must act upward to balance the block’s weight W, and the frictional forces act to the left to prevent the applied force P from moving the block to the right. Close examination of the contacting surfaces between the floor and block reveals how these frictional and normal forces develop, Fig. 8–1c. It can be seen that many microscopic irregularities exist between the two surfaces and, as a result, reactive forces are developed at each point of contact.* As shown, each reactive force contributes both a frictional component and a normal component Equilibrium. The effect of the distributed normal and frictional loadings is indicated by their resultants N and F on the free-body diagram, Fig. 8–1d. Notice that N acts a distance x to the right of the line of action of W, Fig. 8–1d. This location, which coincides with the centroid or geometric center of the normal force distribution in Fig. 8–1b, is necessary in order to balance the “tipping effect” caused by P. For example, if P is applied at a height h from the surface, Fig. 8–1d, then moment equilibrium about point O is satisfied if or x = PhW. Wx = Ph ¢Nn. ¢Fn ¢Rn ¢Fn ¢Nn *Besides mechanical interactions as explained here, which is referred to as a classical approach, a detailed treatment of the nature of frictional forces must also include the effects of temperature, density, cleanliness, and atomic or molecular attraction between the contacting surfaces. See J. Krim, Scientific American, October, 1996. P W (a) Fig. 8–1 P W (b) Nn Fn (c) F1 N1 N2 R1 R2 F2 Fn Rn Nn P W (d) a/2 a/2 h F O N x Resultant Normal and Frictional Forces A B C Regardless of the weight of the rake or shovel that is suspended, the device has been designed so that the small roller holds the handle in equilibrium due to frictional forces that develop at the points of contact, A, B, C.
- 406. 8.1 CHARACTERISTICS OF DRY FRICTION 389 8 W (e) N x Fs Rs Impending motion P Equilibrium h fs Impending Motion. In cases where the surfaces of contact are rather “slippery,” the frictional force F may not be great enough to balance P, and consequently the block will tend to slip. In other words, as P is slowly increased, F correspondingly increases until it attains a certain maximum value called the limiting static frictional force, Fig. 8–1e. When this value is reached, the block is in unstable equilibrium since any further increase in P will cause the block to move. Experimentally, it has been determined that this limiting static frictional force is directly proportional to the resultant normal force N. Expressed mathematically, (8–1) where the constant of proportionality, (mu “sub” s), is called the coefficient of static friction. Thus, when the block is on the verge of sliding, the normal force N and frictional force combine to create a resultant Fig. 8–1e. The angle (phi “sub” s) that makes with N is called the angle of static friction. From the figure, Typical values for are given in Table 8–1. Note that these values can vary since experimental testing was done under variable conditions of roughness and cleanliness of the contacting surfaces. For applications, therefore, it is important that both caution and judgment be exercised when selecting a coefficient of friction for a given set of conditions.When a more accurate calculation of is required, the coefficient of friction should be determined directly by an experiment that involves the two materials to be used. Fs ms fs = tan-1 a Fs N b = tan-1 a msN N b = tan-1 ms Rs fs Rs, Fs ms Fs = msN Fs Fs, Table 8–1 Typical Values for Ms Contact Materials Coefficient of Static Friction 1ms2 Metal on ice 0.03–0.05 Wood on wood 0.30–0.70 Leather on wood 0.20–0.50 Leather on metal 0.30–0.60 Aluminum on aluminum 1.10–1.70
- 407. 390 CHAPTER 8 FRICTION 8 Motion. If the magnitude of P acting on the block is increased so that it becomes slightly greater than the frictional force at the contacting surface will drop to a smaller value called the kinetic frictional force. The block will begin to slide with increasing speed, Fig. 8–2a. As this occurs, the block will “ride” on top of these peaks at the points of contact, as shown in Fig. 8–2b. The continued breakdown of the surface is the dominant mechanism creating kinetic friction. Experiments with sliding blocks indicate that the magnitude of the kinetic friction force is directly proportional to the magnitude of the resultant normal force, expressed mathematically as (8–2) Here the constant of proportionality, is called the coefficient of kinetic friction. Typical values for are approximately 25 percent smaller than those listed in Table 8–1 for As shown in Fig. 8–2a, in this case, the resultant force at the surface of contact, , has a line of action defined by This angle is referred to as the angle of kinetic friction, where By comparison, fs Ú fk. fk = tan-1 a Fk N b = tan-1 a mkN N b = tan-1 mk fk. Rk ms. mk mk, Fk = mkN Fk, Fs, P W (a) N Fk Motion Rk fk (b) F1 N1 N2 R2 R1 F2 Fn Rn Nn Fig. 8–2
- 408. 8.1 CHARACTERISTICS OF DRY FRICTION 391 8 The above effects regarding friction can be summarized by referring to the graph in Fig. 8–3, which shows the variation of the frictional force F versus the applied load P. Here the frictional force is categorized in three different ways: • F is a static frictional force if equilibrium is maintained. • F is a limiting static frictional force when it reaches a maximum value needed to maintain equilibrium. • F is termed a kinetic frictional force when sliding occurs at the contacting surface. Notice also from the graph that for very large values of P or for high speeds, aerodynamic effects will cause and likewise to begin to decrease. Characteristics of Dry Friction. As a result of experiments that pertain to the foregoing discussion, we can state the following rules which apply to bodies subjected to dry friction. • The frictional force acts tangent to the contacting surfaces in a direction opposed to the motion or tendency for motion of one surface relative to another. • The maximum static frictional force that can be developed is independent of the area of contact, provided the normal pressure is not very low nor great enough to severely deform or crush the contacting surfaces of the bodies. • The maximum static frictional force is generally greater than the kinetic frictional force for any two surfaces of contact. However, if one of the bodies is moving with a very low velocity over the surface of another, becomes approximately equal to i.e., • When slipping at the surface of contact is about to occur, the maximum static frictional force is proportional to the normal force, such that • When slipping at the surface of contact is occurring, the kinetic frictional force is proportional to the normal force, such that F k = mkN. F s = msN. ms L mk. F s, F k F s mk F k F k F s F Fs Fk P No motion Motion F P 45 Fig. 8–3
- 409. 392 CHAPTER 8 FRICTION 8 8.2 Problems Involving Dry Friction If a rigid body is in equilibrium when it is subjected to a system of forces that includes the effect of friction, the force system must satisfy not only the equations of equilibrium but also the laws that govern the frictional forces. Types of Friction Problems. In general, there are three types of mechanics problems involving dry friction. They can easily be classified once free-body diagrams are drawn and the total number of unknowns are identified and compared with the total number of available equilibrium equations. No Apparent Impending Motion. Problems in this category are strictly equilibrium problems, which require the number of unknowns to be equal to the number of available equilibrium equations. Once the frictional forces are determined from the solution, however, their numerical values must be checked to be sure they satisfy the inequality otherwise, slipping will occur and the body will not remain in equilibrium. A problem of this type is shown in Fig. 8–4a. Here we must determine the frictional forces at A and C to check if the equilibrium position of the two-member frame can be maintained. If the bars are uniform and have known weights of 100 N each, then the free-body diagrams are as shown in Fig. 8–4b. There are six unknown force components which can be determined strictly from the six equilibrium equations (three for each member). Once and are determined, then the bars will remain in equilibrium provided and are satisfied. Impending Motion at All Points of Contact. In this case the total number of unknowns will equal the total number of available equilibrium equations plus the total number of available frictional equations, When motion is impending at the points of contact, then whereas if the body is slipping, then For example, consider the problem of finding the smallest angle at which the 100-N bar in Fig. 8–5a can be placed against the wall without slipping. The free-body diagram is shown in Fig. 8–5b. Here the five unknowns are determined from the three equilibrium equations and two static frictional equations which apply at both points of contact, so that and . FB = 0.4NB F A = 0.3N A u Fk = mkN. F s = msN; F = mN. F C … 0.5NC F A … 0.3N A NC F C, N A, F A, F … msN; (a) B mC 0.5 mA 0.3 A C (b) Bx By By Bx 100 N 100 N FA FC NA NC Fig. 8–4 A B mB 0.4 mA 0.3 u (a) NB NA FB FA (b) 100 N u Fig. 8–5
- 410. 8.2 PROBLEMS INVOLVING DRY FRICTION 393 8 P (a) A B mC 0.5 mA 0.3 C By Bx 100 N P (b) FC NC By Bx 100 N FA NA Fig. 8–6 P W N F b/2 h x b/2 P W N F b/2 h x b/2 Consider pushing on the uniform crate that has a weight W and sits on the rough surface.As shown on the first free-body diagram, if the magnitude of P is small, the crate will remain in equilibrium.As P increases the crate will either be on the verge of slipping on the surface or if the surface is very rough (large ) then the resultant normal force will shift to the corner, as shown on the second free-body diagram.At this point the crate will begin to tip over.The crate also has a greater chance of tipping if P is applied at a greater height h above the surface, or if its width b is smaller. x = b2, ms 1F = msN2, Impending Motion at Some Points of Contact. Here the number of unknowns will be less than the number of available equilibrium equations plus the number of available frictional equations or conditional equations for tipping. As a result, several possibilities for motion or impending motion will exist and the problem will involve a determination of the kind of motion which actually occurs. For example, consider the two-member frame in Fig. 8–6a. In this problem we wish to determine the horizontal force P needed to cause movement. If each member has a weight of 100 N, then the free-body diagrams are as shown in Fig. 8–6b. There are seven unknowns. For a unique solution we must satisfy the six equilibrium equations (three for each member) and only one of two possible static frictional equations. This means that as P increases it will either cause slipping at A and no slipping at C, so that and or slipping occurs at C and no slipping at A, in which case and The actual situation can be determined by calculating P for each case and then choosing the case for which P is smaller. If in both cases the same value for P is calculated, which in practice would be highly improbable, then slipping at both points occurs simultaneously; i.e., the seven unknowns would satisfy eight equations. F A … 0.3N A. FC = 0.5NC F C … 0.5NC; F A = 0.3N A
- 411. 394 CHAPTER 8 FRICTION 8 Equilibrium Versus Frictional Equations. Whenever we solve problems where the friction force F is to be an “equilibrium force” and satisfies the inequality , then we can assume the sense of direction of F on the free-body diagram. The correct sense is made known after solving the equations of equilibrium for F. If F is a negative scalar the sense of F is the reverse of that which was assumed. This convenience of assuming the sense of F is possible because the equilibrium equations equate to zero the components of vectors acting in the same direction. However, in cases where the frictional equation is used in the solution of a problem, the convenience of assuming the sense of F is lost, since the frictional equation relates only the magnitudes of two perpendicular vectors. Consequently, F must always be shown acting with its correct sense on the free-body diagram, whenever the frictional equation is used for the solution of a problem. F = mN F 6 msN W P FB FA NB NA B A The applied vertical force P on this roll must be large enough to overcome the resistance of friction at the contacting surfaces A and B in order to cause rotation. Procedure for Analysis Equilibrium problems involving dry friction can be solved using the following procedure. Free-Body Diagrams. • Draw the necessary free-body diagrams, and unless it is stated in the problem that impending motion or slipping occurs, always show the frictional forces as unknowns (i.e., do not assume ). • Determine the number of unknowns and compare this with the number of available equilibrium equations. • If there are more unknowns than equations of equilibrium, it will be necessary to apply the frictional equation at some, if not all, points of contact to obtain the extra equations needed for a complete solution. • If the equation is to be used, it will be necessary to show F acting in the correct sense of direction on the free-body diagram. Equations of Equilibrium and Friction. • Apply the equations of equilibrium and the necessary frictional equations (or conditional equations if tipping is possible) and solve for the unknowns. • If the problem involves a three-dimensional force system such that it becomes difficult to obtain the force components or the necessary moment arms, apply the equations of equilibrium using Cartesian vectors. F = mN F = mN
- 412. EXAMPLE 8.1 The uniform crate shown in Fig. 8–7a has a mass of 20 kg. If a force is applied to the crate, determine if it remains in equilibrium. The coefficient of static friction is ms = 0.3. P = 80 N 8.2 PROBLEMS INVOLVING DRY FRICTION 395 8 0.8 m P 80 N 0.2 m 30 (a) Fig. 8–7 P 80 N 0.2 m 30 (b) 196.2 N 0.4 m 0.4 m NC x F O SOLUTION Free-Body Diagram. As shown in Fig. 8–7b, the resultant normal force must act a distance x from the crate’s center line in order to counteract the tipping effect caused by P. There are three unknowns, F, and x, which can be determined strictly from the three equations of equilibrium. Equations of Equilibrium. -80 sin 30° N + NC - 196.2 N = 0 + c©F y = 0; 80 cos 30° N - F = 0 : + ©F x = 0; NC, NC Solving, Since x is negative it indicates the resultant normal force acts (slightly) to the left of the crate’s center line. No tipping will occur since Also, the maximum frictional force which can be developed at the surface of contact is Since the crate will not slip, although it is very close to doing so. F = 69.3 N 6 70.8 N, F max = msNC = 0.31236 N2 = 70.8 N. x 6 0.4 m. x = -0.00908 m = -9.08 mm N C = 236 N F = 69.3 N a 80 sin 30° N10.4 m2 - 80 cos 30° N10.2 m2 + NC1x2 = 0 +©MO = 0;
- 413. 396 CHAPTER 8 FRICTION 8 It is observed that when the bed of the dump truck is raised to an angle of the vending machines will begin to slide off the bed, Fig. 8–8a. Determine the static coefficient of friction between a vending machine and the surface of the truckbed. SOLUTION An idealized model of a vending machine resting on the truckbed is shown in Fig. 8–8b. The dimensions have been measured and the center of gravity has been located. We will assume that the vending machine weighs W. Free-Body Diagram. As shown in Fig. 8–8c, the dimension x is used to locate the position of the resultant normal force N. There are four unknowns, N, F, and x. Equations of Equilibrium. (1) (2) a (3) Since slipping impends at using Eqs. 1 and 2, we have Ans. The angle of is referred to as the angle of repose, and by comparison, it is equal to the angle of static friction, Notice from the calculation that is independent of the weight of the vending machine, and so knowing provides a convenient method for determining the coefficient of static friction. NOTE: From Eq. 3, we find Since indeed the vending machine will slip before it can tip as observed in Fig. 8–8a. 1.17 ft 6 1.5 ft, x = 1.17 ft. u u u = fs. u = 25° ms = tan 25° = 0.466 W sin 25° = ms1W cos 25°2 F s = msN; u = 25°, -W sin 25°12.5 ft2 + W cos 25°1x2 = 0 +©MO = 0; N - W cos 25° = 0 +Q©F y = 0; W sin 25° - F = 0 +R©F x = 0; ms, u = 25° EXAMPLE 8.2 (a) u 25 2.5 ft G 1.5 ft 1.5 ft (b) (c) 2.5 ft G O x 1.5 ft 1.5 ft W 25 N F Fig. 8–8
- 414. 8.2 PROBLEMS INVOLVING DRY FRICTION 397 8 EXAMPLE 8.3 The uniform 10-kg ladder in Fig. 8–9a rests against the smooth wall at B, and the end A rests on the rough horizontal plane for which the coefficient of static friction is . Determine the angle of inclination of the ladder and the normal reaction at B if the ladder is on the verge of slipping. u ms = 0.3 4 m A B A (a) u A (b) NB NA FA (4 m) sin (2 m) cos (2 m) cos 10(9.81) N u u u u Fig. 8–9 SOLUTION Free-Body Diagram. As shown on the free-body diagram, Fig. 8–9b, the frictional force must act to the right since impending motion at A is to the left. Equations of Equilibrium and Friction. Since the ladder is on the verge of slipping, then . By inspection, can be obtained directly. Using this result, . Now can be found. Ans. Finally, the angle can be determined by summing moments about point A. a Ans. u = 59.04° = 59.0° sin u cos u = tan u = 1.6667 (29.43 N)(4 m) sin u - [10(9.81) N](2 m) cos u = 0 +©M A = 0; u NB = 29.43 N = 29.4 N 29.43 N - NB = 0 : + ©F x = 0; NB F A = 0.3(98.1 N) = 29.43 N N A = 98.1 N N A - 10(9.81) N = 0 + c©F y = 0; NA F A = msN A = 0.3N A FA
- 415. 398 CHAPTER 8 FRICTION 8 EXAMPLE 8.4 200 N/m 0.75 m B P 4 m 0.25 m C A (a) 800 N 2 m (b) Ax Ay A 2 m NB 400 N FB 0.75 m 0.25 m P B (c) C 400 N NC FC FB Fig. 8–10 Beam AB is subjected to a uniform load of and is supported at B by post BC, Fig. 8–10a. If the coefficients of static friction at B and C are and determine the force P needed to pull the post out from under the beam. Neglect the weight of the members and the thickness of the beam. SOLUTION Free-Body Diagrams. The free-body diagram of the beam is shown in Fig. 8–10b.Applying we obtain This result is shown on the free-body diagram of the post, Fig. 8–10c. Referring to this member, the four unknowns P, and are determined from the three equations of equilibrium and one frictional equation applied either at B or C. Equations of Equilibrium and Friction. (1) (2) a (3) (Post Slips at B and Rotates about C.) This requires and Using this result and solving Eqs. 1 through 3, we obtain Since slipping at C occurs.Thus the other case of movement must be investigated. (Post Slips at C and Rotates about B.) Here and (4) Solving Eqs. 1 through 4 yields Ans. Obviously, this case occurs first since it requires a smaller value for P. F B = 66.7 N F C = 200 N NC = 400 N P = 267 N FC = 0.5NC FC = mCNC; F B … mBNB FC = 240 N 7 mCNC = 0.51400 N2 = 200 N, NC = 400 N FC = 240 N P = 320 N F B = 0.21400 N2 = 80 N F B = mBNB; FC … mCNC -P10.25 m2 + F B11 m2 = 0 +©MC = 0; NC - 400 N = 0 + c©F y = 0; P - FB - FC = 0 : + ©Fx = 0; NC FC, FB, NB = 400 N. ©MA = 0, mC = 0.5, mB = 0.2 200 Nm
- 416. 8.2 PROBLEMS INVOLVING DRY FRICTION 399 8 EXAMPLE 8.5 Blocks A and B have a mass of 3 kg and 9 kg, respectively, and are connected to the weightless links shown in Fig. 8–11a. Determine the largest vertical force P that can be applied at the pin C without causing any movement. The coefficient of static friction between the blocks and the contacting surfaces is . SOLUTION Free-Body Diagram. The links are two-force members and so the free-body diagrams of pin C and blocks A and B are shown in Fig. 8–11b. Since the horizontal component of tends to move block A to the left, must act to the right. Similarly, must act to the left to oppose the tendency of motion of block B to the right, caused by . There are seven unknowns and six available force equilibrium equations, two for the pin and two for each block, so that only one frictional equation is needed. Equations of Equilibrium and Friction. The force in links AC and BC can be related to P by considering the equilibrium of pin C. Using the result for , for block A, (1) (2) Using the result for , for block B, (3) Movement of the system may be caused by the initial slipping of either block A or block B. If we assume that block A slips first, then (4) Substituting Eqs. 1 and 2 into Eq. 4, Ans. Substituting this result into Eq. 3, we obtain . Since the maximum static frictional force at B is , block B will not slip. Thus, the above assumption is correct. Notice that if the inequality were not satisfied, we would have to assume slipping of block B and then solve for P. (F B)max = msNB = 0.3(88.29 N) = 26.5 N 7 F B F B = 18.4 N P = 31.8 N 0.5774P = 0.3(P + 29.43) F A = msN A = 0.3 N A NB = 88.29 N NB - 9(9.81) N = 0; + c©F y = 0; F B = 0.5774P (0.5774P) - F B = 0; : + ©F x = 0; F BC N A = P + 29.43 N N A - 1.155P cos 30°-3(9.81 N) = 0; + c ©F y = 0; F A = 0.5774P F A - 1.155P sin 30° = 0; : + ©Fx = 0; F AC F BC = 0.5774P 1.155P sin 30° - F BC = 0; : + ©F x = 0; F AC = 1.155P F AC cos 30° - P = 0; + c©F y = 0; FBC FB FA FAC ms = 0.3 A C B (a) P 30 C y x (b) P FAC FA NA FBC 3(9.81)N FAC 1.155 P FBC 0.5774 P FB NB 9(9.81)N 30 30 Fig. 8–11
- 417. 400 CHAPTER 8 FRICTION 8 FUNDAMENTAL PROBLEMS F8–4. If the coefficient of static friction at contact points A and B is , determine the maximum force P that can be applied without causing the 100-kg spool to move. ms = 0.3 F8–3. Determine the maximum force P that can be applied without causing the two 50-kg crates to move. The coefficient of static friction between each crate and the ground is ms = 0.25. F8–2. Determine the minimum force P to prevent the 30-kg rod AB from sliding. The contact surface at B is smooth, whereas the coefficient of static friction between the rod and the wall at A is ms = 0.2. F8–1. If , determine the friction developed between the 50-kg crate and the ground. The coefficient of static friction between the crate and the ground is . ms = 0.3 P = 200 N F8–5. Determine the minimum force P that can be applied without causing movement of the 250-lb crate which has a center of gravity at G.The coefficient of static friction at the floor is . ms = 0.4 4 3 5 P F8–1 3 m A B P 4 m F8–2 B A 30 P F8–3 P 0.6 m 0.9 m B A F8–4 1.5 ft 1.5 ft 2.5 ft 3.5 ft 4.5 ft P A G F8–5
- 418. 8.2 PROBLEMS INVOLVING DRY FRICTION 401 8 PROBLEMS 8–6. The 180-lb man climbs up the ladder and stops at the position shown after he senses that the ladder is on the verge of slipping.Determine the coefficient of static friction between the friction pad at A and ground if the inclination of the ladder is and the wall at B is smooth.The center of gravity for the man is at G. Neglect the weight of the ladder. u = 60° *8–4. If the coefficient of static friction at A is and the collar at B is smooth so it only exerts a horizontal force on the pipe, determine the minimum distance so that the bracket can support the cylinder of any mass without slipping. Neglect the mass of the bracket. x ms = 0.4 •8–5. The 180-lb man climbs up the ladder and stops at the position shown after he senses that the ladder is on the verge of slipping. Determine the inclination of the ladder if the coefficient of static friction between the friction pad A and the ground is .Assume the wall at B is smooth.The center of gravity for the man is at G.Neglect the weight of the ladder. ms = 0.4 u •8–1. Determine the minimum horizontal force P required to hold the crate from sliding down the plane. The crate has a mass of 50 kg and the coefficient of static friction between the crate and the plane is . 8–2. Determine the minimum force P required to push the crate up the plane.The crate has a mass of 50 kg and the coefficient of static friction between the crate and the plane is . 8–3. A horizontal force of is just sufficient to hold the crate from sliding down the plane, and a horizontal force of is required to just push the crate up the plane. Determine the coefficient of static friction between the plane and the crate, and find the mass of the crate. P = 350 N P = 100 N ms = 0.25 ms = 0.25 8–7. The uniform thin pole has a weight of 30 lb and a length of 26 ft. If it is placed against the smooth wall and on the rough floor in the position , will it remain in this position when it is released? The coefficient of static friction is . *8–8. The uniform pole has a weight of 30 lb and a length of 26 ft. Determine the maximum distance d it can be placed from the smooth wall and not slip. The coefficient of static friction between the floor and the pole is . ms = 0.3 ms = 0.3 d = 10 ft P 30 Probs. 8–1/2/3 200 mm x 100 mm B A C G A B 10 ft 3 ft u Probs. 8–5/6 A d B 26 ft Probs. 8–7/8 Prob. 8–4
- 419. 402 CHAPTER 8 FRICTION 8 A B 8 ft 5 ft 5 ft 6 ft P Probs. 8–10/11 A M P B O 125 mm 700 mm 500 mm 300 mm Probs. 8–12/13 A B 0.6 m 0.3 m 60 Prob. 8–14 A B u Prob. 8–9 *8–12. The coefficients of static and kinetic friction between the drum and brake bar are and , respectively. If and determine the horizontal and vertical components of reaction at the pin O. Neglect the weight and thickness of the brake.The drum has a mass of 25 kg. •8–13. The coefficient of static friction between the drum and brake bar is . If the moment , determine the smallest force P that needs to be applied to the brake bar in order to prevent the drum from rotating. Also determine the corresponding horizontal and vertical components of reaction at pin O. Neglect the weight and thickness of the brake bar.The drum has a mass of 25 kg. M = 35 N # m ms = 0.4 P = 85 N M = 50 N # m mk = 0.3 ms = 0.4 8–10. The uniform 20-lb ladder rests on the rough floor for which the coefficient of static friction is and against the smooth wall at B. Determine the horizontal force P the man must exert on the ladder in order to cause it to move. 8–11. The uniform 20-lb ladder rests on the rough floor for which the coefficient of static friction is and against the smooth wall at B. Determine the horizontal force P the man must exert on the ladder in order to cause it to move. ms = 0.4 ms = 0.8 •8–9. If the coefficient of static friction at all contacting surfaces is , determine the inclination at which the identical blocks, each of weight W, begin to slide. u ms 8–14. Determine the minimum coefficient of static friction between the uniform 50-kg spool and the wall so that the spool does not slip.
- 420. 8.2 PROBLEMS INVOLVING DRY FRICTION 403 8 0.1 m G A B 0.4 m P 8–18. The tongs are used to lift the 150-kg crate, whose center of mass is at G. Determine the least coefficient of static friction at the pivot blocks so that the crate can be lifted. *8–16. The 80-lb boy stands on the beam and pulls on the cord with a force large enough to just cause him to slip. If the coefficient of static friction between his shoes and the beam is , determine the reactions at A and B. The beam is uniform and has a weight of 100 lb. Neglect the size of the pulleys and the thickness of the beam. •8–17. The 80-lb boy stands on the beam and pulls with a force of 40 lb. If , determine the frictional force between his shoes and the beam and the reactions at A and B. The beam is uniform and has a weight of 100 lb. Neglect the size of the pulleys and the thickness of the beam. (ms)D = 0.4 (ms)D = 0.4 8–15. The spool has a mass of 200 kg and rests against the wall and on the floor. If the coefficient of static friction at B is , the coefficient of kinetic friction is , and the wall is smooth, determine the friction force developed at B when the vertical force applied to the cable is . P = 800 N (mk)B = 0.2 (ms)B = 0.3 D A C B 5 ft 60 3 ft 12 13 5 4 ft 1 ft Probs. 8–16/17 A u B k 2 lb/ft 275 mm 300 mm 30 500 mm 500 mm A C D F H E B P G Prob. 8–18 8–19. Two blocks A and B have a weight of 10 lb and 6 lb, respectively. They are resting on the incline for which the coefficients of static friction are and . Determine the incline angle for which both blocks begin to slide.Also find the required stretch or compression in the connecting spring for this to occur.The spring has a stiffness of . *8–20. Two blocks A and B have a weight of 10 lb and 6 lb, respectively. They are resting on the incline for which the coefficients of static friction are and . Determine the angle which will cause motion of one of the blocks. What is the friction force under each of the blocks when this occurs? The spring has a stiffness of and is originally unstretched. k = 2 lbft u mB = 0.25 mA = 0.15 k = 2 lbft u mB = 0.25 mA = 0.15 Probs. 8–19/20 Prob. 8–15
- 421. 404 CHAPTER 8 FRICTION 8 B A C D u Prob. 8–21 F 120 N F 120 N Prob. 8–22 8–23. The paper towel dispenser carries two rolls of paper. The one in use is called the stub roll A and the other is the fresh roll B. They weigh 2 lb and 5 lb, respectively. If the coefficients of static friction at the points of contact C and D are and , determine the initial vertical force P that must be applied to the paper on the stub roll in order to pull down a sheet.The stub roll is pinned in the center,whereas the fresh roll is not.Neglect friction at the pin. (ms)D = 0.5 (ms)C = 0.2 8–22. A man attempts to support a stack of books horizontally by applying a compressive force of to the ends of the stack with his hands. If each book has a mass of 0.95 kg, determine the greatest number of books that can be supported in the stack. The coefficient of static friction between the man’s hands and a book is and between any two books . (ms)b = 0.4 (ms)h = 0.6 F = 120 N •8–21. Crates A and B weigh 200 lb and 150 lb, respectively. They are connected together with a cable and placed on the inclined plane. If the angle is gradually increased, determine when the crates begin to slide. The coefficients of static friction between the crates and the plane are and . mB = 0.35 mA = 0.25 u u *8–24. The drum has a weight of 100 lb and rests on the floor for which the coefficient of static friction is . If ft and ft, determine the smallest magnitude of the force P that will cause impending motion of the drum. •8–25. The drum has a weight of 100 lb and rests on the floor for which the coefficient of static friction is . If ft and ft, determine the smallest magnitude of the force P that will cause impending motion of the drum. b = 4 a = 3 ms = 0.5 b = 3 a = 2 ms = 0.6 P 60 3 in. 4 in. 45 A B C D Prob. 8–23 b a P 3 4 5 Probs. 8–24/25
- 422. 8.2 PROBLEMS INVOLVING DRY FRICTION 405 8 •8–29. If the center of gravity of the stacked tables is at G, and the stack weighs 100 lb, determine the smallest force P the boy must push on the stack in order to cause movement. The coefficient of static friction at A and B is . The tables are locked together. ms = 0.3 *8–28. Determine the minimum force P needed to push the two 75-kg cylinders up the incline. The force acts parallel to the plane and the coefficients of static friction of the contacting surfaces are , , and . Each cylinder has a radius of 150 mm. mC = 0.4 mB = 0.25 mA = 0.3 8–26. The refrigerator has a weight of 180 lb and rests on a tile floor for which . If the man pushes horizontally on the refrigerator in the direction shown, determine the smallest magnitude of horizontal force needed to move it. Also, if the man has a weight of 150 lb, determine the smallest coefficient of friction between his shoes and the floor so that he does not slip. 8–27. The refrigerator has a weight of 180 lb and rests on a tile floor for which .Also, the man has a weight of 150 lb and the coefficient of static friction between the floor and his shoes is . If he pushes horizontally on the refrigerator, determine if he can move it. If so, does the refrigerator slip or tip? ms = 0.6 ms = 0.25 ms = 0.25 8–30. The tractor has a weight of 8000 lb with center of gravity at G. Determine if it can push the 550-lb log up the incline.The coefficient of static friction between the log and the ground is , and between the rear wheels of the tractor and the ground . The front wheels are free to roll. Assume the engine can develop enough torque to cause the rear wheels to slip. 8–31. The tractor has a weight of 8000 lb with center of gravity at G. Determine the greatest weight of the log that can be pushed up the incline. The coefficient of static friction between the log and the ground is , and between the rear wheels of the tractor and the ground . The front wheels are free to roll. Assume the engine can develop enough torque to cause the rear wheels to slip. ms œ = 0.7 ms = 0.5 mœ s = 0.8 ms = 0.5 3 ft 3 ft 1.5 ft G A 4 ft Probs. 8–26/27 P A B C 30 Prob. 8–28 G A B 30 3.5 ft 3 ft 2 ft P 2 ft Prob. 8–29 7 ft 3 ft 1.25 ft 2.5 ft 10 A B G C Probs. 8–30/31
- 423. 406 CHAPTER 8 FRICTION 8 A B C 8 m 5 m u Prob. 8–32 L A B u Prob. 8–34 20 in. 3 in. 3 in. 1 in. A C P 30 Prob. 8–33 8–34. The thin rod has a weight W and rests against the floor and wall for which the coefficients of static friction are and , respectively. Determine the smallest value of for which the rod will not move. u mB mA •8–33. A force of is applied perpendicular to the handle of the gooseneck wrecking bar as shown. If the coefficient of static friction between the bar and the wood is , determine the normal force of the tines at A on the upper board.Assume the surface at C is smooth. ms = 0.5 P = 20 lb *8–32. The 50-kg uniform pole is on the verge of slipping at A when . Determine the coefficient of static friction at A. u = 45° •8–37. If the coefficient of static friction between the chain and the inclined plane is , determine the overhang length b so that the chain is on the verge of slipping up the plane.The chain weighs w per unit length. ms = tan u 30 P A 3 in. O u Probs. 8–35/36 b a u Prob. 8–37 8–35. A roll of paper has a uniform weight of 0.75 lb and is suspended from the wire hanger so that it rests against the wall. If the hanger has a negligible weight and the bearing at O can be considered frictionless, determine the force P needed to start turning the roll if . The coefficient of static friction between the wall and the paper is . *8–36. A roll of paper has a uniform weight of 0.75 lb and is suspended from the wire hanger so that it rests against the wall. If the hanger has a negligible weight and the bearing at O can be considered frictionless, determine the minimum force P and the associated angle needed to start turning the roll. The coefficient of static friction between the wall and the paper is ms = 0.25. u ms = 0.25 u = 30°
- 424. 8.2 PROBLEMS INVOLVING DRY FRICTION 407 8 •8–41. The clamp is used to tighten the connection between two concrete drain pipes. Determine the least coefficient of static friction at A or B so that the clamp does not slip regardless of the force in the shaft CD. 8–39. If the coefficient of static friction at B is , determine the largest angle and the minimum coefficient of static friction at A so that the roller remains self-locking, regardless of the magnitude of force P applied to the belt. Neglect the weight of the roller and neglect friction between the belt and the vertical surface. *8–40. If , determine the minimum coefficient of static friction at A and B so that the roller remains self- locking, regardless of the magnitude of force P applied to the belt. Neglect the weight of the roller and neglect friction between the belt and the vertical surface. u = 30° u ms = 0.3 8–38. Determine the maximum height h in meters to which the girl can walk up the slide without supporting herself by the rails or by her left leg.The coefficient of static friction between the girl’s shoes and the slide is . ms = 0.8 8–42. The coefficient of static friction between the 150-kg crate and the ground is , while the coefficient of static friction between the 80-kg man’s shoes and the ground is . Determine if the man can move the crate. 8–43. If the coefficient of static friction between the crate and the ground is , determine the minimum coefficient of static friction between the man’s shoes and the ground so that the man can move the crate. ms = 0.3 ms œ = 0.4 ms = 0.3 y h x y x2 1 –– 3 Prob. 8–38 P A B 30 mm u Probs. 8–39/40 B C D A 100 mm 250 mm Prob. 8–41 30 Probs. 8–42/43
- 425. 408 CHAPTER 8 FRICTION 8 8–47. Block C has a mass of 50 kg and is confined between two walls by smooth rollers. If the block rests on top of the 40-kg spool, determine the minimum cable force P needed to move the spool. The cable is wrapped around the spool’s inner core. The coefficients of static friction at A and B are and . *8–48. Block C has a mass of 50 kg and is confined between two walls by smooth rollers. If the block rests on top of the 40-kg spool, determine the required coefficients of static friction at A and B so that the spool slips at A and B when the magnitude of the applied force is increased to . P = 300 N mB = 0.6 mA = 0.3 •8–45. The 45-kg disk rests on the surface for which the coefficient of static friction is Determine the largest couple moment M that can be applied to the bar without causing motion. 8–46. The 45-kg disk rests on the surface for which the coefficient of static friction is If determine the friction force at A. M = 50 N # m, mA = 0.15. mA = 0.2. *8–44. The 3-Mg rear-wheel-drive skid loader has a center of mass at G. Determine the largest number of crates that can be pushed by the loader if each crate has a mass of 500 kg.The coefficient of static friction between a crate and the ground is , and the coefficient of static friction between the rear wheels of the loader and the ground is . The front wheels are free to roll. Assume that the engine of the loader is powerful enough to generate a torque that will cause the rear wheels to slip. ms œ = 0.5 ms = 0.3 •8–49. The 3-Mg four-wheel-drive truck (SUV) has a center of mass at G. Determine the maximum mass of the log that can be towed by the truck. The coefficient of static friction between the log and the ground is , and the coefficient of static friction between the wheels of the truck and the ground is . Assume that the engine of the truck is powerful enough to generate a torque that will cause all the wheels to slip. 8–50. A 3-Mg front-wheel-drive truck (SUV) has a center of mass at G. Determine the maximum mass of the log that can be towed by the truck. The coefficient of static friction between the log and the ground is , and the coefficient of static friction between the front wheels of the truck and the ground is .The rear wheels are free to roll.Assume that the engine of the truck is powerful enough to generate a torque that will cause the front wheels to slip. ms œ = 0.4 ms = 0.8 ms œ = 0.4 ms = 0.8 0.75 m 0.25 m G 0.3 m B A Prob. 8–44 400 mm 125 mm 300 mm B A C M Probs. 8–45/46 C A B O 0.4 m 0.2 m P Probs. 8–47/48 1.2 m 1.6 m 0.5 m G A B Probs. 8–49/50
- 426. 8.2 PROBLEMS INVOLVING DRY FRICTION 409 8 8–55. If the 75-lb girl is at position d = 4 ft, determine the minimum coefficient of static friction at contact points A and B so that the plank does not slip. Neglect the weight of the plank. *8–56. If the coefficient of static friction at the contact points A and B is , determine the minimum distance d where a 75-lb girl can stand on the plank without causing it to slip. Neglect the weight of the plank. ms = 0.4 ms •8–53. The carpenter slowly pushes the uniform board horizontally over the top of the saw horse. The board has a uniform weight of and the saw horse has a weight of 15 lb and a center of gravity at G. Determine if the saw horse will stay in position, slip, or tip if the board is pushed forward when The coefficients of static friction are shown in the figure. 8–54. The carpenter slowly pushes the uniform board horizontally over the top of the saw horse. The board has a uniform weight of and the saw horse has a weight of 15 lb and a center of gravity at G. Determine if the saw horse will stay in position, slip, or tip if the board is pushed forward when The coefficients of static friction are shown in the figure. d = 14 ft. 3 lbft, d = 10 ft. 3 lbft, 8–51. If the coefficients of static friction at contact points A and B are and respectively, determine the smallest force P that will cause the 150-kg spool to have impending motion. *8–52. If the coefficients of static friction at contact points A and B are and respectively, determine the smallest force P that will cause the 150-kg spool to have impending motion. ms œ = 0.2 ms = 0.4 ms œ = 0.4 ms = 0.3 •8–57. If each box weighs 150 lb, determine the least horizontal force P that the man must exert on the top box in order to cause motion. The coefficient of static friction between the boxes is , and the coefficient of static friction between the box and the floor is . 8–58. If each box weighs 150 lb, determine the least horizontal force P that the man must exert on the top box in order to cause motion. The coefficient of static friction between the boxes is , and the coefficient of static friction between the box and the floor is . ms œ = 0.35 ms = 0.65 ms œ = 0.2 ms = 0.5 P 400 mm 200 mm 150 mm B A Probs. 8–51/52 d G 18 ft 1 ft 1 ft 3 ft m 0.5 m¿ 0.3 m¿ 0.3 Probs. 8–53/54 A G d B 12 ft 45 60 Probs. 8–55/56 3 ft 4.5 ft 5 ft P 4.5 ft A B Probs. 8–57/58
- 427. 410 CHAPTER 8 FRICTION 8 8–62. Blocks A, B, and C have weights of 50 lb, 25 lb, and 15 lb, respectively. Determine the smallest horizontal force P that will cause impending motion. The coefficient of static friction between A and B is , between B and C, , and between block C and the ground, . m¿œ s = 0.35 ms œ = 0.4 ms = 0.3 •8–61. Each of the cylinders has a mass of 50 kg. If the coefficients of static friction at the points of contact are , , , and , determine the smallest couple moment M needed to rotate cylinder E. mD = 0.6 mC = 0.5 mB = 0.5 mA = 0.5 8–59. If the coefficient of static friction between the collars A and B and the rod is , determine the maximum angle for the system to remain in equilibrium, regardless of the weight of cylinder D. Links AC and BC have negligible weight and are connected together at C by a pin. *8–60. If , determine the minimum coefficient of static friction between the collars A and B and the rod required for the system to remain in equilibrium, regardless of the weight of cylinder D. Links AC and BC have negligible weight and are connected together at C by a pin. u = 15° u ms = 0.6 8–63. Determine the smallest force P that will cause impending motion. The crate and wheel have a mass of 50 kg and 25 kg, respectively. The coefficient of static friction between the crate and the ground is , and between the wheel and the ground . *8–64. Determine the smallest force P that will cause impending motion. The crate and wheel have a mass of 50 kg and 25 kg, respectively. The coefficient of static friction between the crate and the ground is , and between the wheel and the ground . ms œ = 0.3 ms = 0.5 ms œ = 0.5 ms = 0.2 D C A B u u 15 15 Probs. 8–59/60 300 mm A D 300 mm E M B C Prob. 8–61 P A B C D Prob. 8–62 300 mm P B C A Probs. 8–63/64
- 428. 8.2 PROBLEMS INVOLVING DRY FRICTION 411 8 CONCEPTUAL PROBLEMS P8–3. The rope is used to tow the refrigerator. Is it best to pull slightly up on the rope as shown, pull horizontally, or pull somewhat downwards? Also, is it best to attach the rope at a high position as shown, or at a lower position? Do an equilibrium analysis to explain your answer. P8–4. The rope is used to tow the refrigerator. In order to prevent yourself from slipping while towing, is it best to pull up as shown, pull horizontally, or pull downwards on the rope? Do an equilibrium analysis to explain your answer. P8–2. The lug nut on the free-turning wheel is to be removed using the wrench. Which is the most effective way to apply force to the wrench? Also, why is it best to keep the car tire on the ground rather than first jacking it up? Explain your answers with an equilibrium analysis. P8–1. Is it more effective to move the load forward at constant velocity with the boom fully extended as shown, or should the boom be fully retracted? Power is supplied to the rear wheels. The front wheels are free to roll. Do an equilibrium analysis to explain your answer. P8–5. Is it easier to tow the load by applying a force along the tow bar when it is in an almost horizontal position as shown, or is it better to pull on the bar when it has a steeper slope? Do an equilibrium analysis to explain your answer. P8–1 P8–2 P8–3/4 P8–5
- 429. 412 CHAPTER 8 FRICTION 8 8.3 Wedges A wedge is a simple machine that is often used to transform an applied force into much larger forces, directed at approximately right angles to the applied force. Wedges also can be used to slightly move or adjust heavy loads. Consider, for example, the wedge shown in Fig. 8–12a, which is used to lift the block by applying a force to the wedge. Free-body diagrams of the block and wedge are shown in Fig. 8–12b. Here we have excluded the weight of the wedge since it is usually small compared to the weight of the block. Also, note that the frictional forces and must oppose the motion of the wedge. Likewise, the frictional force of the wall on the block must act downward so as to oppose the block’s upward motion. The locations of the resultant normal forces are not important in the force analysis since neither the block nor wedge will “tip.” Hence the moment equilibrium equations will not be considered. There are seven unknowns, consisting of the applied force P, needed to cause motion of the wedge, and six normal and frictional forces. The seven available equations consist of four force equilibrium equations, applied to the wedge and block, and three frictional equations, , applied at the surface of contact. If the block is to be lowered, then the frictional forces will all act in a sense opposite to that shown in Fig. 8–12b. Provided the coefficient of friction is very small or the wedge angle is large, then the applied force P must act to the right to hold the block. Otherwise, P may have a reverse sense of direction in order to pull on the wedge to remove it. If P is not applied and friction forces hold the block in place, then the wedge is referred to as self-locking. u F = mN ©Fy = 0 ©Fx = 0, F3 F2 F1 W (a) Impending motion P W u F3 N3 (b) W F2 N2 P F2 N2 F1 N1 u Fig. 8–12 Wedges are often used to adjust the elevation of structural or mechanical parts. Also, they provide stability for objects such as this pipe.
- 430. 8.3 WEDGES 413 8 EXAMPLE 8.6 The uniform stone in Fig. 8–13a has a mass of 500 kg and is held in the horizontal position using a wedge at B. If the coefficient of static friction is at the surfaces of contact, determine the minimum force P needed to remove the wedge.Assume that the stone does not slip at A. ms = 0.3 SOLUTION The minimum force P requires at the surfaces of contact with the wedge. The free-body diagrams of the stone and wedge are shown in Fig. 8–13b. On the wedge the friction force opposes the impending motion, and on the stone at A, since slipping does not occur there. There are five unknowns. Three equilibrium equations for the stone and two for the wedge are available for solution. From the free-body diagram of the stone, a Using this result for the wedge, we have Ans. NOTE: Since P is positive, indeed the wedge must be pulled out. If P were zero, the wedge would remain in place (self-locking) and the frictional forces developed at B and C would satisfy and FC 6 msNC. FB 6 msNB P = 1154.9 N = 1.15 kN P - 0.3(2452.5 N) = 0 2383.1 sin 7° N - 0.312383.1 cos 7° N2 + : + ©F x = 0; NC = 2452.5 N NC - 2383.1 cos 7° N - 0.312383.1 sin 7° N2 = 0 + c©F y = 0; NB = 2383.1 N + 10.3NB sin 7° N211 m2 = 0 -4905 N10.5 m2 + 1NB cos 7° N211 m2 +©MA = 0; FA … msNA, F = msN (a) P 7 B A C 1 m FA 0.3 NB P 7 0.5 m (b) 0.5 m NB NA 7 7 7 NC NB 0.3 NB 0.3 NC 4905 N A Impending motion Fig. 8–13
- 431. r l A B 2pr r A B l (b) B A u (a) Fig. 8–14 414 CHAPTER 8 FRICTION 8 8.4 Frictional Forces on Screws In most cases screws are used as fasteners; however, in many types of machines they are incorporated to transmit power or motion from one part of the machine to another.A square-threaded screw is commonly used for the latter purpose, especially when large forces are applied along its axis. In this section we will analyze the forces acting on square-threaded screws.The analysis of other types of screws, such as the V-thread, is based on these same principles. For analysis, a square-threaded screw, as in Fig. 8–14, can be considered a cylinder having an inclined square ridge or thread wrapped around it. If we unwind the thread by one revolution, as shown in Fig. 8–14b, the slope or the lead angle is determined from . Here l and are the vertical and horizontal distances between A and B, where r is the mean radius of the thread. The distance l is called the lead of the screw and it is equivalent to the distance the screw advances when it turns one revolution. Upward Impending Motion. Let us now consider the case of a square-threaded screw that is subjected to upward impending motion caused by the applied torsional moment M, Fig. 8–15.* A free-body diagram of the entire unraveled thread can be represented as a block as shown in Fig. 8–14a. The force W is the vertical force acting on the thread or the axial force applied to the shaft, Fig. 8–15, and is the resultant horizontal force produced by the couple moment M about the axis of the shaft. The reaction R of the groove on the thread, has both frictional and normal components, where . The angle of static friction is . Applying the force equations of equilibrium along the horizontal and vertical axes, we have R cos (fs + u) - W = 0 + c ©F y = 0; Mr - R sin (fs + u) = 0 : + ©F x = 0; fs = tan-1 (FN) = tan-1 ms F = ms N Mr 2pr u = tan-1 (l2pr) u Eliminating R from these equations, we obtain (8–3) M = rW tan (fs + u) *For applications, M is developed by applying a horizontal force P at a right angle to the end of a lever that would be fixed to the screw. Square-threaded screws find applications on valves, jacks, and vises, where particularly large forces must be developed along the axis of the screw.
- 432. 8.4 FRICTIONAL FORCES ON SCREWS 415 8 Self-Locking Screw. A screw is said to be self-locking if it remains in place under any axial load W when the moment M is removed. For this to occur, the direction of the frictional force must be reversed so that R acts on the other side of N. Here the angle of static friction becomes greater than or equal to , Fig. 8–16d. If , Fig. 8–16b, then R will act vertically to balance W, and the screw will be on the verge of winding downward. Downward Impending Motion. . If a screw is self- locking, a couple moment must be applied to the screw in the opposite direction to wind the screw downward . This causes a reverse horizontal force that pushes the thread down as indicated in Fig. 8–16c. Using the same procedure as before, we obtain (8–4) Downward Impending Motion. . If the screw is not self-locking, it is necessary to apply a moment to prevent the screw from winding downward . Here, a horizontal force is required to push against the thread to prevent it from sliding down the plane, Fig. 8–16d. Thus, the magnitude of the moment required to prevent this unwinding is (8–5) If motion of the screw occurs, Eqs. 8–3, 8–4, and 8–5 can be applied by simply replacing with . fk fs M– = Wr tan (fs - u) M– M–r (fs 6 u ) M– (fs 6 u ) M¿ = rW tan (u - fs) M¿r (fs 7 u) M¿ (fs 7 u) fs = u u fs W h r M Fig. 8–15 W Downward screw motion (u fs) M¿/r n (c) R fs u u W Self-locking screw (u fs) (on the verge of rotating downward) R (b) n u fs u W Downward screw motion (u fs) (d) M–/r R n u u fs Fig. 8–16 W Upward screw motion N F R (a) n Mr u u fs
- 433. 416 CHAPTER 8 FRICTION 8 The turnbuckle shown in Fig. 8–17 has a square thread with a mean radius of 5 mm and a lead of 2 mm. If the coefficient of static friction between the screw and the turnbuckle is determine the moment M that must be applied to draw the end screws closer together. ms = 0.25, EXAMPLE 8.7 Fig. 8–17 M 2 kN 2 kN SOLUTION The moment can be obtained by applying Eq. 8–3. Since friction at two screws must be overcome, this requires (1) Here and Substituting these values into Eq. 1 and solving gives Ans. NOTE: When the moment is removed, the turnbuckle will be self- locking; i.e., it will not unscrew since fs 7 u. = 6374.7 N # mm = 6.37 N # m M = 2[12000 N215 mm2 tan114.04° + 3.64°2] u = tan-1 1l2pr2 = tan-1 12 mm[2p15 mm2]2 = 3.64°. = 14.04°, fs = tan-1 ms = tan-1 10.252 r = 5 mm, W = 2000 N, M = 2[Wr tan1u + f2]
- 434. 8.4 FRICTIONAL FORCES ON SCREWS 417 8 A P B B 15 Prob. 8–65 P 10 A C B 0.5 m Prob. 8–67 P A B 15 Prob. 8–66 PROBLEMS *8–68. The wedge has a negligible weight and a coefficient of static friction with all contacting surfaces. Determine the largest angle so that it is “self-locking.” This requires no slipping for any magnitude of the force P applied to the joint. u ms = 0.35 8–67. Determine the smallest horizontal force P required to lift the 100-kg cylinder. The coefficients of static friction at the contact points A and B are and , respectively; and the coefficient of static friction between the wedge and the ground is . ms = 0.3 (ms)B = 0.2 (ms)A = 0.6 •8–65. Determine the smallest horizontal force P required to pull out wedge A.The crate has a weight of 300 lb and the coefficient of static friction at all contacting surfaces is . Neglect the weight of the wedge. ms = 0.3 8–70. The three stone blocks have weights of , and Determine the smallest horizontal force P that must be applied to block C in order to move this block.The coefficient of static friction between the blocks is and between the floor and each block ms œ = 0.5. ms = 0.3, WC = 500 lb. W B = 150 lb, W A = 600 lb –– 2 –– 2 P u u P Prob. 8–68 •8–69. Determine the smallest horizontal force P required to just move block A to the right if the spring force is and the coefficient of static friction at all contacting surfaces on A is .The sleeve at C is smooth. Neglect the mass of A and B. ms = 0.3 600 N 8–66. Determine the smallest horizontal force P required to lift the 200-kg crate. The coefficient of static friction at all contacting surfaces is . Neglect the mass of the wedge. ms = 0.3 A P B C 45 45 Prob. 8–69 A B C 45 P Prob. 8–70
- 435. 418 CHAPTER 8 FRICTION 8 P 300 mm 450 mm 20 mm A B C F u Probs. 8–71/72 P 30 30 15 Probs. 8–73/74 8–75. If the uniform concrete block has a mass of 500 kg, determine the smallest horizontal force P needed to move the wedge to the left. The coefficient of static friction between the wedge and the concrete and the wedge and the floor is . The coefficient of static friction between the concrete and floor is . ms œ = 0.5 ms = 0.3 •8–73. Determine the smallest vertical force P required to hold the wedge between the two identical cylinders, each having a weight of W. The coefficient of static friction at all contacting surfaces is . 8–74. Determine the smallest vertical force P required to push the wedge between the two identical cylinders, each having a weight of W. The coefficient of static friction at all contacting surfaces is . ms = 0.3 ms = 0.1 8–71. Determine the smallest horizontal force P required to move the wedge to the right. The coefficient of static friction at all contacting surfaces is . Set and . Neglect the weight of the wedge. *8–72. If the horizontal force P is removed, determine the largest angle that will cause the wedge to be self-locking regardless of the magnitude of force F applied to the handle. The coefficient of static friction at all contacting surfaces is . ms = 0.3 u F = 400 N u = 15° ms = 0.3 *8–76. The wedge blocks are used to hold the specimen in a tension testing machine. Determine the largest design angle of the wedges so that the specimen will not slip regardless of the applied load. The coefficients of static friction are at A and at B. Neglect the weight of the blocks. mB = 0.6 mA = 0.1 u A 3 m P 150 mm B 7.5 Prob. 8–75 P A B u u Prob. 8–76
- 436. 1.5 N m ⫺F F Prob. 8–77 8.4 FRICTIONAL FORCES ON SCREWS 419 8 8–79. The jacking mechanism consists of a link that has a square-threaded screw with a mean diameter of 0.5 in. and a lead of 0.20 in., and the coefficient of static friction is . Determine the torque M that should be applied to the screw to start lifting the 6000-lb load acting at the end of member ABC. ms = 0.4 8–78. The device is used to pull the battery cable terminal C from the post of a battery. If the required pulling force is 85 lb, determine the torque M that must be applied to the handle on the screw to tighten it. The screw has square threads, a mean diameter of 0.2 in., a lead of 0.08 in., and the coefficient of static friction is . ms = 0.5 •8–77. The square threaded screw of the clamp has a mean diameter of 14 mm and a lead of 6 mm. If for the threads, and the torque applied to the handle is , determine the compressive force F on the block. 1.5 N # m ms = 0.2 *8–80. Determine the magnitude of the horizontal force P that must be applied to the handle of the bench vise in order to produce a clamping force of 600 N on the block. The single square-threaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction is . •8–81. Determine the clamping force exerted on the block if a force of P = 30 N is applied to the lever of the bench vise. The single square-threaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction is . ms = 0.25 ms = 0.25 C A B M Prob. 8–78 D B C A 7.5 in. 10 in. 15 in. 20 in. 10 in. 6000 lb M Prob. 8–79 100 mm P Probs. 8–80/81
- 437. 420 CHAPTER 8 FRICTION 8 •8–85. If the jack supports the 200-kg crate, determine the horizontal force that must be applied perpendicular to the handle at E to lower the crate. Each single square-threaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction is . 8–86. If the jack is required to lift the 200-kg crate, determine the horizontal force that must be applied perpendicular to the handle at E. Each single square- threaded screw has a mean diameter of 25 mm and a lead of 7.5 mm.The coefficient of static friction is . ms = 0.25 ms = 0.25 *8–84. The clamp provides pressure from several directions on the edges of the board. If the square-threaded screw has a lead of 3 mm, mean radius of 10 mm, and the coefficient of static friction is determine the horizontal force developed on the board at A and the vertical forces developed at B and C if a torque of is applied to the handle to tighten it further.The blocks at B and C are pin connected to the board. M = 1.5 N # m ms = 0.4, 8–82. Determine the required horizontal force that must be applied perpendicular to the handle in order to develop a 900-N clamping force on the pipe. The single square- threaded screw has a mean diameter of 25 mm and a lead of 5 mm.The coefficient of static friction is . Note: The screw is a two-force member since it is contained within pinned collars at A and B. 8–83. If the clamping force on the pipe is 900 N, determine the horizontal force that must be applied perpendicular to the handle in order to loosen the screw. The single square-threaded screw has a mean diameter of 25 mm and a lead of 5 mm.The coefficient of static friction is . Note: The screw is a two-force member since it is contained within pinned collars at A and B. ms = 0.4 ms = 0.4 8–87. The machine part is held in place using the double-end clamp.The bolt at B has square threads with a mean radius of 4 mm and a lead of 2 mm, and the coefficient of static friction with the nut is If a torque of is applied to the nut to tighten it, determine the normal force of the clamp at the smooth contacts A and C. M = 0.4 N # m ms = 0.5. B D E C A 150 mm 200 mm 200 mm Probs. 8–82/83 45 A B C D 45 M Prob. 8–84 C A B D E 100 mm 45 45 45 45 Probs. 8–85/86 260 mm A C B 90 mm Prob. 8–87
- 438. 8.5 FRICTIONAL FORCES ON FLAT BELTS 421 8 8.5 Frictional Forces on Flat Belts Whenever belt drives or band brakes are designed, it is necessary to determine the frictional forces developed between the belt and its contacting surface. In this section we will analyze the frictional forces acting on a flat belt, although the analysis of other types of belts, such as the V-belt, is based on similar principles. Consider the flat belt shown in Fig. 8–18a, which passes over a fixed curved surface. The total angle of belt to surface contact in radians is , and the coefficient of friction between the two surfaces is We wish to determine the tension in the belt, which is needed to pull the belt counterclockwise over the surface, and thereby overcome both the frictional forces at the surface of contact and the tension in the other end of the belt. Obviously, Frictional Analysis. A free-body diagram of the belt segment in contact with the surface is shown in Fig. 8–18b.As shown, the normal and frictional forces, acting at different points along the belt, will vary both in magnitude and direction. Due to this unknown distribution, the analysis of the problem will first require a study of the forces acting on a differential element of the belt. A free-body diagram of an element having a length ds is shown in Fig. 8–18c. Assuming either impending motion or motion of the belt, the magnitude of the frictional force This force opposes the sliding motion of the belt, and so it will increase the magnitude of the tensile force acting in the belt by dT. Applying the two force equations of equilibrium, we have Since is of infinitesimal size, and . Also, the product of the two infinitesimals dT and may be neglected when compared to infinitesimals of the first order.As a result, these two equations become and Eliminating dN yields dT T = m du dN = T du m dN = dT du2 cos(du2) = 1 sin(du2) = du2 du dN - 1T + dT2 sina du 2 b - T sina du 2 b = 0 +Q©Fy = 0; T cosa du 2 b + m dN - 1T + dT2 cosa du 2 b = 0 R+©Fx = 0; dF = m dN. T2 7 T1. T1 T2 m. b Motion or impending motion of belt relative to surface (a) r T2 T1 b u (b) T1 T2 u dF mdN ds (c) T dT T y dN x du 2 du 2 du 2 du 2 Fig. 8–18
- 439. 422 CHAPTER 8 FRICTION 8 Integrating this equation between all the points of contact that the belt makes with the drum, and noting that at and at yields Solving for we obtain (8–6) where T2 = T1emb T2, ln T2 T1 = mb L T2 T1 dT T = m L b 0 du u = b, T = T2 u = 0 T = T1 Motion or impending motion of belt relative to surface r T2 T1 u Flat orV-belts are often used to transmit the torque developed by a motor to a wheel attached to a pump,fan or blower. [ [ T i n f o r m a l t a b l e 0 ] ] T1 = T2, belt tensions; opposes the direction of motion (or impending motion) of the belt measured relative to the surface, while acts in the direction of the relative belt motion (or impending motion); because of friction, T2 7 T1 T2 T1 m = coefficient of static or kinetic friction between the belt and the surface of contact b = angle of belt to surface contact, measured in radians e = base of the natural logarithm 2.718 Á , Note that is independent of the radius of the drum, and instead it is a function of the angle of belt to surface contact, As a result, this equation is valid for flat belts passing over any curved contacting surface. b. T2
- 440. 8.5 FRICTIONAL FORCES ON FLAT BELTS 423 8 EXAMPLE 8.8 The maximum tension that can be developed in the cord shown in Fig.8–19a is 500 N. If the pulley at A is free to rotate and the coefficient of static friction at the fixed drums B and C is determine the largest mass of the cylinder that can be lifted by the cord. ms = 0.25, SOLUTION Lifting the cylinder, which has a weight causes the cord to move counterclockwise over the drums at B and C; hence, the maximum tension in the cord occurs at D. Thus, A section of the cord passing over the drum at B is shown in Fig. 8–19b. Since the angle of contact between the drum and the cord is Using Eq. 8–6, we have Hence, Since the pulley at A is free to rotate, equilibrium requires that the tension in the cord remains the same on both sides of the pulley. The section of the cord passing over the drum at C is shown in Fig. 8–19c.The weight Why? Applying Eq. 8–6, we obtain so that Ans. = 15.7 kg m = W g = 153.9 N 9.81 ms2 W = 153.9 N 277.4 N = We0.25[1342p] T2 = T1emsb ; W 6 277.4 N. T1 = 500 N e0.25[1342p] = 500 N 1.80 = 277.4 N 500 N = T1e0.25[1342p] T2 = T1emsb ; b = 1135°180°2p = 3p4 rad. 180° = p rad F = T2 = 500 N. T2 W = mg, T A (a) C B D 45 45 135 Impending motion B 500 N T1 (b) W mg 277.4 N 135 Impending motion (c) C Fig. 8–19
- 441. F Prob. 8–91 424 CHAPTER 8 FRICTION 8 A B C D m 0.5 mBA 0.6 mAC 0.4 20 Probs. 8–88/89 *8–92. The boat has a weight of 500 lb and is held in position off the side of a ship by the spars at A and B.A man having a weight of 130 lb gets in the boat, wraps a rope around an overhead boom at C, and ties it to the end of the boat as shown. If the boat is disconnected from the spars, determine the minimum number of half turns the rope must make around the boom so that the boat can be safely lowered into the water at constant velocity.Also, what is the normal force between the boat and the man? The coefficient of kinetic friction between the rope and the boom is . Hint:The problem requires that the normal force between the man’s feet and the boat be as small as possible. ms = 0.15 8–90. A cylinder having a mass of 250 kg is to be supported by the cord which wraps over the pipe. Determine the smallest vertical force F needed to support the load if the cord passes (a) once over the pipe, , and (b) two times over the pipe, .Take . ms = 0.2 b = 540° b = 180° 8–91. A cylinder having a mass of 250 kg is to be supported by the cord which wraps over the pipe. Determine the largest vertical force F that can be applied to the cord without moving the cylinder. The cord passes (a) once over the pipe, , and (b) two times over the pipe, .Take . ms = 0.2 b = 540° b = 180° *8–88. Blocks A and B weigh 50 lb and 30 lb, respectively. Using the coefficients of static friction indicated, determine the greatest weight of block D without causing motion. •8–89. Blocks A and B weigh 75 lb each, and D weighs 30 lb. Using the coefficients of static friction indicated, determine the frictional force between blocks A and B and between block A and the floor C. PROBLEMS F Prob. 8–90 A C B Prob. 8–92
- 442. 8.5 FRICTIONAL FORCES ON FLAT BELTS 425 8 8–98. If a force of is applied to the handle of the bell crank, determine the maximum torque M that can be resisted so that the flywheel is not on the verge of rotating clockwise.The coefficient of static friction between the brake band and the rim of the wheel is . ms = 0.3 P = 200 N •8–97. Determine the smallest lever force P needed to prevent the wheel from rotating if it is subjected to a torque of The coefficient of static friction between the belt and the wheel is The wheel is pin connected at its center, B. ms = 0.3. M = 250 N # m. 8–95. A 10-kg cylinder D, which is attached to a small pulley B, is placed on the cord as shown. Determine the smallest angle so that the cord does not slip over the peg at C. The cylinder at E has a mass of 10 kg, and the coefficient of static friction between the cord and the peg is . *8–96. A 10-kg cylinder D, which is attached to a small pulley B, is placed on the cord as shown. Determine the largest angle so that the cord does not slip over the peg at C.The cylinder at E has a mass of 10 kg, and the coefficient of static friction between the cord and the peg is . ms = 0.1 u ms = 0.1 u A Probs. 8–93/94 A B D E C u u Probs. 8–95/96 400 mm 200 mm 750 mm P M B A Prob. 8–97 P 900 mm 400 mm 100 mm 300 mm M O A B C Prob. 8–98 •8–93. The 100-lb boy at A is suspended from the cable that passes over the quarter circular cliff rock. Determine if it is possible for the 185-lb woman to hoist him up; and if this is possible, what smallest force must she exert on the horizontal cable? The coefficient of static friction between the cable and the rock is , and between the shoes of the woman and the ground . 8–94. The 100-lb boy at A is suspended from the cable that passes over the quarter circular cliff rock. What horizontal force must the woman at A exert on the cable in order to let the boy descend at constant velocity? The coefficients of static and kinetic friction between the cable and the rock are and , respectively. mk = 0.35 ms = 0.4 ms œ = 0.8 ms = 0.2
- 443. 426 CHAPTER 8 FRICTION 8 8–103. A 180-lb farmer tries to restrain the cow from escaping by wrapping the rope two turns around the tree trunk as shown. If the cow exerts a force of 250 lb on the rope, determine if the farmer can successfully restrain the cow. The coefficient of static friction between the rope and the tree trunk is , and between the farmer’s shoes and the ground . ms œ = 0.3 ms = 0.15 8–102. The simple band brake is constructed so that the ends of the friction strap are connected to the pin at A and the lever arm at B. If the wheel is subjected to a torque of determine the smallest force P applied to the lever that is required to hold the wheel stationary. The coefficient of static friction between the strap and wheel is ms = 0.5. M = 80 lb # ft, *8–100. Determine the force developed in spring AB in order to hold the wheel from rotating when it is subjected to a couple moment of . The coefficient of static friction between the belt and the rim of the wheel is , and between the belt and peg C, . The pulley at B is free to rotate. •8–101. If the tension in the spring is , determine the largest couple moment that can be applied to the wheel without causing it to rotate. The coefficient of static friction between the belt and the wheel is , and between the belt the peg .The pulley B free to rotate. ms œ = 0.4 ms = 0.2 F AB = 2.5 kN ms œ = 0.4 ms = 0.2 M = 200 N # m C A 200 mm B M 45 Probs. 8–100/101 1.5 ft 3 ft 45 M 80 lb ft 20 1.25 ft A B P O Prob. 8–102 Prob. 8–103 8–99. Show that the frictional relationship between the belt tensions, the coefficient of friction , and the angular contacts and for the V-belt is . T2 = T1embsin(a2) b a m T2 T1 Impending motion b a Prob. 8–99
- 444. 8.5 FRICTIONAL FORCES ON FLAT BELTS 427 8 *8–104. The uniform 50-lb beam is supported by the rope which is attached to the end of the beam, wraps over the rough peg, and is then connected to the 100-lb block. If the coefficient of static friction between the beam and the block, and between the rope and the peg, is determine the maximum distance that the block can be placed from A and still remain in equilibrium. Assume the block will not tip. ms = 0.4, 8–107. The drive pulley B in a video tape recorder is on the verge of slipping when it is subjected to a torque of . If the coefficient of static friction between the tape and the drive wheel and between the tape and the fixed shafts A and C is , determine the tensions and developed in the tape for equilibrium. T2 T1 ms = 0.1 M = 0.005 N # m •8–105. The 80-kg man tries to lower the 150-kg crate using a rope that passes over the rough peg. Determine the least number of full turns in addition to the basic wrap (165°) around the peg to do the job. The coefficients of static friction between the rope and the peg and between the man’s shoes and the ground are and , respectively. 8–106. If the rope wraps three full turns plus the basic wrap (165°) around the peg, determine if the 80-kg man can keep the 300-kg crate from moving. The coefficients of static friction between the rope and the peg and between the man’s shoes and the ground are and , respectively. ms œ = 0.4 ms = 0.1 ms œ = 0.4 ms = 0.1 *8–108. Determine the maximum number of 50-lb packages that can be placed on the belt without causing the belt to slip at the drive wheel A which is rotating with a constant angular velocity. Wheel B is free to rotate. Also, find the corresponding torsional moment M that must be supplied to wheel A. The conveyor belt is pre-tensioned with the 300-lb horizontal force. The coefficient of kinetic friction between the belt and platform P is , and the coefficient of static friction between the belt and the rim of each wheel is . ms = 0.35 mk = 0.2 10 ft 1 ft d A Prob. 8–104 T1 T2 A C B M 5 mN m 10 mm 10 mm 10 mm Prob. 8–107 P 300 lb A P B M 0.5 ft 0.5 ft Prob. 8–108 15 Probs. 8–105/106
- 445. 428 CHAPTER 8 FRICTION 8 8–111. Block A has a weight of 100 lb and rests on a surface for which . If the coefficient of static friction between the cord and the fixed peg at C is , determine the greatest weight of the suspended cylinder B without causing motion. ms = 0.3 ms = 0.25 8–110. Blocks A and B have a mass of 100 kg and 150 kg, respectively. If the coefficient of static friction between A and B and between B and C is and between the ropes and the pegs D and E , determine the smallest force F needed to cause motion of block B if P = 30 N. ms œ = 0.5 ms = 0.25, •8–109. Blocks A and B have a mass of 7 kg and 10 kg, respectively. Using the coefficients of static friction indicated, determine the largest vertical force P which can be applied to the cord without causing motion. *8–112. Block A has a mass of 50 kg and rests on surface B for which . If the coefficient of static friction between the cord and the fixed peg at C is , determine the greatest mass of the suspended cylinder D without causing motion. •8–113. Block A has a mass of 50 kg and rests on surface B for which . If the mass of the suspended cylinder D is 4 kg, determine the frictional force acting on A and check if motion occurs. The coefficient of static friction between the cord and the fixed peg at C is . ms œ = 0.3 ms = 0.25 ms œ = 0.3 ms = 0.25 P 300 mm 400 mm A C D B mD 0.1 mC 0.4 mB 0.4 mA 0.3 Prob. 8–109 B 4 ft 2 ft C 30 A Prob. 8–111 C D A 0.3 m 0.25 m 3 4 5 0.4 m B Probs. 8–112/113 P F D A B E C 45 Prob. 8–110
- 446. 8.6 FRICTIONAL FORCES ON COLLAR BEARINGS, PIVOT BEARINGS, AND DISKS 429 8 *8.6 Frictional Forces on Collar Bearings, Pivot Bearings, and Disks Pivot and collar bearings are commonly used in machines to support an axial load on a rotating shaft. Typical examples are shown in Fig. 8–20. Provided these bearings are not lubricated,or are only partially lubricated, the laws of dry friction may be applied to determine the moment needed to turn the shaft when it supports an axial force. R Pivot bearing (a) M P Collar bearing (b) R1 R2 M P Fig. 8–20 Frictional Analysis. The collar bearing on the shaft shown in Fig. 8–21 is subjected to an axial force P and has a total bearing or contact area Provided the bearing is new and evenly supported, then the normal pressure p on the bearing will be uniformly distributed over this area. Since then p, measured as a force per unit area, is . The moment needed to cause impending rotation of the shaft can be determined from moment equilibrium about the z axis. A differential area element shown in Fig. 8–21, is subjected to both a normal force and an associated frictional force, dF = ms dN = msp dA = msP p1R2 2 - R1 2 2 dA dN = p dA dA = 1r du21dr2, p = Pp1R2 2 - R1 2 2 ©Fz = 0, p1R2 2 - R1 2 2. z r R1 R2 dF dN dA p M P u Fig. 8–21
- 447. 430 CHAPTER 8 FRICTION 8 The normal force does not create a moment about the z axis of the shaft; however, the frictional force does; namely, Integration is needed to compute the applied moment M needed to overcome all the frictional forces.Therefore, for impending rotational motion, Substituting for dF and dA and integrating over the entire bearing area yields or (8–7) The moment developed at the end of the shaft, when it is rotating at constant speed, can be found by substituting for in Eq. 8–7. In the case of a pivot bearing, Fig. 8–20a, then and and Eq. 8–7 reduces to (8–8) Remember that Eqs. 8–7 and 8–8 apply only for bearing surfaces subjected to constant pressure. If the pressure is not uniform, a variation of the pressure as a function of the bearing area must be determined before integrating to obtain the moment. The following example illustrates this concept. M = 2 3 msPR R1 = 0, R2 = R ms mk M = 2 3 msPa R2 3 - R1 3 R2 2 - R1 2 b = msP p1R2 2 - R1 2 2 L R2 R1 r2 dr L 2p 0 du M= L R2 R1 L 2p 0 rc msP p1R2 2 - R1 2 2 d1r du dr2 M - L A r dF = 0 ©Mz = 0; dM = r dF. The motor that turns the disk of this sanding machine develops a torque that must overcome the frictional forces acting on the disk. z p M R1 R2 P Fig. 8–21 (Repeated)
- 448. EXAMPLE 8.9 The uniform bar shown in Fig. 8–22a has a weight of 4 lb. If it is assumed that the normal pressure acting at the contacting surface varies linearly along the length of the bar as shown, determine the couple moment M required to rotate the bar. Assume that the bar’s width is negligible in comparison to its length.The coefficient of static friction is equal to . SOLUTION A free-body diagram of the bar is shown in Fig. 8–22b. The intensity of the distributed load at the center is determined from vertical force equilibrium, Fig. 8–22a. Since at the distributed load expressed as a function of x is The magnitude of the normal force acting on a differential segment of area having a length dx is therefore The magnitude of the frictional force acting on the same element of area is Hence, the moment created by this force about the z axis is The summation of moments about the z axis of the bar is determined by integration, which yields Ans. M = 0.8 lb # ft M = 0.6ax2 - x3 3 b ` 0 2 M - 2 L 2 0 (0.3)(2x - x2 ) dx = 0 ©Mz = 0; dM = x dF = 0.3(2x - x2 )dx dF = ms dN = 0.3(2 - x)dx dN = w dx = (2 - x)dx w = (2 lbft)a1 - x 2 ft b = 2 - x x = 2 ft, w = 0 -4 lb + 2c 1 2 a2 ftbw0 d = 0 w0 = 2 lbft + c ©Fz = 0; 1x = 02 w0 ms = 0.3 8.6 FRICTIONAL FORCES ON COLLAR BEARINGS, PIVOT BEARINGS, AND DISKS 431 8 2 ft 2 ft z M w0 y (a) w w(x) a x 4 lb z (b) y x x dF dx dN dN dx dF M x 4 lb Fig. 8–22
- 449. 432 CHAPTER 8 FRICTION 8 8.7 Frictional Forces on Journal Bearings When a shaft or axle is subjected to lateral loads, a journal bearing is commonly used for support. Provided the bearing is not lubricated, or is only partially lubricated, a reasonable analysis of the frictional resistance on the bearing can be based on the laws of dry friction. Frictional Analysis. A typical journal-bearing support is shown in Fig. 8–23a.As the shaft rotates, the contact point moves up the wall of the bearing to some point A where slipping occurs. If the vertical load acting at the end of the shaft is P, then the bearing reactive force R acting at A will be equal and opposite to P, Fig. 8–23b. The moment needed to maintain constant rotation of the shaft can be found by summing moments about the z axis of the shaft; i.e., or (8–9) where is the angle of kinetic friction defined by In Fig. 8–23c, it is seen that The dashed circle with radius is called the friction circle, and as the shaft rotates,the reaction R will always be tangent to it.If the bearing is partially lubricated, is small, and therefore . Under these conditions, a reasonable approximation to the moment needed to overcome the frictional resistance becomes (8–10) In practice, this type of journal bearing is not suitable for long service since friction between the shaft and bearing will wear down the surfaces. Instead, designers will incorporate “ball bearings” or “rollers” in journal bearings to minimize frictional losses. M L Rrmk sin fk L tan fk L mk mk rf r sin fk = rf. FN = mkNN = mk. tan fk = fk M = Rr sin fk M - 1R sin fk2r = 0 ©Mz = 0; Unwinding the cable from this spool requires overcoming friction from the supporting shaft. A z Rotation (a) Fig. 8–23 M P r A fk fk N R F (b) M P r R (c) rf fk
- 450. EXAMPLE 8.10 The 100-mm-diameter pulley shown in Fig. 8–24a fits loosely on a 10-mm-diameter shaft for which the coefficient of static friction is Determine the minimum tension T in the belt needed to (a) raise the 100-kg block and (b) lower the block. Assume that no slipping occurs between the belt and pulley and neglect the weight of the pulley. ms = 0.4. 8.7 FRICTIONAL FORCES ON JOURNAL BEARINGS 433 8 SOLUTION Part (a). A free-body diagram of the pulley is shown in Fig. 8–24b. When the pulley is subjected to belt tensions of 981 N each, it makes contact with the shaft at point As the tension T is increased, the contact point will move around the shaft to point before motion impends. From the figure, the friction circle has a radius Using the simplification that then so that summing moments about gives a Ans. If a more exact analysis is used, then Thus, the radius of the friction circle would be Therefore, a Ans. Part (b). When the block is lowered, the resultant force R acting on the shaft passes through point as shown in Fig. 8–24c. Summing moments about this point yields a 981 N148 mm2 - T152 mm2 = 0 +©MP3 = 0; T = 1057 N = 1.06 kN 981 N150 mm + 1.86 mm2 - T150 mm - 1.86 mm2 = 0 +©MP2 = 0; 1.86 mm. rf = r sin fs = 5 sin 21.8° = fs = tan-1 0.4 = 21.8°. T = 1063 N = 1.06 kN 981 N152 mm2 - T148 mm2 = 0 +©MP2 = 0; P2 15 mm210.42 = 2 mm, rf L rms = sin fs L tan fs L ms rf = r sin fs. P2 P1. 50 mm r 5 mm 100 kg T (a) Impending motion 52 mm 48 mm 981 N R T P1 P2 rf (b) fs 52 mm 48 mm 981 N R T P3 rf fs (c) Impending motion Fig. 8–24 Ans. NOTE: The difference between raising and lowering the block is thus 157 N. T = 906 N
- 451. 434 CHAPTER 8 FRICTION 8 *8.8 Rolling Resistance When a rigid cylinder rolls at constant velocity along a rigid surface, the normal force exerted by the surface on the cylinder acts perpendicular to the tangent at the point of contact, as shown in Fig. 8–25a. Actually, however,no materials are perfectly rigid,and therefore the reaction of the surface on the cylinder consists of a distribution of normal pressure. For example,consider the cylinder to be made of a very hard material,and the surface on which it rolls to be relatively soft.Due to its weight,the cylinder compresses the surface underneath it, Fig. 8–25b.As the cylinder rolls, the surface material in front of the cylinder retards the motion since it is being deformed, whereas the material in the rear is restored from the deformed state and therefore tends to push the cylinder forward. The normal pressures acting on the cylinder in this manner are represented in Fig.8–25b by their resultant forces and Because the magnitude of the force of deformation, and its horizontal component is always greater than that of restoration, and consequently a horizontal driving force P must be applied to the cylinder to maintain the motion. Fig. 8–25b.* Rolling resistance is caused primarily by this effect, although it is also, to a lesser degree, the result of surface adhesion and relative micro- sliding between the surfaces of contact. Because the actual force P needed to overcome these effects is difficult to determine, a simplified method will be developed here to explain one way engineers have analyzed this phenomenon. To do this, we will consider the resultant of the entire normal pressure, acting on the cylinder, Fig. 8–25c. As shown in Fig. 8–25d, this force acts at an angle with the vertical. To keep the cylinder in equilibrium, i.e., rolling at a constant rate, it is necessary that N be concurrent with the driving force P and the weight W. Summing moments about point A gives Since the deformations are generally very small in relation to the cylinder’s radius, hence, or (8–11) The distance a is termed the coefficient of rolling resistance, which has the dimension of length. For instance, for a wheel rolling on a rail, both of which are made of mild steel. For hardened a L 0.5 mm P L Wa r Wa L Pr cos u L 1; Wa = P1r cos u2. u N = Nd + Nr, Nr, Nd, Nr. Nd *Actually, the deformation force causes energy to be stored in the material as its magnitude is increased, whereas the restoration force as its magnitude is decreased, allows some of this energy to be released. The remaining energy is lost since it is used to heat up the surface, and if the cylinder’s weight is very large, it accounts for permanent deformation of the surface. Work must be done by the horizontal force P to make up for this loss. Nr, Nd (a) r W O N Rigid surface of contact Nd (b) W Soft surface of contact P Nr N Nd Nr (c) (d) r W P A a u N Fig. 8–25
- 452. 8.8 ROLLING RESISTANCE 435 8 steel ball bearings on steel, Experimentally, though, this factor is difficult to measure, since it depends on such parameters as the rate of rotation of the cylinder, the elastic properties of the contacting surfaces, and the surface finish. For this reason, little reliance is placed on the data for determining a.The analysis presented here does, however, indicate why a heavy load (W) offers greater resistance to motion (P) than a light load under the same conditions. Furthermore, since is generally very small compared to the force needed to roll a cylinder over the surface will be much less than that needed to slide it across the surface. It is for this reason that a roller or ball bearings are often used to minimize the frictional resistance between moving parts. mkW, W ar a L 0.1 mm. EXAMPLE 8.11 A 10-kg steel wheel shown in Fig. 8–26a has a radius of 100 mm and rests on an inclined plane made of soft wood. If is increased so that the wheel begins to roll down the incline with constant velocity when determine the coefficient of rolling resistance. u = 1.2°, u SOLUTION As shown on the free-body diagram, Fig. 8–26b, when the wheel has impending motion, the normal reaction N acts at point A defined by the dimension a. Resolving the weight into components parallel and perpendicular to the incline,and summing moments about point A,yields a Solving, we obtain Ans. a = 2.09 mm - 198.1 cos 1.2° N)1a2 + (98.1 sin 1.2° N)1100 cos 1.2° mm2 = 0 +©MA = 0; (b) 1.2 98.1 N 98.1 cos 1.2 N 98.1 sin 1.2 N 100 mm 1.2 O N A a (a) 100 mm u Fig. 8–26 Rolling resistance of railroad wheels on the rails is small since steel is very stiff. By comparison, the rolling resistance of the wheels of a tractor in a wet field is very large.
- 453. 436 CHAPTER 8 FRICTION 8 3 in. 2 in. P M Probs. 8–114/115 6 in. 2 in. M Prob. 8–116 •8–117. The disk clutch is used in standard transmissions of automobiles. If four springs are used to force the two plates A and B together, determine the force in each spring required to transmit a moment of across the plates. The coefficient of static friction between A and B is . ms = 0.3 M = 600 lb # ft *8–116. If the spring exerts a force of 900 lb on the block, determine the torque M required to rotate the shaft. The coefficient of static friction at all contacting surfaces is . ms = 0.3 8–114. The collar bearing uniformly supports an axial force of If the coefficient of static friction is determine the torque M required to overcome friction. 8–115. The collar bearing uniformly supports an axial force of If a torque of is applied to the shaft and causes it to rotate at constant velocity, determine the coefficient of kinetic friction at the surface of contact. M = 3 lb # ft P = 500 lb. ms = 0.3, P = 800 lb. 8–118. If is applied to the handle of the bell crank, determine the maximum torque M the cone clutch can transmit. The coefficient of static friction at the contacting surface is . ms = 0.3 P = 900 N PROBLEMS Fs M 5 in. B 2 in. A M Fs Fs Prob. 8–117 375 mm 200 mm 300 mm 250 mm P M A B C 15 Prob. 8–118
- 454. 8.8 ROLLING RESISTANCE 437 8 •8–121. The shaft is subjected to an axial force P. If the reactive pressure on the conical bearing is uniform, determine the torque M that is just sufficient to rotate the shaft. The coefficient of static friction at the contacting surface is . ms *8–120. The pivot bearing is subjected to a parabolic pressure distribution at its surface of contact. If the coefficient of static friction is , determine the torque M required to overcome friction and turn the shaft if it supports an axial force P. ms 8–119. Because of wearing at the edges, the pivot bearing is subjected to a conical pressure distribution at its surface of contact. Determine the torque M required to overcome friction and turn the shaft, which supports an axial force P. The coefficient of static friction is . For the solution, it is necessary to determine the peak pressure in terms of P and the bearing radius R. p0 ms 8–122. The tractor is used to push the 1500-lb pipe. To do this it must overcome the frictional forces at the ground, caused by sand.Assuming that the sand exerts a pressure on the bottom of the pipe as shown, and the coefficient of static friction between the pipe and the sand is determine the horizontal force required to push the pipe forward.Also, determine the peak pressure p0. ms = 0.3, P M R p0 Prob. 8–119 P p0 p p0 (1 ) r2 –– R2 R r M Prob. 8–120 P M d1 d2 u u Prob. 8–121 15 in. 12 ft p p0 cos u p0 u Prob. 8–122
- 455. 438 CHAPTER 8 FRICTION 8 P M R u Prob. 8–123 P M R2 R1 p0 p p0 R2 r r Prob. 8–124 r P M Prob. 8–125 75 mm P z 60 Probs. 8–126/127 •8–125. The shaft of radius r fits loosely on the journal bearing. If the shaft transmits a vertical force P to the bearing and the coefficient of kinetic friction between the shaft and the bearing is , determine the torque M required to turn the shaft with constant velocity. mk *8–124. Assuming that the variation of pressure at the bottom of the pivot bearing is defined as , determine the torque M needed to overcome friction if the shaft is subjected to an axial force P.The coefficient of static friction is . For the solution, it is necessary to determine in terms of P and the bearing dimensions and . R2 R1 p0 ms p = p01R2r2 8–123. The conical bearing is subjected to a constant pressure distribution at its surface of contact. If the coefficient of static friction is determine the torque M required to overcome friction if the shaft supports an axial force P. ms, 8–126. The pulley is supported by a 25-mm-diameter pin. If the pulley fits loosely on the pin, determine the smallest force P required to raise the bucket. The bucket has a mass of 20 kg and the coefficient of static friction between the pulley and the pin is . Neglect the mass of the pulley and assume that the cable does not slip on the pulley. 8–127. The pulley is supported by a 25-mm-diameter pin. If the pulley fits loosely on the pin, determine the largest force P that can be applied to the rope and yet lower the bucket. The bucket has a mass of 20 kg and the coefficient of static friction between the pulley and the pin is . Neglect the mass of the pulley and assume that the cable does not slip on the pulley. ms = 0.3 ms = 0.3
- 456. 8.8 ROLLING RESISTANCE 439 8 A B 800 mm 600 mm Probs. 8–128/129 A B Probs. 8–130/131 P 10 in. 12 in. 50 lb 45 120 mm P Probs. 8–134/135 *8–132. The 5-kg pulley has a diameter of 240 mm and the axle has a diameter of 40 mm. If the coefficient of kinetic friction between the axle and the pulley is determine the vertical force P on the rope required to lift the 80-kg block at constant velocity. •8–133. Solve Prob. 8–132 if the force P is applied horizontally to the right. mk = 0.15, 8–130. The connecting rod is attached to the piston by a 0.75-in.-diameter pin at B and to the crank shaft by a 2-in.-diameter bearing A. If the piston is moving downwards, and the coefficient of static friction at the contact points is , determine the radius of the friction circle at each connection. 8–131. The connecting rod is attached to the piston by a 20-mm-diameter pin at B and to the crank shaft by a 50-mm-diameter bearing A. If the piston is moving upwards, and the coefficient of static friction at the contact points is , determine the radius of the friction circle at each connection. ms = 0.3 ms = 0.2 *8–128. The cylinders are suspended from the end of the bar which fits loosely into a 40-mm-diameter pin. If A has a mass of 10 kg, determine the required mass of B which is just sufficient to keep the bar from rotating clockwise. The coefficient of static friction between the bar and the pin is . Neglect the mass of the bar. •8–129. The cylinders are suspended from the end of the bar which fits loosely into a 40-mm-diameter pin. If A has a mass of 10 kg, determine the required mass of B which is just sufficient to keep the bar from rotating counterclockwise. The coefficient of static friction between the bar and the pin is . Neglect the mass of the bar. ms = 0.3 ms = 0.3 8–134. The bell crank fits loosely into a 0.5-in-diameter pin. Determine the required force P which is just sufficient to rotate the bell crank clockwise. The coefficient of static friction between the pin and the bell crank is . 8–135. The bell crank fits loosely into a 0.5-in-diameter pin. If P = 41 lb, the bell crank is then on the verge of rotating counterclockwise. Determine the coefficient of static friction between the pin and the bell crank. ms = 0.3 Probs. 8–132/133
- 457. P 3 in. 3 in. 45 440 CHAPTER 8 FRICTION 8 P 250 mm B A 30 Prob. 8–137 300 mm P 30 30 Probs. 8–138/139 P Prob. 8–141 W P r A B Prob. 8–140 1.25 ft P 1.25 ft Prob. 8–142 Prob. 8–136 •8–141. The 1.2-Mg steel beam is moved over a level surface using a series of 30-mm-diameter rollers for which the coefficient of rolling resistance is 0.4 mm at the ground and 0.2 mm at the bottom surface of the beam. Determine the horizontal force P needed to push the beam forward at a constant speed. Hint: Use the result of Prob. 8–140. •8–137. The lawn roller has a mass of 80 kg. If the arm BA is held at an angle of 30° from the horizontal and the coefficient of rolling resistance for the roller is 25 mm, determine the force P needed to push the roller at constant speed. Neglect friction developed at the axle, A, and assume that the resultant force P acting on the handle is applied along arm BA. 8–138. Determine the force P required to overcome rolling resistance and pull the 50-kg roller up the inclined plane with constant velocity. The coefficient of rolling resistance is . 8–139. Determine the force P required to overcome rolling resistance and support the 50-kg roller if it rolls down the inclined plane with constant velocity. The coefficient of rolling resistance is . a = 15 mm a = 15 mm *8–136. The wagon together with the load weighs 150 lb. If the coefficient of rolling resistance is a = 0.03 in., determine the force P required to pull the wagon with constant velocity. *8–140. The cylinder is subjected to a load that has a weight W. If the coefficients of rolling resistance for the cylinder’s top and bottom surfaces are and , respectively, show that a horizontal force having a magnitude of is required to move the load and thereby roll the cylinder forward. Neglect the weight of the cylinder. P = [W(aA + aB)]2r aB aA 8–142. Determine the smallest horizontal force P that must be exerted on the 200-lb block to move it forward.The rollers each weigh 50 lb, and the coefficient of rolling resistance at the top and bottom surfaces is . a = 0.2 in
- 458. CHAPTER REVIEW 441 8 P W Rough surface W N F P W N N Fs ms N Fk mk N Motion motion Impending P P W P W N F Impending slipping F msN P W N F Tipping Dry Friction Frictional forces exist between two rough surfaces of contact. These forces act on a body so as to oppose its motion or tendency of motion. A static frictional force approaches a maximum value of where is the coefficient of static friction. In this case, motion between the contacting surfaces is impending. If slipping occurs, then the friction force remains essentially constant and equal to Here is the coefficient of kinetic friction. The solution of a problem involving friction requires first drawing the free- body diagram of the body. If the unknowns cannot be determined strictly from the equations of equilibrium, and the possibility of slipping occurs, then the friction equation should be applied at the appropriate points of contact in order to complete the solution. It may also be possible for slender objects, like crates, to tip over, and this situation should also be investigated. mk Fk = mkN. ms Fs = msN, CHAPTER REVIEW
- 459. 442 CHAPTER 8 FRICTION 8 Impending motion P W u F3 N3 W F2 N2 P F2 N2 F1 N1 u W r M Motion or impending motion of belt relative to surface r T2 T1 u Wedges Wedges are inclined planes used to increase the application of a force. The two force equilibrium equations are used to relate the forces acting on the wedge. An applied force P must push on the wedge to move it to the right. If the coefficients of friction between the surfaces are large enough, then P can be removed, and the wedge will be self-locking and remain in place. Screws Square-threaded screws are used to move heavy loads. They represent an inclined plane, wrapped around a cylinder. The moment needed to turn a screw depends upon the coefficient of friction and the screw’s lead angle If the coefficient of friction between the surfaces is large enough, then the screw will support the load without tending to turn, i.e., it will be self-locking. u. Flat Belts The force needed to move a flat belt over a rough curved surface depends only on the angle of belt contact, and the coefficient of friction. b, Upward Impending Screw Motion Downward Impending Screw Motion Downward Screw Motion fs 7 u M– = Wr tan1f - us2 u 7 f M¿ = Wr tan1u - fs2 M = Wr tan1u + fs2 T2 7 T1 T2 = T1emb ©Fy = 0 ©Fx = 0
- 460. CHAPTER REVIEW 443 8 z p M R1 R2 P A z Rotation M P r fk N F A r W P a N Collar Bearings and Disks The frictional analysis of a collar bearing or disk requires looking at a differential element of the contact area. The normal force acting on this element is determined from force equilibrium along the shaft, and the moment needed to turn the shaft at a constant rate is determined from moment equilibrium about the shaft’s axis. If the pressure on the surface of a collar bearing is uniform, then integration gives the result shown. Journal Bearings When a moment is applied to a shaft in a nonlubricated or partially lubricated journal bearing, the shaft will tend to roll up the side of the bearing until slipping occurs. This defines the radius of a friction circle, and from it the moment needed to turn the shaft can be determined. Rolling Resistance The resistance of a wheel to rolling over a surface is caused by localized deformation of the two materials in contact.This causes the resultant normal force acting on the rolling body to be inclined so that it provides a component that acts in the opposite direction of the applied force P causing the motion.This effect is characterized using the coefficient of rolling resistance, a, which is determined from experiment. M = Rr sin fk P L Wa r M = 2 3 msPa R2 3 - R1 3 R2 2 - R1 2 b
- 461. 444 CHAPTER 8 FRICTION 8 P s B Drawer 1.25 m 0.3 m Chest A Prob. 8–143 P R A B G 2R –– u p Prob. 8–144 1.5 m 1 m G A B 600 mm 800 mm Probs. 8–145/146 45 60 A B Prob. 8–147 •8–145. The truck has a mass of 1.25 Mg and a center of mass at G. Determine the greatest load it can pull if (a) the truck has rear-wheel drive while the front wheels are free to roll, and (b) the truck has four-wheel drive.The coefficient of static friction between the wheels and the ground is , and between the crate and the ground, it is . 8–146. Solve Prob. 8–145 if the truck and crate are traveling up a 10° incline. ms œ = 0.4 ms = 0.5 *8–144. The semicircular thin hoop of weight W and center of gravity at G is suspended by the small peg at A.A horizontal force P is slowly applied at B. If the hoop begins to slip at A when , determine the coefficient of static friction between the hoop and the peg. u = 30° 8–143. A single force P is applied to the handle of the drawer. If friction is neglected at the bottom and the coefficient of static friction along the sides is , determine the largest spacing s between the symmetrically placed handles so that the drawer does not bind at the corners A and B when the force P is applied to one of the handles. ms = 0.4 8–147. If block A has a mass of 1.5 kg, determine the largest mass of block B without causing motion of the system. The coefficient of static friction between the blocks and inclined planes is . ms = 0.2 REVIEW PROBLEMS
- 462. 8 D 10 10 C B A P P¿ 8000 lb REVIEW PROBLEMS 445 G P h 3 4 h 1 4 h 1 4 h 1 4 Prob. 8–148 2 ft 5 ft 3 ft B A G 2 ft O Probs. 8–149/150 60 20 m u Prob. 8–151 Probs. 8–152/153 8–151. A roofer, having a mass of 70 kg, walks slowly in an upright position down along the surface of a dome that has a radius of curvature of If the coefficient of static friction between his shoes and the dome is determine the angle at which he first begins to slip. u ms = 0.7, r = 20 m. •8–149. The tractor pulls on the fixed tree stump. Determine the torque that must be applied by the engine to the rear wheels to cause them to slip. The front wheels are free to roll. The tractor weighs 3500 lb and has a center of gravity at G. The coefficient of static friction between the rear wheels and the ground is . 8–150. The tractor pulls on the fixed tree stump. If the coefficient of static friction between the rear wheels and the ground is , determine if the rear wheels slip or the front wheels lift off the ground as the engine provides torque to the rear wheels. What is the torque needed to cause this motion? The front wheels are free to roll. The tractor weighs 2500 lb and has a center of gravity at G. ms = 0.6 ms = 0.5 *8–148. The cone has a weight W and center of gravity at G. If a horizontal force P is gradually applied to the string attached to its vertex, determine the maximum coefficient of static friction for slipping to occur. *8–152. Column D is subjected to a vertical load of 8000 lb. It is supported on two identical wedges A and B for which the coefficient of static friction at the contacting surfaces between A and B and between B and C is Determine the force P needed to raise the column and the equilibrium force needed to hold wedge A stationary. The contacting surface between A and D is smooth. •8–153. Column D is subjected to a vertical load of 8000 lb. It is supported on two identical wedges A and B for which the coefficient of static friction at the contacting surfaces between A and B and between B and C is If the forces P and are removed, are the wedges self-locking? The contacting surface between A and D is smooth. P¿ ms = 0.4. P¿ ms = 0.4.
- 463. When a water tank is designed, it is important to be able to determine its center of gravity, calculate its volume and surface area, and reduce three-dimensional distributed loadings caused by the water pressure to their resultants. All of these topics are discussed in this chapter.
- 464. Center of Gravity and Centroid CHAPTER OBJECTIVES • To discuss the concept of the center of gravity, center of mass, and the centroid. • To show how to determine the location of the center of gravity and centroid for a system of discrete particles and a body of arbitrary shape. • To use the theorems of Pappus and Guldinus for finding the surface area and volume for a body having axial symmetry. • To present a method for finding the resultant of a general distributed loading and show how it applies to finding the resultant force of a pressure loading caused by a fluid. 9.1 Center of Gravity, Center of Mass, and the Centroid of a Body In this section we will first show how to locate the center of gravity for a body, and then we will show that the center of mass and the centroid of a body can be developed using this same method. Center of Gravity. A body is composed of an infinite number of particles of differential size, and so if the body is located within a gravitational field, then each of these particles will have a weight dW, Fig. 9–1a. These weights will form an approximately parallel force system, and the resultant of this system is the total weight of the body, which passes through a single point called the center of gravity, G, Fig. 9–1b.* 9 *This is true as long as the gravity field is assumed to have the same magnitude and direction everywhere. That assumption is appropriate for most engineering applications, since gravity does not vary appreciably between, for instance, the bottom and the top of a building.
- 465. 448 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 Using the methods outlined in Sec. 4.8, the weight of the body is the sum of the weights of all of its particles, that is The location of the center of gravity, measured from the y axis, is determined by equating the moment of W about the y axis, Fig. 9–1b, to the sum of the moments of the weights of the particles about this same axis. If dW is located at point ( ), Fig. 9–1a, then Similarly, if moments are summed about the x axis, Finally, imagine that the body is fixed within the coordinate system and this system is rotated about the y axis, Fig. 9–1c. Then the sum of the moments about the y axis gives Therefore, the location of the center of gravity G with respect to the x, y, z axes becomes (9–1) Here are the coordinates of the center of gravity G, Fig. 9–1b. are the coordinates of each particle in the body, Fig. 9–1a. x ' , y ' , z ' x, y, z x = L x ' dW L dW y = L y ' dW L dW z = L z ' dW L dW zW = 1z ' dW (MR)y = ©My; 90° yW = 1y ' dW (MR)x = ©Mx; xW = 1x ' dW (MR)y = ©My; z ' y ' , x ' , W = 1dW + TFR = ©Fz; (a) z z dW y y x x ~ ~ ~ (b) z G y x y x z W (c) z G y x y x z W Fig. 9–1
- 466. 9.1 CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY 449 9 Center of Mass of a Body. In order to study the dynamic response or accelerated motion of a body, it becomes important to locate the body’s center of mass Cm, Fig. 9–2. This location can be determined by substituting dW = g dm into Eqs. 9–1. Since g is constant, it cancels out, and so (9–2) Centroid of a Volume. If the body in Fig. 9–3 is made from a homogeneous material, then its density (rho) will be constant. Therefore, a differential element of volume dV has a mass Substituting this into Eqs. 9–2 and canceling out , we obtain formulas that locate the centroid C or geometric center of the body; namely (9–3) These equations represent a balance of the moments of the volume of the body. Therefore, if the volume possesses two planes of symmetry, then its centroid must lie along the line of intersection of these two planes. For example, the cone in Fig. 9–4 has a centroid that lies on the y axis so that . The location can be found using a single integration by choosing a differential element represented by a thin disk having a thickness dy and radius . Its volume is and its centroid is at , , . z ' = 0 y ' = y x ' = 0 pr2 dy = pz2 dy dV = r = z y — x = z = 0 x = L V x ' dV L V dV y = L V y ' dV L V dV z = L V z ' dV L V dV r dm = r dV. r x = L x ' dm L dm y = L y ' dm L dm z = L z ' dm L dm dm Cm ~ z z y ~ x x ~ y y z x C dV x y z x ~ y y ~ x ~ z z z y x y y y dy r z (0, y, 0) C ~ Fig. 9–4 Fig. 9–3 Fig. 9–2
- 467. 450 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 Centroid of an Area. If an area lies in the x–y plane and is bounded by the curve , as shown in Fig. 9–5a, then its centroid will be in this plane and can be determined from integrals similar to Eqs. 9–3, namely, (9–4) These integrals can be evaluated by performing a single integration if we use a rectangular strip for the differential area element. For example, if a vertical strip is used, Fig. 9–5b, the area of the element is , and its centroid is located at and . If we consider a horizontal strip, Fig. 9–5c, then , and its centroid is located at and . Centroid of a Line. If a line segment (or rod) lies within the x–y plane and it can be described by a thin curve , Fig. 9–6a, then its centroid is determined from (9–5) x = L L x ' dL L L dL y = L L y ' dL L L dL y = f(x) y ' = y x ' = x2 dA = x dy y ' = y2 x ' = x dA = y dx x = L A x ' dA L A dA y = L A y ' dA L A dA y = f(x) y y x x y dx dy x x x y y x (x, y) (x, y) y y 2 x 2 (b) (c) y f(x) y x x y (a) y f(x) y f(x) C ~ ~ ~ ~ Fig. 9–5 Integration must be used to determine the location of the center of gravity of this goal post due to the curvature of the supporting member.
- 468. 9.1 CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY 451 9 Here, the length of the differential element is given by the Pythagorean theorem, , which can also be written in the form or Either one of these expressions can be used; however, for application, the one that will result in a simpler integration should be selected. For example, consider the rod in Fig. 9–6b, defined by .The length of the element is , and since , then .The centroid for this element is located at and . y ' = y x ' = x dL = 21 + (4x22 dx dydx = 4x dL = 21 + 1dydx22 dx y = 2x2 dl = ¢ B a dx dy b 2 + 1 ≤ dy dL = B a dx dy b 2 dy2 + a dy dy b 2 dy2 dl = ¢ B 1 + a dy dx b 2 ≤ dx dL = B a dx dx b 2 dx2 + a dy dx b 2 dx2 dL = 21dx22 + 1dy22 Important Points • The centroid represents the geometric center of a body. This point coincides with the center of mass or the center of gravity only if the material composing the body is uniform or homogeneous. • Formulas used to locate the center of gravity or the centroid simply represent a balance between the sum of moments of all the parts of the system and the moment of the “resultant” for the system. • In some cases the centroid is located at a point that is not on the object, as in the case of a ring, where the centroid is at its center. Also, this point will lie on any axis of symmetry for the body, Fig. 9–7. C dL dL dy dx x y y ~ ~ x O y x (a) y x 2 m 1 m x x y y dx dy y 2x2 ~ ~ (b) C y x Fig. 9–6 Fig. 9–7
- 469. 452 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 Procedure for Analysis The center of gravity or centroid of an object or shape can be determined by single integrations using the following procedure. Differential Element. • Select an appropriate coordinate system, specify the coordinate axes, and then choose a differential element for integration. • For lines the element is represented by a differential line segment of length dL . • For areas the element is generally a rectangle of area dA, having a finite length and differential width. • For volumes the element can be a circular disk of volume dV, having a finite radius and differential thickness. • Locate the element so that it touches the arbitrary point (x, y, z) on the curve that defines the boundary of the shape. Size and Moment Arms. • Express the length dL, area dA, or volume dV of the element in terms of the coordinates describing the curve. • Express the moment arms for the centroid or center of gravity of the element in terms of the coordinates describing the curve. Integrations. • Substitute the formulations for and dL, dA, or dV into the appropriate equations (Eqs. 9–1 through 9–5). • Express the function in the integrand in terms of the same variable as the differential thickness of the element. • The limits of the integral are defined from the two extreme locations of the element’s differential thickness, so that when the elements are “summed” or the integration performed, the entire region is covered.* z ' y ' , x ' , z ' y ' , x ' , *Formulas for integration are given in Appendix A.
- 470. 9.1 CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY 453 9 EXAMPLE 9.1 Locate the centroid of the rod bent into the shape of a parabolic arc as shown in Fig. 9–8. SOLUTION Differential Element. The differential element is shown in Fig. 9–8. It is located on the curve at the arbitrary point (x, y). Area and Moment Arms. The differential element of length dL can be expressed in terms of the differentials dx and dy using the Pythagorean theorem. Since then Therefore, expressing dL in terms of y and dy, we have As shown in Fig. 9–8, the centroid of the element is located at Integrations. Applying Eqs. 9–5, using the formulas in Appendix A to evaluate the integrals, we get Ans. Ans. NOTE: These results for C seem reasonable when they are plotted on Fig. 9–8. y = L L y ' dL L L dL = L 1m 0 y24y2 + 1 dy L 1m 0 24y2 + 1 dy = 0.8484 1.479 = 0.574 m = 0.6063 1.479 = 0.410 m x = L L x ' dL L L dL = L 1m 0 x24y2 + 1 dy L 1m 0 24y2 + 1 dy = L 1m 0 y2 24y2 + 1 dy L 1m 0 24y2 + 1 dy y ' = y. x ' = x, dL = 2(2y22 + 1 dy dxdy = 2y. x = y2 , dL = 2(dx22 + (dy22 = B a dx dy b 2 + 1 dy 1 m ~ C(x, y) y dL 1 m x y y x x O x y2 (x, y) ~ ~ ~ Fig. 9–8
- 471. 454 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 Locate the centroid of the circular wire segment shown in Fig. 9–9. EXAMPLE 9.2 SOLUTION Polar coordinates will be used to solve this problem since the arc is circular. Differential Element. A differential circular arc is selected as shown in the figure.This element intersects the curve at (R, ). Length and Moment Arm. The length of the differential element is and its centroid is located at and Integrations. Applying Eqs. 9–5 and integrating with respect to we obtain Ans. Ans. NOTE: As expected, the two coordinates are numerically the same due to the symmetry of the wire. R2 L p2 0 sin u du R L p2 0 du = 2R p y = L L y ' dL L L dL = L p2 0 1R sin u2R du L p2 0 R du = R2 L p2 0 cos u du R L p2 0 du = 2R p x = L L x ' dL L L dL = L p2 0 1R cos u2R du L p2 0 R du = u, y ' = R sin u. x ' = R cos u dL = R du, u y x d y R sin u ~ ~ C(x, y) (R, u) O R x R cos u dL R d u u u Fig. 9–9
- 472. 9.1 CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY 455 9 EXAMPLE 9.3 Determine the distance measured from the x axis to the centroid of the area of the triangle shown in Fig. 9–10. y y x y h dy y (b x) b x (x,y) (x,y) ~ ~ h b SOLUTION Differential Element. Consider a rectangular element having a thickness dy, and located in an arbitrary position so that it intersects the boundary at (x, y), Fig. 9–10. Area and Moment Arms. The area of the element is and its centroid is located a distance from the x axis. Integration. Applying the second of Eqs. 9–4 and integrating with respect to y yields Ans. NOTE: This result is valid for any shape of triangle. It states that the centroid is located at one-third the height, measured from the base of the triangle. = h 3 y = L A y ' dA L A dA = L h 0 yc b h 1h - y2 dy d L h 0 b h 1h - y2 dy = 1 6 bh2 1 2 bh y ' = y = b h 1h - y2 dy, dA = x dy Fig. 9–10
- 473. 456 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 Locate the centroid for the area of a quarter circle shown in Fig. 9–11. EXAMPLE 9.4 y x d R d x R cos ~ y R sin u ~ R R, u u u u R 3 2 3 2 3 u Fig. 9–11 Ans. Ans. = a 2 3 Rb L p2 0 sin u du L p2 0 du = 4R 3p y = L A y ' dA L A dA = L p2 0 a 2 3 R sin ub R2 2 du L p2 0 R2 2 du = a 2 3 Rb L p2 0 cos u du L p2 0 du = 4R 3p x = L A x ' dA L A dA = L p2 0 a 2 3 R cos ub R2 2 du L p2 0 R2 2 du SOLUTION Differential Element. Polar coordinates will be used, since the boundary is circular.We choose the element in the shape of a triangle, Fig. 9–11. (Actually the shape is a circular sector; however, neglecting higher-order differentials, the element becomes triangular.) The element intersects the curve at point (R, ). Area and Moment Arms. The area of the element is and using the results of Example 9.3, the centroid of the (triangular) element is located at Integrations. Applying Eqs. 9–4 and integrating with respect to we obtain u, y ' = 2 3 R sin u. x ' = 2 3 R cos u, dA = 1 21R21R du2 = R2 2 du u
- 474. 9.1 CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY 457 9 EXAMPLE 9.5 Locate the centroid of the area shown in Fig. 9–12a. SOLUTION I Differential Element. A differential element of thickness dx is shown in Fig. 9–12a. The element intersects the curve at the arbitrary point (x, y), and so it has a height y. Area and Moment Arms. The area of the element is and its centroid is located at Integrations. Applying Eqs.9–4 and integrating with respect to x yields y ' = y2. x ' = x, dA = y dx, Ans. Ans. L 1m 0 1x2 22x2 dx L 1m 0 x2 dx = 0.100 0.333 = 0.3 m y = L A y ' dA L A dA = L 1m 0 1y22y dx L 1m 0 y dx = x = L A x ' dA L A dA = L 1m 0 xy dx L 1m 0 y dx = L 1m 0 x3 dx L 1m 0 x2 dx = 0.250 0.333 = 0.75 m SOLUTION II Differential Element. The differential element of thickness dy is shown in Fig. 9–12b. The element intersects the curve at the arbitrary point (x, y), and so it has a length Area and Moment Arms. The area of the element is and its centroid is located at Integrations. Applying Eqs. 9–4 and integrating with respect to y, we obtain y ' = y x ' = x + a 1 - x 2 b = 1 + x 2 , dA = 11 - x2 dy, 11 - x2. Ans. Ans. = 0.100 0.333 = 0.3 m y = L A y ' dA L A dA = L 1m 0 y11 - x2 dy L 1m 0 11 - x2 dy = L 1m 0 1y - y32 2 dy L 1m 0 11 - 1y2 dy 1 2 L 1m 0 11 - y2 dy L 1m 0 11 - 1y2 dy = 0.250 0.333 = 0.75 m x ' = L A x ' dA L A dA = L 1m 0 [11 + x22]11 - x2 dy L 1m 0 11 - x2 dy = y x2 1 m dx 1 m y x y (a) (x, y) ~ ~ (x, y) x Fig. 9–12 1 m dy 1 m y x y (b) (x, y) ~ ~ (x, y) x y x2 (1 x) NOTE: Plot these results and notice that they seem reasonable. Also, for this problem, elements of thickness dx offer a simpler solution.
- 475. 458 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 1 ft 2 ft y 2 ft y x (a) x x y y 2 2 ft 2 ft dx y y x (b) (x, y) dy x x y y 1 x2 y2 4 1 x2 y2 4 ~ ~ ~ Fig. 9–13 Locate the centroid of the semi-elliptical area shown in Fig. 9–13a. EXAMPLE 9.6 SOLUTION I Differential Element. The rectangular differential element parallel to the y axis shown shaded in Fig. 9–13a will be considered. This element has a thickness of dx and a height of y. Area and Moment Arms. Thus, the area is and its centroid is located at and Integration. Since the area is symmetrical about the y axis, Ans. Applying the second of Eqs. 9–4 with , we have y = B 1- x2 4 x = 0 y ' = y2. x ' = x dA = y dx, Ans. y = L A y ' dA L A dA = L 2 ft -2 ft y 2 (y dx) L 2 ft -2 ft y dx = 1 2 L 2 ft -2 ft a1- x2 4 bdx L 2 ft -2 ftB 1- x2 4 dx = 43 p = 0.424 ft SOLUTION II Differential Element. The shaded rectangular differential element of thickness dy and width 2x, parallel to the x axis, will be considered, Fig. 9–13b. Area and Moment Arms. The area is , and its centroid is at and . Integration. Applying the second of Eqs. 9–4, with , we have x = 231-y2 y ' = y x ' = 0 dA = 2x dy Ans. y = L A y ' dA L A dA = L 1 ft 0 y(2x dy) L 1 ft 0 2x dy = L 1 ft 0 4y31-y2 dy L 1 ft 0 431-y2 dy = 4 3 p ft = 0.424 ft
- 476. 9.1 CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY 459 9 EXAMPLE 9.7 Locate the centroid for the paraboloid of revolution, shown in Fig. 9–14. y 100 mm dy y z z x ~ y y z2 100y 100 mm ~ (0, y, 0) r (o, y, z) Fig. 9–14 SOLUTION Differential Element. An element having the shape of a thin disk is chosen. This element has a thickness dy, it intersects the generating curve at the arbitrary point (0, y, z), and so its radius is Volume and Moment Arm. The volume of the element is and its centroid is located at Integration. Applying the second of Eqs. 9–3 and integrating with respect to y yields y ' = y. 1pz2 2 dy, dV= r = z. Ans. 100p L 100 mm 0 y2 dy 100p L 100 mm 0 y dy = 66.7 mm y = L V y ' dV L V dV = L 100 mm 0 y1pz2 2 dy L 100 mm 0 1pz2 2 dy =
- 477. 460 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 Determine the location of the center of mass of the cylinder shown in Fig. 9–15 if its density varies directly with the distance from its base, i.e., r = 200z kgm3 . EXAMPLE 9.8 y dz z 1 m x 0.5 m z (0,0, z) ~ Fig. 9–15 SOLUTION For reasons of material symmetry, Ans. Differential Element. A disk element of radius 0.5 m and thickness dz is chosen for integration, Fig. 9–15, since the density of the entire element is constant for a given value of z.The element is located along the z axis at the arbitrary point (0, 0, z). Volume and Moment Arm. The volume of the element is and its centroid is located at Integrations. Using an equation similar to the third of Eqs. 9–2 and integrating with respect to z, noting that we have Ans. = L 1 m 0 z2 dz L 1 m 0 z dz = 0.667 m z = L V z ' r dV L V r dV = L 1 m 0 z1200z2Cp10.522 dzD L 1 m 0 1200z2p10.522 dz r = 200z, z ' = z. p10.522 dz, dV= x = y = 0
- 478. 9.1 CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY 461 9 FUNDAMENTAL PROBLEMS F9–4. Locate the center mass of the straight rod if its mass per unit length is given by . m = m0(1 + x2 L2 ) x F9–2. Determine the centroid of the shaded area. (x, y) F9–1. Determine the centroid of the shaded area. (x, y) F9–5. Locate the centroid of the homogeneous solid formed by revolving the shaded area about the axis. y y F9–3. Determine the centroid of the shaded area. y F9–6. Locate the centroid of the homogeneous solid formed by revolving the shaded area about the axis. z z y x y x3 1 m 1 m F9–1 y x 1 m 1 m y x3 F9–2 y x 2 m 1 m 1 m y 2x2 F9–3 y x L F9–4 y x 1 m 0.5 m z z2 y 1 4 F9–5 x z z (12 8y) 1 –– 3 2 ft 1.5 ft 2 ft y F9–6
- 479. 462 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 PROBLEMS 9–3. Determine the distance to the center of mass of the homogeneous rod bent into the shape shown. If the rod has a mass per unit length of , determine the reactions at the fixed support O. 0.5 kgm x 9–2. The uniform rod is bent into the shape of a parabola and has a weight per unit length of . Determine the reactions at the fixed support A. 6 lbft •9–1. Determine the mass and the location of the center of mass of the uniform parabolic-shaped rod. The mass per unit length of the rod is . 2 kgm (x, y) *9–4. Determine the mass and locate the center of mass of the uniform rod. The mass per unit length of the rod is . 3 kgm (x, y) y x 4 m 4 m y2 4x Prob. 9–1 y x 3 ft 3 ft A y2 3x Prob. 9–2 1 m 1 m y x y2 x3 O Prob. 9–3 y x 2 m 4 m y 4 x2 Prob. 9–4
- 480. 9.1 CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY 463 9 *9–8. Determine the area and the centroid of the area. (x, y) 9–6. Determine the location ( , ) of the centroid of the wire. y x •9–5. Determine the mass and the location of the center of mass of the rod if its mass per unit length is . m = m0(1 + xL) x •9–9. Determine the area and the centroid of the area. (x, y) 9–7. Locate the centroid of the circular rod. Express the answer in terms of the radius r and semiarc angle . a x 9–10. Determine the area and the centroid of the area. (x, y) y x L Prob. 9–5 y x y x2 2 ft 4 ft Prob. 9–6 y x C r r – x a a Prob. 9–7 y x 4 m 4 m y2 4x Prob. 9–8 y 1 m 1 m y2 x3 x Prob. 9–9 y x 3 ft 3 ft y x3 1 –– 9 Prob. 9–10
- 481. 464 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 9–14. Determine the area and the centroid of the area. (x, y) *쐍9–12. Locate the centroid of the area. •쐍9–13. Locate the centroid of the area. y x 9–11. Determine the area and the centroid of the area. (x, y) *9–16. Locate the centroid ( , ) of the area. y x 9–15. Determine the area and the centroid of the area. (x, y) y x 2 ab b y2 4ax Prob. 9–11 y x 2 ft x1/2 2x5/3 y Probs. 9–12/13 y x a b xy c2 Prob. 9–14 y x a h y x2 h –– a2 Prob. 9–15 y x 2 m 1 m y 1 – x2 1 – 4 Prob. 9–16
- 482. 9.1 CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY 465 9 *9–20. The plate has a thickness of 0.5 in. and is made of steel having a specific weight of . Determine the horizontal and vertical components of reaction at the pin A and the force in the cord at B. 490 lbft3 9–18. The plate is made of steel having a density of . If the thickness of the plate is 10 mm, determine the horizontal and vertical components of reaction at the pin A and the tension in cable BC. 7850 kgm3 •9–17. Determine the area and the centroid of the area. (x, y) •9–21. Locate the centroid of the shaded area. x 9–19. Determine the location to the centroid C of the upper portion of the cardioid, . r = a(1 - cos u) x x h a y x2 h –– a2 y Prob. 9–17 y A B C x 2 m 4 m y3 2x Prob. 9–18 r r a (1 cos u) C _ x x y u Prob. 9–19 y A B x 3 ft 3 ft y x2 –– 3 y x a ka y 2k(x ) x2 — 2a Prob. 9–21 Prob. 9–20
- 483. 466 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 •9–25. Determine the area and the centroid of the area. (x, y) *9–24. Locate the centroid ( , ) of the area. y x 9–22. Locate the centroid of the area. 9–23. Locate the centroid of the area. y x 9–26. Locate the centroid of the area. 9–27. Locate the centroid of the area. y x y x 2 in. 2 in. y 1 0.5 in. 0.5 in. x Probs. 9–22/23 y x 9 ft 3 ft y 9 x2 Prob. 9–24 y x y y x 3 ft 3 ft x3 –– 9 Prob. 9–25 y x 1 m y x2 1 m y2 x Probs. 9–26/27
- 484. 9.1 CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY 467 9 9–31. Locate the centroid of the area. Hint: Choose elements of thickness dy and length . [(2 - y) - y2 ] 9–30. The steel plate is 0.3 m thick and has a density of . Determine the location of its center of mass. Also determine the horizontal and vertical reactions at the pin and the reaction at the roller support. Hint: The normal force at B is perpendicular to the tangent at B, which is found from tan . u = dydx 7850 kgm3 *9–28. Locate the centroid of the area. •9–29. Locate the centroid of the area. y x *9–32. Locate the centroid of the area. •9–33. Locate the centroid of the area. y x y x h a y xn h –– an Probs. 9–28/29 y A B x 2 m 2 m 2 m y2 2x Prob. 9–30 y x 1 m 1 m 1 m y x 2 y2 x Prob. 9–31 y x 1 ft y 2x 2 ft y2 4x Probs. 9–32/33
- 485. 468 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 •9–37. Locate the centroid of the homogeneous solid formed by revolving the shaded area about the y axis. y 9–35. Locate the centroid of the homogeneous solid formed by revolving the shaded area about the y axis. y 9–34. If the density at any point in the rectangular plate is defined by , where is a constant, determine the mass and locate the center of mass of the plate.The plate has a thickness t. x r0 r = r0(1 + xa) 9–38. Locate the centroid of the homogeneous solid frustum of the paraboloid formed by revolving the shaded area about the z axis. z *9–36. Locate the centroid of the solid. z y a x b –– 2 b –– 2 y x z y2 (z a)2 a2 a Prob. 9–35 y z x a z a 1 a a y ( )2 Prob. 9–36 z y x z2 y3 1 –– 16 2 m 4 m Prob. 9–37 a z (a2 y2) h – a2 h – 2 h – 2 z x y Prob. 9–38 Prob. 9–34
- 486. 9.1 CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY 469 9 •9–41. Determine the mass and locate the center of mass of the hemisphere formed by revolving the shaded area about the y axis.The density at any point in the hemisphere can be defined by , where is a constant. r0 r = r0(1 + ya) y *9–40. Locate the center of mass of the circular cone formed by revolving the shaded area about the y axis. The density at any point in the cone is defined by , where is a constant. r0 r = (r0 h)y y 9–39. Locate the centroid of the homogeneous solid formed by revolving the shaded area about the y axis. y 9–42. Determine the volume and locate the centroid of the homogeneous conical wedge. (y, z) y x z z2 y2 9 3 ft 5 ft 4 ft Prob. 9–39 y x z h a z y a a –– h Prob. 9–40 y x z y2 z2 a2 r Prob. 9–41 z x y a z y a –– h h Prob. 9–42 z y z G x _ r Prob. 9–43 9–43. The hemisphere of radius r is made from a stack of very thin plates such that the density varies with height, , where k is a constant. Determine its mass and the distance to the center of mass G. z r = kz
- 487. 470 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 9.2 Composite Bodies A composite body consists of a series of connected “simpler” shaped bodies,which may be rectangular,triangular,semicircular,etc.Such a body can often be sectioned or divided into its composite parts and, provided the weight and location of the center of gravity of each of these parts are known, we can then eliminate the need for integration to determine the center of gravity for the entire body.The method for doing this follows the same procedure outlined in Sec.9.1.Formulas analogous to Eqs.9–1 result; however,rather than account for an infinite number of differential weights, we have instead a finite number of weights.Therefore, (9–6) Here x = ©x ' W ©W y = ©y ' W ©W z = ©z ' W ©W When the body has a constant density or specific weight, the center of gravity coincides with the centroid of the body.The centroid for composite lines, areas, and volumes can be found using relations analogous to Eqs. 9–6; however, the W’s are replaced by L’s, A’s, and V’s, respectively. Centroids for common shapes of lines, areas, shells, and volumes that often make up a composite body are given in the table on the inside back cover. z y, x, represent the coordinates of the center of gravity G of the composite body. z ' y ' , x ' , represent the coordinates of the center of gravity of each composite part of the body. ©W is the sum of the weights of all the composite parts of the body, or simply the total weight of the body. G In order to determine the force required to tip over this concrete barrier it is first necessary to determine the location of its center of gravity G.Due to symmetry,G will lie on the vertical axis of symmetry.
- 488. 9.2 COMPOSITE BODIES 471 9 Procedure for Analysis The location of the center of gravity of a body or the centroid of a composite geometrical object represented by a line, area, or volume can be determined using the following procedure. Composite Parts. • Using a sketch, divide the body or object into a finite number of composite parts that have simpler shapes. • If a composite body has a hole, or a geometric region having no material, then consider the composite body without the hole and consider the hole as an additional composite part having negative weight or size. Moment Arms. • Establish the coordinate axes on the sketch and determine the coordinates of the center of gravity or centroid of each part. Summations. • Determine by applying the center of gravity equations, Eqs. 9–6, or the analogous centroid equations. • If an object is symmetrical about an axis, the centroid of the object lies on this axis. If desired, the calculations can be arranged in tabular form, as indicated in the following three examples. z y, x, z ' y ' , x ' , The center of gravity of this water tank can be determined by dividing it into composite parts and applying Eqs. 9–6.
- 489. 472 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 Locate the centroid of the wire shown in Fig. 9–16a. SOLUTION Composite Parts. The wire is divided into three segments as shown in Fig. 9–16b. Moment Arms. The location of the centroid for each segment is determined and indicated in the figure. In particular, the centroid of segment is determined either by integration or by using the table on the inside back cover. Summations. For convenience, the calculations can be tabulated as follows: ~ 1 EXAMPLE 9.9 Segment L (mm) x ' (mm) y ' (mm) z ' (mm) x ' L (mm2 ) y ' L (mm2 ) z ' L (mm2 ) 1 p1602 = 188.5 60 -38.2 0 11 310 -7200 0 2 40 0 20 0 0 800 0 3 20 0 40 -10 0 800 -200 ©L = 248.5 ©x ' L = 11 310 ©y ' L = -5600 ©z ' L = -200 Thus, Ans. Ans. Ans. z = ©z ' L ©L = -200 248.5 = -0.805 mm y = ©y ' L ©L = -5600 248.5 = -22.5 mm x = ©x ' L ©L = 11 310 248.5 = 45.5 mm 40 mm 20 mm (a) y z x 60 mm Fig. 9–16 (b) 38.2 mm 20 mm 10 mm 60 mm 20 mm (2) (60) ——— p y x 2 3 1 z
- 490. 9.2 COMPOSITE BODIES 473 9 EXAMPLE 9.10 Locate the centroid of the plate area shown in Fig. 9–17a. (a) y x 1 ft 2 ft 2 ft 1 ft 3 ft Fig. 9–17 y x 1 ft 1.5 ft 1 ft 1.5 ft 1 2 (b) y x 2.5 ft 2 ft 3 SOLUTION Composite Parts. The plate is divided into three segments as shown in Fig. 9–17b. Here the area of the small rectangle is considered “negative” since it must be subtracted from the larger one Moment Arms. The centroid of each segment is located as indicated in the figure. Note that the coordinates of and are negative. Summations. Taking the data from Fig. 9–17b, the calculations are tabulated as follows: ~ 3 ~ 2 x ' ~ 2 . ~ 3 Thus, Ans. Ans. NOTE: If these results are plotted in Fig. 9–17, the location of point C seems reasonable. y = ©y ' A ©A = 14 11.5 = 1.22 ft x = ©x ' A ©A = -4 11.5 = -0.348 ft Segment A (ft2 ) x ' (ft) y ' (ft) x ' A (ft3 ) y ' A (ft3 ) 1 1 2132132 = 4.5 1 1 4.5 4.5 2 132132 = 9 -1.5 1.5 -13.5 13.5 3 -122112 = -2 -2.5 2 5 -4 ©A = 11.5 ©x ' A = -4 ©y ' A = 14
- 491. 474 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 Locate the center of mass of the assembly shown in Fig. 9–18a. The conical frustum has a density of and the hemisphere has a density of There is a 25-mm-radius cylindrical hole in the center of the frustum. SOLUTION Composite Parts. The assembly can be thought of as consisting of four segments as shown in Fig. 9–18b. For the calculations, and must be considered as “negative” segments in order that the four segments, when added together, yield the total composite shape shown in Fig. 9–18a. Moment Arm. Using the table on the inside back cover, the computations for the centroid of each piece are shown in the figure. Summations. Because of symmetry, note that Ans. Since , and g is constant, the third of Eqs. 9–6 becomes The mass of each piece can be computed from and used for the calculations.Also, so that 1 Mgm3 = 10-6 kgmm3 , m = rV z = ©z ' m©m. W = mg x = y = 0 z ' ~ 4 ~ 3 rh = 4 Mgm3 . rc = 8 Mgm3 , EXAMPLE 9.11 Segment m (kg) z ' (mm) z ' m (kg # mm) 1 8110-6 2A1 3 Bp15022 12002 = 4.189 50 209.440 2 4110-6 2A2 3 Bp15023 = 1.047 -18.75 -19.635 3 -8110-6 2A1 3 Bp12522 11002 = -0.524 100 + 25 = 125 -65.450 4 -8110-6 2p12522 11002 = -1.571 50 -78.540 ©m = 3.142 ©z ' m = 45.815 Thus, Ans. z ' = ©z ' m ©m = 45.815 3.142 = 14.6 mm (a) 50 mm 100 mm 25 mm 50 mm x y z Fig. 9–18 200 mm 50 mm 50 mm 50 mm 200 mm 4 1 2 (50) 18.75 mm 8 3 25 mm 4 100 mm 25 mm 100 mm 100 mm 50 mm (b) 3 25 mm 4
- 492. 9.2 COMPOSITE BODIES 475 9 FUNDAMENTAL PROBLEMS F9–10. Locate the centroid of the cross-sectional area. (x, y) F9–8. Locate the centroid of the beam’s cross-sectional area. y F9–7. Locate the centroid of the wire bent in the shape shown. (x, y, z) F9–11. Locate the center of mass of the homogeneous solid block. (x, y, z) x z 400 mm 600 mm 300 mm y F9–7 y x 25 mm 50 mm 300 mm 25 mm 150 mm 150 mm F9–8 y x 400 mm 50 mm 50 mm C 200 mm 50 mm F9–9 x y 4 in. 3 in. C y 0.5 in. 0.5 in. x F9–10 y x z 6 ft 2 ft 4 ft 5 ft 2 ft 3 ft F9–11 y x z 1.8 m 1.5 m 1.5 m 0.5 m 0.5 m 2 m F9–12 F9–9. Locate the centroid of the beam’s cross- sectional area. y F9–12. Determine the center of mass of the homogeneous solid block. (x, y, z)
- 493. 476 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 x y 50 mm 150 mm 100 mm 20 mm Prob. 9–44 x z 400 mm 200 mm y Prob. 9–45 z 6 in. 4 in. y x Prob. 9–46 x y z 4 in. 2 in. 2 in. Prob. 9–47 PROBLEMS 9–46. Locate the centroid ( , , ) of the wire. z y x •9–45. Locate the centroid of the wire. (x, y, z) *9–44. Locate the centroid ( , ) of the uniform wire bent in the shape shown. y x 9–47. Locate the centroid ( , , ) of the wire which is bent in the shape shown. z y x
- 494. 9.2 COMPOSITE BODIES 477 9 3 m 3 m C D B A E y x 3 m Prob. 9–48 9–50. Each of the three members of the frame has a mass per unit length of 6 kg/m. Locate the position ( , ) of the center of mass. Neglect the size of the pins at the joints and the thickness of the members. Also, calculate the reactions at the pin A and roller E. y x •9–49. Locate the centroid of the wire. If the wire is suspended from A, determine the angle segment AB makes with the vertical when the wire is in equilibrium. (x, y) *9–48. The truss is made from seven members, each having a mass per unit length of 6 kg/m. Locate the position ( , ) of the center of mass. Neglect the mass of the gusset plates at the joints. y x 9–51. Locate the centroid of the cross-sectional area of the channel. (x, y) y x A C B 200 mm 200 mm 60 Prob. 9–49 y x A B C D E 4 m 6 m 7 m 4 m Prob. 9–50 x y 9 in. 1 in. 1 in. 22 in. 1 in. Prob. 9–51
- 495. 478 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 x y 3 in. 6 in. 3 in. 27 in. 3 in. 12 in. 12 in. Prob. 9–52 9–54. Locate the centroid of the channel’s cross- sectional area. y •9–53. Locate the centroid of the cross-sectional area of the built-up beam. y *9–52. Locate the centroid of the cross-sectional area of the concrete beam. y 9–55. Locate the distance to the centroid of the member’s cross-sectional area. y y x 6 in. 1 in. 1 in. 1 in. 1 in. 3 in. 3 in. 6 in. Prob. 9–53 2 in. 4 in. 2 in. 12 in. 2 in. C y Prob. 9–54 x y 0.5 in. 6 in. 0.5 in. 1.5 in. 1 in. 3 in. 3 in. Prob. 9–55
- 496. 9.2 COMPOSITE BODIES 479 9 9–58. Locate the centroid of the composite area. x •9–57. The gravity wall is made of concrete. Determine the location ( , ) of the center of mass G for the wall. y x *9–56. Locate the centroid of the cross-sectional area of the built-up beam. y 9–59. Locate the centroid of the composite area. (x, y) y x 1.5 in. 1.5 in. 11.5 in. 1.5 in. 3.5 in. 4in. 1.5 in. 4 in. Prob. 9–56 x y 4 in. 3 in. 3 in. 3 in. Prob. 9–59 3 ft 3 ft 1.5 ft 1 ft y x Prob. 9–60 y 1.2 m x _ x _ y 0.6 m 0.6 m 2.4 m 3 m G 0.4 m Prob. 9–57 x y ri r0 Prob. 9–58 *9–60. Locate the centroid of the composite area. (x, y)
- 497. 480 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 9–63. Locate the centroid of the cross-sectional area of the built-up beam. y 9–62. To determine the location of the center of gravity of the automobile it is first placed in a level position, with the two wheels on one side resting on the scale platform P. In this position the scale records a reading of . Then, one side is elevated to a convenient height c as shown. The new reading on the scale is . If the automobile has a total weight of W, determine the location of its center of gravity G . (x, y) W2 W1 •9–61. Divide the plate into parts, and using the grid for measurement, determine approximately the location ( , ) of the centroid of the plate. y x *9–64. Locate the centroid of the cross-sectional area of the built-up beam. y y x 200 mm 200 mm Prob. 9–61 b P c G – y – x W2 Prob. 9–62 y x 450 mm 150 mm 150 mm 200 mm 20 mm 20 mm Prob. 9–63 200 mm 20 mm 50 mm 150 mm y x 200 mm 300 mm 10 mm 20 mm 20 mm 10 mm Prob. 9–64
- 498. 9.2 COMPOSITE BODIES 481 9 9–67. Uniform blocks having a length L and mass m are stacked one on top of the other, with each block overhanging the other by a distance d, as shown. If the blocks are glued together, so that they will not topple over, determine the location of the center of mass of a pile of n blocks. *9–68. Uniform blocks having a length L and mass m are stacked one on top of the other, with each block overhanging the other by a distance d, as shown. Show that the maximum number of blocks which can be stacked in this manner is . n 6 Ld x y x z G B A 225 mm 150 mm 150 mm 30 mm Prob. 9–65 FA 1129 lb 1168 lb 2297 lb FA 1269 lb 1307 lb 2576 lb FB 975 lb 984 lb 1959 lb A _ x B 9.40 ft 3.0 ft G _ y B G A Prob. 9–66 L d 2d y x Probs. 9–67/68 z y x 60 mm 60 mm 20 mm 20 mm 20 mm 20 mm 60 mm 10 mm dia. holes 80 mm u Prob. 9–60 9–66. The car rests on four scales and in this position the scale readings of both the front and rear tires are shown by and .When the rear wheels are elevated to a height of 3 ft above the front scales, the new readings of the front wheels are also recorded. Use this data to compute the location and to the center of gravity G of the car. The tires each have a diameter of 1.98 ft. y x FB FA •9–65. The composite plate is made from both steel (A) and brass (B) segments. Determine the mass and location of its mass center G. Take and rbr = 8.74 Mgm3 . rst = 7.85 Mgm3 1x, y, z2 •9–69. Locate the center of gravity ( , ) of the sheet- metal bracket if the material is homogeneous and has a constant thickness. If the bracket is resting on the horizontal x–y plane shown, determine the maximum angle of tilt which it can have before it falls over, i.e., begins to rotate about the y axis. u z x
- 499. 482 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 *9–72. Locate the center of mass of the homogeneous block assembly. (x, y, z) 9–71. Major floor loadings in a shop are caused by the weights of the objects shown. Each force acts through its respective center of gravity G. Locate the center of gravity ( , ) of all these components. y x 9–70. Locate the center of mass for the compressor assembly.The locations of the centers of mass of the various components and their masses are indicated and tabulated in the figure.What are the vertical reactions at blocks A and B needed to support the platform? •9–73. Locate the center of mass of the assembly. The hemisphere and the cone are made from materials having densities of and , respectively. 4 Mgm3 8 Mgm3 z x y 1 2 3 4 Instrument panel Filter system Piping assembly Liquid storage Structural framework 230 kg 183 kg 120 kg 85 kg 468 kg 1 2 3 4 5 5 2.30 m 1.80 m 3.15 m 4.83 m 3.26 m A B 2.42 m 2.87 m 1.64 m 1.19m 1.20 m 3.68 m Prob. 9–70 z y G2 G4 G3 G1 x 600 lb 9 ft 7 ft 12 ft 6 ft 8 ft 4 ft 3 ft 5 ft 1500 lb 450 lb 280 lb Prob. 9–71 y z x 150 mm 250 mm 200 mm 150 mm 150 mm 100 mm Prob. 9–72 y z x 100 mm 300 mm Prob. 9–73
- 500. 9.2 COMPOSITE BODIES 483 9 •9–77. Determine the distance to the centroid of the solid which consists of a cylinder with a hole of length bored into its base. 9–78. Determine the distance h to which a hole must be bored into the cylinder so that the center of mass of the assembly is located at . The material has a density of . 8 Mgm3 x = 64 mm h = 50 mm x 9–75. Locate the center of gravity of the homogeneous block assembly having a hemispherical hole. *9–76. Locate the center of gravity of the assembly. The triangular and the rectangular blocks are made from materials having specific weights of and , respectively. 0.1 lbin3 0.25 lbin3 (x, y, z) (x, y, z) 9–74. Locate the center of mass of the assembly. The cylinder and the cone are made from materials having densities of and , respectively. 9 Mgm3 5 Mgm3 z 9–79. The assembly is made from a steel hemisphere, , and an aluminum cylinder, . Determine the mass center of the assembly if the height of the cylinder is . *9–80. The assembly is made from a steel hemisphere, , and an aluminum cylinder, . Determine the height h of the cylinder so that the mass center of the assembly is located at . z = 160 mm ral = 2.70 Mgm3 rst = 7.80 Mgm3 h = 200 mm ral = 2.70 Mgm3 rst = 7.80 Mgm3 y x h 120 mm 40 mm 20 mm Probs. 9–77/78 160 mm h z y x 80 mm z G _ Probs. 9–79/80 z x 0.8 m 0.6 m 0.4 m 0.2 m y Prob. 9–74 y z x 1 in. 3 in. 2.25 in. 2.25 in. 2.5 in. 2.5 in. 1 in. 3 in. Probs. 9–75/76
- 501. 484 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 *9.3 Theorems of Pappus and Guldinus The two theorems of Pappus and Guldinus are used to find the surface area and volume of any body of revolution.They were first developed by Pappus of Alexandria during the fourth century A.D. and then restated at a later time by the Swiss mathematician Paul Guldin or Guldinus (1577–1643). Therefore the first theorem of Pappus and Guldinus states that the area of a surface of revolution equals the product of the length of the generating curve and the distance traveled by the centroid of the curve in generating the surface area. A = surface area of revolution u = angle of revolution measured in radians, u … 2p r = perpendicular distance from the axis of revolution to the centroid of the generating curve L = length of the generating curve The amount of roofing material used on this storage building can be estimated by using the first theorem of Pappus and Guldinus to determine its surface area. 2 r p r L C dL dA r Surface Area. If we revolve a plane curve about an axis that does not intersect the curve we will generate a surface area of revolution. For example, the surface area in Fig. 9–19 is formed by revolving the curve of length L about the horizontal axis.To determine this surface area, we will first consider the differential line element of length dL. If this element is revolved radians about the axis, a ring having a surface area of will be generated.Thus, the surface area of the entire body is Since (Eq. 9–5), then If the curve is revolved only through an angle (radians), then (9–7) where A = urL u A = 2prL. 1r dL = rL A = 2p1r dL. dA = 2pr dL 2p Fig. 9–19
- 502. 9.3 THEOREMS OF PAPPUS AND GULDINUS 485 9 Volume. A volume can be generated by revolving a plane area about an axis that does not intersect the area. For example, if we revolve the shaded area A in Fig. 9–20 about the horizontal axis, it generates the volume shown.This volume can be determined by first revolving the differential element of area dA 2 radians about the axis, so that a ring having the volume is generated.The entire volume is then However, Eq. 9–4, so that . If the area is only revolved through an angle (radians), then (9–8) where V = urA u V = 2prA 1rdA = rA, V = 2p1rdA. dV = 2pr dA p Therefore the second theorem of Pappus and Guldinus states that the volume of a body of revolution equals the product of the generating area and the distance traveled by the centroid of the area in generating the volume. Composite Shapes. We may also apply the above two theorems to lines or areas that are composed of a series of composite parts. In this case the total surface area or volume generated is the addition of the surface areas or volumes generated by each of the composite parts. If the perpendicular distance from the axis of revolution to the centroid of each composite part is then (9–9) and (9–10) Application of the above theorems is illustrated numerically in the following examples. V = u©1r ' A2 A = u©1r ' L2 r ' , dA 2 r C A r r p Fig. 9–20 V = volume of revolution u = angle of revolution measured in radians, u … 2p r = perpendicular distance from the axis of revolution to the centroid of the generating area A = generating area The volume of fertilizer contained within this silo can be determined using the second theorem of Pappus and Guldinus.
- 503. 486 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 Show that the surface area of a sphere is and its volume is V = 4 3 pR3 . A = 4pR2 EXAMPLE 9.12 (b) y x R C 4R 3p y x R C 2R (a) p Fig. 9–21 SOLUTION Surface Area. The surface area of the sphere in Fig. 9–21a is generated by revolving a semicircular arc about the x axis. Using the table on the inside back cover, it is seen that the centroid of this arc is located at a distance from the axis of revolution (x axis). Since the centroid moves through an angle of rad to generate the sphere, then applying Eq. 9–7 we have Ans. Volume. The volume of the sphere is generated by revolving the semicircular area in Fig. 9–21b about the x axis. Using the table on the inside back cover to locate the centroid of the area, i.e., and applying Eq. 9–8, we have Ans. V = 2pa 4R 3p b a 1 2 pR2 b = 4 3 pR3 V = urA; r = 4R3p, A = 2pa 2R p bpR = 4pR2 A = urL; u = 2p r = 2Rp
- 504. 9.3 THEOREMS OF PAPPUS AND GULDINUS 487 9 EXAMPLE 9.13 Determine the surface area and volume of the full solid in Fig. 9–22a. 1 in. 1 in. 2 in. (a) 2.5 in. z Fig. 9–22 (b) z 1 in. 3.5 in. 3 in. 2.5 in. 1 in. 2 in. 1 in. 2 in. 1 in. (c) z 3 in. 2.5 in. ( )(1 in.) 3.1667 in. 2 3 SOLUTION Surface Area. The surface area is generated by revolving the four line segments shown in Fig. 9–22b, radians about the z axis. The distances from the centroid of each segment to the z axis are also shown in the figure.Applying Eq. 9–7, yields Ans. Volume. The volume of the solid is generated by revolving the two area segments shown in Fig. 9–22c, 2 radians about the z axis. The distances from the centroid of each segment to the z axis are also shown in the figure.Applying Eq. 9–10, we have Ans. = 47.6 in3 V=2p©rA=2pE(3.1667 in.)c 1 2 (1 in.)(1 in.) d +(3 in.)[(2 in.)(1 in.)F p = 143 in2 + (3.5 in.)(3 in.) + (3 in.)(1 in.)] A = 2p©rL = 2p[(2.5 in.)(2 in.) + (3 in.)¢ 3(1 in.)2 + (1 in.)2 ≤ 2p
- 505. 488 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 z 1.5 m 2 m 2 m F9–13 1.2 m 0.9 m 1.5 m 1.5 m z F9–14 z 18 in. 15 in. 20 in. 30 in. F9–15 z 2 m 1.5 m 1.5 m F9–16 FUNDAMENTAL PROBLEMS F9–15. Determine the surface area and volume of the solid formed by revolving the shaded area about the z axis. 360° F9–14. Determine the surface area and volume of the solid formed by revolving the shaded area about the z axis. 360° F9–13. Determine the surface area and volume of the solid formed by revolving the shaded area about the z axis. 360° F9–16. Determine the surface area and volume of the solid formed by revolving the shaded area about the z axis. 360°
- 506. 9.3 THEOREMS OF PAPPUS AND GULDINUS 489 9 6 ft 8 ft 8 ft 10 ft Probs. 9–81/82 x y 4 ft 4 ft y2 4x Prob. 9–83 A 1 m z B 1.5 m 3 m Probs. 9–84/85 16 m y x 16 m y 16 (x2 /16) Prob. 9–86 PROBLEMS *9–84. Determine the surface area from A to B of the tank. •9–85. Determine the volume within the thin-walled tank from A to B. 9–83. Determine the volume of the solid formed by revolving the shaded area about the x axis using the second theorem of Pappus–Guldinus.The area and centroid of the shaded area should first be obtained by using integration. y •9–81. The elevated water storage tank has a conical top and hemispherical bottom and is fabricated using thin steel plate. Determine how many square feet of plate is needed to fabricate the tank. 9–82. The elevated water storage tank has a conical top and hemispherical bottom and is fabricated using thin steel plate. Determine the volume within the tank. 9–86. Determine the surface area of the roof of the structure if it is formed by rotating the parabola about the y axis.
- 507. 490 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 2 in. 3 in. z 0.75 in. 0.75 in. 1 in. 0.5 in. Probs. 9–87/88 75 mm 75 mm 75 mm 75 mm 250 mm z 300 mm Prob. 9–89 9–90. Determine the surface area and volume of the solid formed by revolving the shaded area about the z axis. 360° •9–89. Determine the volume of the solid formed by revolving the shaded area about the z axis. 360° 9–87. Determine the surface area of the solid formed by revolving the shaded area about the z axis. *9–88. Determine the volume of the solid formed by revolving the shaded area about the z axis. 360° 360° 9–91. Determine the surface area and volume of the solid formed by revolving the shaded area about the z axis. 360° 1 in. z 2 in. 1 in. Prob. 9–90 z 75 mm 50 mm 400 mm 300 mm 50 mm 75 mm Prob. 9–91
- 508. 9.3 THEOREMS OF PAPPUS AND GULDINUS 491 9 9–94. The thin-wall tank is fabricated from a hemisphere and cylindrical shell. Determine the vertical reactions that each of the four symmetrically placed legs exerts on the floor if the tank contains water which is 12 ft deep in the tank. The specific gravity of water is . Neglect the weight of the tank. 9–95. Determine the approximate amount of paint needed to cover the outside surface of the open tank.Assume that a gallon of paint covers . 400 ft2 62.4 lbft3 •9–93. The hopper is filled to its top with coal. Estimate the volume of coal if the voids (air space) are 35 percent of the volume of the hopper. *9–92. The process tank is used to store liquids during manufacturing. Estimate both the volume of the tank and its surface area.The tank has a flat top and a thin wall. *9–96. Determine the surface area of the tank, which consists of a cylinder and hemispherical cap. •9–97. Determine the volume of the thin-wall tank, which consists of a cylinder and hemispherical cap. 3 m 3 m 6 m 4 m Prob. 9–92 0.2 m 4 m z 1.2 m 1.5 m Prob. 9–93 water surface 8 ft 4 ft 6 ft 8 ft Probs. 9–94/95 8 m 4 m Probs. 9–96/97
- 509. 492 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 •9–101. Determine the outside surface area of the storage tank. 9–102. Determine the volume of the thin-wall storage tank. *9–100. Determine the surface area and volume of the wheel formed by revolving the cross-sectional area about the z axis. 360° 9–98. The water tank AB has a hemispherical top and is fabricated from thin steel plate. Determine the volume within the tank. 9–99. The water tank AB has a hemispherical roof and is fabricated from thin steel plate. If a liter of paint can cover of the tank’s surface, determine how many liters are required to coat the surface of the tank from A to B. 3 m2 9–103. Determine the height h to which liquid should be poured into the conical paper cup so that it contacts half the surface area on the inside of the cup. 15 ft 4 ft 30 ft Probs. 9–101/102 100 mm h 150 mm Prob. 9–103 1.5 m 1.6 m 0.2 m B A 1.6 m Probs. 9–98/99 4 in. 2 in. 1 in. 1 in. 1.5 in. z Prob. 9–100
- 510. 9.4 RESULTANT OF A GENERAL DISTRIBUTED LOADING 493 9 *9.4 Resultant of a General Distributed Loading In Sec. 4.9, we discussed the method used to simplify a two-dimensional distributed loading to a single resultant force acting at a specific point. In this section we will generalize this method to include flat surfaces that have an arbitrary shape and are subjected to a variable load distribution. Consider,for example,the flat plate shown in Fig.9–23a,which is subjected to the loading defined by Pa, where 1 Knowing this function, we can determine the resultant force acting on the plate and its location Fig. 9–23b. Magnitude of Resultant Force. The force dF acting on the differential area of the plate, located at the arbitrary point (x, y), has a magnitude of Notice that p(x, y) the colored differential volume element shown in Fig. 9–23a. The magnitude of is the sum of the differential forces acting over the plate’s entire surface area A.Thus: (9–11) This result indicates that the magnitude of the resultant force is equal to the total volume under the distributed-loading diagram. Location of Resultant Force. The location ( ) of is determined by setting the moments of equal to the moments of all the differential forces dF about the respective y and x axes: From Figs. 9–23a and 9–23b, using Eq. 9–11, this results in FR FR y x, FR = L A p1x, y2 dA = L V dV = V FR = ©F; FR dA = dV, dF = [p1x, y2 Nm2 ]1dA m2 2 = [p1x, y2 dA] N. dA m2 (x, y), FR Pa 1pascal2 = 1 Nm2 . p1x, y2 p = The resultant of a wind loading that is distributed on the front or side walls of this building must be calculated using integration in order to design the framework that holds the building together. x y y x (a) dF p p p(x, y) dA dV Fig. 9–23 x y y x (b) FR (9–12) x = L A xp1x,y2dA L A p1x,y2dA = L V x dV L V dV y = L A yp1x,y2dA L A p1x,y2dA = L V ydV L V dV Hence, the line of action of the resultant force passes through the geometric center or centroid of the volume under the distributed-loading diagram.
- 511. 494 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 *9.5 Fluid Pressure According to Pascal’s law, a fluid at rest creates a pressure p at a point that is the same in all directions.The magnitude of p, measured as a force per unit area, depends on the specific weight or mass density of the fluid and the depth z of the point from the fluid surface.* The relationship can be expressed mathematically as (9–13) where g is the acceleration due to gravity. This equation is valid only for fluids that are assumed incompressible,as in the case of most liquids.Gases are compressible fluids, and since their density changes significantly with both pressure and temperature, Eq. 9–13 cannot be used. To illustrate how Eq. 9–13 is applied, consider the submerged plate shown in Fig. 9–24. Three points on the plate have been specified. Since point B is at depth from the liquid surface, the pressure at this point has a magnitude Likewise, points C and D are both at depth hence, In all cases, the pressure acts normal to the surface area dA located at the specified point. Using Eq. 9–13 and the results of Sec. 9.4, it is possible to determine the resultant force caused by a liquid and specify its location on the surface of a submerged plate. Three different shapes of plates will now be considered. p2 = gz2. z2; p1 = gz1. z1 p = gz = rgz r g z y x b dA dA C z2 z1 Liquid surface dA p1 p2 p2 D B Fig. 9–24 *In particular, for water or since and g = 9.81 ms2 . r = 1000 kgm3 g = rg = 9810 Nm3 g = 62.4 lbft3 ,
- 512. 9.5 FLUID PRESSURE 495 9 Flat Plate of Constant Width. A flat rectangular plate of constant width, which is submerged in a liquid having a specific weight is shown in Fig. 9–25a. Since pressure varies linearly with depth, Eq. 9–13, the distribution of pressure over the plate’s surface is represented by a trapezoidal volume having an intensity of at depth and at depth As noted in Sec. 9.4, the magnitude of the resultant force is equal to the volume of this loading diagram and has a line of action that passes through the volume’s centroid C. Hence, does not act at the centroid of the plate; rather, it acts at point P, called the center of pressure. Since the plate has a constant width, the loading distribution may also be viewed in two dimensions, Fig. 9–25b. Here the loading intensity is measured as force/length and varies linearly from to The magnitude of in this case equals the trapezoidal area, and has a line of action that passes through the area’s centroid C. For numerical applications, the area and location of the centroid for a trapezoid are tabulated on the inside back cover. FR FR w2 = bp2 = bgz2. w1 = bp1 = bgz1 FR FR FR z2. p2 = gz2 z1 p1 = gz1 g, x Liquid surface z2 z1 C P p2 gz2 p1 gz1 (a) y FR z L b 2 b 2 Fig. 9–25 Liquid surface y w2 bp2 z P L (b) FR C y¿ w1 bp1 z2 z1 The walls of the tank must be designed to support the pressure loading of the liquid that is contained within it.
- 513. 496 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 Curved Plate of Constant Width. When a submerged plate of constant width is curved, the pressure acting normal to the plate continually changes both its magnitude and direction, and therefore calculation of the magnitude of and its location P is more difficult than for a flat plate. Three- and two-dimensional views of the loading distribution are shown in Figs. 9–26a and 9–26b, respectively. Although integration can be used to solve this problem, a simpler method exists. This method requires separate calculations for the horizontal and vertical components of For example, the distributed loading acting on the plate can be represented by the equivalent loading shown in Fig. 9–26c. Here the plate supports the weight of liquid contained within the block BDA. This force has a magnitude and acts through the centroid of BDA. In addition, there are the pressure distributions caused by the liquid acting along the vertical and horizontal sides of the block.Along the vertical side AD, the force has a magnitude equal to the area of the trapezoid.It acts through the centroid of this area.The distributed loading along the horizontal side AB is constant since all points lying in this plane are at the same depth from the surface of the liquid. The magnitude of is simply the area of the rectangle. This force acts through the centroid or at the midpoint of AB. Summing these three forces yields Finally, the location of the center of pressure P on the plate is determined by applying which states that the moment of the resultant force about a convenient reference point such as D or B, in Fig. 9–26b, is equal to the sum of the moments of the three forces in Fig. 9–26c about this same point. MR = ©M, FR = ©F = FAD + FAB + Wf. CAB FAB CAD FAD Wf = 1gb21areaBDA2 Wf FR. FR y p1 gz1 Liquid surface z x z1 L z2 FR p2 gz2 b C P (a) Fig. 9–26 Liquid surface y w2 bp2 C FR w1 bp1 B z P D (b) B CAB FAB A z1 z2 y w1 bp1 z CAD FAD w1 bp2 Wf CBDA Liquid surface D (c)
- 514. 9.5 FLUID PRESSURE 497 9 Flat Plate of Variable Width. The pressure distribution acting on the surface of a submerged plate having a variable width is shown in Fig. 9–27. If we consider the force dF acting on the differential area strip dA, parallel to the x axis, then its magnitude is . Since the depth of dA is z, the pressure on the element is . Therefore, and so the resultant force becomes If the depth to the centroid of the area is , Fig. 9–27, then, . Substituting, we have (9–14) In other words, the magnitude of the resultant force acting on any flat plate is equal to the product of the area A of the plate and the pressure at the depth of the area’s centroid As discussed in Sec. 9.4, this force is also equivalent to the volume under the pressure distribution. Realize that its line of action passes through the centroid C of this volume and intersects the plate at the center of pressure P, Fig. 9–27. Notice that the location of does not coincide with the location of P. C¿ C¿. p = gz FR = gzA 1z dA = zA z C¿ FR = 1 dF = g1 z dA dF = (gz)dA p = gz dF = p dA y x y¿ Liquid surface z FR p gz dy¿ dA dF C¿ P x z C Fig. 9–27 The resultant force of the water pressure and its location on the elliptical back plate of this tank truck must be determined by integration.
- 515. 498 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 Determine the magnitude and location of the resultant hydrostatic force acting on the submerged rectangular plate AB shown in Fig. 9–28a.The plate has a width of 1.5 m; SOLUTION I The water pressures at depths A and B are Since the plate has a constant width, the pressure loading can be viewed in two dimensions as shown in Fig. 9–28b. The intensities of the load at A and B are From the table on the inside back cover, the magnitude of the resultant force created by this distributed load is Ans. This force acts through the centroid of this area, Ans. measured upward from B, Fig. 9–31b. SOLUTION II The same results can be obtained by considering two components of , defined by the triangle and rectangle shown in Fig. 9–28c. Each force acts through its associated centroid and has a magnitude of Hence, Ans. The location of is determined by summing moments about B, Fig. 9–28b and c, i.e., c Ans. NOTE: Using Eq. 9–14, the resultant force can be calculated as FR = gzA = (9810 Nm3 )(3.5 m)(3 m)(1.5 m) = 154.5 kN. h = 1.29 m 1154.52h = 88.311.52 + 66.2112 +1MR2B = ©MB; FR FR = FRe + Ft = 88.3 + 66.2 = 154.5 kN Ft = 1 2144.15 kNm213 m2 = 66.2 kN FRe = 129.43 kNm213 m2 = 88.3 kN FR h = 1 3 a 2129.432 + 73.58 29.43 + 73.58 b132 = 1.29 m FR = area of a trapezoid = 1 2132129.4 + 73.62 = 154.5 kN FR wB = bpB = 11.5 m2149.05 kPa2 = 73.58 kNm wA = bpA = 11.5 m2119.62 kPa2 = 29.43 kNm pB = rwgzB = 11000 kgm3 219.81 ms2 215 m2 = 49.05 kPa pA = rwgzA = 11000 kgm3 219.81 ms2 212 m2 = 19.62 kPa rw = 1000 kgm3 . EXAMPLE 9.14 2 m 3 m 1.5 m A B (a) Fig. 9–28 (b) 2 m 3 m A B h FR wB 73.58 kN/m wA 29.43 kN/m (c) 2 m 3 m A B Ft 1 m 44.15 kN/m 29.43 kN/m FRe 1.5 m
- 516. 9.5 FLUID PRESSURE 499 9 EXAMPLE 9.15 Determine the magnitude of the resultant hydrostatic force acting on the surface of a seawall shaped in the form of a parabola as shown in Fig. 9–29a.The wall is 5 m long; rw = 1020 kgm3 . 3 m 1 m (a) Fig. 9–29 Fh wB 150.1 kN/m C Fv A B (b) SOLUTION The horizontal and vertical components of the resultant force will be calculated, Fig. 9–29b. Since then Thus, The area of the parabolic sector ABC can be determined using the table on the inside back cover. Hence, the weight of water within this 5 m long region is The resultant force is therefore Ans. = 231 kN FR = 2Fh 2 + Fv 2 = 21225.1 kN22 + 150.0 kN22 = 11020 kgm3 219.81 ms2 215 m2C1 311 m213 m2D = 50.0 kN Fv = 1rwgb21areaABC2 Fh = 1 213 m21150.1 kNm2 = 225.1 kN wB = bpB = 5 m130.02 kPa2 = 150.1 kNm pB = rwgzB = 11020 kgm3 219.81 ms2 213 m2 = 30.02 kPa
- 517. 500 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 1 m 1 m (a) E Fig. 9–30 Determine the magnitude and location of the resultant force acting on the triangular end plates of the water trough shown in Fig. 9–30a; rw = 1000 kgm3 . EXAMPLE 9.16 SOLUTION The pressure distribution acting on the end plate E is shown in Fig.9–30b. The magnitude of the resultant force is equal to the volume of this loading distribution.We will solve the problem by integration. Choosing the differential volume element shown in the figure, we have The equation of line AB is Hence, substituting and integrating with respect to z from to yields Ans. This resultant passes through the centroid of the volume. Because of symmetry, Ans. Since for the volume element, then z ' = z x = 0 = 9810 L 1 m 0 1z - z2 2 dz = 1635 N = 1.64 kN F = V = L V dV = L 1 m 0 119 6202z[0.511 - z2] dz z = 1 m z = 0 x = 0.511 - z2 dF = dV = p dA = rwgz12x dz2 = 19 620zx dz 0.5 m y x z 1 m z dz A B (b) O 2x dF Ans. NOTE: We can also determine the resultant force by applying Eq. 9–14, FR = gzA = (9810 Nm3 )(1 3)(1 m)[1 2(1 m)(1 m)] = 1.64 kN. = 0.5 m 9810 L 1 m 0 1z2 - z3 2 dz 1635 z = L V z ' dV L V dV = L 1 m 0 z119 6202z[0.511 - z2] dz 1635 =
- 518. 9.5 FLUID PRESSURE 501 9 FUNDAMENTAL PROBLEMS F9–20. Determine the magnitude of the hydrostatic force acting on gate AB, which has a width of 2 m. Water has a density of . r = 1 Mgm3 F9–18. Determine the magnitude of the hydrostatic force acting on gate AB, which has a width of 4 ft. The specific weight of water is . g = 62.4 lbft3 F9–17. Determine the magnitude of the hydrostatic force acting per meter length of the wall. Water has a density of . r = 1 Mgm3 F9–21. Determine the magnitude of the hydrostatic force acting on gate AB, which has a width of 2 ft. The specific weight of water is . g = 62.4 lbft3 A B 4 ft 3 ft F9–18 B A 2 m 1.5 m F9–19 B A 2 m 3 m F9–20 6 m F9–17 B A 3 ft 4 ft 6 ft F9–21 F9–19. Determine the magnitude of the hydrostatic force acting on gate AB, which has a width of 1.5 m. Water has a density of . r = 1 Mgm3
- 519. 502 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 PROBLEMS 9–106. The symmetric concrete “gravity” dam is held in place by its own weight. If the density of concrete is , and water has a density of , determine the smallest distance d at its base that will prevent the dam from overturning about its end A.The dam has a width of 8 m. rw = 1.0 Mgm3 rc = 2.5 Mgm3 •9–105. The concrete “gravity” dam is held in place by its own weight. If the density of concrete is , and water has a density of , determine the smallest dimension d that will prevent the dam from overturning about its end A. rw = 1.0 Mgm3 rc = 2.5 Mgm3 *9–104. The tank is used to store a liquid having a specific weight of . If it is filled to the top, determine the magnitude of the force the liquid exerts on each of its two sides ABDC and BDFE. 80 lbft3 9–107. The tank is used to store a liquid having a specific weight of . If the tank is full, determine the magnitude of the hydrostatic force on plates CDEF and ABDC. 60 lbft3 A B E C D F 12 ft 8 ft 4 ft 6 ft 6 ft Prob. 9–104 A 6 m d Prob. 9–105 A d 1.5 m 9 m Prob. 9–106 x B A D E y z 5 ft 2 ft 2 ft 1.5 ft 1.5 ft 1.5 ft 1.5 ft C F Prob. 9–107
- 520. 9.5 FLUID PRESSURE 503 9 *9–112. Determine the magnitude of the hydrostatic force acting per foot of length on the seawall. . gw = 62.4 lbft3 9–110. Determine the magnitude of the hydrostatic force acting on the glass window if it is circular, A. The specific weight of seawater is . 9–111. Determine the magnitude and location of the resultant hydrostatic force acting on the glass window if it is elliptical, B. The specific weight of seawater is . gw = 63.6 lbft3 gw = 63.6 lbft3 *9–108. The circular steel plate A is used to seal the opening on the water storage tank. Determine the magnitude of the resultant hydrostatic force that acts on it. The density of water is . •9–109. The elliptical steel plate B is used to seal the opening on the water storage tank. Determine the magnitude of the resultant hydrostatic force that acts on it. The density of water is . rw = 1 Mgm3 rw = 1 Mgm3 •9–113. If segment AB of gate ABC is long enough, the gate will be on the verge of opening. Determine the length L of this segment in order for this to occur. The gate is hinged at B and has a width of 1 m. The density of water is . 9–114. If L = 2 m, determine the force the gate ABC exerts on the smooth stopper at C. The gate is hinged at B, free at A, and is 1 m wide.The density of water is . rw = 1 Mgm3 rw = 1 Mgm3 45 1 m 2 m 0.5 m 0.5 m 1 m A B 1 m Probs. 9–108/109 4 ft 0.5 ft 0.5 ft 1 ft 1 ft 1 ft A B Probs. 9–110/111 x y 8 ft 2 ft y 2x2 Prob. 9–112 C B A L 2 m 4 m Probs. 9–113/114
- 521. 504 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 9–118. The concrete gravity dam is designed so that it is held in position by its own weight. Determine the minimum dimension x so that the factor of safety against overturning about point A of the dam is 2.The factor of safety is defined as the ratio of the stabilizing moment divided by the overturning moment. The densities of concrete and water are and , respectively. Assume that the dam does not slide. rw = 1 Mgm3 rconc = 2.40 Mgm3 •9–117. The concrete gravity dam is designed so that it is held in position by its own weight. Determine the factor of safety against overturning about point A if . The factor of safety is defined as the ratio of the stabilizing moment divided by the overturning moment. The densities of concrete and water are and , respectively.Assume that the dam does not slide. rw = 1 Mgm3 rconc = 2.40 Mgm3 x = 2 m 9–115. Determine the mass of the counterweight A if the 1-m-wide gate is on the verge of opening when the water is at the level shown. The gate is hinged at B and held by the smooth stop at C.The density of water is . *9–116. If the mass of the counterweight at A is 6500 kg, determine the force the gate exerts on the smooth stop at C. The gate is hinged at B and is 1-m wide. The density of water is . rw = 1 Mgm3 rw = 1 Mgm3 9–119. The underwater tunnel in the aquatic center is fabricated from a transparent polycarbonate material formed in the shape of a parabola. Determine the magnitude of the hydrostatic force that acts per meter length along the surface AB of the tunnel. The density of the water is . rw = 1000 kg/m3 A B C 1 m 45 2 m 2 m Probs. 9–115/116 y x x 3 –– 2 y x2 6 m 2 m A Probs. 9–118 y x x 3 –– 2 y x2 6 m 2 m A Probs. 9–117 y x 2 m 2 m 2 m 4 m y 4 x2 A B Prob. 9–119
- 522. CHAPTER REVIEW 505 9 CHAPTER REVIEW Center of Gravity and Centroid The center of gravity G represents a point where the weight of the body can be considered concentrated. The distance from an axis to this point can be determined from a balance of moments, which requires that the moment of the weight of all the particles of the body about this axis must equal the moment of the entire weight of the body about the axis. The center of mass will coincide with the center of gravity provided the acceleration of gravity is constant. The centroid is the location of the geometric center for the body. It is determined in a similar manner, using a moment balance of geometric elements such as line, area, or volume segments. For bodies having a continuous shape, moments are summed (integrated) using differential elements. The center of mass will coincide with the centroid provided the material is homogeneous, i.e., the density of the material is the same throughout. The centroid will always lie on an axis of symmetry. z = L z ' dW L dW y = L y ' dW L dW x = L x ' dW L dW x = L V x ' dV L V dV y = L V y ' dV L V dV z = L V z ' dV L V dV x = L A x ' dA L A dA y = L A y ' dA L A dA z = L A z ' dA L A dA x = L L x ' dL L L dL y = L L y ' dL L L dL z = L L z ' dL L L dL G dV ~ z z y ~ x x ~ y y z x W dW C y x
- 523. 506 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 Composite Body If the body is a composite of several shapes, each having a known location for its center of gravity or centroid, then the location of the center of gravity or centroid of the body can be determined from a discrete summation using its composite parts. z = ©z ' W ©W y = ©y ' W ©W x = ©x ' W ©W x y z Theorems of Pappus and Guldinus The theorems of Pappus and Guldinus can be used to determine the surface area and volume of a body of revolution. The surface area equals the product of the length of the generating curve and the distance traveled by the centroid of the curve needed to generate the area. The volume of the body equals the product of the generating area and the distance traveled by the centroid of this area needed to generate the volume. A = urL V = urA
- 524. CHAPTER REVIEW 507 9 General Distributed Loading The magnitude of the resultant force is equal to the total volume under the distributed-loading diagram. The line of action of the resultant force passes through the geometric center or centroid of this volume. y = L V y dV L V dV x = L V x dV L V dV FR = L A p1x, y2 dA = L V dV Fluid Pressure The pressure developed by a liquid at a point on a submerged surface depends upon the depth of the point and the density of the liquid in accordance with Pascal’s law, This pressure will create a linear distribution of loading on a flat vertical or inclined surface. If the surface is horizontal, then the loading will be uniform. In any case, the resultants of these loadings can be determined by finding the volume under the loading curve or using , where is the depth to the centroid of the plate’s area. The line of action of the resultant force passes through the centroid of the volume of the loading diagram and acts at a point P on the plate called the center of pressure. z FR = gzA p = rgh = gh. x y y x dF p dV dA p p(x, y) Liquid surface P FR
- 525. 508 CHAPTER 9 CENTER OF GRAVITY AND CENTROID 9 REVIEW PROBLEMS 9–123. Locate the centroid of the solid. z 9–122. Locate the centroid of the beam’s cross-sectional area. y *9–120. Locate the centroid of the shaded area. •9–121. Locate the centroid of the shaded area. y x *9–124. The steel plate is 0.3 m thick and has a density of Determine the location of its center of mass. Also compute the reactions at the pin and roller support. 7850 kgm3 . y x 1 in. 1 in. 4 in. 1 in. y x2 Probs. 9–120/121 100 mm 25 mm 25 mm x 25 mm y 50 mm 50 mm y 75 mm 75 mm C Prob. 9–122 z x 2a y y2 a a – z – 2 a Prob. 9–123 A B x y y2 2x y x 2 m 2 m 2 m Prob 9.124
- 526. REVIEW PROBLEMS 509 9 9–127. Locate the centroid of the shaded area. y 9–126. Determine the location ( , ) of the centroid for the structural shape. Neglect the thickness of the member. y x •9–125. Locate the centroid ( , ) of the area. y x *9–128. The load over the plate varies linearly along the sides of the plate such that . Determine the resultant force and its position ( , ) on the plate. y x p = 2 3 [x(4 - y)] kPa y x 3 in. 1 in. 3 in. 6 in. Prob. 9–125 1.5 in. 1.5 in. 1.5 in. 1.5 in. 1 in. 1 in. 3 in. x y Prob. 9–126 x y a — 2 a — 2 a a a Prob. 9–127 p 3 m 4 m y x 8 kPa Prob. 9–128 p x y 6 m 5 m 100 Pa 300 Pa Prob. 9–129 •9–129. The pressure loading on the plate is described by the function . Determine the magnitude of the resultant force and coordinates of the point where the line of action of the force intersects the plate. p = 5-240(x + 1) + 3406 Pa
- 527. The design of a structural member, such as a beam or column, requires calculation of its cross-sectional moment of inertia. In this chapter, we will discuss how this is done.
- 528. Moments of Inertia CHAPTER OBJECTIVES • To develop a method for determining the moment of inertia for an area. • To introduce the product of inertia and show how to determine the maximum and minimum moments of inertia for an area. • To discuss the mass moment of inertia. 10.1 Definition of Moments of Inertia for Areas Whenever a distributed loading acts perpendicular to an area and its intensity varies linearly, the computation of the moment of the loading distribution about an axis will involve a quantity called the moment of inertia of the area. For example, consider the plate in Fig. 10–1, which is subjected to a fluid pressure p. As discussed in Sec. 9.5, this pressure p varies linearly with depth, such that , where is the specific weight of the fluid. Thus, the force acting on the differential area of the plate is . The moment of this force about the axis is therefore , and so integrating over the entire area of the plate yields .The integral is called the moment of inertia of the area about the axis. Integrals of this form often arise in formulas used in fluid mechanics, mechanics of materials, structural mechanics, and mechanical design, and so the engineer needs to be familiar with the methods used for their computation. x Ix 1y 2 dA M = g1y2 dA dM dM = y dF = gy2 dA x dF = p dA = (g y)dA dA g p = gy 10 y x z y dF dA p gy Fig. 10–1
- 529. 512 CHAPTER 10 MOMENTS OF INERTIA 10 Moment of Inertia. By definition, the moments of inertia of a differential area dA about the x and y axes are and respectively, Fig. 10–2. For the entire area A the moments of inertia are determined by integration; i.e., (10–1) We can also formulate this quantity for dA about the “pole” O or z axis, Fig. 10–2. This is referred to as the polar moment of inertia. It is defined as where r is the perpendicular distance from the pole (z axis) to the element dA. For the entire area the polar moment of inertia is (10–2) This relation between and is possible since Fig. 10–2. From the above formulations it is seen that and will always be positive since they involve the product of distance squared and area. Furthermore, the units for moment of inertia involve length raised to the fourth power, e.g., or 10.2 Parallel-Axis Theorem for an Area The parallel-axis theorem can be used to find the moment of inertia of an area about any axis that is parallel to an axis passing through the centroid and about which the moment of inertia is known.To develop this theorem, we will consider finding the moment of inertia of the shaded area shown in Fig. 10–3 about the x axis.To start, we choose a differential element dA located at an arbitrary distance from the centroidal axis. If the distance between the parallel x and axes is then the moment of inertia of dA about the x axis is . For the entire area, = L A y¿2 dA + 2dy L A y¿ dA + dy 2 L A dA Ix = L A 1y¿ + dy22 dA dIx = 1y¿ + dy22 dA dy, x¿ x¿ y¿ in4 . ft4 , mm4 , m4 , JO Iy, Ix, r2 = x2 + y2 , Iy Ix, JO JO = L A r2 dA = Ix + Iy dJO = r2 dA, Ix = L A y 2 dA Iy = L A x 2 dA dIy = x2 dA, dIx = y2 dA O x y y x r dA A Fig. 10–2 O x y d dx dy x¿ y x¿ y¿ dA C Fig. 10–3
- 530. 10.3 RADIUS OF GYRATION OF AN AREA 513 10 The first integral represents the moment of inertia of the area about the centroidal axis, The second integral is zero since the axis passes through the area’s centroid C; i.e., since Since the third integral represents the total area A, the final result is therefore (10–3) A similar expression can be written for i.e., (10–4) And finally, for the polar moment of inertia, since and , we have (10–5) The form of each of these three equations states that the moment of inertia for an area about an axis is equal to its moment of inertia about a parallel axis passing through the area’s centroid plus the product of the area and the square of the perpendicular distance between the axes. 10.3 Radius of Gyration of an Area The radius of gyration of an area about an axis has units of length and is a quantity that is often used for the design of columns in structural mechanics.Provided the areas and moments of inertia are known,the radii of gyration are determined from the formulas (10–6) The form of these equations is easily remembered since it is similar to that for finding the moment of inertia for a differential area about an axis. For example, whereas for a differential area, dIx = y2 dA. Ix = kx 2 A; kO = D JO A ky = D Iy A kx = D Ix A JO = JC + Ad2 d2 = d2 x + d2 y JC = Ix¿ + Iy¿ Iy = Iy¿ + Adx 2 Iy; Ix = Ix¿ + Ady 2 y¿ = 0. 1y¿ dA = y¿1dA = 0 x¿ Ix¿. In order to predict the strength and deflection of this beam, it is necessary to calculate the moment of inertia of the beam’s cross-sectional area.
- 531. 514 CHAPTER 10 MOMENTS OF INERTIA 10 y (a) y x dy x (x, y) y f(x) dA x (b) y x y dx (x, y) dA y f(x) Procedure for Analysis In most cases the moment of inertia can be determined using a single integration. The following procedure shows two ways in which this can be done. • If the curve defining the boundary of the area is expressed as , then select a rectangular differential element such that it has a finite length and differential width. • The element should be located so that it intersects the curve at the arbitrary point (x, y). Case 1 • Orient the element so that its length is parallel to the axis about which the moment of inertia is computed.This situation occurs when the rectangular element shown in Fig. 10–4a is used to determine for the area.Here the entire element is at a distance y from the x axis since it has a thickness .Thus .To find ,the element is oriented as shown in Fig. 10–4b. This element lies at the same distance x from the y axis so that . Case 2 • The length of the element can be oriented perpendicular to the axis about which the moment of inertia is computed; however, Eq. 10–1 does not apply since all points on the element will not lie at the same moment-arm distance from the axis. For example, if the rectangular element in Fig. 10–4a is used to determine , it will first be necessary to calculate the moment of inertia of the element about an axis parallel to the y axis that passes through the element’s centroid, and then determine the moment of inertia of the element about the y axis using the parallel-axis theorem. Integration of this result will yield . See Examples 10.2 and 10.3. Iy Iy Iy = 1x 2 dA Iy Ix = 1y 2 dA dy Ix y = f(x) Fig. 10–4
- 532. 10.3 RADIUS OF GYRATION OF AN AREA 515 10 EXAMPLE 10.1 Determine the moment of inertia for the rectangular area shown in Fig. 10–5 with respect to (a) the centroidal axis, (b) the axis passing through the base of the rectangle, and (c) the pole or axis perpendicular to the plane and passing through the centroid C. SOLUTION (CASE 1) Part (a). The differential element shown in Fig. 10–5 is chosen for integration. Because of its location and orientation, the entire element is at a distance from the axis. Here it is necessary to integrate from to Since then Ans. Part (b). The moment of inertia about an axis passing through the base of the rectangle can be obtained by using the above result of part (a) and applying the parallel-axis theorem, Eq. 10–3. Ans. Part (c). To obtain the polar moment of inertia about point C, we must first obtain which may be found by interchanging the dimensions b and h in the result of part (a), i.e., Using Eq. 10–2, the polar moment of inertia about C is therefore Ans. JC = Ix¿ + Iy¿ = 1 12 bh1h2 + b2 2 Iy¿ = 1 12 hb3 Iy¿, = 1 12 bh3 + bha h 2 b 2 = 1 3 bh3 Ixb = Ix¿ + Ady 2 Ix¿ = 1 12 bh3 Ix¿ = L A yœ2 dA = L h2 -h2 yœ2 1b dy¿2 = b L h2 -h2 yœ2 dyœ dA = b dy¿, y¿ = h2. y¿ = -h2 x¿ y¿ x¿–y¿ z¿ xb x¿ x¿ y¿ y¿ xb C dy¿ b 2 b 2 h 2 h 2 Fig. 10–5
- 533. 516 CHAPTER 10 MOMENTS OF INERTIA 10 x y 200 mm 100 mm y x dy y2 400x (a) (100 – x) x y 200 mm x y 100 mm dx x¿ y2 400x (b) y ~ y –– 2 Fig. 10–6 Determine the moment of inertia for the shaded area shown in Fig. 10–6a about the x axis. SOLUTION I (CASE 1) A differential element of area that is parallel to the x axis, as shown in Fig. 10–6a, is chosen for integration. Since this element has a thickness dy and intersects the curve at the arbitrary point (x, y), its area is Furthermore, the element lies at the same distance y from the x axis. Hence, integrating with respect to y, from to yields y = 200 mm, y = 0 dA = 1100 - x2 dy. EXAMPLE 10.2 Ans. = 1071106 2 mm4 = L 200 mm 0 y2 a100 - y2 400 b dy = L 200 mm 0 a100y2 - y4 400 bdy Ix = L A y2 dA = L 200 mm 0 y2 1100 - x2 dy SOLUTION II (CASE 2) A differential element parallel to the y axis, as shown in Fig. 10–6b, is chosen for integration. It intersects the curve at the arbitrary point (x, y). In this case, all points of the element do not lie at the same distance from the x axis, and therefore the parallel-axis theorem must be used to determine the moment of inertia of the element with respect to this axis. For a rectangle having a base b and height h, the moment of inertia about its centroidal axis has been determined in part (a) of Example 10.1.There it was found that For the differential element shown in Fig. 10–6b, and and thus Since the centroid of the element is from the x axis, the moment of inertia of the element about this axis is (This result can also be concluded from part (b) of Example 10.1.) Integrating with respect to x, from to yields Ans. = 1071106 2 mm4 Ix = L dIx = L 100 mm 0 1 3 y3 dx = L 100 mm 0 1 3 1400x232 dx x = 100 mm, x = 0 dIx = dIx¿ + dA y '2 = 1 12 dx y3 + y dxa y 2 b 2 = 1 3 y3 dx y ' = y2 dIx¿ = 1 12 dx y3 . h = y, b = dx Ix¿ = 1 12 bh3 .
- 534. 10.3 RADIUS OF GYRATION OF AN AREA 517 10 EXAMPLE 10.3 Determine the moment of inertia with respect to the x axis for the circular area shown in Fig. 10–7a. SOLUTION I (CASE 1) Using the differential element shown in Fig. 10–7a, since we have Ans. SOLUTION II (CASE 2) When the differential element shown in Fig. 10–7b is chosen, the centroid for the element happens to lie on the x axis, and since for a rectangle, we have Integrating with respect to x yields Ans. NOTE: By comparison, Solution I requires much less computation. Therefore, if an integral using a particular element appears difficult to evaluate, try solving the problem using an element oriented in the other direction. Ix = L a -a 2 3 1a2 - x2 232 dx = pa4 4 = 2 3 y3 dx dIx = 1 12 dx12y23 Ix¿ = 1 12 bh3 = L a -a y2 A22a2 - y2 B dy = pa4 4 Ix = L A y2 dA = L A y2 12x2 dy dA = 2x dy, x y y x x dy (x, y) (x, y) x2 y2 a2 (a) O a O x y a (x, y) (x, y) dx y y (b) (x, y) ~ ~ x2 y2 a2 Fig. 10–7
- 535. 518 CHAPTER 10 MOMENTS OF INERTIA 10 y x 1 m 1 m y3 x2 y x 1 m 1 m y3 x2 F10–2 y x 1 m 1 m y3 x2 y x 1 m 1 m y3 x2 F10–4 F10–3. Determine the moment of inertia of the shaded area about the y axis. F10–2. Determine the moment of inertia of the shaded area about the x axis. F10–1. Determine the moment of inertia of the shaded area about the x axis. F10–4. Determine the moment of inertia of the shaded area about the y axis. FUNDAMENTAL PROBLEMS F10–1 F10–3
- 536. 10.3 RADIUS OF GYRATION OF AN AREA 519 10 y x 2 m 2 m y 0.25 x3 Probs. 10–1/2 •10–5. Determine the moment of inertia of the area about the axis. 10–6. Determine the moment of inertia of the area about the axis. y x 10–3. Determine the moment of inertia of the area about the axis. *10–4. Determine the moment of inertia of the area about the axis. y x •10–1. Determine the moment of inertia of the area about the axis. 10–2. Determine the moment of inertia of the area about the axis. y x 10–7. Determine the moment of inertia of the area about the axis. *10–8. Determine the moment of inertia of the area about the axis. •10–9. Determine the polar moment of inertia of the area about the axis passing through point . O z y x PROBLEMS y x y2 x3 1 m 1 m Probs. 10–3/4 y x y2 2x 2 m 2 m Probs. 10–5/6 y x O y 2x4 2 m 1 m Probs. 10–7/8/9
- 537. 520 CHAPTER 10 MOMENTS OF INERTIA 10 y x 2 in. 8 in. y x3 Probs. 10–10/11 10–14. Determine the moment of inertia of the area about the x axis. Solve the problem in two ways, using rectangular differential elements: (a) having a thickness of dx, and (b) having a thickness of dy. 10–15. Determine the moment of inertia of the area about the y axis. Solve the problem in two ways, using rectangular differential elements: (a) having a thickness of dx, and (b) having a thickness of dy. *10–12. Determine the moment of inertia of the area about the x axis. •10–13. Determine the moment of inertia of the area about the y axis. 10–10. Determine the moment of inertia of the area about the x axis. 10–11. Determine the moment of inertia of the area about the y axis. x y 1 in. 2 in. y 2 – 2x 3 Probs. 10–12/13 1 in. 1 in. 4 in. y 4 – 4x2 x y Probs. 10–14/15 k *10–16. Determine the moment of inertia of the triangular area about the x axis. •10–17. Determine the moment of inertia of the triangular area about the y axis. y (b x) h –– b y x b h Probs. 10–16/17
- 538. 10.3 RADIUS OF GYRATION OF AN AREA 521 10 10–22. Determine the moment of inertia of the area about the x axis. 10–23. Determine the moment of inertia of the area about the y axis. *10–20. Determine the moment of inertia of the area about the x axis. •10–21. Determine the moment of inertia of the area about the y axis. 10–18. Determine the moment of inertia of the area about the x axis. 10–19. Determine the moment of inertia of the area about the y axis. *10–24. Determine the moment of inertia of the area about the axis. •10–25. Determine the moment of inertia of the area about the axis. 10–26. Determine the polar moment of inertia of the area about the axis passing through point O. z y x x y b h y x2 h — b2 Probs. 10–18/19 y x y3 x 2 in. 8 in. Probs. 10–20/21 y x y 2 cos ( x) –– 8 2 in. 4 in. 4 in. π Probs. 10–22/23 y x x2 y2 r2 r0 0 Probs. 10–24/25/26
- 539. For design or analysis of this Tee beam, engineers must be able to locate the centroid of its cross-sectional area, and then find the moment of inertia of this area about the centroidal axis. 522 CHAPTER 10 MOMENTS OF INERTIA 10 10.4 Moments of Inertia for Composite Areas A composite area consists of a series of connected “simpler” parts or shapes, such as rectangles, triangles, and circles. Provided the moment of inertia of each of these parts is known or can be determined about a common axis, then the moment of inertia for the composite area about this axis equals the algebraic sum of the moments of inertia of all its parts. Procedure for Analysis The moment of inertia for a composite area about a reference axis can be determined using the following procedure. Composite Parts. • Using a sketch, divide the area into its composite parts and indicate the perpendicular distance from the centroid of each part to the reference axis. Parallel-Axis Theorem. • If the centroidal axis for each part does not coincide with the reference axis, the parallel-axis theorem, should be used to determine the moment of inertia of the part about the reference axis. For the calculation of , use the table on the inside back cover. Summation. • The moment of inertia of the entire area about the reference axis is determined by summing the results of its composite parts about this axis. • If a composite part has a “hole,” its moment of inertia is found by “subtracting” the moment of inertia of the hole from the moment of inertia of the entire part including the hole. I I = I + Ad2 ,
- 540. 10.4 MOMENTS OF INERTIA FOR COMPOSITE AREAS 523 10 x 100 mm 75 mm 75 mm 25 mm (a) x 100 mm 75 mm 75 mm 25 mm – (b) Fig. 10–8 EXAMPLE 10.4 Determine the moment of inertia of the area shown in Fig. 10–8a about the x axis. SOLUTION Composite Parts. The area can be obtained by subtracting the circle from the rectangle shown in Fig. 10–8b. The centroid of each area is located in the figure. Parallel-Axis Theorem. The moments of inertia about the x axis are determined using the parallel-axis theorem and the data in the table on the inside back cover. Circle Rectangle Summation. The moment of inertia for the area is therefore Ans. = 1011106 2 mm4 Ix = -11.41106 2 + 112.51106 2 = 1 12 11002115023 + 110021150217522 = 112.51106 2 mm4 Ix = Ixœ + Ady 2 = 1 4 p12524 + p12522 17522 = 11.41106 2 mm4 Ix = Ixœ + Ady 2
- 541. 524 CHAPTER 10 MOMENTS OF INERTIA 10 Determine the moments of inertia for the cross-sectional area of the member shown in Fig. 10–9a about the x and y centroidal axes. SOLUTION Composite Parts. The cross section can be subdivided into the three rectangular areas A, B, and D shown in Fig. 10–9b. For the calculation, the centroid of each of these rectangles is located in the figure. Parallel-Axis Theorem. From the table on the inside back cover, or Example 10.1, the moment of inertia of a rectangle about its centroidal axis is Hence, using the parallel-axis theorem for rectangles A and D, the calculations are as follows: Rectangles A and D Rectangle B Summation. The moments of inertia for the entire cross section are thus Ans. Ans. = 5.601109 2 mm4 Iy = 2[1.901109 2] + 1.801109 2 = 2.901109 2 mm4 Ix = 2[1.4251109 2] + 0.051109 2 Iy = 1 12 11002160023 = 1.801109 2 mm4 Ix = 1 12 16002110023 = 0.051109 2 mm4 = 1.901109 2 mm4 Iy = Iy¿ + Adx 2 = 1 12 13002110023 + 1100213002125022 = 1.4251109 2 mm4 Ix = Ix¿ + Ady 2 = 1 12 11002130023 + 1100213002120022 I = 1 12 bh3 . EXAMPLE 10.5 100 mm 400 mm 100 mm 100 mm 600 mm 400 mm x y (a) C 100 mm 100 mm x y 300 mm 300 mm 200 mm 250 mm 200 mm (b) A B D 250 mm Fig. 10–9
- 542. 10.4 MOMENTS OF INERTIA FOR COMPOSITE AREAS 525 10 FUNDAMENTAL PROBLEMS F10–7. Determine the moment of inertia of the cross- sectional area of the channel with respect to the y axis. F10–6. Determine the moment of inertia of the beam’s cross-sectional area about the centroidal x and y axes. F10–5. Determine the moment of inertia of the beam’s cross-sectional area about the centroidal x and y axes. F10–8. Determine the moment of inertia of the cross- sectional area of the T-beam with respect to the axis passing through the centroid of the cross section. x¿ 300 mm 200 mm 30 mm 30 mm 30 mm 30 mm x y F10–6 x y 50 mm 50 mm 300 mm 50 mm 200 mm F10–7 30 mm 150 mm 150 mm 30 mm y x¿ F10–8 F10–5 200 mm 150 mm 150 mm 200 mm 50 mm 50 mm x y
- 543. 526 CHAPTER 10 MOMENTS OF INERTIA 10 2 in. 4 in. 1 in. 1 in. C x¿ x y y 6 in. Probs. 10–27/28/29 PROBLEMS *10–32. Determine the moment of inertia of the composite area about the axis. •10–33. Determine the moment of inertia of the composite area about the axis. y x 10–30. Determine the moment of inertia of the beam’s cross-sectional area about the axis. 10–31. Determine the moment of inertia of the beam’s cross-sectional area about the axis. y x 10–27. Determine the distance to the centroid of the beam’s cross-sectional area; then find the moment of inertia about the axis. *10–28. Determine the moment of inertia of the beam’s cross-sectional area about the x axis. •10–29. Determine the moment of inertia of the beam’s cross-sectional area about the y axis. x¿ y 10–34. Determine the distance to the centroid of the beam’s cross-sectional area; then determine the moment of inertia about the axis. 10–35. Determine the moment of inertia of the beam’s cross-sectional area about the y axis. x¿ y y x 15 mm 15 mm 60 mm 60 mm 100 mm 100 mm 50 mm 50 mm 15 mm 15 mm Probs. 10–30/31 y x 150 mm 300 mm 150 mm 100 mm 100 mm 75 mm Probs. 10–32/33 x x¿ C y 50 mm 50 mm 75 mm 25 mm 25 mm 75 mm 100 mm _ y 25 mm 25 mm 100 mm Probs. 10–34/35
- 544. 10.4 MOMENTS OF INERTIA FOR COMPOSITE AREAS 527 10 •10–41. Determine the moment of inertia of the beam’s cross-sectional area about the axis. 10–42. Determine the moment of inertia of the beam’s cross-sectional area about the axis. y x 10–38. Determine the distance to the centroid of the beam’s cross-sectional area; then find the moment of inertia about the axis. 10–39. Determine the moment of inertia of the beam’s cross-sectional area about the x axis. *10–40. Determine the moment of inertia of the beam’s cross-sectional area about the y axis. x¿ y *10–36. Locate the centroid of the composite area, then determine the moment of inertia of this area about the centroidal axis. •10–37. Determine the moment of inertia of the composite area about the centroidal axis. y x¿ y 10–43. Locate the centroid of the cross-sectional area for the angle. Then find the moment of inertia about the centroidal axis. *10–44. Locate the centroid of the cross-sectional area for the angle. Then find the moment of inertia about the centroidal axis. y¿ Iy¿ x x¿ Ix¿ y y 1 in. 1 in. 2 in. 3 in. 5 in. x¿ x y 3 in. C Probs. 10–36/37 300 mm 100 mm 200 mm 50 mm 50 mm y C x y x¿ Probs. 10–38/39/40 y 50 mm 50 mm 15 mm 115 mm 115 mm 7.5 mm x 15 mm Probs. 10–41/42 6 in. 2 in. 6 in. x 2 in. C x¿ y¿ y – x – y Probs. 10–43/44
- 545. 528 CHAPTER 10 MOMENTS OF INERTIA 10 •10–49. Determine the moment of inertia of the section.The origin of coordinates is at the centroid C. 10–50. Determine the moment of inertia of the section. The origin of coordinates is at the centroid C. Iy¿ Ix¿ 10–47. Determine the moment of inertia of the composite area about the centroidal axis. *10–48. Locate the centroid of the composite area, then determine the moment of inertia of this area about the axis. x¿ y y •10–45. Determine the moment of inertia of the composite area about the axis. 10–46. Determine the moment of inertia of the composite area about the axis. y x 10–51. Determine the beam’s moment of inertia about the centroidal axis. *10–52. Determine the beam’s moment of inertia about the centroidal axis. y Iy x Ix y x 150 mm 150 mm 150 mm 150 mm Probs. 10–45/46 x x¿ y C 400 mm 240 mm 50 mm 150 mm 150 mm 50 mm 50 mm y Probs. 10–47/48 200 mm 600 mm 20 mm C y¿ x¿ 200 mm 20 mm 20 mm Probs. 10–49/50 y x 50 mm 50 mm 100 mm 15 mm 15 mm 10 mm 100 mm C Probs. 10–51/52
- 546. 10.4 MOMENTS OF INERTIA FOR COMPOSITE AREAS 529 10 •10–57. Determine the moment of inertia of the beam’s cross-sectional area about the axis. 10–58. Determine the moment of inertia of the beam’s cross-sectional area about the axis. y x 10–55. Determine the moment of inertia of the cross- sectional area about the axis. *10–56. Locate the centroid of the beam’s cross- sectional area, and then determine the moment of inertia of the area about the centroidal axis. y¿ x x •10–53. Locate the centroid of the channel’s cross- sectional area, then determine the moment of inertia of the area about the centroidal axis. 10–54. Determine the moment of inertia of the area of the channel about the axis. y x¿ y 10–59. Determine the moment of inertia of the beam’s cross-sectional area with respect to the axis passing through the centroid C of the cross section. . y = 104.3 mm x¿ 6 in. 0.5 in. 0.5 in. 0.5 in. 6.5 in. 6.5 in. y C x¿ x y Probs. 10–53/54 100 mm 10 mm 10 mm 180 mm x y¿ y C 100 mm 10 mm x Probs. 10–55/56 y 100 mm 12 mm 125 mm 75 mm 12 mm 75 mm x 12 mm 25 mm 125 mm 12 mm Probs. 10–57/58 x¿ C A B – y 150 mm 15 mm 35 mm 50 mm Prob. 10–59
- 547. 530 CHAPTER 10 MOMENTS OF INERTIA 10 *10.5 Product of Inertia for an Area It will be shown in the next section that the property of an area, called the product of inertia, is required in order to determine the maximum and minimum moments of inertia for the area.These maximum and minimum values are important properties needed for designing structural and mechanical members such as beams, columns, and shafts. The product of inertia of the area in Fig. 10–10 with respect to the and axes is defined as (10–7) If the element of area chosen has a differential size in two directions, as shown in Fig. 10–10, a double integration must be performed to evaluate Most often, however, it is easier to choose an element having a differential size or thickness in only one direction in which case the evaluation requires only a single integration (see Example 10.6). Like the moment of inertia, the product of inertia has units of length raised to the fourth power, e.g., or However, since x or y may be negative, the product of inertia may either be positive, negative, or zero, depending on the location and orientation of the coordinate axes. For example, the product of inertia for an area will be zero if either the x or y axis is an axis of symmetry for the area, as in Fig. 10–11. Here every element dA located at point (x, y) has a corresponding element dA located at (x, ). Since the products of inertia for these elements are, respectively, xy dA and the algebraic sum or integration of all the elements that are chosen in this way will cancel each other. Consequently, the product of inertia for the total area becomes zero. It also follows from the definition of that the “sign” of this quantity depends on the quadrant where the area is located. As shown in Fig. 10–12, if the area is rotated from one quadrant to another, the sign of will change. Ixy Ixy -xy dA, -y Ixy in4 . ft4 , mm4 m4 , Ixy. Ixy = L A xy dA y x x y x y A dA Fig. 10–10 x y x y y dA dA Fig. 10–11 The effectiveness of this beam to resist bending can be determined once its moments of inertia and its product of inertia are known.
- 548. 10.5 PRODUCT OF INERTIA FOR AN AREA 531 10 Parallel-Axis Theorem. Consider the shaded area shown in Fig. 10–13, where and represent a set of axes passing through the centroid of the area, and x and y represent a corresponding set of parallel axes. Since the product of inertia of dA with respect to the x and y axes is then for the entire area, The first term on the right represents the product of inertia for the area with respect to the centroidal axes, The integrals in the second and third terms are zero since the moments of the area are taken about the centroidal axis. Realizing that the fourth integral represents the entire area A, the parallel-axis theorem for the product of inertia becomes (10–8) It is important that the algebraic signs for and be maintained when applying this equation. dy dx Ixy = Ix¿y¿ + Adxdy Ix¿y¿. = L A x¿y¿ dA + dx L A y¿ dA + dy L A x¿ dA + dxdy L A dA Ixy = L A 1x¿ + dx21y¿ + dy2 dA dIxy = 1x¿ + dx21y¿ + dy2 dA, y¿ x¿ x y y y x x x y y x Ixy xy dA Ixy xy dA Ixy xy dA Ixy xy dA Fig. 10–12 x y x¿ y¿ dx dy C dA y¿ x¿ Fig. 10–13
- 549. 532 CHAPTER 10 MOMENTS OF INERTIA 10 x y h b (a) x y h b (x, y) dx y (b) (x, y) ~ ~ y x h b x y h b (x, y) dy y (b x) x (c) (x, y) ~ ~ y x h b Fig. 10–14 Determine the product of inertia for the triangle shown in Fig. 10–14a. SOLUTION I A differential element that has a thickness dx, as shown in Fig. 10–14b, has an area The product of inertia of this element with respect to the x and y axes is determined using the parallel-axis theorem. where and locate the centroid of the element or the origin of the axes. (See Fig. 10–13.) Since due to symmetry, and then Integrating with respect to x from to yields Ans. SOLUTION II The differential element that has a thickness dy,as shown in Fig.10–14c, can also be used. Its area is . The centroid is located at point so the product of inertia of the element becomes Integrating with respect to y from to yields Ans. Ixy = 1 2 L h 0 yab2 - b2 h2 y2 b dy = b2 h2 8 y = h y = 0 = ab - b h yb dyc b + 1bh2y 2 dy = 1 2 yab2 - b2 h2 y2 b dy = 0 + 1b - x2 dya b + x 2 by dIxy = dIx¿y¿ + dA x ' y ' y ' = y, x ' = x + 1b - x22 = 1b + x22, dA = 1b - x2 dy Ixy = h2 2b2 L b 0 x3 dx = b2 h2 8 x = b x = 0 = h2 2b2 x3 dx dIxy = 0 + 1y dx2xa y 2 b = a h b x dxbxa h 2b xb y ' = y2, x ' = x, dIx¿y¿ = 0, y¿ x¿, y ' x ' dIxy = dIx¿y¿ + dA x ' y ' dA = y dx. Ixy EXAMPLE 10.6
- 550. 10.5 PRODUCT OF INERTIA FOR AN AREA 533 10 100 mm 400 mm 100 mm 100 mm 600 mm 400 mm x y (a) C 100 mm 100 mm x y 300 mm 300 mm 200 mm 250 mm 200 mm (b) A B D 250 mm Fig. 10–15 EXAMPLE 10.7 Determine the product of inertia for the cross-sectional area of the member shown in Fig. 10–15a, about the x and y centroidal axes. SOLUTION As in Example 10.5, the cross section can be subdivided into three composite rectangular areas A, B, and D, Fig. 10–15b.The coordinates for the centroid of each of these rectangles are shown in the figure. Due to symmetry, the product of inertia of each rectangle is zero about a set of axes that passes through the centroid of each rectangle. Using the parallel-axis theorem, we have Rectangle A Rectangle B Rectangle D The product of inertia for the entire cross section is therefore Ans. NOTE: This negative result is due to the fact that rectangles A and D have centroids located with negative x and negative y coordinates, respectively. Ixy = -1.501109 2 + 0 - 1.501109 2 = -3.001109 2 mm4 = 0 + 1300211002125021-2002 = -1.501109 2 mm4 Ixy = Ix¿y¿ + Adxdy = 0 + 0 = 0 Ixy = Ix¿y¿ + Adxdy = 0 + 13002110021-250212002 = -1.501109 2 mm4 Ixy = Ix¿y¿ + Adxdy y¿ x¿,
- 551. 534 CHAPTER 10 MOMENTS OF INERTIA 10 *10.6 Moments of Inertia for an Area about Inclined Axes In structural and mechanical design, it is sometimes necessary to calculate the moments and product of inertia and for an area with respect to a set of inclined u and axes when the values for and are known.To do this we will use transformation equations which relate the x, y and u, coordinates. From Fig. 10–16, these equations are With these equations, the moments and product of inertia of dA about the u and axes become Expanding each expression and integrating, realizing that and we obtain Using the trigonometric identities and we can simplify the above expressions, in which case (10–9) Notice that if the first and second equations are added together, we can show that the polar moment of inertia about the z axis passing through point O is, as expected, independent of the orientation of the u and axes; i.e., JO = Iu + Iv = Ix + Iy v Iu = Ix + Iy 2 + Ix - Iy 2 cos 2u - Ixy sin 2u Iv = Ix + Iy 2 - Ix - Iy 2 cos 2u + Ixy sin 2u Iuv = Ix - Iy 2 sin 2u + Ixy cos 2u = cos2 u - sin2 u cos 2 u sin 2 u = 2 sin u cos u Iuv = Ix sin u cos u - Iy sin u cos u + Ixy1cos2 u - sin2 u2 Iv = Ix sin2 u + Iy cos2 u + 2Ixy sin u cos u Iu = Ix cos2 u + Iy sin2 u - 2Ixy sin u cos u Ixy = 1xy dA, Iy = 1x2 dA, Ix = 1y2 dA, dIuv = uv dA = 1x cos u + y sin u21y cos u - x sin u2 dA dIv = u2 dA = 1x cos u + y sin u22 dA dIu = v2 dA = 1y cos u - x sin u22 dA v v = y cos u - x sin u u = x cos u + y sin u v Ixy Iy, Ix, u, v Iuv Iv, Iu, x y O u v A dA x y x cos u u v y cos u y sin u x sin u u u u Fig. 10–16
- 552. 10.6 MOMENTS OF INERTIA FOR AN AREA ABOUT INCLINED AXES 535 10 Principal Moments of Inertia. Equations 10–9 show that and depend on the angle of inclination, of the u, axes. We will now determine the orientation of these axes about which the moments of inertia for the area are maximum and minimum.This particular set of axes is called the principal axes of the area, and the corresponding moments of inertia with respect to these axes are called the principal moments of inertia. In general, there is a set of principal axes for every chosen origin O. However, for structural and mechanical design, the origin O is located at the centroid of the area. The angle which defines the orientation of the principal axes can be found by differentiating the first of Eqs. 10–9 with respect to and setting the result equal to zero.Thus, Therefore, at (10–10) The two roots and of this equation are 90° apart, and so they each specify the inclination of one of the principal axes. In order to substitute them into Eq. 10–9, we must first find the sine and cosine of and This can be done using these ratios from the triangles shown in Fig. 10–17, which are based on Eq. 10–10. Substituting each of the sine and cosine ratios into the first or second of Eqs. 10–9 and simplifying, we obtain (10–11) Depending on the sign chosen, this result gives the maximum or minimum moment of inertia for the area. Furthermore, if the above trigonometric relations for and are substituted into the third of Eqs. 10–9, it can be shown that that is, the product of inertia with respect to the principal axes is zero. Since it was indicated in Sec. 10.6 that the product of inertia is zero with respect to any symmetrical axis, it therefore follows that any symmetrical axis represents a principal axis of inertia for the area. Iuv = 0; up2 up1 Imax min = Ix + Iy 2 ; C a Ix - Iy 2 b 2 + Ixy 2 2up2 . 2up1 up2 up1 tan 2up = -Ixy 1Ix - Iy22 u = up, dIu du = -2a Ix - Iy 2 b sin 2u - 2Ixy cos 2u = 0 u v u, Iuv Iv, Iu, 2up2 2up1 Ix Iy 2 ( ) Ix Iy 2 ( ) Ixy Ixy Ix Iy 2 ( ) 2 I2 xy Fig. 10–17
- 553. 536 CHAPTER 10 MOMENTS OF INERTIA 10 Determine the principal moments of inertia and the orientation of the principal axes for the cross-sectional area of the member shown in Fig. 10–18a with respect to an axis passing through the centroid. SOLUTION The moments and product of inertia of the cross section with respect to the x, y axes have been determined in Examples 10.5 and 10.7. The results are Using Eq. 10–10, the angles of inclination of the principal axes u and are Thus, by inspection of Fig. 10–18b, Ans. The principal moments of inertia with respect to these axes are determined from Eq. 10–11. Hence, or Ans. NOTE: The maximum moment of inertia, occurs with respect to the u axis since by inspection most of the cross- sectional area is farthest away from this axis. Or, stated in another manner, occurs about the u axis since this axis is located within of the y axis, which has the larger value of Also, this can be concluded by substituting the data with into the first of Eqs. 10–9 and solving for . Iu u = 57.1° I 1Iy 7 Ix2. ;45° Imax Imax = 7.541109 2 mm4 , Imax = 7.541109 2 mm4 Imin = 0.9601109 2 mm4 Imax min = 4.251109 2 ; 3.291109 2 ; C c 2.901109 2 - 5.601109 2 2 d 2 + [-3.001109 2]2 = 2.901109 2 + 5.601109 2 2 Imax min = Ix + Iy 2 ; C a Ix - Iy 2 b 2 + Ixy 2 up2 = -32.9° and up1 = 57.1° 2up = -65.8° and 114.2° tan 2up = -Ixy 1Ix - Iy22 = -[-3.001109 2] [2.901109 2 - 5.601109 2]2 = -2.22 v Ix = 2.901109 2 mm4 Iy = 5.601109 2 mm4 Ixy = -3.001109 2 mm4 EXAMPLE 10.8 100 mm 400 mm 100 mm 100 mm 600 mm 400 mm x y (a) C x y (b) C u v up1 57.1 up2 32.9 Fig. 10–18
- 554. 10.7 MOHR’S CIRCLE FOR MOMENTS OF INERTIA 537 10 *10.7 Mohr’s Circle for Moments of Inertia Equations 10–9 to 10–11 have a graphical solution that is convenient to use and generally easy to remember. Squaring the first and third of Eqs. 10–9 and adding, it is found that Here and are known constants. Thus, the above equation may be written in compact form as When this equation is plotted on a set of axes that represent the respective moment of inertia and the product of inertia, as shown in Fig. 10–19, the resulting graph represents a circle of radius and having its center located at point (a, 0), where The circle so constructed is called Mohr’s circle, named after the German engineer Otto Mohr (1835–1918). a = 1Ix + Iy22. R = C a Ix - Iy 2 b 2 + Ixy 2 1Iu - a22 + Iuv 2 = R2 Ixy Iy, Ix, aIu - Ix + Iy 2 b 2 + Iuv 2 = a Ix - Iy 2 b 2 + Ixy 2 x y u v up1 Axis for minor principal moment of inertia, Imin Axis for major principal moment of inertia, Imax (a) P I O Imax Imin A (b) 2up 1 Ixy Ixy Ix R Ix Iy 2 2 I2 xy Ix Iy 2 Ix Iy 2 Fig. 10–19
- 555. 538 CHAPTER 10 MOMENTS OF INERTIA 10 Using trigonometry, the above procedure can be verified to be in accordance with the equations developed in Sec. 10.6. Procedure for Analysis The main purpose in using Mohr’s circle here is to have a convenient means for finding the principal moments of inertia for an area.The following procedure provides a method for doing this. Determine . • Establish the x, y axes and determine and Fig. 10–19a. Construct the Circle. • Construct a rectangular coordinate system such that the abscissa represents the moment of inertia I, and the ordinate represents the product of inertia Fig. 10–19b. • Determine the center of the circle, O, which is located at a distance from the origin, and plot the reference point A having coordinates ( ). Remember, is always positive, whereas can be either positive or negative. • Connect the reference point A with the center of the circle and determine the distance OA by trigonometry. This distance represents the radius of the circle, Fig. 10–19b. Finally, draw the circle. Principal Moments of Inertia. • The points where the circle intersects the axis give the values of the principal moments of inertia and Notice that, as expected, the product of inertia will be zero at these points, Fig. 10–19b. Principal Axes. • To find the orientation of the major principal axis, use trigonometry to find the angle measured from the radius OA to the positive I axis, Fig. 10–19b. This angle represents twice the angle from the x axis to the axis of maximum moment of inertia Fig. 10–19a. Both the angle on the circle, and the angle must be measured in the same sense, as shown in Fig. 10–19. The axis for minimum moment of inertia is perpendicular to the axis for Imax. Imin up1 , 2up1 , Imax, 2up1 , Imax. Imin I Ixy Ix Ixy Ix, 1Ix + Iy22 Ixy, Ixy, Iy, Ix, Ix , Iy , and Ixy x y u v up1 Axis for minor principal moment of inertia, Imin Axis for major principal moment of inertia, Imax (a) P I O Imax Imin A (b) 2up 1 Ixy Ixy Ix R Ix Iy 2 2 I2 xy Ix Iy 2 Ix Iy 2 Fig. 10–19
- 556. 10.7 MOHR’S CIRCLE FOR MOMENTS OF INERTIA 539 10 EXAMPLE 10.9 Using Mohr’s circle, determine the principal moments of inertia and the orientation of the major principal axes for the cross-sectional area of the member shown in Fig. 10–20a, with respect to an axis passing through the centroid. Ixy (109 ) mm4 I (109 ) mm4 O (b) 4.25 2.90 1.35 3.00 A (2.90, 3.00) B O (c) A (2.90, 3.00) Imin 0.960 Imax 7.54 2up1 3.29 Ixy (109 ) mm4 I (109 ) mm4 x y C u v up1 57.1 (d) Fig. 10–20 SOLUTION Determine . The moments and product of inertia have been determined in Examples 10.5 and 10.7 with respect to the x, y axes shown in Fig. 10–20a. The results are and Construct the Circle. The I and axes are shown in Fig. 10–20b.The center of the circle,O,lies at a distance from the origin.When the reference point or is connected to point O, the radius OA is determined from the triangle OBA using the Pythagorean theorem. The circle is constructed in Fig. 10–20c. Principal Moments of Inertia. The circle intersects the I axis at points (7.54, 0) and (0.960, 0). Hence, Ans. Ans. Principal Axes. As shown in Fig. 10–20c, the angle is determined from the circle by measuring counterclockwise from OA to the direction of the positive I axis. Hence, The principal axis for is therefore oriented at an angle measured counterclockwise, from the positive x axis to the positive u axis. The axis is perpendicular to this axis. The results are shown in Fig. 10–20d. v up1 = 57.1°, Imax = 7.541109 2 mm4 2up1 = 180° - sin-1 a ƒBA ƒ ƒOAƒ b = 180° - sin-1 a 3.00 3.29 b = 114.2° 2up1 Imin = (4.25 - 3.29)109 = 0.9601109 2 mm4 Imax = (4.25 + 3.29)109 = 7.541109 2 mm4 OA = 2(1.3522 + (-3.0022 = 3.29 A(2.90,-3.00) A(Ix, Ixy) 1Ix+Iy22 = 12.90 + 5.6022 = 4.25 Ixy Ixy = -3.001109 2 mm4 . Iy = 5.601109 2 mm4 , Ix = 2.901109 2 mm4 , Ix, Iy, Ixy 100 mm 400 mm 100 mm 100 mm 600 mm 400 mm x y (a) C
- 557. y a b x 1 x 2 –– a 2 y 2 –– b 2 Prob. 10–62 y x 2 m 3 m 8y x3 2x2 4x Prob. 10–65 y x y 2x2 2 in. 1 in. 540 CHAPTER 10 MOMENTS OF INERTIA 10 *10–64. Determine the product of inertia of the area with respect to the and axes. y x 10–62. Determine the product of inertia of the quarter elliptical area with respect to the and axes. y x *10–60. Determine the product of inertia of the parabolic area with respect to the x and y axes. •10–61. Determine the product of inertia of the right half of the parabolic area in Prob. 10–60, bounded by the lines . and . x = 0 y = 2 in Ixy •10–65. Determine the product of inertia of the area with respect to the and axes. y x PROBLEMS Probs. 10–60/61 y x 8 in. 2 in. y3 x Prob. 10–63 y x y x –– 4 4 in. 4 in. (x 8) Prob. 10–64 y x 2 m 1 m y2 1 0.5x Prob. 10–66 10–63. Determine the product of inertia for the area with respect to the x and y axes. 10–66. Determine the product of inertia for the area with respect to the x and y axes.
- 558. 10.7 MOHR’S CIRCLE FOR MOMENTS OF INERTIA 541 10 10–70. Determine the product of inertia of the composite area with respect to the and axes. y x *10–68. Determine the product of inertia for the area of the ellipse with respect to the x and y axes. 10–67. Determine the product of inertia for the area with respect to the x and y axes. 10–71. Determine the product of inertia of the cross- sectional area with respect to the x and y axes that have their origin located at the centroid C. •10–69. Determine the product of inertia for the parabolic area with respect to the x and y axes. y x y3 x b h3 h b Prob. 10–67 y x 4 in. 2 in. x2 4y2 16 Prob. 10–68 y 4 in. 2 in. x y2 x Prob. 10–69 1.5 in. y x 2 in. 2 in. 2 in. 2 in. Prob. 10–70 4 in. 4 in. x y 5 in. 1 in. 1 in. 3.5 in. 0.5 in. C Prob. 10–71
- 559. 542 CHAPTER 10 MOMENTS OF INERTIA 10 10–74. Determine the product of inertia for the beam’s cross-sectional area with respect to the x and y axes that have their origin located at the centroid C. *10–72. Determine the product of inertia for the beam’s cross-sectional area with respect to the x and y axes that have their origin located at the centroid C. x y 5 mm 30 mm 5 mm 50 mm 7.5 mm C 17.5 mm Prob. 10–72 x y 300 mm 100 mm 10 mm 10 mm 10 mm Prob. 10–73 1 in. 5 in. 5 in. 5 in. 1 in. C 5 in. x y 1 in. 0.5 in. Prob. 10–74 y x u x 200 mm 200 mm 175 mm 20 mm 20 mm 20 mm C 60 v Prob. 10–75 •10–73. Determine the product of inertia of the beam’s cross-sectional area with respect to the x and y axes. 10–75. Locate the centroid of the beam’s cross-sectional area and then determine the moments of inertia and the product of inertia of this area with respect to the and axes.The axes have their origin at the centroid C. v u x
- 560. 10.7 MOHR’S CIRCLE FOR MOMENTS OF INERTIA 543 10 10–78. Determine the moments of inertia and the product of inertia of the beam’s cross-sectional area with respect to the and axes. v u •10–77. Determine the product of inertia of the beam’s cross-sectional area with respect to the centroidal and axes. y x *10–76. Locate the centroid ( , ) of the beam’s cross- sectional area, and then determine the product of inertia of this area with respect to the centroidal and axes. y¿ x¿ y x 10–79. Locate the centroid of the beam’s cross-sectional area and then determine the moments of inertia and the product of inertia of this area with respect to the and axes. v u y x¿ y¿ x y 300 mm 200 mm 10 mm 10 mm C y x 10 mm 100 mm Prob. 10–76 x C 150 mm 100 mm 100 mm 10 mm 10 mm 10 mm y 150 mm 5 mm Prob. 10–77 3 in. 1.5 in. 3 in. y u x 1.5 in. C v 30 Prob. 10–78 y x u 8 in. 4 in. 0.5 in. 0.5 in. 4.5 in. 0.5 in. y 4.5 in. C v 60 Prob. 10–79
- 561. 544 CHAPTER 10 MOMENTS OF INERTIA 10 10–82. Locate the centroid of the beam’s cross-sectional area and then determine the moments of inertia of this area and the product of inertia with respect to the and axes. The axes have their origin at the centroid C. v u y •10–81. Determine the orientation of the principal axes, which have their origin at centroid C of the beam’s cross- sectional area.Also, find the principal moments of inertia. *10–80. Locate the centroid and of the cross-sectional area and then determine the orientation of the principal axes, which have their origin at the centroid C of the area. Also, find the principal moments of inertia. y x 10–83. Solve Prob. 10–75 using Mohr’s circle. *10–84. Solve Prob. 10–78 using Mohr’s circle. •10–85. Solve Prob. 10–79 using Mohr’s circle. 10–86. Solve Prob. 10–80 using Mohr’s circle. 10–87. Solve Prob. 10–81 using Mohr’s circle. *10–88. Solve Prob. 10–82 using Mohr’s circle. y x 6 in. 0.5 in. 6 in. y x 0.5 in. C Prob. 10–80 y C x 100 mm 100 mm 20 mm 20 mm 20 mm 150 mm 150 mm Prob. 10–81 200 mm 25 mm y u C x y 60 75 mm 75 mm 25 mm 25 mm v Prob. 10–82
- 562. 10.8 MASS MOMENT OF INERTIA 545 10 10.8 Mass Moment of Inertia The mass moment of inertia of a body is a measure of the body’s resistance to angular acceleration. Since it is used in dynamics to study rotational motion, methods for its calculation will now be discussed.* Consider the rigid body shown in Fig. 10–21. We define the mass moment of inertia of the body about the z axis as (10–12) Here r is the perpendicular distance from the axis to the arbitrary element dm. Since the formulation involves r, the value of I is unique for each axis about which it is computed.The axis which is generally chosen, however, passes through the body’s mass center G. Common units used for its measurement are or If the body consists of material having a density , then Fig. 10–22a. Substituting this into Eq. 10–12, the body’s moment of inertia is then computed using volume elements for integration; i.e. (10–13) For most applications, will be a constant, and so this term may be factored out of the integral, and the integration is then purely a function of geometry. (10–14) I = r L V r2 dV r I = L V r2 r dV dm = r dV , r slug # ft2 . kg # m2 I = L m r2 dm *Another property of the body which measures the symmetry of the body’s mass with respect to a coordinate system is the mass product of inertia. This property most often applies to the three-dimensional motion of a body and is discussed in Engineering Mechanics: Dynamics (Chapter 21). r dm z Fig. 10–21 z y x dm rdV (x, y, z) (a) Fig. 10–22
- 563. 546 CHAPTER 10 MOMENTS OF INERTIA 10 y z (x, y) (b) z x y dy (c) z y x z (x,y) dz y Fig. 10–22 Procedure for Analysis If a body is symmetrical with respect to an axis, as in Fig. 10–22, then its mass moment of inertia about the axis can be determined by using a single integration. Shell and disk elements are used for this purpose. Shell Element. • If a shell element having a height z, radius y, and thickness dy is chosen for integration, Fig. 10–22b, then its volume is • This element can be used in Eq. 10–13 or 10–14 for determining the moment of inertia of the body about the z axis since the entire element, due to its “thinness,” lies at the same perpendicular distance from the z axis (see Example 10.10). Disk Element. • If a disk element having a radius y and a thickness dz is chosen for integration, Fig. 10–22c, then its volume is • In this case the element is finite in the radial direction, and consequently its points do not all lie at the same radial distance r from the z axis.As a result, Eqs. 10–13 or 10–14 cannot be used to determine Instead, to perform the integration using this element, it is first necessary to determine the moment of inertia of the element about the z axis and then integrate this result (see Example 10.11). Iz. dV = 1py2 2 dz. r = y Iz dV = 12py21z2 dy.
- 564. 10.8 MASS MOMENT OF INERTIA 547 10 EXAMPLE 10.10 Determine the mass moment of inertia of the cylinder shown in Fig. 10–23a about the z axis.The density of the material, is constant. r, y z x R O (a) h 2 h 2 z r dr y x (b) O h 2 h 2 Fig. 10–23 SOLUTION Shell Element. This problem will be solved using the shell element in Fig. 10–23b and thus only a single integration is required. The volume of the element is and so its mass is Since the entire element lies at the same distance r from the z axis, the moment of inertia of the element is Integrating over the entire cylinder yields Since the mass of the cylinder is then Ans. Iz = 1 2 mR2 m = L m dm = r2ph L R 0 r dr = rphR2 Iz = L m r2 dm = r2ph L R 0 r3 dr = rp 2 R4 h dIz = r2 dm = r2phr3 dr dm = r dV = r12phr dr2. dV = 12pr21h2 dr,
- 565. 548 CHAPTER 10 MOMENTS OF INERTIA 10 A solid is formed by revolving the shaded area shown in Fig. 10–24a about the y axis. If the density of the material is determine the mass moment of inertia about the y axis. 5 slugft3 , EXAMPLE 10.11 SOLUTION Disk Element. The moment of inertia will be determined using this disk element, as shown in Fig. 10–24b. Here the element intersects the curve at the arbitrary point (x, y) and has a mass Although all points on the element are not located at the same distance from the y axis, it is still possible to determine the moment of inertia of the element about the y axis. In the previous example it was shown that the moment of inertia of a homogeneous cylinder about its longitudinal axis is where m and R are the mass and radius of the cylinder. Since the height of the cylinder is not involved in this formula, we can also use this result for a disk.Thus, for the disk element in Fig. 10–24b, we have Substituting and integrating with respect to y, from to yields the moment of inertia for the entire solid. Ans. Iy = 5p 2 L 1ft 0 x4 dy = 5p 2 L 1ft 0 y8 dy = 0.873 slug # ft2 y = 1 ft, y = 0 r = 5 slugft3 , x = y2 , dIy = 1 2 1dm2x2 = 1 2 [r1px2 2 dy]x2 I = 1 2 mR2 , dIy dm = r dV = r1px2 2 dy x 1 ft y2 x (a) y 1 ft y 1 ft x 1 ft y dy (x, y) (b) Fig. 10–24
- 566. 10.8 MASS MOMENT OF INERTIA 549 10 Parallel-Axis Theorem. If the moment of inertia of the body about an axis passing through the body’s mass center is known, then the moment of inertia about any other parallel axis can be determined by using the parallel-axis theorem.To derive this theorem, consider the body shown in Fig. 10–25. The axis passes through the mass center G, whereas the corresponding parallel z axis lies at a constant distance d away. Selecting the differential element of mass dm, which is located at point ( ), and using the Pythagorean theorem, the moment of inertia of the body about the z axis is Since the first integral represents The second integral is equal to zero, since the axis passes through the body’s mass center, i.e., since Finally, the third integral is the total mass m of the body. Hence, the moment of inertia about the z axis becomes (10–15) where I = IG + md2 x = 0. 1x¿ dm = x1dm = 0 z¿ IG. r¿2 = x¿2 + y¿2 , = L m 1x¿2 + y¿2 2 dm + 2d L m x¿ dm + d2 L m dm I = L m r2 dm = L m [1d + x¿22 + y¿2 ] dm r2 = 1d + x¿22 + y¿2 , y¿ x¿, z¿ IG = moment of inertia about the axis passing through the mass center G z¿ m = mass of the body d = distance between the parallel axes y¿ x¿ z z¿ y¿ r¿ x¿ d r dm A G Fig. 10–25
- 567. 550 CHAPTER 10 MOMENTS OF INERTIA 10 Radius of Gyration. Occasionally, the moment of inertia of a body about a specified axis is reported in handbooks using the radius of gyration, k. This value has units of length, and when it and the body’s mass m are known, the moment of inertia can be determined from the equation (10–16) Note the similarity between the definition of k in this formula and r in the equation which defines the moment of inertia of a differential element of mass dm of the body about an axis. Composite Bodies. If a body is constructed from a number of simple shapes such as disks, spheres, and rods, the moment of inertia of the body about any axis z can be determined by adding algebraically the moments of inertia of all the composite shapes computed about the same axis. Algebraic addition is necessary since a composite part must be considered as a negative quantity if it has already been included within another part—as in the case of a “hole” subtracted from a solid plate. Also, the parallel-axis theorem is needed for the calculations if the center of mass of each composite part does not lie on the z axis. In this regard, formulas for the mass moment of inertia of some common shapes, such as disks, spheres, and rods, are given in the table on the inside back cover. dI = r2 dm, I = mk2 or k = A I m This flywheel, which operates a metal cutter,has a large moment of inertia about its center. Once it begins rotating it is difficult to stop it and therefore a uniform motion can be effectively transferred to the cutting blade.
- 568. 10.8 MASS MOMENT OF INERTIA 551 10 O 0.25 m 0.125 m G (a) Thickness 0.01 m 0.25 m G G – 0.125 m (b) Fig. 10–26 EXAMPLE 10.12 If the plate shown in Fig. 10–26a has a density of and a thickness of 10 mm, determine its mass moment of inertia about an axis perpendicular to the page and passing through the pin at O. 8000 kgm3 SOLUTION The plate consists of two composite parts, the 250-mm-radius disk minus a 125-mm-radius disk, Fig. 10–26b.The moment of inertia about O can be determined by finding the moment of inertia of each of these parts about O and then algebraically adding the results. The computations are performed by using the parallel-axis theorem in conjunction with the data listed in the table on the inside back cover. Disk. The moment of inertia of a disk about an axis perpendicular to the plane of the disk and passing through G is The mass center of both disks is 0.25 m from point O.Thus, Hole. For the smaller disk (hole), we have The moment of inertia of the plate about the pin is therefore Ans. = 1.20 kg # m2 = 1.473 kg # m2 - 0.276 kg # m2 IO = 1IO2d - 1IO2h = 0.276 kg # m2 = 1 213.93 kg210.125 m22 + 13.93 kg210.25 m22 1IO2h = 1 2 mhrh 2 + mhd2 mh = rhV h = 8000 kgm3 [p10.125 m22 10.01 m2] = 3.93 kg = 1.473 kg # m2 = 1 2115.71 kg210.25 m22 + 115.71 kg210.25 m22 1IO2d = 1 2 mdrd 2 + mdd2 md = rdV d = 8000 kgm3 [p10.25 m22 10.01 m2] = 15.71 kg IG = 1 2 mr2 .
- 569. 552 CHAPTER 10 MOMENTS OF INERTIA 10 The pendulum in Fig. 10–27 consists of two thin rods each having a weight of 10 lb. Determine the pendulum’s mass moment of inertia about an axis passing through (a) the pin at O, and (b) the mass center G of the pendulum. SOLUTION Part (a). Using the table on the inside back cover, the moment of inertia of rod OA about an axis perpendicular to the page and passing through the end point O of the rod is Hence, Realize that this same value may be computed using and the parallel-axis theorem; i.e., For rod BC we have The moment of inertia of the pendulum about O is therefore Ans. Part (b). The mass center G will be located relative to the pin at O. Assuming this distance to be Fig. 10–27, and using the formula for determining the mass center, we have The moment of inertia may be computed in the same manner as which requires successive applications of the parallel-axis theorem in order to transfer the moments of inertia of rods OA and BC to G.A more direct solution, however, involves applying the parallel-axis theorem using the result for determined above; i.e., Ans. IG = 0.362 slug # ft2 1.76 slug # ft2 = IG + a 20 lb 32.2 fts2 b11.50 ft22 IO = IG + md2 ; IO IO, IG y = ©y ' m ©m = 111032.22 + 211032.22 11032.22 + 11032.22 = 1.50 ft y, IO = 0.414 + 1.346 = 1.76 slug # ft2 = 1.346 slug # ft2 1IBC2O = 1 12 ml2 + md2 = 1 12 a 10 lb 32.2 fts2 b12 ft22 + 10 lb 32.2 fts2 12 ft22 = 0.414 slug # ft2 1IOA2O = 1 12 ml2 + md2 = 1 12 a 10 lb 32.2 fts2 b12 ft22 + 10 lb 32.2 fts2 11 ft22 IG = 1 12 ml2 1IOA2O = 1 3 ml2 = 1 3 a 10 lb 32.2 fts2 b12 ft22 = 0.414 slug # ft2 IO = 1 3 ml2 . EXAMPLE 10.13 2 ft y – O G A B C 1 ft 1 ft Fig. 10–27
- 570. x y l z Prob. 10–91 10.8 MASS MOMENT OF INERTIA 553 10 PROBLEMS 10–91. Determine the mass moment of inertia of the slender rod. The rod is made of material having a variable density , where is constant. The cross- sectional area of the rod is . Express the result in terms of the mass m of the rod. A r0 r = r0(1 + xl) Iy 10–90. Determine the mass moment of inertia of the right circular cone and express the result in terms of the total mass m of the cone.The cone has a constant density . r Ix •10–89. Determine the mass moment of inertia of the cone formed by revolving the shaded area around the axis. The density of the material is . Express the result in terms of the mass of the cone. m r z Iz •10–93. The paraboloid is formed by revolving the shaded area around the x axis. Determine the radius of gyration . The density of the material is . r = 5 Mgm3 kx z z (r0 y) h –– y h x r0 r0 Prob. 10–89 h y x r r – h x y Prob. 10–90 z y2 x y z 1 4 2 m 1 m Prob. 10–92 *10–92. Determine the mass moment of inertia of the solid formed by revolving the shaded area around the axis. The density of the material is . Express the result in terms of the mass of the solid. m r y Iy y x 100 mm y2 50 x 200 mm Prob. 10–93
- 571. 554 CHAPTER 10 MOMENTS OF INERTIA 10 y a b z x 1 y2 –– a 2 z 2 –– b 2 Prob. 10–94 y x 2b b – a x b y a b Prob. 10–95 y3 9x 3 in. x 3 in. y Prob. 10–96 2 m 4 m z2 8y z y x Prob. 10–97 *10–96. The solid is formed by revolving the shaded area around the y axis. Determine the radius of gyration The specific weight of the material is g = 380 lbft3 . ky. 10–95. The frustum is formed by rotating the shaded area around the x axis. Determine the moment of inertia and express the result in terms of the total mass m of the frustum.The material has a constant density . r Ix 10–94. Determine the mass moment of inertia of the solid formed by revolving the shaded area around the axis. The density of the material is . Express the result in terms of the mass of the semi-ellipsoid. m r y Iy •10–97. Determine the mass moment of inertia of the solid formed by revolving the shaded area around the axis. The density of the material is . r = 7.85 Mgm3 z Iz
- 572. 10.8 MASS MOMENT OF INERTIA 555 10 *10–100. Determine the mass moment of inertia of the pendulum about an axis perpendicular to the page and passing through point O.The slender rod has a mass of 10 kg and the sphere has a mass of 15 kg. 10–99. Determine the mass moment of inertia of the solid formed by revolving the shaded area around the axis. The total mass of the solid is . 1500 kg y Iy 10–98. Determine the mass moment of inertia of the solid formed by revolving the shaded area around the axis. The solid is made of a homogeneous material that weighs 400 lb. z Iz •10–101. The pendulum consists of a disk having a mass of 6 kg and slender rods AB and DC which have a mass per unit length of . Determine the length L of DC so that the center of mass is at the bearing O. What is the moment of inertia of the assembly about an axis perpendicular to the page and passing through point O? 2 kgm 4 ft 8 ft y x z y 3 –– 2 z Prob. 10–98 y x z 4 m 2 m z2 y3 1 –– 16 O Prob. 10–99 450 mm A O B 100 mm Prob. 10–100 O 0.2 m L A B C D 0.8 m 0.5 m Prob. 10–101
- 573. 556 CHAPTER 10 MOMENTS OF INERTIA 10 •10–105. The pendulum consists of the 3-kg slender rod and the 5-kg thin plate. Determine the location of the center of mass G of the pendulum; then find the mass moment of inertia of the pendulum about an axis perpendicular to the page and passing through G. y 10–103. The thin plate has a mass per unit area of . Determine its mass moment of inertia about the y axis. *10–104. The thin plate has a mass per unit area of . Determine its mass moment of inertia about the z axis. 10 kgm2 10 kgm2 10–102. Determine the mass moment of inertia of the 2-kg bent rod about the z axis. 10–106. The cone and cylinder assembly is made of homogeneous material having a density of . Determine its mass moment of inertia about the axis. z 7.85 Mgm3 300 mm 300 mm z y x Prob. 10–102 200 mm 200 mm 200 mm 200 mm 200 mm 200 mm 200 mm 200 mm z y x 100 mm 100 mm Probs. 10–103/104 G 2 m 1 m 0.5 m y O Prob. 10–105 300 mm 300 mm z x y 150 mm 150 mm Prob. 10–106
- 574. 10.8 MASS MOMENT OF INERTIA 557 10 10–110. Determine the mass moment of inertia of the thin plate about an axis perpendicular to the page and passing through point O. The material has a mass per unit area of . 20 kgm2 •10–109. If the large ring, small ring and each of the spokes weigh 100 lb,15 lb,and 20 lb,respectively,determine the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A. 10–107. Determine the mass moment of inertia of the overhung crank about the x axis. The material is steel having a density of . *10–108. Determine the mass moment of inertia of the overhung crank about the axis. The material is steel having a density of . r = 7.85 Mgm3 x¿ r = 7.85 Mgm3 10–111. Determine the mass moment of inertia of the thin plate about an axis perpendicular to the page and passing through point O. The material has a mass per unit area of . 20 kgm2 90 mm 50 mm 20 mm 20 mm 20 mm x x¿ 50 mm 30 mm 30 mm 30 mm 180 mm Probs. 10–107/108 A O 1 ft 4 ft Prob. 10–109 400 mm 150 mm 400 mm O 50 mm 50 mm 150 mm 150 mm 150 mm Prob. 10–110 200 mm 200 mm O 200 mm Prob. 10–111
- 575. 558 CHAPTER 10 MOMENTS OF INERTIA 10 CHAPTER REVIEW Area Moment of Inertia The area moment of inertia represents the second moment of the area about an axis. It is frequently used in formulas related to the strength and stability of structural members or mechanical elements. If the area shape is irregular but can be described mathematically, then a differential element must be selected and integration over the entire area must be performed to determine the moment of inertia. Ix = L A y2 dA Parallel-Axis Theorem If the moment of inertia for an area is known about a centroidal axis, then its moment of inertia about a parallel axis can be determined using the parallel-axis theorem. Composite Area If an area is a composite of common shapes, as found on the inside back cover, then its moment of inertia is equal to the algebraic sum of the moments of inertia of each of its parts. I = I + Ad2 Product of Inertia The product of inertia of an area is used in formulas to determine the orientation of an axis about which the moment of inertia for the area is a maximum or minimum. Ixy = L A xy dA If the product of inertia for an area is known with respect to its centroidal axes, then its value can be determined with respect to any x, y axes using the parallel-axis theorem for the product of inertia. y¿ x¿, Ixy = Ix¿y¿ + Adxdy Iy = L A x2 dA x y y x dx y f(x) dA d C A I I x x – O x y d dx dy x¿ y x¿ y¿ dA C
- 576. CHAPTER REVIEW 559 10 Principal Moments of Inertia Provided the moments of inertia, and and the product of inertia, are known, then the transformation formulas, or Mohr’s circle, can be used to determine the maximum and minimum or principal moments of inertia for the area, as well as finding the orientation of the principal axes of inertia. Ixy, Iy, Ix tan 2up = -Ixy 1Ix - Iy22 Imax min = Ix + Iy 2 ; C a Ix - Iy 2 b 2 + Ixy 2 Mass Moment of Inertia The mass moment of inertia is a property of a body that measures its resistance to a change in its rotation. It is defined as the “second moment” of the mass elements of the body about an axis. I = r L V r2 dV For homogeneous bodies having axial symmetry, the mass moment of inertia can be determined by a single integration, using a disk or shell element. The mass moment of inertia of a composite body is determined by using tabular values of its composite shapes found on the inside back cover, along with the parallel-axis theorem. I = IG + md2 I = L m r2 dm r dm z z y x z (x,y) dz y y z (x, y) z x y dy
- 577. 560 CHAPTER 10 MOMENTS OF INERTIA 10 C x y d 2 d 2 d 2 d 2 60 60 Probs. 10–112/113 a a a a a –– 2 y – x y x Prob. 10–114 REVIEW PROBLEMS 10–115. Determine the moment of inertia of the beam’s cross-sectional area with respect to the axis passing through the centroid C. x¿ 10–114. Determine the moment of inertia of the beam’s cross-sectional area about the x axis. *10–112. Determine the moment of inertia of the beam’s cross-sectional area about the x axis which passes through the centroid C. •10–113. Determine the moment of inertia of the beam’s cross-sectional area about the y axis which passes through the centroid C. *10–116. Determine the product of inertia for the angle’s cross-sectional area with respect to the and axes having their origin located at the centroid C. Assume all corners to be right angles. y¿ x¿ 0.5 in. 0.5 in. 4 in. 2.5 in. C x¿ 0.5 in. _ y Prob. 10–115 C 57.37 mm x¿ y¿ 200 mm 20 mm 57.37 mm 200 mm 20 mm Prob. 10–116
- 578. REVIEW PROBLEMS 561 10 *10–120. The pendulum consists of the slender rod OA, which has a mass per unit length of . The thin disk has a mass per unit area of . Determine the distance to the center of mass G of the pendulum; then calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G. y 12 kgm2 3 kgm 10–119. Determine the moment of inertia of the area about the x axis. Then, using the parallel-axis theorem, find the moment of inertia about the axis that passes through the centroid C of the area. . y = 120 mm x¿ •10–117. Determine the moment of inertia of the area about the y axis. 10–118. Determine the moment of inertia of the area about the x axis. •10–121. Determine the product of inertia of the area with respect to the x and y axes. y 4y 4 – x2 1 ft x 2 ft Probs. 10–117/118 1 ––– 200 200 mm 200 mm y x x¿ – y C y x2 Prob. 10–119 G 1.5 m A y O 0.3 m 0.1 m Prob. 10–120 y x3 y 1 m 1 m x Prob. 10–121
- 579. Equilibrium and stability of this articulated crane boom as a function of the boom position can be analyzed using methods based on work and energy, which are explained in this chapter.
- 580. Virtual Work CHAPTER OBJECTIVES • To introduce the principle of virtual work and show how it applies to finding the equilibrium configuration of a system of pin-connected members. • To establish the potential-energy function and use the potential- energy method to investigate the type of equilibrium or stability of a rigid body or system of pin-connected members. 11.1 Definition of Work The principle of virtual work was proposed by the Swiss mathematician Jean Bernoulli in the eighteenth century.It provides an alternative method for solving problems involving the equilibrium of a particle, a rigid body, or a system of connected rigid bodies. Before we discuss this principle, however, we must first define the work produced by a force and by a couple moment. 11
- 581. 564 CHAPTER 11 VIRTUAL WORK 11 Work of a Force. A force does work when it undergoes a displacement in the direction of its line of action. Consider, for example, the force F in Fig. 11–1a that undergoes a differential displacement . If is the angle between the force and the displacement, then the component of F in the direction of the displacement is . And so the work produced by F is Notice that this expression is also the product of the force F and the component of displacement in the direction of the force, , Fig. 11–1b. If we use the definition of the dot product (Eq. 2–14) the work can also be written as As the above equations indicate, work is a scalar, and like other scalar quantities, it has a magnitude that can either be positive or negative. In the SI system, the unit of work is a joule (J), which is the work produced by a 1-N force that displaces through a distance of 1 m in the direction of the force .The unit of work in the FPS system is the foot-pound , which is the work produced by a 1-lb force that displaces through a distance of 1 ft in the direction of the force. The moment of a force has this same combination of units; however, the concepts of moment and work are in no way related. A moment is a vector quantity, whereas work is a scalar. Work of a Couple Moment. The rotation of a couple moment also produces work. Consider the rigid body in Fig. 11–2, which is acted upon by the couple forces F and –F that produce a couple moment M having a magnitude . When the body undergoes the differential displacement shown, points and move and to their final positions and , respectively. Since , this movement can be thought of as a translation , where A and B move to and , and a rotation about , where the body rotates through the angle about A.The couple forces do no work during the translation because each force undergoes the same amount of displacement in opposite directions, thus canceling out the work. During rotation, however, F is displaced , and so it does work . Since , the work of the couple moment M is therefore If M and have the same sense, the work is positive; however, if they have the opposite sense, the work will be negative. du dU = Mdu M = Fr dU = F dr– = F r du dr– = r du drA du A¿ B– A¿ drA drB = drA + dr¿ B¿ A¿ drB drA B A M = Fr (ft # lb) (1 J = 1 N # m) dU = F # dr dr cos u dU = F dr cos u F cos u u dr F dr F cos (a) u u F dr dr cos u (b) u Fig. 11–1 F –F A A¿ B– dr¿ drA drA drB B¿ B r du Fig. 11–2
- 582. 11.2 PRINCIPLE OF VIRTUAL WORK 565 11 Virtual Work. The definitions of the work of a force and a couple have been presented in terms of actual movements expressed by differential displacements having magnitudes of dr and Consider now an imaginary or virtual movement of a body in static equilibrium, which indicates a displacement or rotation that is assumed and does not actually exist.These movements are first-order differential quantities and will be denoted by the symbols and (delta r and delta ), respectively. The virtual work done by a force having a virtual displacement is (11–1) Similarly, when a couple undergoes a virtual rotation in the plane of the couple forces, the virtual work is (11–2) 11.2 Principle of Virtual Work The principle of virtual work states that if a body is in equilibrium, then the algebraic sum of the virtual work done by all the forces and couple moments acting on the body, is zero for any virtual displacement of the body.Thus, (11–3) For example, consider the free-body diagram of the particle (ball) that rests on the floor, Fig. 11–3. If we “imagine” the ball to be displaced downwards a virtual amount then the weight does positive virtual work, and the normal force does negative virtual work, For equilibrium the total virtual work must be zero, so that Since then as required by applying . ©Fy = 0 N = W dy Z 0, dU= Wdy -N dy = 1W-N2dy = 0. -N dy. W dy, dy, dU = 0 dU = M du du dU = F cos u dr dr u du dr du. W N dy Fig. 11–3
- 583. 566 CHAPTER 11 VIRTUAL WORK 11 In a similar manner, we can also apply the virtual-work equation to a rigid body subjected to a coplanar force system. Here, separate virtual translations in the x and y directions and a virtual rotation about an axis perpendicular to the x–y plane that passes through an arbitrary point O, will correspond to the three equilibrium equations, and When writing these equations, it is not necessary to include the work done by the internal forces acting within the body since a rigid body does not deform when subjected to an external loading, and furthermore, when the body moves through a virtual displacement, the internal forces occur in equal but opposite collinear pairs, so that the corresponding work done by each pair of forces will cancel. To demonstrate an application, consider the simply supported beam in Fig. 11–4a. When the beam is given a virtual rotation about point B, Fig. 11–4b, the only forces that do work are P and Since and the virtual work equation for this case is . Since then Excluding notice that the terms in parentheses actually represent the application of . As seen from the above two examples, no added advantage is gained by solving particle and rigid-body equilibrium problems using the principle of virtual work. This is because for each application of the virtual-work equation, the virtual displacement, common to every term, factors out, leaving an equation that could have been obtained in a more direct manner by simply applying an equation of equilibrium. ©MB = 0 du, Ay = P2. du Z 0, dU = Ay1l du2 - P1l22 du = 1Ayl - Pl22 du = 0 dy¿ = 1l22 du, dy = l du Ay. du ©MO = 0. ©Fy = 0, ©Fx = 0, dU = 0 A (a) B P l –– 2 l –– 2 (b) By Ay Bx P du l –– 2 l –– 2 dy dy¿ Fig. 11–4
- 584. 11.3 PRINCIPLE OF VIRTUAL WORK FOR A SYSTEM OF CONNECTED RIGID BODIES 567 11 11.3 Principle of Virtual Work for a System of Connected Rigid Bodies The method of virtual work is particularly effective for solving equilibrium problems that involve a system of several connected rigid bodies, such as the ones shown in Fig. 11–5. Each of these systems is said to have only one degree of freedom since the arrangement of the links can be completely specified using only one coordinate . In other words, with this single coordinate and the length of the members, we can locate the position of the forces F and P. In this text, we will only consider the application of the principle of virtual work to systems containing one degree of freedom*. Because they are less complicated, they will serve as a way to approach the solution of more complex problems involving systems with many degrees of freedom. The procedure for solving problems involving a system of frictionless connected rigid bodies follows. u Important Points • A force does work when it moves through a displacement in the direction of the force. A couple moment does work when it moves through a collinear rotation. Specifically, positive work is done when the force or couple moment and its displacement have the same sense of direction. • The principle of virtual work is generally used to determine the equilibrium configuration for a system of multiply connected members. • A virtual displacement is imaginary; i.e., it does not really happen. It is a differential displacement that is given in the positive direction of a position coordinate. • Forces or couple moments that do not virtually displace do no virtual work. *This method of applying the principle of virtual work is sometimes called the method of virtual displacements because a virtual displacement is applied, resulting in the calculation of a real force. Although it is not used here, we can also apply the principle of virtual work as a method of virtual forces.This method is often used to apply a virtual force and then determine the displacements of points on deformable bodies. See R. C. Hibbeler, Mechanics of Materials, 7th edition, Pearson/Prentice Hall, 2007. This scissors lift has one degree of freedom. Without the need for dismembering the mechanism, the force in the hydraulic cylinder required to provide the lift can be determined directly by using the principle of virtual work. AB A B F l l P F l l P u u u u Fig. 11–5
- 585. 568 CHAPTER 11 VIRTUAL WORK 11 Procedure for Analysis Free-Body Diagram. • Draw the free-body diagram of the entire system of connected bodies and define the coordinate q. • Sketch the “deflected position” of the system on the free-body diagram when the system undergoes a positive virtual displacement Virtual Displacements. • Indicate position coordinates each measured from a fixed point on the free-body diagram. These coordinates are directed to the forces that do work. • Each of these coordinate axes should be parallel to the line of action of the force to which it is directed, so that the virtual work along the coordinate axis can be calculated. • Relate each of the position coordinates to the coordinate q; then differentiate these expressions in order to express each virtual displacement in terms of Virtual-Work Equation. • Write the virtual-work equation for the system assuming that, whether possible or not, each position coordinate undergoes a positive virtual displacement If a force or couple moment is in the same direction as the positive virtual displacement, the work is positive. Otherwise, it is negative. • Express the work of each force and couple moment in the equation in terms of • Factor out this common displacement from all the terms, and solve for the unknown force, couple moment, or equilibrium position q. dq. ds. s dq. ds s s, dq.
- 586. 11.3 PRINCIPLE OF VIRTUAL WORK FOR A SYSTEM OF CONNECTED RIGID BODIES 569 11 EXAMPLE 11.1 Determine the angle for equilibrium of the two-member linkage shown in Fig. 11–6a. Each member has a mass of 10 kg. SOLUTION Free-Body Diagram. The system has only one degree of freedom since the location of both links can be specified by the single coordinate As shown on the free-body diagram in Fig. 11–6b, when has a positive (clockwise) virtual rotation only the force F and the two 98.1-N weights do work. (The reactive forces and are fixed, and does not displace along its line of action.) Virtual Displacements. If the origin of coordinates is established at the fixed pin support D, then the position of F and W can be specified by the position coordinates and . In order to determine the work, note that, as required, these coordinates are parallel to the lines of action of their associated forces. Expressing these position coordinates in terms of and taking the derivatives yields (1) (2) It is seen by the signs of these equations,and indicated in Fig.11–6b,that an increase in (i.e., ) causes a decrease in and an increase in Virtual-Work Equation. If the virtual displacements and were both positive, then the forces W and F would do positive work since the forces and their corresponding displacements would have the same sense. Hence, the virtual-work equation for the displacement is (3) Substituting Eqs. 1 and 2 into Eq. 3 in order to relate the virtual displacements to the common virtual displacement yields Notice that the “negative work” done by F (force in the opposite sense to displacement) has actually been accounted for in the above equation by the “negative sign” of Eq. 1. Factoring out the common displacement and solving for noting that yields Ans. NOTE: If this problem had been solved using the equations of equilibrium, it would be necessary to dismember the links and apply three scalar equations to each link. The principle of virtual work, by means of calculus, has eliminated this task so that the answer is obtained directly. u = tan-1 98.1 50 = 63.0° 198.1 cos u - 50 sin u2 du = 0 du Z 0, u, du 98.110.5 cos u du2 + 98.110.5 cos u du2 + 251-2 sin u du2 = 0 du W dyw + W dyw + F dxB = 0 dU = 0; du dyw dxB yw. xB du u dyw = 0.5 cos u du m yw = 1 211 sin u2 m dxB = -2 sin u du m xB = 211 cos u2 m u yw xB By Dy Dx du, u 1q =2 u. u 1 m 1 m D u B F 25 N C (a) D B F 25 N (b) W 98.1 N W 98.1 N By Dy Dx dxB du u dyw dyw yw xB Fig. 11–6
- 587. 570 CHAPTER 11 VIRTUAL WORK 11 Determine the required force P in Fig. 11–7a, needed to maintain equilibrium of the scissors linkage when . The spring is unstretched when . Neglect the mass of the links. SOLUTION Free-Body Diagram. Only and P do work when undergoes a positive virtual displacement , Fig. 11–7b. For the arbitrary position , the spring is stretched , so that Virtual Displacements. The position coordinates, and , measured from the fixed point A, are used to locate and . These coordinates are parallel to the line of action of their corresponding forces. Expressing and in terms of the angle using trigonometry, Differentiating, we obtain the virtual displacements of points B and D. (1) (2) Virtual-Work Equation. Force P does positive work since it acts in the positive sense of its virtual displacement.The spring force does negative work since it acts opposite to its positive virtual displacement.Thus, the virtual-work equation becomes Since , then this equation requires When , Ans. P = 500 sin 60° - 250 = 183 N u = 60° P = 500 sin u - 250 cos u du Z 0 [0.9P + 225 - 450 sin u] cos u du = 0 - [1500 sin u - 750] (0.3 cos u du) + P (0.9 cos u du) = 0 -FsdxB+PdxD = 0 dU = 0; Fs dxD = 0.9 cos u du dxB = 0.3 cos u du xD = 3[(0.3 m) sin u] = (0.9 m) sin u xB = (0.3 m) sin u u xD xB P Fs xD xB = (1500 sin u - 750) N Fs = ks = 5000 N/m [(0.3 m) sin u - (0.3 m) sin 30°] (0.3 m) sin u - (0.3 m) sin 30° u du u Fs u = 30° u = 60° EXAMPLE 11.2 A B k 5 kN/m (a) C E D G 0.3m 0.3m 0.3m 0.3m u u P B (b) Gx Ax Ay Fs xD xB dxD dxB P u du Fig. 11–7
- 588. 11.3 PRINCIPLE OF VIRTUAL WORK FOR A SYSTEM OF CONNECTED RIGID BODIES 571 11 EXAMPLE 11.3 If the box in Fig. 11–8a has a mass of 10 kg, determine the couple moment M needed to maintain equilibrium when . Neglect the mass of the members. u = 60° D C A B M 0.45 m 0.45 m (a) 0.2 m 0.4 m u u 0.45 m C b A M (b) yE yE Bx Dx By Dy 10(9.81) N u d u d u u d Fig. 11–8 SOLUTION Free-Body Diagram. When undergoes a positive virtual displacement , only the couple moment M and the weight of the box do work, Fig. 11–8b. Virtual Displacements. The position coordinate , measured from the fixed point B, locates the weight, . Here, where b is a constant distance. Differentiating this equation, we obtain (1) Virtual-Work Equation. The virtual-work equation becomes Substituting Eq. 1 into this equation Since , then Since it is required that , then Ans. M = 44.145 cos 60° = 22.1 N # m u = 60° M - 44.145 cos u = 0 du Z 0 du(M - 44.145 cos u) = 0 Mdu - 10(9.81) N(0.45 m cos u du) = 0 Mdu - [10(9.81) N]dyE = 0 dU = 0; dyE = 0.45 m cos u du yE = (0.45 m) sin u + b 10(9.81) N yE du u
- 589. 572 CHAPTER 11 VIRTUAL WORK 11 The mechanism in Fig. 11–9a supports the 50-lb cylinder. Determine the angle for equilibrium if the spring has an unstretched length of 2 ft when . Neglect the mass of the members. SOLUTION Free-Body Diagram. When the mechanism undergoes a positive virtual displacement , Fig. 11–9b, only and the 50-lb force do work. Since the final length of the spring is , then Virtual Displacements. The position coordinates and are established from the fixed point A to locate at and at . The coordinate also measured from A, specifies the position of the 50-lb force at B. The coordinates can be expressed in terms of using trigonometry. Differentiating, we obtain the virtual displacements of points , , and as (1) (2) (3) Virtual-Work Equation. The virtual-work equation is written as if all virtual displacements are positive, thus Since , then Solving by trial and error, Ans. u = 34.9° 800 sin u cos u - 800 sin u + 100 cos u = 0 du Z 0 du(800 sin u cos u - 800 sin u + 100 cos u) = 0 -(400 - 400 cos u)(-1 sin u du) = 0 (400 - 400 cos u)(-3 sin u du) + 50(2 cos u du) FsdxE + 50dyB - FsdxD = 0 dU = 0; dyB = 2 cos u du dxE = -3 sin u du dxD = -1 sin u du B E D yB = (2 ft) sin u xE = 3[(1 ft) cos u] = (3 ft) cos u xD = (1 ft) cos u u yB, E D Fs xE xD Fs = ks = (200 lb/ft)(2 ft - 2 ft cos u) = (400 - 400 cos u) lb 2(1 ft cos u) Fs du u = 0° u EXAMPLE 11.4 C D E k 200 lb/ft B A 1 ft 1 ft 1 ft (a) 1 ft u u E D 50 lb (b) Ax Ay Cy Fs Fs xD yB xE dxE dxD dyB u du Fig. 11–9
- 590. 11.3 PRINCIPLE OF VIRTUAL WORK FOR A SYSTEM OF CONNECTED RIGID BODIES 573 11 FUNDAMENTAL PROBLEMS F11–4. The linkage is subjected to a force of . Determine the angle for equilibrium. The spring is unstretched at . Neglect the mass of the links. u = 60° u P = 6 kN F11–2. Determine the magnitude of force P required to hold the 50-kg smooth rod in equilibrium at u = 60°. F11–1. Determine the required magnitude of force P to maintain equilibrium of the linkage at Each link has a mass of 20 kg. u = 60°. F11–5. Determine the angle where the 50-kg bar is in equilibrium.The spring is unstretched at u = 60°. u F11–3. The linkage is subjected to a force of Determine the angle for equilibrium. The spring is unstretched when . Neglect the mass of the links. u = 0° u P = 2 kN. F11–6. The scissors linkage is subjected to a force of Determine the angle for equilibrium. The spring is unstretched at . Neglect the mass of the links. u = 0° u P = 150 N. 0.9 m k 20 kN/m 0.9 m A B C P 6 kN u F11–4 1.5 m 1.5 m A B C P u u F11–1 D k 15 kN/m A B C 0.6 m 0.6 m 0.6 m P 2 kN u u F11–3 5 m A B k 600 N/m u F11–5 5 m P u A B F11–2 C 0.3 m 0.3 m P 150 N A B k 15 kN/m u F11–6
- 591. 574 CHAPTER 11 VIRTUAL WORK 11 PROBLEMS 11–3. The “Nuremberg scissors” is subjected to a horizontal force of . Determine the angle for equilibrium. The spring has a stiffness of and is unstretched when . *11–4. The “Nuremberg scissors” is subjected to a horizontal force of . Determine the stiffness k of the spring for equilibrium when . The spring is unstretched when . u = 15° u = 60° P = 600 N u = 15° k = 15 kNm u P = 600 N 11–2. The uniform rod OA has a weight of 10 lb. When the rod is in a vertical position, , the spring is unstretched. Determine the angle for equilibrium if the end of the spring wraps around the periphery of the disk as the disk turns. u u = 0° •11–1. The 200-kg crate is on the lift table at the position . Determine the force in the hydraulic cylinder AD for equilibrium. Neglect the mass of the lift table’s components. u = 30° •11–5. Determine the force developed in the spring required to keep the 10 lb uniform rod AB in equilibrium when . u = 35° P 200 mm 200 mm A C D E B k u Probs. 11–3/4 A u B k 15 lb/ft 6 ft M = 10 lb ft Prob. 11–5 A B C D E H I F 1.2 m 1.2 m u Prob. 11–1 O u A k 30 lb/ft 2 ft 0.5 ft Prob. 11–2
- 592. 11.3 PRINCIPLE OF VIRTUAL WORK FOR A SYSTEM OF CONNECTED RIGID BODIES 575 11 •11–9. If a force is applied to the lever arm of the toggle press, determine the clamping force developed in the block when . Neglect the weight of the block. u = 45° P = 100 N 11–7. The pin-connected mechanism is constrained at A by a pin and at B by a roller. If , determine the angle for equilibrium. The spring is unstretched when . Neglect the weight of the members. *11–8. The pin-connected mechanism is constrained by a pin at A and a roller at B. Determine the force P that must be applied to the roller to hold the mechanism in equilibrium when . The spring is unstretched when . Neglect the weight of the members. u = 45° u = 30° u = 45° u P = 10 lb 11–6. If a force of is applied to the handle of the mechanism,determine the force the screw exerts on the cork of the bottle.The screw is attached to the pin at A and passes through the collar that is attached to the bottle neck at B. P = 5 lb 11–10. When the forces are applied to the handles of the bottle opener, determine the pulling force developed on the cork. 3 in. D B A u 30° P 5 lb Prob. 11–6 A C B D E F 90 mm 90 mm 15 mm 15 mm 5 N 5 N P P Prob. 11–10 0.5 ft B P A u 0.5 ft 0.5 ft k 50 lb/ft Probs. 11–7/8 200 mm 200 mm 500 mm B C D E F A P u u Prob. 11–9
- 593. 576 CHAPTER 11 VIRTUAL WORK 11 11–14. The truck is weighed on the highway inspection scale. If a known mass m is placed a distance s from the fulcrum B of the scale, determine the mass of the truck if its center of gravity is located at a distance d from point C. When the scale is empty, the weight of the lever ABC balances the scale CDE. mt •11–13. Determine the angles for equilibrium of the 4-lb disk using the principle of virtual work. Neglect the weight of the rod.The spring is unstretched when and always remains in the vertical position due to the roller guide. u = 0° u 11–11. If the spring has a stiffness k and an unstretched length , determine the force P when the mechanism is in the position shown. Neglect the weight of the members. *11–12. Solve Prob. 11–11 if the force P is applied vertically downward at B. l0 11–15. The assembly is used for exercise. It consists of four pin-connected bars, each of length L, and a spring of stiffness k and unstretched length . If horizontal forces are applied to the handles so that is slowly decreased, determine the angle at which the magnitude of P becomes a maximum. u u a (6 2L) P l k B u C A l Probs. 11–11/12 k 50 lb/ft A B C u 3 ft 1 ft Prob. 11–13 s a C B D E A m a d Prob. 11–14 L L u u L L D –P k P B C A Prob. 11–15
- 594. 11.3 PRINCIPLE OF VIRTUAL WORK FOR A SYSTEM OF CONNECTED RIGID BODIES 577 11 11–19. The spring is unstretched when and has a stiffness of . Determine the angle for equilibrium if each of the cylinders weighs 50 lb. Neglect the weight of the members.The spring remains horizontal at all times due to the roller. u k = 1000 lbft u = 45° 11–18. If a vertical force of is applied to the handle of the toggle clamp, determine the clamping force exerted on the pipe. P = 50 N *11–16. A 5-kg uniform serving table is supported on each side by pairs of two identical links, and , and springs . If the bowl has a mass of , determine the angle where the table is in equilibrium. The springs each have a stiffness of and are unstretched when . Neglect the mass of the links. •11–17. A 5-kg uniform serving table is supported on each side by two pairs of identical links, and , and springs . If the bowl has a mass of and is in equilibrium when , determine the stiffness of each spring.The springs are unstretched when . Neglect the mass of the links. u = 90° k u = 45° 1 kg CE CD AB u = 90° k = 200 Nm u 1 kg CE CD AB *11–20. The machine shown is used for forming metal plates. It consists of two toggles ABC and DEF, which are operated by the hydraulic cylinder. The toggles push the moveable bar G forward, pressing the plate into the cavity. If the force which the plate exerts on the head is , determine the force F in the hydraulic cylinder when . u = 30° P = 8 kN A B C P 50 N D 300 mm 500 mm 100 mm 150 mm u 45 Prob. 11–18 D E k A B C 4 ft 4 ft 2 ft 2 ft u u Prob. 11–19 A C k 250 mm 250 mm 150 mm 150 mm B D E u u Probs. 11–16/17 200 mm 200 mm 200 mm 200 mm A B H E u u D F F –F C G P 30 plate Prob. 11–20
- 595. 578 CHAPTER 11 VIRTUAL WORK 11 *11–24. Determine the magnitude of the couple moment required to support the 20-kg cylinder in the configuration shown. The smooth peg at can slide freely within the slot. Neglect the mass of the members. B M 11–22. Determine the weight of block required to balance the differential lever when the 20-lb load F is placed on the pan.The lever is in balance when the load and block are not on the lever.Take . 11–23. If the load weighs 20 lb and the block weighs 2 lb, determine its position for equilibrium of the differential lever. The lever is in balance when the load and block are not on the lever. x G F x = 12 in G •11–21. The vent plate is supported at B by a pin.If it weighs 15 lb and has a center of gravity at G, determine the stiffness k of the spring so that the plate remains in equilibrium at .The spring is unstretched when . u = 0° u = 30° •11–25. The crankshaft is subjected to a torque of . Determine the vertical compressive force F applied to the piston for equilibrium when . u = 60° M = 50 lb # ft 0.5 ft 1 ft A k u B G C 4 ft Prob. 11–21 A u B M 3 in. 5 in. F Prob. 11–25 2.5 m 1 m 1 m A B C D E M u 30 4 in. 4 in. x A B C G E D 2 in. F Probs. 11–22/23 Prob. 11–24
- 596. 11.4 CONSERVATIVE FORCES 579 11 *11.4 Conservative Forces If the work of a force only depends upon its initial and final positions, and is independent of the path it travels, then the force is referred to as a conservative force.The weight of a body and the force of a spring are two examples of conservative forces. Weight. Consider a block of weight that travels along the path in Fig. 11–10a. When it is displaced up the path by an amount , then the work is or , as shown in Fig. 11–10b. In this case, the work is negative since acts in the opposite sense of . Thus, if the block moves from to , through the vertical displacement , the work is The weight of a body is therefore a conservative force, since the work done by the weight depends only on the vertical displacement of the body, and is independent of the path along which the body travels. Spring Force. Now consider the linearly elastic spring in Fig. 11–11, which undergoes a displacement ds. The work done by the spring force on the block is . The work is negative because acts in the opposite sense to that of . Thus, the work of when the block is displaced from to is Here the work depends only on the spring’s initial and final positions, and , measured from the spring’s unstretched position. Since this result is independent of the path taken by the block as it moves, then a spring force is also a conservative force. s2 s1 U = - 3 s2 s1 ks ds = - a1 2 ks2 2 - 1 2 ks1 2 b s = s2 s = s1 Fs ds Fs dU = -Fs ds = -ks ds U = - 3 h 0 W dy = -Wh h B A dy W dU = -W(dr cos u) = -Wdy dU = W # dr dr W y h dr A B W s W (a) Undeformed position s ds Fs Fig. 11–11 dr dy dr cos u W (b) u Fig. 11–10
- 597. 580 CHAPTER 11 VIRTUAL WORK 11 Friction. In contrast to a conservative force, consider the force of friction exerted on a sliding body by a fixed surface. The work done by the frictional force depends on the path; the longer the path, the greater the work. Consequently, frictional forces are nonconservative, and most of the work done by them is dissipated from the body in the form of heat. *11.5 Potential Energy When a conservative force acts on a body, it gives the body the capacity to do work. This capacity, measured as potential energy, depends on the location of the body relative to a fixed reference position or datum. Gravitational Potential Energy. If a body is located a distance y above a fixed horizontal reference or datum as in Fig. 11–12, the weight of the body has positive gravitational potential energy since W has the capacity of doing positive work when the body is moved back down to the datum. Likewise, if the body is located a distance y below the datum, is negative since the weight does negative work when the body is moved back up to the datum.At the datum, Measuring y as positive upward, the gravitational potential energy of the body’s weight W is therefore (11–4) Elastic Potential Energy. When a spring is either elongated or compressed by an amount s from its unstretched position (the datum), the energy stored in the spring is called elastic potential energy. It is determined from (11–5) This energy is always a positive quantity since the spring force acting on the attached body does positive work on the body as the force returns the body to the spring’s unstretched position, Fig. 11–13. V e = 1 2 ks2 Vg = Wy Vg = 0. V g V g y y Datum Vg Wy Vg Wy Vg 0 W W Fig. 11–12 Fs Fs s s Undeformed position Undeformed position Ve ks2 1 2 Fig. 11–13
- 598. 11.5 POTENTIAL ENERGY 581 11 Potential Function. In the general case, if a body is subjected to both gravitational and elastic forces, the potential energy or potential function V of the body can be expressed as the algebraic sum (11–6) where measurement of V depends on the location of the body with respect to a selected datum in accordance with Eqs. 11–4 and 11–5. In particular, if a system of frictionless connected rigid bodies has a single degree of freedom, such that its vertical position from the datum is defined by the coordinate q, then the potential function for the system can be expressed as The work done by all the weight and spring forces acting on the system in moving it from to , is measured by the difference in V; i.e., (11–7) For example, the potential function for a system consisting of a block of weight W supported by a spring, as in Fig. 11–14, can be expressed in terms of the coordinate measured from a fixed datum located at the unstretched length of the spring. Here (11–8) If the block moves from to then applying Eq. 11–7 the work of W and is U1-2 = V1y12 - V1y22 = -W(y1 - y2) + 1 2 ky1 2 - 1 2 ky2 2 Fs y2, y1 = -Wy + 1 2 ky2 V = Vg + V e 1q =2 y, U1-2 = V1q12 - V1q22 q2 q1 V = V1q2. V = V g + Ve y2 y1 y Datum W k (a) Fig. 11–14
- 599. 582 CHAPTER 11 VIRTUAL WORK 11 *11.6 Potential-Energy Criterion for Equilibrium If a frictionless connected system has one degree of freedom, and its position is defined by the coordinate q, then if it displaces from q to Eq. 11–7 becomes or If the system is in equilibrium and undergoes a virtual displacement rather than an actual displacement dq, then the above equation becomes However, the principle of virtual work requires that and therefore, , and so we can write . Since , this expression becomes (11–9) Hence, when a frictionless connected system of rigid bodies is in equilibrium, the first derivative of its potential function is zero. For example, using Eq. 11–8 we can determine the equilibrium position for the spring and block in Fig. 11–14a. We have Hence, the equilibrium position is Of course, this same result can be obtained by applying to the forces acting on the free-body diagram of the block, Fig. 11–14b. ©Fy = 0 yeq = W k y = yeq dV dy = -W + ky = 0 dV dq = 0 dq Z 0 dV = (dVdq)dq = 0 dV = 0 dU = 0, dU = -dV. dq, dU = -dV dU = V1q2 - V1q + dq2 q + dq, y2 y1 y Datum W k (a) Fig. 11–14 W Fs kyeq (b)
- 600. 11.7 STABILITY OF EQUILIBRIUM CONFIGURATION 583 11 *11.7 Stability of Equilibrium Configuration The potential function V of a system can also be used to investigate the stability of the equilibrium configuration, which is classified as stable, neutral, or unstable. Stable Equilibrium. A system is said to be stable if a system has a tendency to return to its original position when a small displacement is given to the system. The potential energy of the system in this case is at its minimum.A simple example is shown in Fig. 11–15a.When the disk is given a small displacement, its center of gravity G will always move (rotate) back to its equilibrium position, which is at the lowest point of its path.This is where the potential energy of the disk is at its minimum. Neutral Equilibrium. A system is said to be in neutral equilibrium if the system still remains in equilibrium when the system is given a small displacement away from its original position. In this case, the potential energy of the system is constant. Neutral equilibrium is shown in Fig. 11–15b, where a disk is pinned at G. Each time the disk is rotated, a new equilibrium position is established and the potential energy remains unchanged. Unstable Equilibrium. A system is said to be unstable if it has a tendency to be displaced further away from its original equilibrium position when it is given a small displacement. The potential energy of the system in this case is a maximum. An unstable equilibrium position of the disk is shown in Fig. 11–15c. Here the disk will rotate away from its equilibrium position when its center of gravity is slightly displaced. At this highest point, its potential energy is at a maximum. Stable equilibrium Unstable equilibrium Neutral equilibrium (a) (b) (c) G G G The counterweight at balances the weight of the deck of this simple lift bridge. By applying the method of potential energy we can study the stability of the structure for various equilibrium positions of the deck. B A A B Fig. 11–15
- 601. 584 CHAPTER 11 VIRTUAL WORK 11 One-Degree-of-Freedom System. If a system has only one degree of freedom, and its position is defined by the coordinate q, then the potential function V for the system in terms of q can be plotted, Fig. 11–16. Provided the system is in equilibrium, then , which represents the slope of this function, must be equal to zero. An investigation of stability at the equilibrium configuration therefore requires that the second derivative of the potential function be evaluated. If is greater than zero, Fig. 11-16a, the potential energy of the system will be a minimum. This indicates that the equilibrium configuration is stable.Thus, (11–10) If is less than zero, Fig. 11-16b, the potential energy of the system will be a maximum. This indicates an unstable equilibrium configuration.Thus, (11–11) Finally, if is equal to zero, it will be necessary to investigate the higher order derivatives to determine the stability. The equilibrium configuration will be stable if the first non-zero derivative is of an even order and it is positive. Likewise, the equilibrium will be unstable if this first non-zero derivative is odd or if it is even and negative. If all the higher order derivatives are zero, the system is said to be in neutral equilibrium, Fig 11-16c.Thus, (11–12) This condition occurs only if the potential-energy function for the system is constant at or around the neighborhood of . qeq dV dq = d2 V dq2 = d3 V dq3 = Á = 0 neutral equilibrium d2 Vdq2 d2 V dq2 6 0 unstable equilibrium dV dq = 0, d2 Vdq2 d2 V dq2 7 0 stable equilibrium dV dq = 0, d2 Vdq2 dV/dq V q qeq d2V dq2 0 Neutral equilibrium (c) dV dq 0 Fig. 11–16 V q qeq d2 V dq2 0 Stable equilibrium (a) dV dq 0 V q qeq d2 V dq2 0 Unstable equilibrium (b) dV dq 0 During high winds and when going around a curve,these sugar-cane trucks can become unstable and tip over since their center of gravity is high off the road when they are fully loaded.
- 602. 11.7 STABILITY OF EQUILIBRIUM CONFIGURATION 585 11 Procedure for Analysis Using potential-energy methods, the equilibrium positions and the stability of a body or a system of connected bodies having a single degree of freedom can be obtained by applying the following procedure. Potential Function. • Sketch the system so that it is in the arbitrary position specified by the coordinate q. • Establish a horizontal datum through a fixed point* and express the gravitational potential energy in terms of the weight W of each member and its vertical distance y from the datum, • Express the elastic potential energy of the system in terms of the stretch or compression, s, of any connecting spring, • Formulate the potential function and express the position coordinates y and s in terms of the single coordinate q. Equilibrium Position. • The equilibrium position of the system is determined by taking the first derivative of V and setting it equal to zero, Stability. • Stability at the equilibrium position is determined by evaluating the second or higher-order derivatives of V. • If the second derivative is greater than zero, the system is stable; if all derivatives are equal to zero, the system is in neutral equilibrium; and if the second derivative is less than zero, the system is unstable. *The location of the datum is arbitrary, since only the changes or differentials of V are required for investigation of the equilibrium position and its stability. dVdq = 0. V = V g + V e Ve = 1 2 ks2 . V e Vg = Wy. V g
- 603. 586 CHAPTER 11 VIRTUAL WORK 11 The uniform link shown in Fig. 11–17a has a mass of 10 kg. If the spring is unstretched when determine the angle for equilibrium and investigate the stability at the equilibrium position. SOLUTION Potential Function. The datum is established at the bottom of the link, Fig. 11–17b. When the link is located in the arbitrary position the spring increases its potential energy by stretching and the weight decreases its potential energy. Hence, Since or and , then Equilibrium Position. The first derivative of V is or This equation is satisfied provided Ans. Ans. Stability. The second derivative of V is Substituting values for the constants, with and yields Ans. = -29.4 6 0 1unstable equilibrium at u = 0°2 d2 V du2 ` u=0° = 20010.622 1cos 0° - cos 0°2 - 1019.81210.62 2 cos 0° u = 53.8°, u = 0° = kl2 1cos u - cos 2u2 - Wl 2 cos u d2 V du2 = kl2 11 - cos u2 cos u + kl2 sin u sin u - Wl 2 cos u u = cos-1 a1 - W 2kl b = cos-1 c1 - 1019.812 21200210.62 d = 53.8° sin u = 0 u = 0° lckl11 - cos u2 - W 2 d sin u = 0 dV du = kl2 11 - cos u2 sin u - Wl 2 sin u = 0 V = 1 2 kl2 11 - cos u22 + Wa l 2 cos ub y = (l2) cos u s = l11 - cos u2, l = s + l cos u V = Ve + V g = 1 2 ks2 + Wy u, u u = 0°, EXAMPLE 11.5 l 0.6 m A k 200 N/m B (a) u s cos u l — 2 l — 2 l W W l — 2 k Datum (b) F ks u u y Fig 11–17 Ans. = 46.9 7 0 1stable equilibrium at u = 53.8°2 d2 V du2 ` u=53.8° = 20010.622 1cos53.8° - cos107.6°2 - 1019.81210.62 2 cos53.8°
- 604. 11.7 STABILITY OF EQUILIBRIUM CONFIGURATION 587 11 EXAMPLE 11.6 If the spring AD in Fig. 11–18a has a stiffness of 18 kN/m and is unstretched when , determine the angle for equilibrium.The load has a mass of 1.5 Mg. Investigate the stability at the equilibrium position. SOLUTION Potential Energy. The gravitational potential energy for the load with respect to the fixed datum, shown in Fig. 11–18b, is u u = 60° Vg = mgy = 1500(9.81) N[(4 m) sin u + h] = 58 860 sin u + 14 715h where h is a constant distance. From the geometry of the system, the elongation of the spring when the load is on the platform is . Thus, the elastic potential energy of the system is The potential energy function for the system is therefore (1) Equilibrium. When the system is in equilibrium, Since Solving by trial and error, Ans. Stability. Taking the second derivative of Eq. 1, Substituting yields Ans. And for , Ans. d2 V du2 = 64 073 7 0 Stable u = 45.51° d2 V du2 = -60 409 6 0 Unstable u = 28.18° d2 V du2 = -58 860 sin u - 288 000 cos 2u + 144 000 cos u u = 45.51° u = 28.18° and 58 860 cos u - 144 000 sin 2u + 144 000 sin u = 0 sin 2u = 2 sin u cos u, 58 860 cos u - 288 000 sin u cos u + 144 000 sin u = 0 dV du = 58 860 cos u + 18 000(4 cos u - 2)(-4 sin u) = 0 V = Vg + V e = 58 860 sin u + 14 715h + 9000(4 cos u - 2)2 Ve = 1 2 ks2 = 1 2(18 000 N/m)(4 m cos u - 2 m)2 = 9000(4 cos u - 2)2 s = (4 m) cos u - (4 m) cos 60° = (4 m) cos u - 2 m (a) 2 m 2 m A C E B D G k 18 kN/m u u 2 m 2 m A C E B D G k 18 kN/m (b) 4 m cosu (4 m)sinu h y Datum u u Fig 11–18
- 605. 588 CHAPTER 11 VIRTUAL WORK 11 The uniform block having a mass m rests on the top surface of the half cylinder, Fig. 11–19a. Show that this is a condition of unstable equilibrium if SOLUTION Potential Function. The datum is established at the base of the cylinder, Fig. 11–19b. If the block is displaced by an amount from the equilibrium position, the potential function is From Fig. 11–18b, Thus, Equilibrium Position. Note that satisfies this equation. Stability. Taking the second derivative of V yields At Since all the constants are positive, the block is in unstable equilibrium provided because then d2 Vdu2 6 0. h 7 2R, d2 V du2 ` u=0° = -mga h 2 - Rb u = 0°, d2 V du2 = mga- h 2 cos u + R cos u - Ru sin ub u = 0° = mga- h 2 sin u + Ru cos ub = 0 dV du = mgc - aR + h 2 b sin u + R sin u + Ru cos ud = 0 V = mgc aR + h 2 b cos u + Ru sin ud y = aR + h 2 b cos u + Ru sin u = 0 + mgy V = V e + Vg u h 7 2R. EXAMPLE 11.7 h m R (a) y W mg R (b) ) cos u (R Ru sin u Ru h — 2 h — 2 Datum u u Fig 11–19
- 606. 11.7 STABILITY OF EQUILIBRIUM CONFIGURATION 589 11 11–31. If the springs at A and C have an unstretched length of 10 in. while the spring at B has an unstretched length of 12 in., determine the height h of the platform when the system is in equilibrium. Investigate the stability of this equilibrium configuration. The package and the platform have a total weight of 150 lb. 11–30. The spring has a stiffness and is unstretched when .If the mechanism is in equilibrium when determine the weight of cylinder D. Neglect the weight of the members. Rod AB remains horizontal at all times since the collar can slide freely along the vertical guide. u = 60°, u = 45° k = 600 lbft 11–26. If the potential energy for a conservative one- degree-of-freedom system is expressed by the relation , where x is given in feet, determine the equilibrium positions and investigate the stability at each position. 11–27. If the potential energy for a conservative one- degree-of-freedom system is expressed by the relation , , determine the equilibrium positions and investigate the stability at each position. *11–28. If the potential energy for a conservative one- degree-of-freedom system is expressed by the relation , where y is given in meters, determine the equilibrium positions and investigate the stability at each position. •11–29. The 2-Mg bridge, with center of mass at point G, is lifted by two beams CD, located at each side of the bridge. If the 2-Mg counterweight E is attached to the beams as shown, determine the angle for equilibrium. Neglect the weight of the beams and the tie rods. u V = (3y3 + 2y2 - 4y + 50) J 0° … u … 90° V = (24 sin u + 10 cos 2u) ft # lb V = (4x3 - x2 - 3x + 10) ft # lb PROBLEMS D A G C B E 2.5 m 2.5 m 2 m 5 m 0.3 m 2 m u u Prob. 11–29 A k B C D 5 ft u Prob. 11–30 h A B C k1 20 lb/in. k1 20 lb/in. k2 30 lb/in. Prob. 11–31
- 607. 590 CHAPTER 11 VIRTUAL WORK 11 11–35. Determine the angles for equilibrium of the 200-lb cylinder and investigate the stability of each position. The spring has a stiffness of and an unstretched length of 0.75 ft. k = 300 lbft u 11–34. If a 10-kg load I is placed on the pan, determine the position x of the 0.75-kg block H for equilibrium.The scale is in balance when the weight and the load are not on the scale. •11–33. A 5-kg uniform serving table is supported on each side by pairs of two identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg, determine the angle where the table is in equilibrium. The springs each have a stiffness of and are unstretched when . Neglect the mass of the links. u = 90° k = 200 Nm u *11–36. Determine the angles for equilibrium of the 50-kg cylinder and investigate the stability of each position. The spring is uncompressed when u = 60°. u B C F I H E D A 100 mm 100 mm 100 mm 50 mm x Prob. 11–34 A B C E k D 3 ft 1.5 ft u u Prob. 11–35 A B C 1 m 1 m u k 900 N/m Prob. 11–36 A C k 250 mm 250 mm 150 mm 150 mm B D E u u Prob. 11–33 *11–32. The spring is unstretched when and has a stiffness of . Determine the angle for equilibrium if each of the cylinders weighs 50 lb. Neglect the weight of the members. u k = 1000 lbft u = 45° D E k A B C 4 ft 4 ft 2 ft 2 ft u u Prob. 11–32
- 608. 11.7 STABILITY OF EQUILIBRIUM CONFIGURATION 591 11 11–39. The uniform link AB has a mass of 3 kg and is pin connected at both of its ends.The rod BD, having negligible weight, passes through a swivel block at C. If the spring has a stiffness of and is unstretched when , determine the angle for equilibrium and investigate the stability at the equilibrium position. Neglect the size of the swivel block. u u = 0° k = 100 Nm 11–38. The uniform rod OA weighs 20 lb, and when the rod is in the vertical position, the spring is unstretched. Determine the position for equilibrium. Investigate the stability at the equilibrium position. u •11–37. If the mechanism is in equilibrium when determine the mass of the bar BC.The spring has a stiffness of and is uncompressed when . Neglect the mass of the links. u = 0° k = 2 kNm u = 30°, *11–40. The truck has a mass of 20 Mg and a mass center at G. Determine the steepest grade along which it can park without overturning and investigate the stability in this position. u 600 mm 450 mm B C D H F A k 2 kN/m u u Prob. 11–37 3 ft k 2 lb/in. A O 1 ft u Prob. 11–38 k 100 N/m 400 mm 400 mm D C B A u Prob. 11–39 G u 3.5 m 1.5 m 1.5 m Prob. 11–40
- 609. 592 CHAPTER 11 VIRTUAL WORK 11 11–43. Determine the height h of the cone in terms of the radius r of the hemisphere so that the assembly is in neutral equilibrium. Both the cone and the hemisphere are made from the same material. 11–42. The cap has a hemispherical bottom and a mass m. Determine the position h of the center of mass G so that the cup is in neutral equilibrium. •11–41. The cylinder is made of two materials such that it has a mass of m and a center of gravity at point G. Show that when G lies above the centroid C of the cylinder, the equilibrium is unstable. *11–44. A homogeneous block rests on top of the cylindrical surface. Derive the relationship between the radius of the cylinder, r, and the dimension of the block, b, for stable equilibrium. Hint: Establish the potential energy function for a small angle , i.e., approximate , and . cos u L 1 - u2 2 sin u L 0 u h r G Prob. 11–42 h r Prob. 11–43 b r b Prob. 11–44 C G a r Prob. 11–41
- 610. 11.7 STABILITY OF EQUILIBRIUM CONFIGURATION 593 11 *11–48. The assembly shown consists of a semicircular cylinder and a triangular prism. If the prism weighs 8 lb and the cylinder weighs 2 lb, investigate the stability when the assembly is resting in the equilibrium position. 11–46. The assembly shown consists of a semicylinder and a rectangular block. If the block weighs 8 lb and the semicylinder weighs 2 lb, investigate the stability when the assembly is resting in the equilibrium position. Set 11–47. The 2-lb semicylinder supports the block which has a specific weight of . Determine the height h of the block which will produce neutral equilibrium in the position shown. g = 80 lbft3 h = 4 in. •11–45. The homogeneous cone has a conical cavity cut into it as shown. Determine the depth d of the cavity in terms of h so that the cone balances on the pivot and remains in neutral equilibrium. •11–49. A conical hole is drilled into the bottom of the cylinder, and it is then supported on the fulcrum at A. Determine the minimum distance d in order for it to remain in stable equilibrium. 4 in. 6 in. 8 in. Prob. 11–48 d A r h Prob. 11–49 r d h Prob. 11–45 h 4 in. 10 in. Probs. 11–46/47
- 611. 594 CHAPTER 11 VIRTUAL WORK 11 CHAPTER REVIEW Principle of Virtual Work The forces on a body will do virtual work when the body undergoes an imaginary differential displacement or rotation. For equilibrium, the sum of the virtual work done by all the forces acting on the body must be equal to zero for any virtual displacement. This is referred to as the principle of virtual work, and it is useful for finding the equilibrium configuration for a mechanism or a reactive force acting on a series of connected members. If the system of connected members has one degree of freedom, then its position can be specified by one independent coordinate such as To apply the principle of virtual work, it is first necessary to use position coordinates to locate all the forces and moments on the mechanism that will do work when the mechanism undergoes a virtual movement du. u. The coordinates are related to the independent coordinate and then these expressions are differentiated in order to relate the virtual coordinate displacements to the virtual displacement . Finally, the equation of virtual work is written for the mechanism in terms of the common virtual displacement , and then it is set equal to zero. By factoring out of the equation, it is then possible to determine either the unknown force or couple moment, or the equilibrium position u. du du du u dU = 0 du–virtual rotation dy, dy¿–virtual displacements By Ay Bx P du dy dy¿ F l l P F l l P u u u u
- 612. CHAPTER REVIEW 595 11 Potential-Energy Criterion for Equilibrium When a system is subjected only to conservative forces, such as weight and spring forces, then the equilibrium configuration can be determined using the potential-energy function V for the system. V = V g + Ve = -Wy + 1 2 ky2 The potential-energy function is established by expressing the weight and spring potential energy for the system in terms of the independent coordinate q. Once the potential-energy function is formulated, its first derivative is set equal to zero. The solution yields the equilibrium position for the system. qeq dV dq = 0 The stability of the system can be investigated by taking the second derivative of V. dV dq = d2 V dq2 = d3 V dq3 = Á = 0 neutral equilibrium dV dq = 0, d2 V dq2 6 0 unstable equilibrium dV dq = 0, d2 V dq2 7 0 stable equilibrium y2 y1 y Datum W k (a)
- 613. 596 CHAPTER 11 VIRTUAL WORK 11 REVIEW PROBLEMS *11–52. The uniform links AB and BC each weigh 2 lb and the cylinder weighs 20 lb. Determine the horizontal force P required to hold the mechanism at . The spring has an unstretched length of 6 in. u = 45° 11–51. The uniform rod has a weight W. Determine the angle for equilibrium. The spring is uncompressed when . Neglect the weight of the rollers. u = 90° u 11–50. The punch press consists of the ram R, connecting rod AB, and a flywheel. If a torque of is applied to the flywheel, determine the force F applied at the ram to hold the rod in the position . u = 60° M = 50 N # m •11–53. The spring attached to the mechanism has an unstretched length when . Determine the position for equilibrium and investigate the stability of the mechanism at this position. Disk A is pin connected to the frame at B and has a weight of 20 lb. u u = 90° 1.25 ft 1.25 ft A B C u u u u k 16 lb/ft Prob. 11–53 F 0.1m M B R A u 0.4 m Prob. 11–50 k A B L u Prob. 11–51 P 10 in. B A u C 10 in. k = 2 lb/in. Prob. 11–52
- 614. REVIEW PROBLEMS 597 11 *11–56. The uniform rod AB has a weight of 10 lb. If the spring DC is unstretched when , determine the angle for equilibrium using the principle of virtual work. The spring always remains in the horizontal position due to the roller guide at D. •11–57. Solve Prob. 11–56 using the principle of potential energy. Investigate the stability of the rod when it is in the equilibrium position. u u = 0° 11–55. The uniform bar AB weighs 100 lb. If both springs DE and BC are unstretched when , determine the angle for equilibrium using the principle of potential energy. Investigate the stability at the equilibrium position. Both springs always remain in the horizontal position due to the roller guides at C and E. u u = 90° 11–54. Determine the force P that must be applied to the cord wrapped around the drum at C which is necessary to lift the bucket having a mass m. Note that as the bucket is lifted, the pulley rolls on a cord that winds up on shaft B and unwinds from shaft A. 11–58. Determine the height h of block B so that the rod is in neutral equilibrium. The springs are unstretched when the rod is in the vertical position.The block has a weight W. P c C B A b a Prob. 11–54 B A k k l h Prob. 11–58 A u k 2 lb/in. k 4 lb/in. 2 ft 4 ft D B C E Prob. 11–55 A k 50 lb/ft 1 ft 2 ft C u B D Probs. 11–56/57
- 615. 598 Mathematical Review and Expressions Geometry and Trigonometry Review The angles in Fig. A–1 are equal between the transverse and two parallel lines. u APPENDIX A 180⬚ ⫺ u u u u u u Fig. A–1 u u u u Fig. A–2 For a line and its normal, the angles in Fig.A–2 are equal. u
- 616. APPENDIX A MATHEMATICAL REVIEW AND EXPRESSIONS 599 A s r r u Fig. A–3 For the circle in Fig. A–3 so that when rad then the circumference is Also, since rad, then The area of the circle is A = pr2 . u 1rad2 = 1p180°2u°. 180° = p s = 2pr. u = 360° = 2p s = ur, The sides of a similar triangle can be obtained by proportion as in Fig.A–4, where For the right triangle in Fig.A–5, the Pythagorean theorem is The trigonometric functions are This is easily remembered as “soh, cah, toa”, i.e., the sine is the opposite over the hypotenuse, etc. The other trigonometric functions follow from this. cot u = 1 tan u = a o sec u = 1 cos u = h a csc u = 1 sin u = h o tan u = o a cos u = a h sin u = o h h = 21o22 + 1a22 a A = b B = c C . a b c A B C Fig. A–4 a (adjacent) o (opposite) h (hypotenuse) u Fig. A–5
- 617. 600 APPENDIX A MATHEMATICAL REVIEW AND EXPRESSIONS A Trigonometric Identities Quadratic Formula Hyperbolic Functions tanh x = sinh x cosh x cosh x = ex + e-x 2 , sinh x = ex - e-x 2 , then x = -b ; 2b2 - 4ac 2a If ax2 + bx + c = 0, 1 + tan2 u = sec2 u 1 + cot2 u = csc2 u tan u = sin u cos u sin u = ; A 1 - cos 2u 2 cos u = ; A 1 + cos 2u 2 , cos 2u = cos2 u - sin2 u cos1u ; f2 = cos u cos f sin u sin f sin 2u = 2 sin u cos u sin1u ; f2 = sin u cos f ; cos u sin f sin2 u + cos2 u = 1 Power-Series Expansions Derivatives d dx 1csc u2 = -csc u cot u du dx d dx 1sec u2 = tan u sec u du dx d dx 1cosh u2 = sinh u du dx d dx 1cot u2 = -csc2 u du dx d dx 1sinh u2 = cosh u du dx d dx a u v b = v du dx - u dv dx v2 d dx 1tan u2 = sec2 u du dx d dx 1uv2 = u dv dx + v du dx d dx 1cos u2 = -sin u du dx d dx 1un 2 = nun-1 du dx d dx 1sin u2 = cos u du dx cosh x = 1 + x2 2! + Á sinh x = x + x3 3! + Á , cos x = 1 - x2 2! + Á sin x = x - x3 3! + Á ,
- 618. L cosh x dx = sinh x + C L sinh x dx = cosh x + C L xeax dx = eax a2 1ax - 12 + C L eax dx = 1 a eax + C L x2 cos1ax2 dx = 2x a2 cos1ax2 + a2 x2 - 2 a3 sin1ax2 + C L x cos1ax2 dx = 1 a2 cos1ax2 + x a sin1ax2 + C L cos x dx = sin x + C L sin x dx = -cos x + C c 6 0 = 1 1-c sin-1 a -2cx - b 2b2 - 4ac b + C, c 7 0 x1c + b 21c d + C, L dx 2a + bx + cx2 = 1 1c lnc 2a + bx + cx2 + L x dx 2x2 ; a2 = 2x2 ; a2 + C L dx 2a + bx = 22a + bx b + C a2 8 x2x2 ; a2 - a4 8 lnAx + 2x2 ; a2 B + C L x2 2x2 ; a2 dx = x 4 21x2 ; a2 23 L x2x2 ; a2 dx = 1 3 21x2 ; a2 23 + C APPENDIX A MATHEMATICAL REVIEW AND EXPRESSIONS 601 A Integrals 1 2 cx2x2 ; a2 ; a2 lnAx + 2x2 ; a2 B d + C L 2x2 ; a2 dx = a 7 0 + a2 sin-1 x a b + C, + a2 8 ax2a2 - x2 L x2 2a2 - x2 dx = - x 4 21a2 - x2 23 L x2a2 - x2 dx = - 1 3 21a2 - x2 23 + C a 7 0 L 2a2 - x2 dx = 1 2 cx2a2 - x2 + a2 sin-1 x a d + C, 218a2 - 12abx + 15b2 x2 221a + bx23 105b3 + C L x2 2a + bx dx = L x2a + bx dx = -212a - 3bx221a + bx23 15b2 + C L 2a + bx dx = 2 3b 21a + bx23 + C ab 7 0 L x2 dx a + bx2 = x b - a b2ab tan-1 x2ab a + C, L x dx a + bx2 = 1 2b ln1bx2 + a2 + C ab 6 0 L dx a + bx2 = 1 22-ba lnc a + x2-ab a - x2-ab d + C, L dx a + bx = 1 b ln1a + bx2 + C n Z -1 L xn dx = xn+1 n + 1 + C,
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- 620. F2–9. Ans. Ans. F2–10. Ans. Ans. F2–11. Ans. Ans. F2–12. Ans. Ans. F2–13. Ans. Ans. Ans. g = cos-1 A -37.5 75 B = 120° b = cos-1 A45.93 75 B = 52.2° a = cos-1 A45.93 75 B = 52.2° Fz = -75 sin 30° = -37.5 lb Fy = 75 cos 30° cos 45° = 45.93 lb Fx = 75 cos 30° sin 45° = 45.93 lb u = 39.8° FR = 31.2 kN (FR)y = 15A3 5 B + 20 - 15A3 5 B = 20 kN c (FR)x = 15A4 5 B + 0 + 15A4 5 B = 24 kN : F = 62.5 lb tan u = 0.2547 u = 14.29° = 14.3°a -(80 lb) sin45° = F sin u - A4 5 B(90 lb) + c(FR)y = ©Fy; (800 lb) cos 45° = F cos u + 50 lb - A3 5 B90 lb : + (FR)x = ©Fx; F = 236 N tan u = 0.6190 u = 31.76° = 31.8°a 0 = F sin u + A12 13B(325 N) - (600 N)sin 45° + c(FR)y =©Fy; 750 N = F cos u + A5 13B(325 N) + (600 N)cos45° + :(FR)x = ©Fx; u = 180° + f = 180° + 78.68° = 259° f = tan-1 A 1230 lb 246.22 lb B = 78.68° FR = 2(246.22 lb)2 + (1230 lb)2 = 1254 lb = -1230 lb (FR)y = -(700 lb) sin 30° - 400 lb - A4 5 B (600 lb) + c(FR)y = ©Fy; = -246.22 lb (F R)x = - (700 lb) cos 30° + 0 + A3 5 B (600 lb) + :(FR)x = ©Fx; Fundamental Problems Partial Solutions And Answers Chapter 2 F2–1. Ans. Ans. F2–2. Ans. F2–3. Ans. Ans. F2–4. Ans. Ans. F2–5. Ans. Ans. F2–6. Ans. Ans. F2–7. Ans. Ans. Ans. Ans. Ans. F2–8. Ans. Ans. u = tan-1 350 446.4 = 38.1°a FR = 2(446.4)2 + 3502 = 567 N FRy = 400 sin 30° + 250A3 5 B = 350 N FRx = 300 + 400 cos 30° - 250A4 5 B = 446.4 N (F3)y = A4 5 B600 N = 480 N (F3)x = A3 5 B600 N = 360 N (F2)y = (450 N) sin 45° = 318 N (F2)x = -(450 N) cos 45° = -318 N (F1)x = 0 (F1)y = 300 N Fv sin 45° = 6 sin 105° Fv = 4.39 kN F sin 30° = 6 sin 105° F = 3.11 kN F AC = 636 lb F AC sin 45° = 450 sin 30° F AB = 869 lb F AB sin 105° = 450 sin 30° Fv sin 30° = 30 sin 105° ; Fv = 15.5 lb Fu sin 45° = 30 sin 105° ; Fu = 22.0 lb f = a - 30° = 73.90° - 30° = 43.9° a = 73.90° sin a 800 = sin 60° 721.11 ; = 721.11 N = 721 N FR = 26002 + 8002 - 2(600)(800) cos 60° = 666 N FR = 22002 + 5002 - 2(200)(500) cos 140° u = 45° + f = 45° + 58.49° = 103° f = 58.49° sin f 6 kN = sin 105° 6.798 kN , = 6.798 kN = 6.80 kN F R = 2(2 kN)2 + (6 kN)2 - 2(2 kN)(6 kN) cos 105° 603
- 621. 604 PARTIAL SOLUTIONS AND ANSWERS F2–14. Require . Ans. F2–15. Ans. F2–16. Ans. F2–17. Ans. F2–18. Ans. F2–19. Ans. Ans. Ans. F2–20. Ans. Ans. Ans. F2–21. Ans. = 5180i + 270j - 540k6 N = (630 N)A2 7 i + 3 7 j - 6 7 kB FB = FBuB rB = 52i + 3j - 6k6 m u = 180° - 131.8° = 48.2° a = cos-1 A -4 ft 6 ft B = 131.8° rAB = 2(-4 ft)2 + (2 ft)2 + (4 ft)2 = 6 ft rAB = 5-4i + 2j + 4k6 ft a = 132°, b = 48.2°, g = 70.5° rAB = 2(-6 m)2 + (6 m)2 + (3 m)2 = 9 m rAB = 5-6i + 6j + 3k6 m FR = F1 + F2 = 5490i + 683j - 266k6 lb = 5489.90i + 282.84j - 565.69k6 lb + (800 lb) sin 45° (-k) + [(800 lb) cos 45°] sin 30°j F2 = [(800 lb) cos 45°] cos 30° i = 5400j + 300k6 lb F1 = A4 5 B(500 lb) j + A3 5 B (500 lb)k F2 = 5265i - 459j + 530k6 N Fy = (530.33 N) sin 60° = 459.3 N Fx = (530.33 N) cos 60° = 265.1 N F¿ = (750 N) cos 45° = 530.33 N Fz = (750 N) sin 45° = 530.33 N F = 5-21.2i + 28.3j + 35.4k6 lb Fy = A4 5 B(35.36 lb) = 28.28 lb Fx = A3 5 B (35.36 lb) = 21.21 lb F¿ = (50 lb) cos 45° = 35.36 lb Fz = (50 lb) sin 45° = 35.36 lb = 5250i - 354j - 250k6 N F = FuF = (500 N)(0.5i - 0.7071j - 0.5k) a = 60° cos2 a + cos2 135° + cos2 120° = 1 = 5-250i - 354j + 250k6 N F = FuF = (500 N)(-0.5i - 0.7071j + 0.5k) b = 135° cos b = 21 - cos2 120° - cos2 60° = ;0.7071 F2–22. F2–23. Ans. F2–24. Ans. F2–25. Ans. F2–26. Ans. F2–27. Ans. F2–28. FOA = FOA uOA = 5231i + 96.2j6 N FOA = F # uOA = 250 N F = FuF = [650j] N uOA = 12 13 i + 5 13 j cos u = 5 13; u = 67.4° uOA # j = uOA (1) cos u uOA = 12 13 i + 5 13 j u = cos-1 (uAB # uF) = 68.9° uF = 4 5 i - 3 5 j uAB = -3 5 j + 4 5 k u = cos-1 (uAO # uF) = 57.7° uF = -0.5345i + 0.8018j + 0.2673k uAO = -1 3 i + 2 3 j - 2 3 k FR = FB + FC = 5-620i + 610j -540k6 lb = 5-420i + 210j - 140k6 lb = (490 lb)A -6 7 i + 3 7 j - 2 7 kB FC = FCuC = 5-200i + 400j - 400k6 lb = (600 lb)A -1 3 i + 2 3 j - 2 3 kB FB = FBuB = 1.18 kN FR = 2(480 N)2 + (-60 N)2 + (-1080 N)2 = 5120i + 180j - 360k6 N = (420 N) A2 7 i + 3 7 j - 6 7 kB FC = FCuC = 5360i - 240j - 720k6 N = (840 N)A3 7 i - 2 7 j - 6 7 kB FB = FBuB = 5-400i + 700j - 400k6 N F = FuAB = 900NA -4 9 i + 7 9 j - 4 9 kB
- 622. FUNDAMENTAL PROBLEMS 605 F2–29. Ans. F2–30. Ans. Ans. Chapter 3 F3–1. Ans. Ans. F3–2. Ans. F3–3. Ans. F3–4. Ans. F3–5. Ans. mA = 20 kg + c ©Fy = 0; (392.4 N)sin 30° - mA(9.81) = 0 l0 = 0.283 m Fsp = k(l - l0); 43.35 = 200(0.5 - l0) Fsp = 43.35 N +Q©Fx = 0; 4 5 (Fsp) - 5(9.81) sin 45° = 0 T = 40.9 N u = tan-1 A0.15 m 0.2 m B = 36.87° + c©Fy = 0; 2T sin u - 49.05 N = 0 f = u : + ©Fx = 0; T cos u - T cos f = 0 LABC = 2A 5 ft cos 13.5° B = 10.3 ft u = 13.5° + c©Fy = 0; -2(1500) sin u + 700 = 0 FAC = 518 lb FAB = 478 lb + c©Fy = 0; 3 5 FAC + FAB sin 30° - 550 = 0 + : ©F x = 0; 4 5F AC - F AB cos 30° = 0 = 401 lb (FA)per = 2(600 lb)2 - (446.41 lb)2 (FA)proj = F # uA = 446.41 lb = 446 lb uA = -2 3 i + 2 3 j + 1 3 k = 5-150i + 259.81j + 519.62k6 lb + [(600 lb) sin 60°] k + [(600 lb) cos 60°] cos 30° j F = [(-600 lb) cos 60°] sin 30°i (FAO)proj = F # uAO = 244 N = - 0.5547j - 0.8321k uAO = 5-4 j- 6 k6m 2(-4 m)2 + (-6 m)2 = 5219.78i + 54.94j - 329.67k6 N F = (400 N) 54 i+1 j-6 k6m 2(4 m)2 + (1 m)2 + (-6 m)2 F3–6. Ans. Ans. Ans. Ans. F3–7. (1) (2) (3) Ans. Ans. Ans. F3–8. Ans. Ans. Ans. F3–9. Ans. Ans. Ans. F3–10. Ans. Ans. Ans. FAB = 138.60 lb = 139 lb ©F x = 0; F AB - 0.25(202.92) - 0.5(175.74) = 0 FAC = 202.92 lb = 203 lb FAD = 175.74 lb = 176 lb ©Fz = 0; 0.8660F AC + 0.7071F AD - 300 = 0 ©F y = 0; 0.4330F AC - 0.5F AD = 0 = -0.5FAD i - 0.5FAD j + 0.7071FAD k FAD = FAD5cos 120° i + cos 120° j + cos 45°k6 = -0.25F AC i + 0.4330F AC j + 0.8660F AC k + cos 60° cos 30° j + sin 60° k6 FAC = FAC 5-cos 60° sin 30° i FAC = 646.41 N = 646 N 1 3 (900) + 692.82 sin 30° - FAC = 0 ©Fx = 0; FAB = 692.82 N = 693 N FAB cos 30° - 2 3 (900) = 0 ©Fy = 0; FAD = 900 N 2 3FAD - 600 = 0 ©Fz = 0; FAD = FADa rAD rAD b = 1 3 FADi - 2 3 FAD j + 2 3 F AD k FAB = 506.25 N = 506 N ©Fx = 0; FAB - 843.75A3 5 B = 0 FAC = 843.75 N = 844 N ©Fy = 0; FACA4 5 B - 1125A3 5 B = 0 FAD = 1125 N = 1.125 kN ©Fz = 0; FADA4 5 B - 900 = 0 F2 = 879 N F1 = 466 N F3 = 776 N ©Fz = 0; A4 5 BF3 + A3 5 BF1 - 900 N = 0 ©Fy = 0; A4 5 BF1 - C A3 5 BF3D A4 5 B = 0 ©Fx = 0; CA3 5 BF3D A3 5B + 600 N - F2 = 0 u = 21.9° TCD = 395 N + c©Fy = 0; TCD sin u - 15(9.81) N = 0 + :©Fx = 0; TCD cos u - 366.11 N = 0 TBC = 366.11 N = 366 N + :©Fx = 0; TBC - 379.03 N cos 15° = 0 TAB = 379.03 N = 379 N + c©Fy = 0; TAB sin 15° - 10(9.81) N = 0
- 623. 606 PARTIAL SOLUTIONS AND ANSWERS F4–8. a b F4–9. a Ans. F4–10. Ans. or Ans. F4–11. Ans. or Ans. F4–12. Ans. = 5485i - 1000j + 1020k6 lb # ft (MR)O = rA * FR = 3 i j k 4 5 3 -100 130 175 3 = 5-100i + 130j + 175k6 lb + (75 + 100)k6 lb = 5(100 - 200)i + (-120 + 250)j FR = F1 + F2 = 5200j - 400k6 lb # ft MO = rB * F = 3 i j k 1 4 2 80 -80 -40 3 = 5200j - 400k6 lb # ft MO = rC * F = 3 i j k 5 0 0 80 -80 -40 3 = 580i - 80j - 40k6 lb = 120 lbB 54 i -4 j- 2 k6 ft 2(4 ft)2 + (-4 ft)2 + (-2 ft)2 R F = FuBC = 5-1200k6 N # m MO = rOB * F = 54i6 m * 5400i - 300j6 N = 5-1200k6 N # m MO = rOA * F = 53j6 m * 5400i - 300j6 N F = FuAB = 500 NA4 5 i - 3 5 jB = 5400i - 300j6N = 2.60 kip # ft + (200 lb)(6 cos 30° ft) - (300 sin 30° lb)(6 cos 30° ft) (MR)O = (300 cos 30° lb)(6 ft + 6 sin 30° ft) +(MR)O = ©Fd; = -268 N # m = 268 N # m - [(600 N) sin 60°](0.425 m) - [(600 N) cos 60°](0.25 m) - C A4 5 B500 ND(0.25 m) (MR)O = C A3 5 B500 ND(0.425 m) +(MR)O = ©Fd; F3–11. (1) (2) (3) Ans. Ans. Ans. Chapter 4 F4–1. a Ans. F4–2. a b Ans. F4–3. a Ans. F4–4. a Ans. F4–5. c Ans. F4–6. a Ans. F4–7. a Ans. = 1254 N # m = 1.25 kN # m - (300N)[(2.5 m) sin 45°] + (500 N)[3 m + (2.5 m) cos 45°] (MR)O = -(600 N)(1 m) +(MR)O = ©Fd; = 1.06 kN # m - 500 cos 45° (3 sin 45°) +MO = 500 sin 45° (3 + 3 cos 45°) = 11.2 N # m - 50 cos 60°(0.2 sin 45°) +MO = 50 sin 60° (0.1 + 0.2 cos 45° + 0.1) = 3.07 kip # ft +MO = (600 lb)(4 ft + (3 ft)cos 45° - 1 ft) = 36.7 N # m - [(300 N) cos 30°][(0.3 m) sin 45°] +MO = [(300 N) sin 30°][0.4m + (0.3 m) cos 45°] = -460 N # m = 460 N # m +MO = - A4 5 B(100 N)(2 m)- A3 5 B(100 N)(5 m) = 2.49 kip # ft +MO = 600 sin 50° (5) + 600 cos 50° (0.5) FD = 346.15 lb = 346 lb FC = 1.5(162 lb) = 242 lb FB = 162 lb ©Fz = 0; 2 7 FB + 3 7 FC - 150 = 0 ©Fy = 0; 3 7 FB - 2 7 FC = 0 ©Fx = 0; -6 7 FB - 6 7 FC + FD = 0 W = 5-150k6 lb FD = FDi = -6 7 FCi - 2 7FCj + 3 7 FC k = FCB 5-6i - 2j + 3k6 ft 2(-6 ft)2 + (-2 ft)2 + (3ft)2 R FC = FCa rAC rAC b = -6 7FBi + 3 7 FBj + 2 7 FBk = FBB 5-6i + 3j + 2k6 ft 2(-6 ft)2 + (3 ft)2 + (2 ft)2 R FB = FBa rAB rAB b
- 624. FUNDAMENTAL PROBLEMS 607 F4–13. Ans. F4–14. Ans. F4–15. F4–16. Ans. F4–17. Ans. F4–18. Ans. Ans. Ans. = -160 N#m Mz = 240 N(2 m) - 320 N(2 m) = -120 N#m My = 300 N(2 m) - 240 N(3 m) = -360 N # m Mx = 300 N(2 m) - 320 N(3 m) Fz = (500 N)A3 5 B = 300 N Fy = C A4 5 B500 ND A4 5 B = 320 N Fx = C A4 5 B500 ND A3 5 B = 240 N MAB = MABuAB = 53.2i - 2.4j6 lb#ft = -4 lb#ft = 4 i j k -0.8 0.6 0 0 0 2 50 -40 20 4 MAB = uAB #(rAC * F) uAB = rAB rAB = 5-4i + 3j6ft 2(-4 ft)2 + (3 ft)2 = -0.8i + 0.6j = 210 N # m Mp = j#(rA * F) = 3 0 1 0 -3 -4 2 30 -20 50 3 = 17.4 N # m MO = i#(rA * F) = 3 1 0 0 0 0.3 0.25 -100 100 141.42 3 = 5-100i + 100j + 141.42k6 N + (200 N) cos 60° j + (200 N) cos 45° k F = (200 N) cos 120° i = -72 N # m MOA = uOA #(rAB * F) = 3 0.6 0.8 0 0 0 -0.2 300 -200 150 3 uOA = rA rA = 50.3i + 0.4j6 m 2(0.3 m)2 + (0.4 m)2 = 20 N # m Mx = i#(rOB*F) = 3 1 0 0 0.3 0.4 -0.2 300 -200 150 3 F4–19. c Ans. Also, c Ans. F4–20. a Ans. F4–21. a Ans. F4–22. a b F4–23. Ans. F4–24. Ans. Also, Ans. = {108j + 144k} N#m = 3 i j k 0 0 0.3 0 -360 270 3 + 3 i j k 0.4 0 0.3 0 360 -270 3 Mc = (rA * FA) + (rB * FB) = {108j + 144k} N#m Mc = rAB * FB = 3 i j k 0.4 0 0 0 360 -270 3 = 5360j - 270k6 N FB = A4 5 B(450 N)j - A3 5 B(450 N) k (Mc)R = {-20i - 40j + 100k} lb#ft (Mc)R = ©Mc; = 5180i - 240j6 lb#ft (Mc)3 = (Mc)3 u3 = (300 lb#ft)A1.5 2.5 i - 2 2.5 jB = 5-250k6 lb#ft (Mc)2 = (Mc)2u2 = (250 lb#ft)(-k) = 5-200i + 200j + 350k6 lb#ft = (450 lb#ft)A- 2 4.5 i + 2 45 j + 3.5 4.5 kB (Mc)1 = (Mc)1u1 u3 = 1.5 2.5 i - 2 2.5 j u2 = -k = - 2 4.5 i + 2 4.5 j + 3.5 4.5 k u1 = r1 r1 = [-2i + 2j + 3.5k] ft 2(-2 ft)2 + (2 ft)2 + (3.5 ft)2 = 20 kN # m +MC = 10A3 5 B(2) - 10A4 5 B(4) = -20 kN # m F = 2.33 kN -1.5 kN#m = (2 kN)(0.3 m) -F(0.9 m) +(MB)R = ©MB = 2600 lb # ft +MCR = 300(4) + 200(4) + 150(4) = 740 N # m +MCR = 300(5) - 400(2) + 200(0.2) + 200(0.2) = 740 N # m +MCR = ©MA = 400(3) - 400(5) + 300(5)
- 625. 608 PARTIAL SOLUTIONS AND ANSWERS F4–29. Ans. Ans. F4–30. Ans. Ans. F4–31. Ans. c Ans. F4–32. Ans. a Ans. d = 3.12 ft 163.30(d) = 200(3) - 100A4 5B(6) + 50 cos 30°(9) +(MR)A = ©MA; u = tan-1 A163.30 85 B = 62.5° a FR = 2852 + 163.302 = 184 lb = 163.30 lbc (F R)y = 200 + 50 cos 30° - 100A4 5B + c(FR)y = ©Fy; (FR)x = 100A3 5 B + 50 sin 30° = 85 lb : : + (FR)x = ©Fx; x = 6 ft 1250(x) = 500(3) + 250(6) + 500(9) FRx = ©MO; + = 1250 lb + TFR = ©Fy; FR = 500 + 250 + 500 = {-105i - 48 j + 80k} N#m + 3 i j k 0 0.5 0.3 -160 0 -120 3 + (-75i) (MR )O = (0.3k)*(-100j) FR = {-160i - 100j - 120k} N Mc = 5-75i6 N#m = 5-160i - 120k6 N F2 = (200 N)B 5-0.4i - 0.3k6 m 2(-0.4 m)2 + (-0.3 m)2 R F1 = 5-100j6 N = 5-650i + 375k6N#m = 3 i j k -1.5 2 1 -300 150 200 3 + 3 i j k 0 2 0 0 0 -450 3 (MR)O = rOB * F1 + rOA * F2 (MR)O = ©M; = 5-1.5i + 2j + 1k6 m rOB = (-1.5 - 0)i + (2 - 0)j + (1 - 0)k rOA = (2 - 0)j = 52j6 m = 5-300i + 150j - 250k6 N = (-300i + 150j + 200k) + (-450k) FR = F1 + F2 FR = ©F; F4–25. Ans. Ans. c Ans. F4–26. Ans. Ans. c Ans. F4–27. Ans. Ans. a b Ans. F4–28. Ans. Ans. a b Ans. = -640 = 640 lb#ft (MR)A = 100A4 5 B(1) - 100A3 5 B(6) - 150A4 5 B(3) +(MR)A = ©MA; u = tan-1 A180 60 B = 71.6° c F R = 2602 + 1802 = 189.74 lb = 190 lb = -180 lb = 180 lb T (F R)y = -150A4 5 B - 100A3 5 B + c(FR)y = ©Fy; (FR)x = 150A3 5 B + 50 - 100A4 5 B = 60 lb : + : (FR)x = ©Fx; = 960 N # m = -959.57 N # m (MR)A = 300 - 900 cos30° (0.75) - 300(2.25) (MR)A = ©MA; + u = tan-1 A1079.42 450 B = 67.4° c = 1169.47 N = 1.17 kN FR = 24502 + 1079.422 = -1079.42 N = 1079.42 N T (FR)y = -900 cos 30° - 300 + c(FR)y = ©Fy; (FR)x = 900 sin 30° = 450 N : : + (FR)x = ©Fx; = 470 N # m MAR = 30(3) + 3 5 (50)(6) + 200 MAR = ©MA; + u = tan-1 A100 40 B = 68.2° c FR = 2(40)2 + (100)2 = 108 N = 100 N + TFRy = ©Fy; FRy = 40 + 30 + 3 5 (50) FRx = 4 5 (50) = 40 N + : FR x = ©Fx ; MRA = 210 lb # ft MAR = 3 5 (100)(4) - 4 5 (100)(6) + 150(3) MAR = ©MA; + u = tan-1 A 70 140 B = 26.6° d FR = 21402 + 702 = 157 lb + TFRy = ©Fy; FRy = 150 - 4 5 (100) = 70 lb + ; FRx = ©Fx; FRx = 200 - 3 5 (100) = 140 lb
- 626. FUNDAMENTAL PROBLEMS 609 F4–33. Ans. Ans. a Ans. F4–34. Ans. Ans. a Ans. F4–35. Ans. Ans. Ans. F4–36. Ans. Ans. Ans. x = 0.667 m 600x = 100(3) + 100(3) + 200(2) - 200(3) MRy = ©My; y = -0.667 m -600y = 200(1) + 200(1) + 100(3) - 100(3) MRx = ©Mx; = 600 N FR = 200 + 200 + 100 + 100 + TFR = ©Fz; x = 2.125 m MRy = ©My; 800x = 500(4) - 100(3) y = 4.50 m MRx = ©Mx; -800y = -400(4) - 500(4) = 800 N + TFR = ©Fz; FR = 400 + 500 - 100 d = 0.2 m - C A3 5 B5kND(4 m) - C A4 5 B5 kND(2 m) 5 kN(d) = 8 kN(3 m) - 6 kN(0.5 m) +(MR)A = ©MA; u = tan-1 A10 kN 5 kN B = 63.4°d FR = 252 + 102 = 11.2 kN = -10 kN = 10 kNT (FR)y = -6 kN - A4 5 B 5 kN + c(FR)y = ©Fy; = -5 kN = 5 kN ; (FR)x = A3 5 B 5 kN - 8 kN + :(FR)x = ©Fx; d = 0.909 m -11(d) = -20(2) - 15A4 5 B(2) + 15A3 5 B(6) +(MR)A = ©MA; u = tan-1 A11 12 B = 42.5° c FR = 2122 + 112 = 16.3 kN (FR)y = -20 + 15A3 5 B = -11 kN = 11 kN T + c(FR)y = ©Fy; (FR)x = 15A4 5 B = 12 kN : : + (FR)x = ©Fx; F4–37. Ans. a Ans. F4–38. Ans. c Ans. F4–39. Ans. a Ans. F4–40. Ans. c Ans. F4–41. Ans. a Ans. F4–42. c x = L xw(x) dx L w(x) dx = L 4 0 2.5x4 dx 160 = 3.20 m MAR = ©MA; + FR = L w(x) dx = L 4 0 2.5x3 dx = 160 N d = 2.59 m -24.75(d) = - 1 2 (3)(4.5)(1.5) - 3(6)(3) +(MR)A = ©MA; FR = 24.75 kN T -FR = - 1 2 (3)(4.5) - 3(6) + cFR = ©Fy; d = 5.03 ft 1550d = C1 2 (50)(6)D(4) + [150(6)](3) + 500(9) MAR = ©MA; + = 1550 lb FR = 1 2 (50)(6) + 150(6) + 500 + TFR = ©Fy; d = 1 m -27(d) = 1 2 (6)(3)(1) - 1 2 (6)(6)(2) +(MR)A = ©MA; FR = 27 kN T -FR = - 1 2 (6)(3) - 1 2 (6)(6) + cFR = ©Fy; d = 8.36 ft 1650d = C1 2 (6)(150)D(4) + [8(150)](10) MAR = ©MA; + FR = 1 2 (6)(150) + 8(150) = 1650 lb d = 1.25 m - 9(3)(1.5) - 3(1.5)(3.75) -40.5(d) = 6(1.5)(0.75) +(MR)A = ©MA; FR = 40.5 kN T -FR = -6(1.5) - 9(3) - 3(1.5) + cFR = ©Fy;
- 627. 610 PARTIAL SOLUTIONS AND ANSWERS F5–5. a Ans. Ans. Ans. F5–6. Ans. a Ans. Ans. F5–7. Ans. F5–8. Ans. Ans. Ans. Ans. Ans. TBC = 352.5 N TBC + 660 N + 487.5 N - 900 N - 600 N = 0 ©Fz = 0; Dy = 0 ©Fy = 0; Dx = 0 ©Fx = 0; Dz = 487.5 N Dz(0.8 m) - 600 N(0.5 m) - 900 N(0.1 m) = 0 ©Mx = 0; FA = 660 N 600 N(0.2 m) + 900 N(0.6 m) - FA(1 m) = 0 ©My = 0; TB = 250 lb, TC = 100 lb TA = 350 lb, -TB(4) - TC(4) + 500(2) + 200(2) = 0 ©My = 0; TA(3) + TC(3) - 500(1.5) - 200(3) = 0 ©Mx = 0; TA + TB + TC - 200 - 500 = 0 ©Fz = 0; NB = 327 N - (250 N) cos 60° = 0 NB - 577.4 N + (433.0 N)cos 30° + c©Fy = 0; NA = 577.4 N = 577 N + [(250 N) cos 30°](0.6 m) = 0 -NA sin 30°(0.15 m) - 433.0 N(0.2 m) +©MB = 0; NC = 433.0 N = 433 N NC sin 30° - (250 N) sin 60° = 0 + :©Fx = 0; FA = 93.5 N + (151.71 N) sin 60° - 25(9.81) N = 0 FA + (78.53 N) sin 15° + c©Fy = 0; TAB = 78.53 N = 78.5 N TAB cos 15° - (151.71 N) cos 60° = 0 + :©Fx = 0; NC = 151.71 N = 152 N NC(0.7 m) - [25(9.81) N] (0.5 m) cos 30° = 0 +©MA = 0; Chapter 5 F5–1. Ans. c Ans. Ans. F5–2. a Ans. Ans. Ans. F5–3. a Ans. Ans. Ans. F5–4. Ans. Ans. a Ans. MA = 3.90 kN # m - 400 sin 30°(4.5) - 400 cos 30°(3 sin 60°) = 0 MA - 200(2.5) - 200(3.5) - 200(4.5) +©MA = 0; Ay = 800 N Ay - 200 - 200 - 200 - 400 sin 30° = 0 + c©Fy = 0; Ax = 346 N -Ax + 400 cos 30° = 0 : + ©Fx = 0; Ay = 5.49 kN Ay + 8.047 kN - (5 kN) sin 45° - 10 kN = 0 + c©Fy = 0; Ax = 3.54 kN (5 kN) cos 45° - Ax = 0 + :©Fx = 0; NB = 8.047 kN = 8.05 kN - 5 kN(4 m) = 0 - 10 kN[2 m + (6 m) cos 45°] NB[6 m + (6 m) cos 45°] +©MA = 0; Ay = -4 kN = 4 kN T Ay + (11.31 kN) sin 45° - 4 kN = 0 + c©Fy = 0; Ax = -8 kN = 8 kN ; + :©Fx = 0; Ax + (11.31 kN) cos 45° = 0 FCD = 11.31 kN = 11.3 kN FCD sin 45°(1.5 m) - 4 kN(3 m) = 0 +©MA = 0; Ay = 140 lb + c©Fy = 0; Ay + 260 - 500A4 5 B = 0 By = 260 lb ©MA = 0; By(10) - 500A4 5 B(5) - 600 = 0 + Ax = 300 lb : + ©Fx = 0; -Ax + 500A3 5 B = 0
- 628. FUNDAMENTAL PROBLEMS 611 F5–9. Ans. Ans. Ans. Ans. Ans. Ans. F5–10. Ans. Ans. Ans. Ans. Az + 1350 N + (-1800 N) - 450 N = 0 ©Fz = 0; Cz = 1350 N Bz = -1800 N -Cz - 0.6Bz + 270 = 0 - Bz(0.6 m) + 450 N(0.6 m) = 0 -Cz(0.6 m + 0.4 m) ©My = 0; 1.2Cz + 0.6Bz - 540 = 0 - 450 N(0.6 m + 0.6 m) = 0 ©Mx = 0; Cz(0.6 m + 0.6 m) + Bz(0.6 m) Ay = 0 Ay + 0 = 0 ©Fy = 0; Cy = 0 Cy(0.4 m + 0.6 m) = 0 ©Mz = 0; Bx = 0 ©Fx = 0; Az = 333.3 N Az - 933.3N + 600 N = 0 ©Fz =0; Ax = 500 N 1400 N + (-900 N)-Ax = 0 ©Fx = 0; Bx = 1400 N + (-400 N)(0.6 m) = 0 -(-900 N)(1.2 m) -Bx (0.6 m) + ©Mz = 0; Bz = -933.3 N + (-400 N)(0.4 m) = 0 Bz (0.6 m) + 600 N (1.2 m) ©Mx = 0; Cx = -900 N -Cx (0.4 m) - 600 N ( 0.6 m) = 0 ©My = 0; Cy = -400 N 400 N + Cy = 0; ©Fy = 0; F5–12. Ans. Ans. Ans. Ans. Ans. Chapter 6 F6–1. Joint A. Ans. Ans. Joint B. Ans. Ans. Joint D. Ans. F6–2. Joint D: Ans. Ans. Ans. F6–3. Joint A: Ans. Joint C: Ans. FDC = 400 lb (C) + c©Fy = 0; -FDC + 400 = 0; FAE = 667 lb (C) + c©Fy = 0; - 3 5 FAE + 400 = 0 Ax = 0, Ay = Cy = 400 lb FBC = 500 lb (T), FAC = FAB = 0 FAD = 400 lb (C) : + ©Fx = 0; -FAD + 4 5 (500) = 0 FCD = 500 lb (T) + c©Fy = 0; 3 5 FCD - 300 = 0; FCD = 318.20 lb = 318 lb (T) FCD cos 45° + (318.20 lb) cos 45° - 450 lb = 0 + :©Fx = 0; + c©Fy = 0; FBD = 0 FBC = 225 lb (T) + :©Fx = 0; FBC - 225 lb = 0 FAB = 225 lb (T) + :©Fx = 0; FAB - (318.20 lb) cos 45° = 0 FAD = 318.20 lb = 318 lb (C) + c©Fy = 0; 225 lb - FAD sin 45° = 0 Az = 40 lb (MA)x = 240 lb # ft ©Mz = 0; (MA)z = 0 ©My = 0; 3FBC - 80(1.5) = 0 FBC = 40 lb ©Mx = 0; (MA)x + 6FBC - 80(6) = 0 ©Fz = 0; Az + FBC - 80 = 0 ©Fy = 0; Ay = 0 ©Fx = 0; Ax = 0 F5–11. Ans. Ans. Ans. Ans. Ans. Ans. FDB = -6.75 kN ©Fz = 0; FDB + 9 - 9 + 6.75 = 0 ©Fx = 0; Ax + 6 - 6 = 0 Ax = 0 Az = 6.75 kN ©My = 0; 9(4) - Az (4) - 6(1.5) = 0 FCF = 6 kN ©Mz = 0; FCF(3) - 6(3) = 0 FCE = 9 kN ©Mx = 0; -9(3) + FCE(3) = 0 ©Fy = 0; Ay = 0 Ans. Az = 900 N
- 629. 612 PARTIAL SOLUTIONS AND ANSWERS F6–8. Ans. a Ans. Ans. F6–9. a From the geometry of the truss, . a Ans. a Ans. Ans. F6–10. From the geometry of the truss, a Ans. a Ans. a Ans. F6–11. From the geometry of the truss, . The location of G can be found using similar triangles. x = 2 m 4 m = 2 m + x 1 m 2 m = 2 m 2 m + x f = tan-1 (3 m2 m) = 56.31° u = tan-1 (1 m2 m) = 26.57° FBC = 346.41 lb = 346 lb (T) 300 lb(9 ft) - 300 lb(3 ft) - FBC(9 ft)tan 30° = 0 +©MF = 0; FCF = 346.41 lb = 346 lb (T) 300 lb(6 ft) - FCF sin 60° (6 ft) = 0 +©MD = 0; FEF = -600 lb = 600 lb (C) FEF sin 30°(6 ft) + 300 lb(6 ft) = 0 +©MC = 0; tan f = (9 ft) tan 30° 3 ft = 1.732 f = 60° FKD = 8.01 kN (T) 33.33 kN - 40 kN + FKD sin 56.31° = 0 + c©Fy = 0; FKJ = 66.7 kN (C) +©MD = 0; 33.33 kN(6 m) - FKJ(3 m) = 0 FCD = 62.2 kN (T) 33.33 kN(8 m) - 40 kN(2 m) - FCD(3 m) = 0 +©MK = 0; f = tan-1 (3 m2 m) = 56.31° Gy = 33.33 kN - 30 kN(4 m) - 40 kN(6 m) = 0 Gy(12 m) - 20 kN(2 m) +©MA = 0; FLK = 62.2 kN (C) FLK - 62.22 kN = 0 + :©Fx = 0; FCD = 62.22 kN = 62.2 kN (T) 33.33 kN(8 m) - 40 kN(2 m) - FCD(3 m) = 0 + ©MK = 0; FKC = 6.67 kN (C) FKC + 33.33 kN - 40 kN = 0 + c©Fy = 0; F6–4. Joint C. Joint B. The smaller value of P is chosen, Ans. F6–5. Ans. Ans. Ans. Ans. F6–6. Joint C. Ans. Ans. Joint D. Ans. Ans. Joint B. Ans. Ans. Joint A. Ans. F6–7. Ans. a Ans. a Ans. FBC = 2200 lb (C) +©MF = 0; FBC(4) - 600(4) - 800(8) = 0 FFE = 800 lb (T) ©MC = 0; FFE(4) - 800(4) = 0 + FCF = 1980 lb (T) + c©Fy = 0; FCF sin 45° - 600 - 800 = 0 FAE = 340 lb (C) + c©Fy = 0; 340.19 lb - FAE = 0 FAB = 450 lb (T) + :©Fx = 0; 450 lb - FAB = 0 c©Fy = 0; FBE sin f = 0 FBE = 0 FDE = 519.62 lb = 520 lb (C) +R©Fx¿ = 0; FDE - 519.62 lb = 0 +Q©Fy¿ = 0; FBD cos 30° = 0 FBD = 0 FBC = 450 lb (T) + :©Fx = 0; (519.62 lb) cos 30° - FBC = 0 FCD = 519.62 lb = 520 lb (C) + c©Fy = 0; 259.81 lb - FCD sin 30° = 0 FDE = 0 FAE = 0 FCD = 0 FCB = 0 P = 2.598 kN = 2.60 kN P = 2.598 kN FAC = FBC = 0.5774P = 1.5 kN P = 6.928 kN FAB = 0.2887P = 2 kN FAB = 0.2887P (T) 0.5774P cos 60° - FAB = 0 + :©Fx = 0; FAC = FBC = F = P 2 cos 30° = 0.5774P (C) + c©Fy = 0; 2F cos 30° - P = 0
- 630. FUNDAMENTAL PROBLEMS 613 a Ans. a Ans. a Ans. F6–12. a Ans. a Ans. a Ans. F6–13. Ans. F6–14. a Ans. Ans. F6–15. a Ans. Ans. = 575 N FA = 2(353.55 N)2 + (453.55 N)2 Ay = 453.55 N + c©Fy = 0; Ay - 100 N - (500 N) cos 45° = 0 Ax = 353.55 N : + ©Fx = 0; (500 N) sin 45° - Ax = 0 NB = 500 N +©MA = 0; 100 N(250 mm) - NB(50 mm) = 0 Cy = 467 lb + c ©Fy = 0; Cy + 4 5 (541.67) - 400 - 500 = 0 Cx = 325 lb : + ©Fx = 0; -Cx + 3 5 (541.67) = 0 FAB = 541.67 lb - A4 5 B(FAB)(9) + 400(6) + 500(3) = 0 ©MC = 0; + P = 20 lb + c©Fy = 0; 3P - 60 = 0 FJI = 0 - 900 lb(12 ft) - 1600 lb(9 ft) = 0 FJI cos 45°(12 ft) + 1200 lb(21 ft) +©MC = 0; FHI = 900 lb (C) 1200 lb(21 ft) - 1600 lb(9 ft) - FHI(12 ft) = 0 +©MD = 0; FDC = 1900 lb (C) FDC(12 ft) + 1200 lb(9 ft) - 1600 lb(21 ft) = 0 +©MH = 0; FGD = 2.253 kN = 2.25 kN (T) - FGD sin 56.31°(4 m) = 0 15 kN(4 m) - 26.25 kN(2 m) +©MO = 0; FGF = 29.3 kN (C) 26.25 kN(2 m) - FGF cos 26.57°(2 m) = 0 +©MD = 0; FCD = 25 kN (T) 26.25 kN(4 m) - 15 kN(2 m) - FCD(3 m) = 0 +©MG = 0; F6–16. a Ans. Ans. F6–17. Plate A: Plate B: Ans. F6–18. Pulley C: Beam: Ans. a Ans. Chapter 7 F7–1. a Ans. Ans. a Ans. F7–2. a Ans. Ans. a Ans. MC = -22.5 kN # m +©MC = 0; MC + 30 - 5(1.5) = 0 VC = 5 kN + c©Fy = 0; 5 - VC = 0 : + ©Fx = 0; NC = 0 Ay = 5 kN +©MB = 0; 30 - 10(1.5) - Ay(3) = 0 MC = 18.75 kN # m +©MC = 0; 13.75(3) - 15(1.5) - MC = 0 VC = 1.25 kN + c©Fy = 0; VC + 13.75 - 15 = 0 : + ©Fx = 0; NC = 0 By = 13.75 kN +©MA = 0; By(6) - 10(1.5) - 15(4.5) = 0 x = 0.333 m ©MA = 0; 2(1) - 6(x) = 0 + P = 2 kN + c©Fy = 0; 2P + P - 6 = 0 + c©Fy = 0; T - 2P = 0; T = 2P T = 32.5 lb, NAB = 35 lb + c©Fy = 0; 2T - NAB - 30 = 0 + c©Fy = 0; 2T + NAB - 100 = 0 Cy = 400 N + c©Fy = 0; -Cy + 1131.37 sin 45° - 400 = 0 Cx = 800 N : + ©Fx = 0; -Cx + 1131.37 cos 45° = 0 FAB = 1131.37 N + 800 + 400(2) = 0 FAB cos 45°(1) - FAB sin 45°(3) ©MC = 0; +
- 631. 614 PARTIAL SOLUTIONS AND ANSWERS F7–8. a F7–9. a F7–10. a F7–11. Region a Region a F7–12. Region a Region a F7–13. x = 3, V = -18, M = -32; x = 2+ , V = -18, M = -14; x = 1+ , V = -10, M = -4; x = 0, V = -4, M = 0; M = A4(6 - x)B kN # m +©MO = 0; 4(6 - x) - M = 0 + c©Fy = 0; V + 4 = 0 V = -4 kN 3 m 6 x … 6 m M = 12 kN # m +©MO = 0; M - 12 = 0 + c©Fy = 0; V = 0 0 … x 6 3 m M = A5(6 - x)B kN # m +©MO = 0; 5(6 - x) - M = 0 V = -5 kN + c©Fy = 0; V + 5 = 0 0 6 x … 6 m M = (-5x) kN # m +©MO = 0; M + 5x = 0 V = -5 kN + c©Fy = 0; -V - 5 = 0 3 … x 6 3 m M = (-2x) kN # m +©MO = 0; M + 2x = 0 V = -2 kN + c©Fy = 0; -V - 2 = 0 M = - A1 3 x3 B kN # m +©MO = 0; M + 1 2 (2x)(x)Ax 3 B = 0 V = - Ax2 B kN + c©Fy = 0; -V - 1 2 (2x)(x) = 0 M|x=9 ft = 15 - 92 = -66 kip # ft V|x=9 ft = -2(9) = -18 kip M = A15 - x2 B kip # ft +©MO = 0; M + 2xAx 2 B - 15 = 0 V = (-2x) kip + c©Fy = 0; -V - 2x = 0 F7–3. a Ans. a Ans. a Ans. F7–4. a Ans. Ans. a Ans. F7–5. a Ans. Ans. a Ans. F7–6. a Ans. Ans. a Ans. F7–7. a M = (6x - 18) kN # m +©MO = 0; M + 18 - 6x = 0 + c©Fy = 0; 6 - V = 0 V = 6 kN MC = 22.5 kN # m +©MC = 0; 16.5(3) - 6(3)(1.5) - MC = 0 VC = 1.50 kN + c©Fy = 0; VC + 16.5 - 6(3) = 0 + :©Fx = 0; NC = 0 By = 16.5 kN By(6) - 1 2 (6)(3)(2) - 6(3)(4.5) = 0 +©MA = 0; MC = 27 kN # m +©MC = 0; 13.5(3) - 1 2 (9)(3)(1) - MC = 0 VC = 0 + c©Fy = 0; VC + 13.5 - 1 2 (9)(3) = 0 : + ©Fx = 0; NC = 0 By = 13.5 kN +©MA = 0; By(6) - 1 2 (9)(6)(3) = 0 MC = 24.75 kN # m 23.25(1.5) - 9(1.5)(0.75) - MC = 0 +©MC = 0; VC = -9.75 kN + c©Fy = 0; VC + 23.25 - 9(1.5) = 0 : + ©Fx = 0; NC = 0 By = 23.25 kN +©MA = 0; By(6) - 12(1.5) - 9(3)(4.5) = 0 MC = -27 kip # ft +©MC = 0; -MC - 6(4.5) = 0 VC = 6 kip + c©Fy = 0; VC - 6 = 0 : + ©Fx = 0; NC = 0 By = 6 kip +©MA = 0; 3(6)(3) - By(9) = 0 : + ©Fx = 0; Bx = 0
- 632. FUNDAMENTAL PROBLEMS 615 F7–14. F7–15. F7–16. F7–17. F7–18. Chapter 8 F8–1. , therefore Ans. F8–2. a Ans. F8–3. Crate A T = 122.62 N : + ©Fx = 0; T - 0.25(490.5) = 0 NA = 490.5 N + c©Fy = 0; NA - 50(9.81) = 0 P = 154.89 N = 155 N + :©Fx = 0; P - 154.89 = 0 NA = 154.89 N NA(3) + 0.2NA(4) - 30(9.81)(2) = 0 +©MB = 0; F = 160 N F 6 Fmax = msN = 0.3(610.5) = 183.15 N F = 160 N + :©Fx = 0; F - 200A4 5 B = 0 N = 610.5N + c©Fy = 0; N - 50(9.81) - 200A3 5 B = 0 x = 6, V = -13.5, M = 0 x = 3, V = 0, M = 27; x = 0, V = 13.5, M = 0; x = 6, V = -9; M = 0 x = 3. V = 0, M = 9; x = 0, V = 9, M = 0; x = 6, V = 0, M = 0 x = 4.5+ , V = 9, M = -6.75; x = 1.5+ , V = 0, M = -6.75; x = 0, V = 0, M = 0; x = 18, V = -10, M = 0 x = 12+ , V = -10, M = 60; x = 6+ , V = 2, M = 48; x = 0, V = 8, M = 0; x = 3, V = 6, M = 0; x = 1.5, V = 6, M = -9; x = 0, V = 18, M = -27; Crate B Ans. F8–4. a Ans. F8–5. If slipping occurs: If tipping occurs: a Ans. Chapter 9 F9–1. Ans. Ans. F9–2. Ans. Ans. = 0.286 m y = L A y dA L A dA = L 1 m 0 1 2 x3 Ax3 dxB L 1 m 0 x 3 dx = 0.8 m x = L A x dA L A dA = L 1m 0 x(x3 dx) L 1 m 0 x 3 dx y = L A y dA L A dA = L 1 m 0 y4/3 dy L 1 m 0 y1/3 dy = 0.571 m x = L A x dA L A dA = 1 2 L 1 m 0 y2/3 dy L 1 m 0 y1/3 dy = 0.4 m P = 83.3 lb +©MA = 0; -P(4.5) + 250(1.5) = 0 P = 100 lb : + ©Fx = 0; P - 0.4(250) = 0 NC = 250 lb + c©Fy = 0; NC - 250 lb = 0 P = 343 N NA = 175.70 N NB = 585.67 N - 0.3 NA (0.9) = 0 P(0.6) + NB(0.9) - 0.3NB(0.9) +©MO = 0; NB + 0.3NA + P - 100(9.81) = 0 + c©Fy = 0; + :©Fx = 0; NA - 0.3NB = 0 P = 247 N P cos30° - 0.25(490.5 - 0.5 P) - 122.62 = 0 : + ©Fx = 0; NB = 490.5 - 0.5P NB + P sin 30° - 50(9.81) = 0 + c©Fy = 0;
- 633. 616 PARTIAL SOLUTIONS AND ANSWERS F9–9. Ans. F9–10. Ans. Ans. F9–11. Ans. Ans. Ans. F9–12. Ans. Ans. Ans. F9–13. Ans. Ans. F9–14. Ans. Ans. = 22.6 m3 = 2pC1.8A1 2 B(0.9)(1.2) + 1.95(0.9)(1.5)D V = 2p©rA = 77.5 m2 = 2pC1.952(0.9)2 + (1.2)2 + 2.4(1.5) + 1.95(0.9) + 1.5(2.7)D A = 2p©rL = 18.8 m3 = 2pC0.75(1.5)(2) + 0.5A1 2 B(1.5)(2)D V = 2p©rA = 37.7 m2 = 2pC0.75(1.5) + 1.5(2) + 0.752(1.5)2 + (2)2 D A = 2p©rL z = ©z V ©V = 2.835 3.6 = 0.7875 m y = ©y V ©V = 5.00625 3.6 = 1.39 m = 0.391 m = 0.25[0.5(2.5)(1.8)] + 0.25B 1 2 (1.5)(1.8)(0.5)R + B 1 2 (1.5)(1.8)(0.5)R 0.5(2.5)(1.8) + 1 2 (1.5)(1.8)(0.5) + 1 2 (1.5)(1.8)(0.5) x = ©x V ©V = 2.67 ft z = ©z V ©V = 3[2(7)(6)] + 1.5[4(2)(3)] 2(7)(6) + 4(2)(3) = 2.94 ft y = ©y V ©V = 3.5[2(7)(6)] + 1[4(2)(3)] 2(7)(6) + 4(2)(3) = 1.67 ft x = ©x V ©V = 1[2(7)(6)] + 4[4(2)(3)] 2(7)(6) + 4(2)(3) = 1.33 in. y = ©y A ©A = 2[4(0.5)] + 0.25[(0.5)(2.5)] 4(0.5) + (0.5)(2.5) = 0.827 in. x = ©x A ©A = 0.25[4(0.5)] + 1.75[0.5(2.5)] 4(0.5) + 0.5(2.5) = 162.5 mm y = ©y A ©A = 100[2(200)(50)] + 225[50(400)] 2(200)(50) + 50(400) F9–3. Ans. F9–4. Ans. F9–5. Ans. F9–6. Ans. F9–7. Ans. Ans. Ans. F9–8. Ans. = 237.5 mm y = ©y A ©A = 150[300(50)] + 325[50(300)] 300(50) + 50(300) = -61.5 mm = 0(300) + 0(600) + (-200)(400) 300 + 600 + 400 z = ©z L ©L = 323 mm = 0(300) + 300(600) + 600(400) 300 + 600 + 400 y = ©y L ©L = 265 mm = 150(300) + 300(600) + 300(400) 300 + 600 + 400 x = ©x L ©L = 0.786 ft z = L V z dV L V dV = L 2 ft 0 zc 9p 64 (4 - z)2 dzd L 2 ft 0 9p 64 (4 - z)2 dz = 0.667 m y = L V y dV L V dV = L 1 m 0 y¢ p 4 ydy≤ L 1 m 0 p 4 y dy = 9 16 L x = L m x dm L m dm = L L 0 xBm0¢1 + x2 L2 ≤dxR L L 0 m0¢1 + x2 L2 ≤dx = 1.2 m y = L A y dA L A dA = L 2 m 0 ya2a y1/2 22 b bdy L 2 m 0 2a y1/2 22 bdy
- 634. FUNDAMENTAL PROBLEMS 617 F9–15. Ans. Ans. F9–16. Ans. Ans. F9–17. Ans. F9–18. Ans. F9–19. Ans. F9–20. Ans. F9–21. Ans. Chapter 10 F10–1. Ans. = 0.111 m4 Ix = L A y2 dA = L 1 m 0 y2 C A1 - y3/2 BdyD = 4.99 kip FR = 1 2 (748.8 + 1248)A 2(3)2 + (4)2 B wB = gw hB b = 62.4(10)(2) = 1248 lbft wA = gwhA b = 62.4(6)(2) = 748.8 lbft FR = 1 2 (58.86 + 98.1)(2) = 157 kN = 98.1 kNm wB = rwghBb = 1000(9.81)(5)(2) = 58.86 kNm wA = rwghAb = 1000(9.81)(3)(2) = 36.8 kN FR = 1 2 (29.43)A 2(1.5)2 + (2)2 B = 29.43 kNm wb = rwghBb = 1000(9.81)(2)(1.5) FR = 998.4(3) = 3.00 kip wb = gw hb = 62.4 (4)(4) = 998.4 lbft FR = 1 2 (58.76)(6) = 176.58 kN = 177 kN = 58.86 kNm wb = rwghb = 1000(9.81)(6)(1) = 21.2 m3 = 2pC4(1.5) 3p ApA1.52 B 4 B + 0.75(1.5)(2)D V = 2p©rA = 40.1 m2 = 2pC2(1.5) p Ap(1.5) 2 B + 1.5(2) + 0.75(1.5)D A = 2p©rL = 45 710 in.3 = 2pC7.5(15)(38) + 20A1 2 B(15)(20)D V = 2p©rA = 8765 in.2 = 2pC7.5(15) + 15(18) + 22.52152 + 202 + 15(30)D A = 2p©rL F10–2. Ans. F10–3. Ans. F10–4. Ans. F10–5. Ans. Ans. F10–6. Ans. Ans. F10–7. Ans. F10–8. Ans. Chapter 11 F11–1. Ans. P = 98.1 cot u|u=60° = 56.6 N (294.3 cos u - 3P sin u)du = 0 dU = 0; 2WdyG + PdxC = 0 xC = 2(1.5) cos u dxC = -3 sin u du yG = 0.75 sin u dyG = 0.75 cos u du = 25.1 (106 ) mm4 + C 1 12(30)(150)3 + 30(150)(105 - 65)2 D = C 1 12 (150)(30)3 + (150)(30)(60 - 15)2 D Ix¿ = ©(I + Ad2 ) y = ©y A ©A = 15(150)(30) + 105(30)(150) 150(30) + 30(150) = 60 mm = 69.8 (106 ) mm4 + C 1 12(300)A503 B + 0D Iy = 2C 1 12(50)A2003 B + 0D = 463A106 B mm4 Iy = 1 12 (200)A3603 B - 1 12(140)A3003 B = 171A106 B mm4 Ix = 1 12 (360)A2003 B - 1 12 (300)A1403 B = 183A106 B mm4 + 2C 1 12(50)A1503 B + (150)(50)(100)2 D Iy = C 1 12 (450)A503 B + 0D = 383A106 B mm4 Ix = C 1 12 (50)A4503 B + 0D + C 1 12(300)A503 B + 0D = 0.0606 m4 Iy = L A x2 dA = L 1 m 0 x2 C(1 - x2/3 ) dxD Iy = L A x2 dA = L 1 m 0 x2 Ax2/3 Bdx = 0.273 m4 Ix = L A y2 dA = L 1 m 0 y2 Ay3/2 dyB = 0.222 m4
- 635. 618 PARTIAL SOLUTIONS AND ANSWERS Ans. Ans. F11–5. Ans. F11–6. Ans. u = 20.9° + 5400 sin u cos u)du = 0 (135 cos u - 5400 sin u dU = 0; PdxC + FspdyB = 0 yB = 2[0.3 cos u] dyB = -0.6 sin u du xC = 3[0.3 sin u] dxC = 0.9 cos u du Fsp = 15 000 (0.6 - 0.6 cos u) u = 56.33° = 56.3° - 1226.25 cos u)du = 0 (15 000 sin u cos u - 7500 sin u dU = 0; -WdyG + A -FspdxAB = 0 xA = 5 cos u dxC = -5 sin u du yG = 2.5 sin u dyG = 2.5 cos u du u = 54.31° = 54.3° 64800 cos u - 37 800 = 0 sin u = 0 u = 0° sin u (64 800 cos u - 37 800)du = 0 F11–2. Ans. F11–3. Ans. Ans. F11–4. - 36A103 B(cos u - 0.5)(-1.8 sin u du) = 0 6A103 B(-0.9 sin u du) dU = 0; PdxB + A -Fsp dxCB = 0 xC = 2(0.9 cos u) dxC = -1.8 sin udu xB = 0.9 cos u dxB = -0.9 sin udu u = 77.16° = 77.2° -5400 cos u + 1200 = 0 sin u = 0 u = 0° - 2000(-0.6 sin u du) = 0 -9A103 B sin u (0.6 cos u du) dU = 0; -FspdxB + (-PdyC) = 0 yC = 0.6 cos u dyC = -0.6 sin u du xB = 0.6 sin u dxB = 0.6 cos u du P = 245.25 cot u|u=60° = 142 N (5P sin u - 1226.25 cos u)du = 0 dU = 0; -PdxA + (-WdyG) = 0 yG = 2.5 sin u dyG = 2.5 cos u du xA = 5 cos u dxA = -5 sin u du
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- 637. Chapter 2 2–1. 2–2. 2–3. 2–5. 2–6. 2–7. 2–9. 2–10. 2–11. 2–13. 2–14. 2–15. 2–17. 2–18. 2–19. 2–21. u = 60° F2 = 6 sin 30° = 3 kN F1 = 6 cos 30° = 5.20 kN u = 70.0° FA = 893 N FB = 325 N FB = 346 N FA = 774 N F2u sin 75° = 150 sin 75° , F2u = 150 N F2v sin 30° = 150 sin 75° , F2v = 77.6 N f = 38.3° FAB = 621 lb u = 53.5° Fy sin 70° = 360 sin 80° Fy = 344 N -Fx¿ sin 30° = 360 sin 80° Fx¿ = -183 N u = 60° FR = 400 N FR = 10.4 kN u = 54.9° f = 3.16° u¿ = 33.16° sin u¿ 6 = sin 100° 10.80 FR = 282 + 62 - 2(8)(6) cos 100° = 10.8 kN FR = 3.92 kN u = 78.6° Fv = 260 lb Fu = 150 lb Fv = 283 lb Fu sin 105° = 200 sin 30° Fu = 386 lb u = 30.6° T = 6.57 kN f = 17.5° FR = 10.5 kN f = 3.05° sin a 8 = sin 75° 8.669 a = 63.05° FR = 262 + 82 - 2(6)(8) cos 75° = 8.67 kN Answers to Selected Problems 620 Chapter 1 1–1. a. b. c. d. 1–2. a. N b. MN/m c. N/s2 d. MN/s 1–3. a. b. c. 1–5. a. GN/s b. Gg/N c. GN/(kg · s) 1–6. a. b. c. 1–7. a. b. c. d. 1–9 1–10. a. b. c. 1–11. a. b. c. 1–13. a. b. c. 1–14. 1–15. 1–17. a. b. c. 1–18. a. b. c. 1–19. 1–21. 26.9 mm # kg/N F = 7.41 mN 0.064 km3 25 mm2 0.04 MN2 m = 6.12 Gg m = 15.3 Mg m = 2.04 g rw = 1.00 Mg/m3 2.71 Mg/m3 1.27 mm/s 70.7 kN/m3 27.1 N # m 18.8 GN/m 0.911 kg # s 0.447 kg # m/N W = 44.1 kN W = 4.90 mN W = 98.1 N 1 ATM = 101 kPa 1 Pa = 20.9(10-3 ) lb/ft2 mm = me = 3.65 Gg Wm = 5.89 MN We = 35.8 MN 3.65 Gg 5.63 mg 56.8 km 45.3 MN 5.32 m 35.3 kN 0.431 g 2.77 Mg 4.56 kN 55.6 s 4.66 m
- 638. ANSWERS TO SELECTED PROBLEMS 621 2–22. 2–23. 2–25. 2–26. 2–27. 2–29. 2–30. 2–31. 2–33. 2–34. 2–35. 2–37. 2–38. 2–39. 2–41. 2–42. 2–43. 2–45. 2–46. 2–47. u = 54.3° FA = 686 N u = 63.7° F3 = 1.20F1 FR = 140 lb F1 = 420 lb FR = F1 cos f + 240 - 100 0 = F1 sin f - 180 - 240 u = 39.6° FR = 463 lb f = 14.8° FR = 839 N u = 68.6° FB = 960 N 1500 = 700 cos 30° + FB sin u 0 = 700 sin 30° - FB cos u u = 29.1° F1 = 275 N FR = 13.2 kN F2 = 12.9 kN f = 47.3° F2 = 4.71 kN -3 = -3.464 + F2 sin f - 3 5.196 = -2 + F2 cos f + 4 Fy = -162 lb Fx = 67.3 lb f = 42.4° F1 = 731 N u = 44.6° FR = 2499.622 + 493.012 = 702 N u = 16.2° F = 97.4 lb Fmin = 235 lb u = 10.9° FB = 600 sin 30° = 300 N FA = 600 cos 30° = 520 N u = 60° FA = 8.66 kN FB = 5.00 kN FB = 7.07 kN FA = 3.66 kN FR = 2F cos Au 2 B FR = 2(F)2 + (F)2 - 2(F)(F) cos (180° - u) f = u 2 F sin f = F sin(u - f) f = 98.5° FR = 8.09 kN FR = 4.33 kN F2 = 2.50 kN u = 90° 2–49. 2–50. 2–51. 2–53. , 2–54. 2–55. 2–57. 2–58. 2–59. 2–61. 2–62. 2–63. 2–65. g = 144° b = 107° a = 59.8° F2 = 66.4 lb + (F2z - 45.96)k} -100k = {(F2x - 33.40)i + (F2y + 19.28)j Fz = 283 N Fy = 200 N Fx = -200 N g = 77.7° b = 68.0° a = 25.5° FR = 754 lb = 5200i + 283j - 200k6 N + 400 cos 120°k F2 = 400 cos 60°i + 400 cos 45°j = 5480i + 360k6 N F1 = 600A4 5 B(+i)+0 j+600A3 5 B(+k) F2 = 5424i + 300j - 300k6 N F1 = 5-159i + 276j + 318k6 N F1 = 434 N u = 67.0° F3 = {-100j} N F2 = { 350i } N F1 = {F1 cos u i + F1 sin u j} N FR = 7.87 kN F = 2.03 kN + 2(7 - F sin 45°)(-sin 45°) = 0 2FR dFR dF = 2(-4.1244 - F cos 45°)(-cos 45°) FR 2 = (-4.1244 - F cos 45°)2 + (7-F sin 45°)2 u = 38.3° FR = 161 lb F2 = 88.1 lb u = 103° FR = 380 N F1 = 57.8 N 2FR dFR dF1 = 2F1 - 115.69 = 0 FR 2 = F1 2 - 115.69F1 + 147 600 FR = 2(0.5F1 + 300)2 + (0.8660F1 - 240)2 u = 16.4° FR = 391 N 1.22 kN … P … 3.17 kN u = 202° FR = 2(-103.05)2 + (-42.57)2 = 111 lb
- 639. 622 ANSWERS TO SELECTED PROBLEMS 2–82. 2–83. 2–85. 2–86. 2–87. 2–89. 2–90. 2–91. 2–93. 2–94. 2–95. g = 128° b = 131° a = 63.9° F = {59.4i - 88.2j - 83.2k} lb F = 52.1 lb g = 180° b = 90° a = 90° FR = 150 lb = {33.3 j - 49.9 k} lb FC = 60 (4 j - 6 k) 2(4)2 + (-6)2 FB = {-28.8 i - 16.6 j - 49.9 k} lb = {28.8 i - 16.6 j - 49.9 k} lb FA = 60 (4 cos 30° i - 4 sin 30° j - 6 k) 2(4 cos 30°)2 + (-4 sin 30°)2 + (-6)2 g = 144° b = 125° a = 82.4° FR = 1.38 kN FR = 822 N g = 162° b = 83.3° a = 72.8° g = 137° b = 84.0° a = 47.4° FR = 26502 + 1002 + (-700)2 = 960 lb FC = 5250i + 500j - 500k6 lb FB = 5400i - 400j - 200k6 lb z = 5.35 m rAB = 7 m rAB = 5-3i + 6j + 2k6 m g2 = 144° b2 = 74.5° a2 = 122° = 32.4 lb F2 = 2(-17.10)2 + (8.68)2 + (-26.17)2 g = 27.5° b = 63.7° a = 97.5° F3 = 166 N Fy = 0.523 kN F = 2.02 kN 2–66. 2–67. 2–69. 2–70. 2–71. 2–73. 2–74. 2–75. 2–77. 2–78. 2–79. 2–81. Fz = 0.776 kN Fy = 2.60 kN Fx = 1.28 kN a = 64.67° g3 = 77.0° b3 = 98.4° a3 = 15.5° F3 = 9.58 kN b = 52.5° FR = 754 N g = 53.1° a = 121° g2 = 75.0° b2 = 119° a2 = 147° F2 = 180 N F2 cos g2 = 46.59 F2 cos b2 = -86.93 F2 cos a2 = -150.57 g1 = 66.4° b1 = 53.1° a1 = 90° g1 = 66.4° b1 = 53.1° a1 = 45.6° g = 64.0° b = 85.1° a = 26.6° FR = 2(550)2 + (52.1)2 + (270)2 = 615 N gR = 103° bR = 13.3° aR = 86.8° FR = 718 lb a = 121° b = 52.7° g = 53.0° F = 882 N a = 131° b = 70.5° g = 47.5° F = 1.15 kN F2 (cos2 a + cos2 b + cos2 g) = 1 333 518.08 + (F cos ai + F cos bj + F cosgk) = (459.28i + 265.17j - 530.33k) -300i + 650j + 250k F2 = {90i - 127j + 90k} lb F1 = {14.0j - 48.0k} lb g = 140° b = 71.3° a = 124°
- 640. ANSWERS TO SELECTED PROBLEMS 623 2–97. 2–98. 2–99. 2–101. 2–102. 2–103. 2–105. 2–106. 2–107. 2–109. 2–110. F = {143i + 248j - 201k} lb g = 2.81° b = 92.6° a = 88.8° FR = 18.5 kN FB = {0.970i - 1.68j + 7.76k} kN + (0 - 0.75 cos 30°)j + (3 - 0)k rB = [0 - (-0.75 sin 30°)]i FC = {0.857 i + 0.857 j + 4.85 k} kN + [0 - (-0.75 cos 45°)]j + (3 - 0)k rC = [0 - (-0.75 sin 45°)]i FA = {-1.46 i + 5.82 k} kN = {-0.75 i + 0j + 3 k} m rA = (0 - 0.75)i + (0 - 0)j + (3 - 0)k F = {-6.61i - 3.73j + 9.29k} lb F = 105 lb g = 180° b = 90° a = 90° FR = 240 lb FD = 5-30i - 20j - 60k6 lb FC = 5-30i + 20j - 60k6 lb FB = 530i + 20j - 60k6 lb FA = 530i - 20j - 60k6 lb FA = FB = FC = 326 lb g = 180° b = 90° a = 90° FR = 1.24 kip z = 16 ft y = 18 ft x = 24 ft u = F F = -120 170 i - 90 170 j - 80 170 k FR = 3.46 kN FB = 2.42 kN FC = 1.62 kN FB = {53.2i - 79.8j - 146k} N FA = {-43.5i + 174j - 174k} N FC = {159i + 183j - 59.7k} N FA = {285j - 93.0k} N + (0 - 0.750) k} m rCD = {[-0.5 - (-2.5)] i + [0 - (-2.299)] j + (0 - 0.750) k} m rAB = {(0 - 0)i + [0 - (-2.299)] j 2–113. 2–114. 2–115. 2–117. 2–118. 2–119. 2–121. 2–122. 2–123. 2–125. 2–126. 2–127. 2–129. 2–130. 2–131. 2–133. 2–134. 2–135. 2–137. 2–138. 2–139. u = 52.7° FAB = 215 lb u = 85.2° Fr = 178 N u = 143° rBC = {6i + 4j - 2k} ft rBA = {-3i} ft u = 52.7° FR = 215 lb u = 100° FR = 178 N (F1)F2 = 5.44 lb uF2 = cos 135°i + cos 60°j + cos 60°k uF1 = cos 30° sin 30°i + cos 30° cos 30°j - sin 30°k Fy = 260 N Fx = -75 N FAC = 45.5 lb Fx = 47.8 lb u = 34.2° rAC = {-15i - 8j + 12k} ft rAB = {-15i + 3j + 8k} ft u = 97.3° (F1)F2 = 50.6 N f = 65.8° uOA = 1 3 i + 2 3 j - 2 3 k uOD = -sin 30°i + cos 30°j (FBC)⬜ = 316 N (FBC)|| = 245 N FAC = 5293j + 219k6 lb FAC = 366 lb (FAC)z = -569 lb uAC = 0.1581i + 0.2739j - 0.9487k F2 = 373 N F1 = 333 N FBC = {32i - 32j} N FBC = 45.2 N |Proj F2| = 71.6 N u1 = cos 120° i + cos 60° j + cos 45° k (FED)⬜ = 498 N (FED)|| = 334 N rBC = 5.39 m (FAO)⬜ = 2(56)2 - (46.86)2 = 30.7 N + 16A-2 7 B = 46.9 N (F AO)|| = (24)A3 7 B + (-48)A -6 7 B
- 641. 624 ANSWERS TO SELECTED PROBLEMS 3–26. 3–27. 3–29. 3–30. 3–31. 3–33. 3–34. 3–35. 3–37. 3–38. 3–39. 3–41. 3–42. 3–43. 3–45. 3–46. 3–47. 3–49. 3–50. 3–51. FAD = 1.42 kip FAC = 0.914 kip FAB = 1.47 kip FAD = 1.70 kip FAC = 0.744 kip FAB = 1.37 kip W = 375 lb FAC = 225 lb FAD = 450 lb 2 3FAB + 1 3FAC - W = 0 1 3FAB - 2 3FAC = 0 -2 3FAB - 2 3FAC + FAD = 0 FBD = 3.64 kN FCB = 2.52 kN FAB = 2.52 kN m = 102 kg FAB = FAC = 1.96 kN FAD = 2.94 kN 1 3FAD - 981 = 0 - FAC+2 3FAD = 0 FAB-2 3FAD = 0 y = 6.59 m m = 2.37 kg m = 15.6 kg -2(107.1) cos 44.4° + m(9.81) = 0 -150 + 2 T sin u = 0 WE = 18.3 lb k = 6.80 lb/in. d = 7.13 in. -TAC + Fs cos u = 0 l = 19.1 in. P = 147 N FR = 40.8 N, (B and C) FR = 14.9 N, (A and D) T = 28.9 N 2 (T cos 30°) - 50 = 0 F = 39.3 lb T = 53.1 lb W = 51.0 lb u = 78.7° 100 cos u = WA 5 13 B WF = 123 lb u = 2.95° FBC = 57.1 lb FCD = 65.9 lb FBA = 80.7 lb 2–141. 2–142. 2–143. Chapter 3 3–1. 3–2. , 3–3. 3–5. 3–6. 3–7. 3–9. 3–10. 3–11 3–13. 3–14. 3–15. 3–17. 3–18. 3–19. 3–21. Joint D, , Joint B, , 3–22. 3–23. , 3–25. Joint E, Joint B, W = 57.7 lb 1.3957W cos 30° - 0.8723WA3 5 B - FBA = 0 FED cos 30° - FEBA3 5 B = 0 l¿ = 2.66 ft 40 = 50(212 - l¿) u = 35.0° m = 48.2 kg FBC + 8.7954m cos 45° - 12.4386m cos 30° = 0 ©Fx = 0 FCD cos 30° - FBD cos 45° = 0 ©Fx = 0 d = 2.42 m FAB = 98.6 N FAC = 267 N FCA = 42.6 N u = 64.3° FCB = 85.2 N FCB cos u - FCA cos 30° = 0 m = 8.56 kg xAB = 0.467 m xAC = 0.793 m u = 40.9° WC = 240 lb WC cos 30° - 275 cos u = 0 u = 70.1° T = 7.66 kN F = 5.40 kN T = 7.20 kN W = 412 lb FAC = 294.63 lb FAB cos 45° - FACA3 5 B = 0 TBD = 32.6 kN TBC = 22.3 kN u = 36.3°, T = 14.3 kN T = 13.3 kN, F2 = 10.2 kN FBC = 15.2 kN, FBD = 21.5 kN FAB = 29.4 kN y = 0.841 m FBC = 2.90 kN FBC = 3.40 kN FBA sin 30° - 200(9.81) = 0 FBA = 3.92 kN FE = {-194i + 291k} N FC = {-324i - 130j + 195k} N FB = {-324i + 130j + 195k} N Proj F = 48.0 N Fv = 98.7 N 250 sin 120° = Fu sin 40° Fu = 186 N
- 642. ANSWERS TO SELECTED PROBLEMS 625 3–53. 3–54. 3–55. 3–57. 3–58. 3–59. 3–61. 3–62. 3–63. 3–65. 3–66. 3–67. 3–69. -0.2236 FOC + 100 = 0 -0.4472 FOA - 0.2236 FOB u = 0° -0.3873 FOB + 0.3873 FOC + 100 sin u = 0 FAB = FAC = FAD = 375 lb d = 1.64 ft z = 173 mm 3F£ z 20.52 + z2 ≥ - 100(9.81) = 0 FAB£ 0.5 20.52 + z2 ≥ - 2CF£ 0.5 sin 30° 20.52 + z2 ≥ S = 0 F AD£ 0.5 cos 30° 20.52 + z2 ≥ - F AC£ 0.5 cos 30° 20.52 + z2 ≥ = 0 z = 2.07 ft F = 831 lb z = 2.51 ft y = 0.374 ft d = 3.61 m FAC = FAD = 260 N FAB = 520 N (FAB)z + 3 14FAB + 3 14FAB - 490.5 = 0 (FAB)x - 3 7FAB - 3 7FAB = 0 m = 90.3 kg FAD = 415 N FAC = 35.6 N FAB = 831 N m = 2.62 Mg - 12 14FB - 12 14FC - 12 14FD + W = 0 - 6 14FB - 4 14FC + 6 14FD = 0 4 14FB - 6 14FC - 4 14FD = 0 FAD = 708.5 N FAC = 763 N FAB = 1.31 kN FAD = 750 N FAC = 606 N FAB = 1.21 kN FD = 6.32 kN FC = 10.4 kN FB = 19.2 kN 0.6402 FB - 0.4432 FC - 0.4364 FD - 4905 = 0 0.7682 FB - 0.8865 FC - 0.8729 FD = 0 0.1330 FC - 0.2182 FD = 0 3–70. 3–71. 3–73. 3–74. 3–75. 3–77. 3–78. 3–79. Chapter 4 4–5. 4–6. b 4–7. 4–9. 4–10. d d 4–11. b 4–13. 4–14. c 4–15. b (MR)A = 2.09 N # m F = 23.7 lb + MA = 123 lb # in.b 0 = 36 cos u + 18 sin u, u = 117° When MA = 0, u = 26.6°, (MA)max = 40.2 kN # m dMA du = -36 sin u + 18 cos u = 0 MA = (36 cos u + 18 sin u) kN # m MA = 38.2 kN # m MO = 520 N # m MO = 120 N # m F = 27.6 lb -500 = -F cos 30°(18) - F sin 30°(5) u = 64.0° MA = 7.21 kN # m F = 39.8 lb 30 (cos 45°) (18) = FA4 5 B (12) F3 = 238 lb F2 = 311 lb F1 = 0 FCA = FCB = 198 lb FCD = 625 lb F3 = 357 lb F2 = 280 lb F1 = 400 lb F1 cos 60° - 200 = 0 800 A4 5 B + F1 cos 135° - F3 = 0 F2 + F1 cos 60° - 800A3 5 B = 0 g3 = 119° b3 = 148° a3 = 77.2° P = 639 lb FAC = 85.8 N FAO = 319 N FAB = 110 N F = 0.850 mN 1.699(10)-3 cos 60° - F = 0 F1 = 4.31 kN u = 4.69° u = 11.5° FOB = FOC = 74.5 lb FOA = 149 lb 0.8944 FOA - 0.8944 FOB - 0.8944 FOC = 0
- 643. 626 ANSWERS TO SELECTED PROBLEMS 4–49. 4–50. 4–51. 4–53. 4–54. 4–55. 4–57. 4–58. 4–59. 4–61. 4–62. 4–63. 4–65. 4–66. 4–67. 4–69. 4–70. 4–71. 4–73 4–74. 4–75. b 4–77. 4–78. 4–79. 4–81. c P = 70.7 N - P sin 15° (0.3) - P cos 15°(0.3) = 15 +MR = 100 cos 30° (0.3) + 100 sin 30° (0.3) u = 56.1° F = 111 N F = 133 N F¿ = 33.3 N (Mc)R = 260 lb # ft F = 625 N M3 = 300 N # m 0 = 424.26 cos 45° - M3 M2 = 424 N # m F = 20.2 N Mx = 14.8 N # m W = 56.8 lb MOA = uOA # rOB * W = uOA # rOB * W (Ma)2 = 8 lb # in. (Ma)1 = 30 lb # in. M = 282 lb # in. My = 282 lb # ft rAC = -6 cos 15° iœ + 3 jœ + 6 sin 15° k uy = -sin 30° iœ + cos 30° jœ My¿ = 464 lb # ft F = 162 lb = uCD # rDB * F = -432 lb # ft MCD = uCD # rCA * F F = 771 N Mx = 73.0 N # m My = 0.828 N # m rOB = {0.2 cos 45°i - 0.2 sin 45°k} m MAC = {11.5i + 8.64j} lb # ft Mz = 36.0 lb # ft My = 4.00 lb # ft Mx = 15.0 lb # ft Mz = 15.5 N # m r = 0.25 sin 30° i + 0.25 cos 30° j u = k MAF = 59.33i + 9.33j - 4.67k6 N # m g = 20.6° b = 110° a = 95.2° MO = 4.27 N # m MB = rBC * F = {10i + 0.750j - 1.56k} kN # m uF = b b b = rCA * rCB 4–17. b d Since , the gate will rotate counterclockwise. 4–18. 4–19. 4–21. a. d 4–22. a d 4–23. 4–25. 4–26. b b 4–27. b 4–29. a d a d 4–30. a d 4–31. a d 4–33. c 4–34. 4–35. 4–37. 4–38. 4–39. 4–41. 4–42. 4–43. 4–45. 4–46. 4–47. d = 1.15 m z = 3 m y = 1 m MB = {10.6i + 13.1j + 29.2k} N # m = {-5.39i + 13.1j + 11.4k} N # m MA = rAC * F (MB)O = 518i + 7.5j + 30 k6 N # m (MA)O = 5-18i + 9j - 3k6 N # m MO = 5-720i + 720j6 N # m MO = rOC * FC = {1080i + 720j}N # m MO = rOA * FC = {1080i + 720j} N # m (MR)O = {200i - 180j + 30k} lb # ft MO = 590i - 130j - 60k6 lb # ft MO = rOA * F1 = {110i - 50j + 90k}lb # ft F = 84.3 N F = 115 N u = 33.6° +(MO)max = 80.0 kN # m Maximum moment, OB ⬜ BA +MA = 7.71 N # m +MA = 195 lb # ft + MC = 141 lb # ft +MB = 40 cos 25°(2.5) = 90.6 lb # ft FC = 82.2 N ; MA = 73.9 N # m (MA)2 = 140 lb # in. (MA)1 = 118 lb # in. F = 191 lb 1500 = F sin 23.15°(20) sin u 10 = sin 105° 24.57 u = 23.15° BC = 24.57 ft umin = 146° Mmin = 0 umax = 56.3° Mmax = 1.44 kN # m +MA = 1200 sin u + 800 cos u u = 56.3° MA = 1.44 kN # m MA = 4002(3)2 + (2)2 MP = (537.5 cos u + 75 sin u) lb # ft FA = 28.9 lb (MFB )C (MFA )C (MFB )C = 260 lb # ft = -162 lb # ft = 162 lb # ft (MFA ) C = -30A3 5 B(9)
- 644. ANSWERS TO SELECTED PROBLEMS 627 4–82. For minimum P require 4–83. 4–85. a. b b. b 4–86. b 4–87. 4–89. a. a b b. a b 4–90. a. a b b. a b 4–91. 4–93. 4–94. 4–95. 4–97. 4–98. 4–99. 4–101. 4–102. 4–103. 4–105. b MRA = 34.8 kN # m u = 77.8° FR = 2 1.252 + 5.7992 = 5.93 kN F2 = 150 lb F1 = 200 lb g = 90° b = 63.4° a = 153° (MC)R = 224 N # m M1 = M2 = 287 lb # ft M3 = 318 lb # ft 0 = 1 3M3 - 106.7 0 = M1 - 2 3M3 - 75 0 = -M2 + 2 3M3 + 75 d = 342 mm MR = {-12.1i - 10.0j - 17.3k} N # m F = 15.4 N MC = F (1.5) (MR)y¿ = 29.8 kip # ft (MR)x¿ = 4.84 kip # ft F = 98.1 N g = 90° b = 101° a = 11.3° Mc = 40.8 N # m M c = r AB * F = rBA * -F g = 136° b = 61.3° a = 120° (Mc)R = 1.04 kN # m +MC = 53.4 lb # ft +MC = 53.4 lb # ft +MC = -53.4 lb # ft = 53.4 lb # ft = 53.4 lb # ft +MC = 40 cos 30°(4)-60A4 5 B(4) F = 14.2 kN # m (Mc)R = 5.20 kN # m MR = 9.69 kN # m MR = 9.69 kN # m - 2 sin 30°(0.3) - 2 cos 30°(3.3) - 8 cos 45°(3.3) MR = 8 cos 45°(1.8) + 8 sin 45°(0.3) + 2 cos 30°(1.8) N = 26.0 N P = 49.5 N u = 45° 4–106. d 4–107. d 4–109. d 4–110. b 4–111. b 4–113. 4–114. 4–115. 4–117. 4–118. d 4–119. 4–121. 4–122. 4–123. 4–125. 4–126. d = 4.62 ft u = 49.8°c FR = 65.9 lb d = 2.10 ft u = 49.8°c FR = 2(42.5)2 + (50.31)2 = 65.9 lb d = 0.824 ft u = 42.6°a FR = 197 lb d = 5.24 ft u = 42.6°a FR = 197 lb d = 6.10 ft f = 23.6° u = 6.35° FR = 2(100)2 + (898.2)2 = 904 lb d = 9.26 ft FR = 10.75 kip T d = 13.7 ft MRA = 99.5 kip # ft FR = 10.75 kip T = {36.0i - 26.1j + 12.2k} kN # m MRO = r1 * F1 + r2 * F2 FR = {0.232i + 5.06j + 12.4k} kN F2 = {-1.768i + 3.062j + 3.536k} kN MRO = {1.30 i + 3.30 j - 0.450 k} N # m FR = {6 i - 1 j - 14 k} N MRO = {-15i + 225j} N # m FR = {-210k} N = 5-6i + 12j6 kN # m (MR)O = rOB * FB + rOC * FD FR = 52i - 10k6 kN (MR)O = 438 N # m u = 49.4°c FR = 461 N (MR)A = 239 kN # m u = 84.3°d FR = 50.2 kN (MR)A = 441 N # m u = 10.6°b FR = 2533.012 + 1002 = 542 N MR O = 214 lb # in. u = 78.4°a FR = 29.9 lb MRB = 11.6 kN # m u = 77.8°d FR = 5.93 kN
- 645. 628 ANSWERS TO SELECTED PROBLEMS 4–150. 4–151. 4–153. 4–154. 4–155. b 4–157. 4–158. 4–159. 4–161. 4–162. d 4–163. 4–165. 4–166. 4–167. 4–169. a. b. 4–170. 4–171. 4–173. Mz = k # (rBA * F) = k # (rOA * F) = -4.03 N # m MRP = {-240i + 720j + 960k} lb # ft FR = {-80i - 80j + 40k} lb F = 992 N MC = {-5i + 8.75 j} N # m MC = rOB * (25 k) + rOA * (-25 k) MC = {-5i + 8.75 j} N # m MC = r AB * (25 k) MA = {-59.7i - 159k} N # m MA = 2.89 kip # ft b = {298i + 15.lj - 200k} lb # in MO = rOA * F d = 5.54 ft MRA = 533 lb # ft FR = 533 lb T FR = 223 lb c (dFR)x = 62.5(1 + cos u) sin u du x¿ = 2.40 ft FR = 53.3 lb wmax = 18 lb/ft x = 1.60 ft FR = 53.3 lb x = 14.6 ft 80640x = 34560(6) + L x 0 (x + 12) wdx FR = 80.6 kip c MRA = 2.20 kip # ft FR = 577 lb, u = 47.5°c x = 1 m FR = 10.7 kN T h = 1.60 m z = L 4 m 0 c A20z 3 2 B (103 )ddz L 4 m 0 A20z 1 2 B (103 )dz z = L z 0 zwdz L z 0 wdz FR = 107 kN ; x = 0.268 ft FR = 7 lb a = 9.75 ft b = 4.50 ft MRO = { - 194 j - 54 k } N # m - A0.1 + 1 3(1.2)B (108) k 4–127. 4–129. 4–130. 4–131. 4–133. 4–134. 4–135. 4–137. 4–138. 4–139. 4–141. 4–142. 4–143. 4–145. 4–146. 4–147. 4–149. MRO = - A1 + 2 3 (1.2)B (108) j FR = {-108 i} N w2 = 282 lb/ft w1 = 190 lb/ft d = 11.3 ft FR = 3.90 kip c x = 5 12L -1 2w0L(x) = -1 2w0AL 2 B AL 6 B -1 2w0AL 2 B A2 3LB FR = 1 2w0L T x = 3.4 m FR = 30 kN T x = 1.20 m FR = 75 kN T y = 2.06 m MR = 3.07 kN # m x = 1.16 m uFR = -0.5051 i + 0.3030 j + 0.8081 k FR = 990 N MW = -1003 lb # ft x = 3.52 ft y = 0.138 ft FR = 808 lb FR = 48.7 kN FA = 18.0 kN FB = 16.7 kN x = 3.85 mm y = 82.7 mm -26(y) = 6(650) + 5(750)-7(600)-8(700) FR = 26 kN FA = 30 kN FB = 20 kN FR = 190 kN x = 3.54 m y = 3.68 m FR = 215 kN FC = 223 lb FB = 163 lb 0 = 200(1.5 cos 45°) - FB (1.5 cos 30°) FC = 600 N FD = 500 N y = 7.29 m x = 6.43 m FR = 140 kN x = 5.71 m y = 7.14 m -140y = -50(3)-30(11)-40(13) FR = 140 kNT d = 0.827 m u = 10.6°b FR = 542 N
- 646. ANSWERS TO SELECTED PROBLEMS 629 Chapter 5 5–1. W is the effect of gravity (weight) on the paper roll. NA and NB are the smooth blade reactions on the paper roll. 5–2. NA force of plane on roller. Bx, By force of pin on member. 5–3. W is the effect of gravity (weight) on the dumpster. Ay and Ax are the reactions of the pin A on the dumpster. FBC is the reaction of the hydraulic cylinder BC on the dumpster. 5–5. Cy and Cx are the reactions of pin C on the truss. TAB is the tension of cable AB on the truss. 3 kN and 4 kN force are the effect of external applied forces on the truss. 5–6. W is the effect of gravity (weight) on the boom. Ay and Ax are the reactions of pin A on the boom. TBC is the force reaction of cable BC on the boom. The 1250 lb force is the suspended load reaction on the boom. 5–7. Ax, Ay, NB forces of cylinder on wrench. 5–9. NA, NB, NC forces of wood on bar. 10 lb forces of hand on bar. 5–10. Cx, Cy forces of pin on drum. FAB forces of pawl on drum gear. 500 lb forces of cable on drum. 5–11. 5–13. 5–14. 5–15. 5–17. NB = 12.2 lb NA = 23.7 lb - 5.77(3.464) = 0 10 cos 30°(13 - 1.732) - NA(5 - 1.732) NC = 5.77 lb Ay = 20 lb Ax = 140 lb NB = 140 lb Ay = 6.15 kip Ax = 10.2 kip TBC = 11.1 kip Cy = 4.05 kN Cx = 5.11 kN TAB = 5.89 kN TAB cos 30°(2) + TAB sin 30°(4) - 3(2) - 4(4) = 0 NA = 425 N NB = 245 N 5–18. 5–19. 5–21. 5–22. 5–23. 5–25. 5–26. 5–27. 5–29. 5–30. 5–31. 5–33. 5–34. 5–35. 5–37. wA = 1.44 kN/m wB = 1.11 kN/m -490.5 (3.15) + 1 2 wB (0.3) (9.25) = 0 w = 267 lb/ft d = 6 ft Ay = 600 N Ax = 0 NB = 1.04 kN Cy = 4.38 kN Cx = 32 kN x = 5.22 m 40 000A3 5 B(4) + 40 000A4 5 B(0.2) - 2000(9.81)(x) = 0 Ay = 46.9 lb Ax = 1.42 kip F = 93.75 lb Ax = 1.51 kip NB = 1.60 kip Ay = 50 lb FA = 2(3433.5d)2 + (4578d - 6867)2 FBC = 5722.5d FBCA4 5 B(1.5) - 700(9.81)(d) = 0 Cx = 2.66 kN Cy = 6.56 kN FAB = 0.864 kN By = 95.4 N Bx = 34.0 N FCD = 131 N Ax = 150 lb Ay = 300 lb NB = 150 lb NB(3) - 300(1.5) = 0 Ay = 118 N Ax = 105 N NC = 213 N Ay = 1.97 kip Ax = 3.21 kip FB = 4.19 kip Ay = 87.7 kN Ax = 20.8 kN T = 34.62 kN TA3 5 B(3) + TA4 5 B(1) - 60(1) - 30 = 0 (NA)s = 100 lb, (NB)s = 20 lb (NA)r = 98.6 lb, (NB)r = 21.4 lb Cy = 722 lb Cx = 333 lb FAB = 401 lb
- 647. 630 ANSWERS TO SELECTED PROBLEMS 5–61. 5–62. 5–63. 5–65. 5–66. , 5–67. 5–69. 5–70. 5–71. 5–73. 5–74. Ax = 0 FBD = 150 lb FEF = 100 lb FCD = 0 Ay = 0 Ax = 0 TCD = 43.5 N TCD + 373.21 + 333.33 - 350 - 200 - 200 = 0 Az = 333 N NB = 373 N NB (3) - 200(3) - 200(3 sin 60°) = 0 Az = 62.5 lb Ay = 0 Ax = 0 Ez = 562.5 lb Ex = 0 FDC = 375 lb Az = 100 N Ay = 0 Ax = 66.7 N TBD = TCD = 117 N Ax = 475 N Ax + 25 - 500 = 0 Bx = 25 N Az = 125 N Bz = 1.125 kN Cz = 250 N Cz (0.9 + 0.9) - 900(0.9) + 600(0.6) = 0 Cy = 450 N RF = 13.7 kip RE = 22.6 kip RD = 22.6 kip x = 0.667 m y = 0.667 m TAB = 0.75 kN TEF = 2.25 kN TCD = 3 kN TCD (2)-6(1) = 0 NB = 332 N NA = 213 N NC = 289 N a = 3(4 r2 l) 2 3 - 4 r2 Ay = 48.8 N Ax = 108 N F = 50.6 N 95.35 sin 45°(300)-F(400) = 0 5–38. 5–39. 5–41. 5–42. 5–43. 5–45. 5–46. 5–47. 5–49. 5–50. 5–51. 5–53. 5–54. 5–55. 5–57. For disk E: For disk D: 5–58. 5–59. a = 10.4° NC = 143 lb NA = 262 lb Pmax = 210 lb NC = 141 lb NB = 9.18 lb NA = 250 lb NAA4 5 B - N¿¢ 224 5 ≤ = 0 -P + N¿¢ 224 5 ≤ = 0 a = 1.02° k = 11.2 lb/ft u = 12.8° FC (6 cos u) - FA (6 cos u) = 0 T = 29.2 kN u = 63.4° NB = 24.9 kN NA = 17.3 kN F = 5.20 kN Pmin = 395 N u = 33.6° For Pmin ; dP du = 0 - P cos u(0.5) - P sin u (0.3317) = 0 50(9.81) sin 20° (0.5) + 50(9.81) cos 20°(0.3317) FA = 432 lb FB = 0 FC = 432 lb W = 5.34 kip NB = 1.15 kip NA = 1.85 kip - NA(2.2 + 1.4 + 8.4) = 0 2500(1.4 + 8.4) - 500(15 cos 30° - 8.4) T = 9.08 lb MB = 227 N # m Ax = 900 N NB = 1.27 kN Ax = 825 lb NB = 825 lb NB (4 sin 30°)-300(1)-450(3) Ay = 750 lb Ax = 353 N Ay = 300 N u = 23.1° Ax = 398 N Ay = 300 N k = 1.33 kNm
- 648. ANSWERS TO SELECTED PROBLEMS 631 5–75. 5–77. 5–78. 5–79. 5–81. 5–82. 5–83. 5–85. 5–86. 5–87. 5–89. 5–90. F = 354 N Ay = 1.80 kN Ax = 3.60 kN NB = 5.09 kN 600(6) + 600(4) + 600(2) - N B cos 45°(2) = 0 P = 0.5 W d = r 2 d = r 2 A1 + W P B - P(d + r cos 60°) = 0 ©MAB = 0; TC (r + r cos 60°) - W(r cos 60°) FBC = 105 lb MAz = -720 lb # ft MAy = 0 MAx = -300 lb # ft Ay = -10 lb Ax = 130 lb FBC = 175 lb MAz = 0 MAy = 165 lb # ft Az = 0 Ay = -27.5 lb Ax = -27.5 lb FCB = 67.4 lb -55(3) + ¢ 6 254 ≤FCB (3) = 0 Ax + ¢ 3 254 ≤FCB = 0 Az = 16.7 kN Ay = 5.00 kN Ax = 0 TB = 16.7 kN TCD = 0.0398W TEF = 0.570W TAB = 1.14W TEF = 0.583W d = 0.550L TEF(L) - WAL 2 B -0.75WAL 2 - d cos 45°B = 0 MAz = 0 MAx = 0 Az = 600 lb Ay = 0 Ax = 0 F = 900 lb Az = 100 lb Ay = 0 5–91. 5–93. 5–94. 5–95. Chapter 6 6–1. 6–2. 6–3. 6–5. - (196.2 + 302.47) cos 26.57° = 0 Joint E: FEC cos 36.87° FBE = 196 N (C) Joint B: FBC - 332.45 = 0 FBC = 332 N (T) FAB = 332 N (T) FAE = 372 N (C) Joint A: FAEa 1 25 b - 166.22 = 0 FDE = 1.60 kip (C) FDC = 1.13 kip (T) FBC = 800 lb (T) FBD = 0 FAB = 800 lb (T) FAD = 1.13 kip (C) FDE = 1.60 kip (C) FDC = 1.41 kip (T) FBC = 600 lb (T) FBD = 400 lb (C) FAB = 600 lb (T) FAD = 849 lb (C) FEA = 2.10 kN (T) FEB = 1.27 kN (C) Joint E: 900 - FEB sin 45° = 0 FCB = 1.34 kN (C) FCE = 0 Joint C: -FCE cos 26.57° = 0 FDE = 1.20 kN (T) FDC = 1.34 kN (C) Joint D: 600 - FDC sin 26.57° = 0 Az = 40 lb By = 0 Ax = 136 lb Bx = -35.7 lb Bz = 40 lb P = 100 lb FD = 982 N Dy = -507.66 N T = 1.01 kN By = 16.6 kip Bx = 0.5 kip Ay = 7.36 kip 5(14) + 7(6) + 0.5(6) - 2(6) - Ay(14) = 0 By = 5.00 kN Bx = 5.20 kN NA = 8.00 kN
- 649. 632 ANSWERS TO SELECTED PROBLEMS 6–15. 6–17. 6–18. 6–19. 6–21. 6–22. 6–23. 6–25. Joint E: 1.4142 P sin 45° - P - FEB sin 45° = 0 Joint F: FFE - 1.4142 P sin 45° = 0 Joint D: FDC - 1.4142 P cos 45° = 0 Joint A: 1.4142 P cos 45° - FAB = 0 FEA = 286 N (C) FBA = 202 N (T) FBE = 118 N (T) FDE = 286 N (C) FDB = 118 N (T) FCB = 202 N (T) FCD = 286 N (C) FBA = 450 N (T) FEB = 70.7 N (T) FEA = 636 N (C) FDE = 500 N (C) FDB = 70.7 N (C) FCB = 550 N (T) FCD = 778 N (C) m = 1.80 Mg Joint A: FAG - 1.414 W sin 45° = 0 Joint D: FDC sin 45° + FDE cos 30.25° - W = 0 FDC = 79.2 lb (C) FBD = 55 lb (T) FBC = 63.3 lb (T) FAB = 63.3 lb (T) FAD = 154 lb (C) FEA = 55 lb (C) FFE = 60 lb (T) FED = 60 lb (T) FFA = 75 lb (C) FDC = 250 lb (T) FBD = 0 FBC = 200 lb (C) FAB = 200 lb (C) FAD = 1250 lb (C) FEA = 0 FED = 1200 lb (T) FFE = 1200 lb (T) FFA = 1500 lb (C) P = 1.50 kN (controls) - 0.8333P cos 73.74° = 0 Joint D: FDE - 0.8333P - P cos 53.13° Joint B: 0.8333PA4 5 B - FBCA4 5 B = 0 - FAB = 0 Joint A: 0.8333P cos 73.74° + P cos 53.13° P = 2000 lb 6–6. 6–7. 6–9. 6–10. 6–11. 6–13. 6–14. FFD = 1768 lb (C) FEF = 1768 lb (T) FED = 1250 lb (C) FAF = 0 FAE = 2450 lb (C) FGF = 2500 lb (T) FGD = 1768 lb (C) FCD = 1250 lb (C) FCG = 1768 lb (T) FBC = 2450 lb (C) FBG = 0 P2 = 135 lb Joint B: FBC - 2.60 P2 sin 22.62° = 0 Joint D: 2.60 P2 cos 22.62° - FDC = 0 Joint A: FAC sin u = 0 FDC = FDE = 825 lb (C) FBC = FBA = 708 lb (C) FFC = FFE = 333 lb (T) FDF = 400 lb (C) FBG = FGC = FGA = 0 FBC = FCD = 667 lb (C) FAB = FDE = 667 lb (C) FEF = 0 FBG = FCG = FAG = FDF = FCF = (controls) 3P = 600 lb P = 200 lb 2P = 800 lb P = 400 lb Joint C: 3P - NC = 0 Joint B: FBD sin 45° - 1.4142 P sin 45° = 0 Joint E: FED - 2P = 0 Joint F: FFB cos 45° - 1.4142 P cos 45° = 0 Joint A: FAF sin 45° - P = 0 FBA = 12.0 kN (T) FBE = 4.00 kN (C) FDB = 4.00 kN (T) FDE = 6.93 kN (C) FCD = 6.93 kN (C) FCB = 8.00 kN (T) FBA = 5.00 kN (T) FBE = 2.00 kN (C) FDB = 2.00 kN (T) FDE = 2.60 kN (C) FCD = 2.60 kN (C) FCB = 3.00 kN (T) FDC = 582 N (T) FED = 929 N (C) FEC = 558 N (T)
- 650. ANSWERS TO SELECTED PROBLEMS 633 6–26. 6–27. 6–29. 6–30. 6–31. 6–33. 6–34. 6–35. FHI = 21.1 kN (C) FFI = 7.21 kN (T) FEF = 12.9 kN (T) FCJ = 1.60 kN (C) FCD = 12 kN (T) FJK = 11.1 kN (C) FHB = 21.2 kN (C) FHI = 35.0 kN (C) FBC = 50.0 kN (T) FBC (4) + 20(4) + 30(8) - 65.0(8) = 0 Ax = 0 Ay = 65.0 kN FHC = 180 lb (C) FBC = 130 lb (T) FBH = 255 lb (T) 336° … u … 347° 127° … u … 196° P = 1.25 kN - 2.00P = 0 Joint F: FFD + 2c1.863Pa 0.5 2 1.25 b d - FBFa 0.5 2 1.25 b - FBDa 0.5 2 1.25 b = 0 Joint B: 2.404Pa 1.5 2 3.25 b - P Joint A: FAF - 2.404Pa 1.5 2 3.25 b = 0 FCD = 0.471P (C) FEC = 1.41P (T) FBD = 1.49P (C) FBF = 1.41P (T) FAC = 1.49P (C) FAE = 1.67P (T) FAB = 0.471P (C) FFD = 1.67P (T) FFE = 0.667P (T) FBA = 722 lb (T) FBE = 297 lb (T) FDE = 780 lb (C) FDB = 0 FCB = 720 lb (T) FCD = 780 lb (C) P = 1.06 kN 1.4142 P = 1.5 P = 1 kN (controls) Joint C: FCB = P (C) 6–37. 6–38. 6–39. 6–41. 6–42. AB, BC, CD, DE, HI, and GI are all zero-force members. 6–43. AB, BC, CD, DE, HI, and GI are all zero-force members. 6–45. 6–46. 6–47. 6–49. 6–50. FEF = 9.38 kN (T) FDE = 15.6 kN (C), FDF = 12.5 kN (T), FCF = 3.12 kN (C), FCD = 9.38 kN (C), FCG = 3.12 kN (T), FFG = 11.2 kN (T), FBC = 13.1 kN (C), FBG = 17.5 kN (T), FAB = 21.9 kN (C), FAG = 13.1 kN (T), FBC = 15 kN (T) FKC = 7.50 kN (C) FKJ = 18.0 kN (C) FKJ sin 33.69°(4) + 5(2) + 3(4) - 15.5(4) = 0 Ay = 15.5 kN Ax = 0 FFD = FFC = 0 FGF = 1.53 kN (T) FCD = 1.92 kN (C) FCH = 1.92 kN 1T2 FBC = 3.25 kN 1C2 FLD = 424 lb (T) FCD = 2600 lb (T) FKL = 3800 lb (C) FKL (8) + 1000(8) - 900(8) - 1300(24) = 0 NA = 1300 lb FGF = 5.625 kN (T) FJE = 9.38 kN (C) FCG = 9.00 kN (T) FIC = 5.62 kN (C) FBG = 200 lb (C) FHG = 420 lb (C) FBC = 495 lb (T) 240(8) - FBC cos 14.04°(4) = 0 Ax = 100 lb Ay = 240 lb FEH = 29.2 kN (T) FED = 100 kN (C) FGH = 76.7 kN (T) FDC = 125 kN (C) FHC = 100 kN (T) FHI = 42.5 kN (T) FCF = 0.770 kN (T) FCD = 8.47 kN (C) FFG = 8.08 kN (T) 7.333 (4.5) - 8 (1.5) - FFG(3 sin 60°) = 0 Ey = 7.333 kN
- 651. 634 ANSWERS TO SELECTED PROBLEMS 6–62. 6–63. 6–65. are lying in the same plane. are lying in the same plane. 6–66. 6–67. 6–69. Apply the force equation of equilibrium along the y axis of each pulley 6–70. 6–71. 6–73. 6–74. 6–75. 6–77. Cy = 273.6 lb Ax = 92.3 lb Cx = 100 lb By = 449 lb Ax = 0 Ay = 5.00 kN MA = 30.0 kN # m By = 15.0 kN Cy = 5.00 kN Ax = 161 lb Cx = 90 lb Cy = 161 lb Ay = 60 lb MC = 1.25 kN # m Cy = 1.30 kN Cx = 795 N Ay = 795 N Ax = 795 N NB = 1125 N NB (0.8) - 900 = 0 FA = P = 25.0 lb FB = 60.0 lb P = 25.0 lb P = 5 lb P = 18.9 N 2P + 2R + 2T - 50(9.81) = 0 P = 12.5 lb FFD = 0 FGE = 505 lb (C) FGD = 157 lb (T) FED cos u = 0 FED = 0 Joint E: FEG, FEC, and FEB FFE cos u = 0 FFE = 0 Joint F: FFG, FFD, and FFC F = 170 N FBE = FBC = 141 N (T) FBD = 707 N (C) FAB = 583 N (C) FAE = FAC = 220 N (T) FEF = 525 lb (C) FDF = 1230 lb (T) Joint F: FBF = 225 lb (T) FCF = 0 FCD = 650 lb (C) Joint C: FCB = 0 6–51. 6–53. 6–54. 6–55. 6–57. 6–58. 6–59. 6–61. Fz = 700 lb Fy = 650 lb Fx = 150 lb Ex = 550 lb Cy = 650 lb Dx = 100 lb FAB = 3.46 kN (C) FED = 3.46 kN (T) FCD = 2.31 kN (T) FCF = 0 FBE = 4.16 kN (T) FDF = 4.16 kN (C) FBC = 1.15 kN (C) FEC = 295 N (C) FAC = 221 N (T) FBC = 148 N (T) Joint C: FBC - 1 2 7.25 (397.5) = 0 FCD = 397 N (C) FBD = 186 N (T) FAD = 343 N (T) + 1 2 7.25 F C D - 200 = 0 Joint D: - 1 3 FAD + 5 2 31.25 FBD FDB = 544 lb (C) FAB = FAD = 424 lb (T) FCB = 344 lb (C) FCD = 406 lb (T) FCA = 1000 lb (C) FDB = 50 lb (T) FAD = FAB = 354 lb (C) FCD = 333 lb (T) FCB = 667 lb (C) FCA = 833 lb (T) FDE = 2.13 kip (T) FJI = 2.13 kip (C) 1.60(40) - FJI (30) = 0 Gy = 1.60 kip FFC = 6.25 kN (C) FDC = 18.8 kN (C), FDF = 25.0 kN (T) FED = 31.2 kN (C), FEF = 18.8 kN (T) FGC = 6.25 kN (T), FGF = 22.5 kN (T) FBC = 26.2 kN (C), FBG = 35.0 kN (T) FAB = 43.8 kN (C), FAG = 26.2 kN (T)
- 652. ANSWERS TO SELECTED PROBLEMS 635 6–78. 6–79. 6–81. 6–82. 6–83. 6–85. Member AB, Member EFG, Member CDI, 6–86. 6–87. 6–89. 6–90. 6–91. NB = 7.05 kN Cy = 7.05 kN NA = 4.60 kN Dy = 1000 lb Dx = 945 lb Ey = 500 lb Ex = 945 lb Ay = 130 lb Ax = 52.6 lb By = 130 lb Bx = 97.4 lb Member AB: FBD = 162.4 lb FBD = 2.60 kN FFB = 1.94 kN mL = 106 kg ms = 1.71 kg FED = 158.9 N FBG = 264.9 N Cy = 833 N Cx = 1.33 kN Ay = 1.17 kN Ax = 167 N ND = 1.05 kN Ay = 2.94 kN Ax = 12.7 kN NC = 12.7 kN Ey = 75 kip Ex = 0 Segment DEF: Fy = 135 kip Ay = 75 kip Ax = 0 Segment ABC: Cy = 135 kip Dy = 30 kip Dx = 0 Segment BD: By = 30 kip Ay = 100 N Ax = 333 N ND = 333 N Cy = 300 N Cx = 300 N Ax = 300 N Ay = 300 N MA = 359 lb # ft Ay = 186 lb 6–93. 6–94. 6–95. 6–97. 6–98. 6–99. 6–101. Member ABC Member CD 6–102. 6–103. 6–105. Member BC Member ACD 6–106. 6–107. 6–109. Clamp Handle 6–110. 6–111. WC = 0.812W NA = 284 N FBE = 2719.69 N F = 370 N Cx = 1175 N NC = 87.5 lb F = 87.5 lb NC = 350 lb F = 175 lb FAD = 3.43 kip FAB = 3.08 kip FAC = 2.51 kip Bx = 2.98 kN Ax = 2.98 kN Ay = 235 N Cx = 2.98 kN By = 549 N Cy = 1.33 kN ME = 500 N # m Ey = 417 N Ex = 0 Ay = 183 N FABC = 319 N FCD = 1.01 kN Ax = 695 N Dx = 695 N Dy = 245 N Ay = 245 N F = 5.07 kN M = 2.43 kN # m NB - NC = 49.5 N 80 - NG cos 36.03° - NC cos 36.03° = 0 F = 562.5 N Ws = 3.35 lb Ax = 2.00 kip Dy = 1.84 kip Member DB: Dx = 1.82 kip Member ABC: Ay = 700 lb Pulley E: T = 350 lb
- 653. 636 ANSWERS TO SELECTED PROBLEMS 6–127. 6–129. 6–130. 6–131. 6–133. Member AC: Member AC: Member CB: 6–134. 6–135. Ey = 5.69 kip Ex = 8.31 kip Ay = 0.308 kip Ax = 8.31 kip P = kL 2 tan u sin u (2 - csc u) By = 97.4 N Bx = 97.4 N Ay = 397 N Ax = 117 N Cy = 97.4 N Cx = 402.6 N FDF = 424 lb (T) FCF = 300 lb (C) FCD = 500 lb (C) FEF = 300 lb (C) FDE = 0 FAE = 367 lb (C) FAD = 0 FAC = 972 lb (T) FAB = 300 lb (C) FBE = 500 lb (T) FBC = 0 FBF = 0 FDF = 424 lb (T) FCF = 300 lb (C) FCD = 300 lb (C) FEF = 300 lb (C) FDE = 0 FAE = 667 lb (C) FAD = 333 lb (T) FAC = 583 lb (T) FAB = 300 lb (C) FBE = 500 lb (T) FBC = 0 FBF = 0 Joint A: FAE = 8.00 kN (T) Joint B: FBA = 17.9 kN (C) Joint D: FDE = 8.00 kN (T) FCD = 8.00 kN (T) Joint C: FCB = 17.9 kN (C) FDB = FBE = 0 FB = 133 lb MCz = 0 MCy = -429 N # m Cz = 125 N Cy = 61.9 N 6–113. 6–114. 6–115. 6–117. 6–118. 6–119. 6–121. 6–122. 6–123. 6–125. 6–126. Cx = 47.3 N Ay = 115 N Ax = 172 N Az = 0 By = -13.3 lb Bx = -30 lb Bz = 0 Bz + 6 9(270) - 180 = 0 FDE = 270 lb -6 9 FDE(3) + 180(3) = 0 Bz = Dz = 283 N By = Dy = 283 N Bx = Dx = 42.5 N P = 283 N W3 = 75 lb W2 = 21 lb W1 = 3 lb M = 4PL sin2 u sin f [cos(f - u)] Nc = 4P sin2 u sin f Cy = 53.3 lb Cx = 413 lb Bx = 333 lb By = 133 lb Ay = 80 lb Ax = 80 lb NC = 15.0 lb Ay = 0 Ax = 120 lb MD = 2.66 kN # m Dy = 1.96 kN Dx = 0 Cy = 7.01 kN Cx = 2.17 kN FAB = 9.23 kN lAB = 861.21 mm, LCAB = 76.41°, NE = 187 N FBC = 15.4 kN (C) FIJ = 9.06 kN (T) W1 = b aW Wx 12b + 3c(4b) + W¢1 - x 3b + 3 4c ≤(b) - W1(a) = 0 ©MA = 0; FCD(c) - Wx A3b + 3 4cB A1 4cB = 0 ©ME = 0; W(x) - NBA3b + 3 4cB = 0
- 654. ANSWERS TO SELECTED PROBLEMS 637 Chapter 7 7–1. 7–2. 7–3. 7–5. 7–6. 7–7. 7–9. . 7–10. 7–11. MD = 13.5 kN # m V D = 1 kN ND = 0 MC = 9.375 kN # m V C = 3.25 kN NC = 0 MC = 1.5 kN # m V C = 0 NC = 0 MC = -480 lb # in MC + 80(6) = 0 V C = 0 NC = -80 lb NC + 80 = 0 MC = - 5 48w0L2 V C = 3w0L 8 NC = 0 MC = 9 kN # m V C = -1 kN NC = 0 MC = -144 N # m V C = -96 N NC = 400 N Ay = 96 N Ax = 400 N MC = 9750 lb # ft V C = -125 lb NC = -1804 lb MD = -600 lb # ft V D = 300 lb ND = 0 MC = -857 lb # ft V C = -386 lb NC = 0 MD = 48.0 kip # ft V D = -1.00 kip ND = 0 MC = 56.0 kip # ft V C = -1.00 kip NC = 0 Ax = 0 Ay = 7.00 kip By = 1.00 kip 7–13. 7–14. 7–15. 7–17. 7–18. 7–19. 7–21. 7–22. 7–23. ME = 0 NE = 86.0 N V E = 0 MD = 19.0 N # m V D = 26.0 N ND = 0 MD = 42.5 kN # m V D = -10.6 kN ND = 0 MG = 1160 lb # ft V G = -580 lb NG = 0 MF = 1040 lb # ft V F = 20 lb NF = 0 Ay = 520 lb Ax = 0 Ey = 580 lb Dy = 540 lb FBC = 560 lb Dx = 0 a = 2 3L ME = -4.875 kN # m V E = 3.75 kN NE = 4 kN MD = -18 kN # m V D = -9 kN ND = 4 kN a b = 1 4 Ay = w 6b(2a + b) (b - a) MC = 800 lb # ft V C = 0 NC = 0 MD = -1.60 kip # ft V D = 800 lb ND = 0 ME = 1000 N # m V E = 500 N NE = -1.48 kN MD = 500 N # m V D = 0 ND = 1.26 kN Member BC: Bx = 1258.33 N Member AB: By = 500 N
- 655. 638 ANSWERS TO SELECTED PROBLEMS 7–38. 7–39. 7–41. 7–42. 7–43. 7–45. 7–46. 7–47. 7–49. 7–50. 7–51. 7–53. at x = 3.87 ft V = 0 V = 25 - 1.667x2 0 … x 6 9 ft Mmax = 0.866 kN # m x = 1.732 m M = 25(100x - 5x2 - 6) V = 250(10 - x) M = -7.5x + 75 V = -7.5 5 m 6 x 6 10 m M = 2.5x - x2 V = 2.5 - 2x 0 … x 6 5 m M = 759 N # m x = 1.75 m M = w0L2 16 x = L2 M = 9 128w0L2 x = A3 8 BL M0 = 44 kN # m For Mmax = M02, M0 = 44 kN # m For Vmax = M0L, M0 = 45 kN # m M = -10 kN # m x = 0, V = 4 kN x = 12+ , V = -333, M = 0 x = 8+ , V = -833, M = 1333 M|x=4 = 12 kN # m M = {36 - 6x} kN # m V = -6kN 4 m 6 x … 6 m M = {3x} kN # m V = 3kN 0 … x 6 4 m (ME)z = -26.8 kN # m (ME)y = -43.5 kN # m (ME)x = 0 (V E)z = -87.0 kN (V E)y = 53.6 kN (NE)x = 0 (MD)z = 26.2 kN # m (MD)y = 87.0 kN # m (MD)x = 49.2 kN # m (V D)x = 0 (ND)y = -65.6 kN (V D)x = 116 kN 7–25. Use top segment of frame. 7–26. 7–27. 7–29. Beam reaction 7–30. 7–31. 7–33. 7–34. 7–35. 7–37. (MC)z = 675 N # m TC = 30 N # m (MC)x = -825 N # m (V C)z = -550 N (V C)x = 450 N (NC)y = 0 Bx = 900 N Bz = 550 N (MC)z = -750 lb # ft (MC)y = -1.20 kip # ft (MC)x = 1.40 kip # ft (V C)z = 700 lb (V C)x = -150 lb (NC)y = -350 lb (MC)z = -178 lb # ft (MC)y = 72.0 lb # ft (MC)x = 20.0 lb # ft (V C)z = 10.0 lb (V C)x = 104 lb (NC)y = 0 MD = 8.89 N # m V D = 37.5 N ND = -29.4 N By = 37.5 N Bx = 29.39 N MD = 1.06 kip # ft V D = 1.06 kip ND = 844 lb MC = -844 lb # ft V C = -844 lb NC = 1.75 kip MC = -17.8 kip # ft R = 700 lb MC = 382 N # m V C = 0 NC = -1.91 kN MC = wL2 8 cos u V C = 0 NC = -wL 2 csc u ME = 1140 lb # ft V E = 120 lb NE = 360 lb MD = 900 lb # ft V D = 0 ND = 200 lb
- 656. ANSWERS TO SELECTED PROBLEMS 639 7–54. 7–55. 7–57. 7–58. 7–59. 7–61. 7–62. 7–63. 7–65. 7–66. 7–67. 7–69. 7–70. x = AL 3 B + , V = -P, M = PL x = 4+ , V = -12.5, M = 10 x = 2- , V = 7.5, M = 15 x = 4+ , V = 6, M = -22 x = 2+ , V = -14.5, M = 7 x = 6- , V = -5, M = -10 x = 2+ , V = 5, M = -10 x = 6, V = -625, M = 1250 x = 4- +, V = 275, M = 1900 x = 2- , V = 675, M = 1350 Mz = 0 My = 8.00 lb # ft Mx = {2y2 - 24y + 64.0} lb # ft V z = {24.0 - 4y} lb V x = 0 M = - pgr2 0 12L2 c(L + x)4 - L3 (4x + L)d V = pgr2 0 3L2 c(L + x)3 - L3 d M|x=7.5 ft = 2250 lb # ft x = 6 ft M = {3000x - 250x2 - 6750} lb # ft V = {3000 - 500x} lb w0 = 8.52 kN/m w0 = 21.8 lb/ft M|x=3 m = -18 kN # m V|x=3 m+ = 12 kN V|x=3 m- = -10 kN M = 5-2(6 - x)2 6 kN # m V = 524 - 4x6 kN 3 m 6 x … 6 m M = E -2 9x3 - 4xF kN # m V = E -2 3x2 - 4F kN 0 … x 6 3 m M = E - 1 18(24 - x)3 F kip # ft V = E1 6(24 - x)2 F kip M = E48.0x - x3 18 - 576F kip # ft V = E48.0 - x2 6 F kip w = 22.2 lb/ft M = -180 V = 0 9 ft 6 x 6 13.5 ft Mmax = 64.5 lb # ft M = 25x - 0.5556x3 7–71. 7–73. 7–74. 7–75. 7–77. 7–78. 7–79. 7–81. 7–82. 7–83. 7–85. Use . 7–86. 7–87. 7–89. Entire cable 7–90. 7–91. 7–93. 7–94. 7–95. yD = 7.04 ft yB = 8.67 ft Tmax = 157 N yB = 2.43 m Tmax = TDE = 8.17 kN TCD = 4.60 kN TBC = 4.53 kN TAB = 6.05 kN yB = 3.53 m xB = 3.98 ft P = 72.0 lb L = 15.7 ft Joint D: TCD = 43.7 lb Joint A: TAC = 74.7 lb TBD = 78.2 lb x = 900, V = -487, M = 350 x = 300, V = 722, M = 277 x = 6, V = 2.5, M = 0 x = 3+ , V = 11.5, M = -21 x = 6+ , V = 4 w, M = -120 m w = 2 kip/ft w = 5 kip/ft Mmax = -6w w = 2 kip/ft V max = 4w x = 3, V = -12, M = 12 x = L- , V = -2wL 3 , M = -wL2 6 x = 18, V = -3.625, M = 0 x = 9+ , V = -1.375, M = 25.9 x = 9- , V = 0.625, M = 25.9 x = 0, V = 5.12, M = 0 x = 6, V = -900, M = -3000 x = 14.1, V = 0, M = 334 x = 8- , V = 1017, M = -1267 x = 5- , V = -225, M = -300 x = 1+ , V = 175, M = -200 x = 1.5- , V = 250, M = 712.5 x = 3+ , V = 15, M = -7.50 x = 1+ , V = -9.17, M = -1.17 x = 1, V = -3.84, M = 0 x = 0.8- , V = 0.16, M = 0.708 x = 0, V = 1.76, M = 0 x = 0.2+ , V = 96.7, M = -31 x = A2L 3 B + , V = -2P, M = A2 3 B PL
- 657. 640 ANSWERS TO SELECTED PROBLEMS 7–118. 7–119. 7–121. 7–122. 7–123. 7–125. 7–126. 7–127. Chapter 8 8–1. 8–2. 8–3. 8–5. u = 52.0° 180(10 cos u) - 0.4(180)(10 sin u) - 180(3) = 0 ms = 0.256 P = 474 N N = 494.94 N P = 140 N P cos 30° + 0.25N - 50(9.81) sin 30° = 0 h = 93.75 ft l = 238 ft s = 18.2 ft ME = 86.6 lb # ft V E = 28.9 lb NE = 0 ND = FCD = -86.6 lb V D = MD = 0 FCD = 86.6 lb M = -300 - 200y N = -150 lb V = 200 lb 0 … y … 2 ft M = 150 cos u + 200 sin u - 150 N = 150 cos u + 200 sin u V = 150 sin u - 200 cos u 0° … u … 180° M = {16.0 - 2.71x - 0.0981x2 } kN # m V = { - 0.196x - 2.71} kN 2 m 6 x … 5 m M = {5.29x - 0.0981x2 } kN # m V = {5.29 - 0.196x} kN 0 … x 6 2 m M = {27.0 - 4.50x} kN # m V = -4.50 kN 3 m 6 x … 6 m M = {1.50x} kN # m V = 1.50 kN 0 … x 6 3 m Ay = 1.50 kN FCD = 6.364 kN Tmax = 76.7 lb a = 0.366L MD = -54.9 N # m V D = -220 N ND = -220 N Segment CD ME = 112.5 N # m 7–97. 7–98. 7–99. 7–101. 7–102. 7–103. 7–105. 7–106. 7–107. 7–109. 7–111. 7–113. 7–114. 7–115. 7–117. V E = 0 NE = 80.4 N Segment CE FBC = 310.58 N x = 5+ , V = -1.14, M = 2.29 x = 2- , V = 4.86, M = 9.71 h = 10.6 ft Total length = 55.6 ft h = 1.47 m y = 135.92Ccosh 7.3575(10-3 )x - 1D dy dx = sinh 7.3575(10-3 )x L = 15.5 m Tmax = 1.60 kN y = 23.5[cosh 0.0425x - 1] m FH = 1153.41 N L = 45 = 2e FH 49.05 sinha 49.05 FH (20)b f y = FH 49.05 ccosha 49.05 FH xb - 1d m L = 51.3 m Tmax = 5.36 kN Tmax = 48.7 kip y = 46.0(10-6 )x3 + 0.176x w0 = 77.8 kNm y = 150 m at x = -(1000 - x0) y = 75 m at x = x0 y = w0 4FH x2 dy dx = w0 2FH x L = 13.4 ft h = 2.68 ft 4.42 kip w0 = 264 lb/ft 10 = w0 2 FH (25 - x)2 15 = w0 2 FH x2 w0 = 0.846 kN/m P = 71.4 lb xB = 4.36 ft Joint C: 30 - 2xB 2(xB - 3)2 + 64 TBC = 102 Joint B: 13xB - 15 2(xB - 3)2 + 64 TBC = 200
- 658. ANSWERS TO SELECTED PROBLEMS 641 8–6. 8–7. Yes, the pole will remain stationary. 8–9. 8–10. 8–11. 8–13. 8–14. 8–15. 8–17. Boy does not slip. 8–18. 8–19. 8–21. 8–22. 8–23. 8–25. Assume 8–26. 8–27. The man is capable of moving the refrigerator. The refrigerator slips. 8–29. 8–30. Tractor can move log. 8–31. 8–33. The bar will not slip. 8–34. 8–35. 8–37. 8–38. h = 0.48 m b = 2a sin u N = wa cos u P = 0.127 lb u = tan-1 a 1 - mAmB 2mA b NA = 130 lb FA = 17.32 lb W = 836 lb NA = 12.9 N NB = 72.4 N P = 29.5 N ms œ = 0.300 P = 45.0 lb P = 100 lb x = 1.44 ft 6 1.5 ft N = 160 lb P = 100 lb P = 0.990 lb n = 12 FCD = 8.23 lb u = 16.3° NB = 150 cos u NA = 200 cos u u = 10.6° x = 0.184 ft ms = 0.595 By = 228 lb Bx = 34.6 lb Ay = 468 lb FD = 36.9 lb ND = 95.38 lb FB = 200 N ms = 0.577 P = 350 N NB = 700 N FB = 280 N P = 1 lb P = 15 lb d = 13.4 ft 30 (13 cos u) - 9 (26 sin u) = 0 ms = 0.231 8–39. 8–41. 8–42. He can move the crate. 8–43. 8–45. 8–46. 8–47. 8–49. 8–50. 8–51. 8–53. Slipping of board on saw horse . Slipping at ground . Tipping . The saw horse will start to slip. 8–54. The saw horse will start to slip. 8–55. 8–57. 8–58. 8–59. 8–61. 8–62. 8–63. 8–65. 8–66. 8–67. 8–69. 8–70. 8–71. P = 574 N All blocks slip at the same time; P = 625 lb P = 1.29 kN NC = 600 N NA = 1212.18 N P = 863 N P = 1.98 N P = 90.7 lb NB = 82.57 lb NC = 275.23 lb P = 49.0 N P = 45 lb FB = 37.73 N NB = 679.15 N NA = 150.92 N M = 90.6 N # m ND = 188.65 N NC = 377.31 N u = 16.0° P = 90 lb F¿ = 60 lb N¿ = 150 lb P = 60 lb ms = 0.304 Px = 21.2 lb Px = 19.08 lb Px = 24.3 lb N = 48.6 lb P = 1.02 kN ml = 800 kg ml = 1500 kg Nl = 9.81ml T = 11 772 N P = 589 N FA = 71.4 N M = 77.3 N # m By = 110.4 N Bx = 110.4 N NA = 551.8 N ms œ = 0.376 ms = 0.4 NA = 0.9285 FCA FA = 0.3714 FCA ms = 0.3 u = 33.4°
- 659. 642 ANSWERS TO SELECTED PROBLEMS 8–102. 8–103. Since the man will not slip, and he will successfully restrain the cow. 8–105. Thus, the required number of full turns is 8–106. The man can hold the crate in equilibrium. 8–107. 8–109. For motion to occur, block A will have to slip. 8–110. 8–111. 8–113. No tipping occurs. 8–114. 8–115. 8–117. Apply Eq. 8–7. 8–118. 8–119. 8–121. 8–122. 8–123. 8–125. 8–126. 8–127. 8–129. mB = 13.1 kg fs = 16.699° P = 179 N P = 215 N M = a mk 21 + mk 2 bpr sin fk = mk 21 + mk 2 tan fk = mk M = 2msPR 3 cos u F = 573 lb p0 = 0.442 psi M = msP 3 cos u a d2 3 - d1 2 d2 2 - d1 2 b A = p 4 cos u(d2 2 - d1 2 ) N = P cos u M = ms PR 2 M = 270 N # m Fsp = 1.62 kip mk = 0.0568 M = 304 lb # in. x = 0.00697 m 6 0.125 m NA = 478.4 N FA = 16.2 N T = 20.19 N W = 39.5 lb F = 2.49 kN FB = T = 36.79 N P = 223 N T2 = 1.59 N T1 = 1.85 N n = 2 b = (2n + 0.9167)p rad T = 486.55 N N = 314.82 N F 6 Fmax = 54 lb, P = 17.1 lb M = 216 N # m (ms)req = 0.3 8–73. 8–74. 8–75. 8–77. 8–78. 8–79. 8–81. 8–82. 8–83. 8–85. 8–86. 8–87. 8–89. 8–90. 8–91. 8–93. Yes, just barely. 8–94. 8–95. 8–97. 8–98. 8–101. T1 = 688.83 T2 = 1767.77 N TC = 150.00 N TA = 616.67 N M = 187 N # m P = 42.3 N F¿ = 19.53P F = 4.75P u = 24.2° T1 = 57.7 lb F = 136.9 lb N = 185 lb F = 16.2 kN F = 4.60 kN F = 372 N F = 1.31 kN FB = 38.5 lb NB = 65.8 lb FC = 13.7 lb TB = 13.678 lb NA = 42.6 N NC = 123 N F = 174 N F = 74.0 N fs = 14.036° u = 5.455° FAB = 1962 N FBD = 1387.34 N F = 1387.34 N FCA = FCB F = 49.2 N F = 71.4 N F = 678 N fs = 14.036° u = 5.455° M = 145 lb # ft M = 5.69 lb # in F = 620 N fs = 11.310° u = 7.768° P = 1.80 kN P = 1.38W P = 0.0329W FB = 0.05240W NB = 1.1435W NA = 0.5240W
- 660. ANSWERS TO SELECTED PROBLEMS 643 8–130. 8–131. 8–133. 8–134. 8–135. 8–137. 8–138. 8–139. 8–141. 8–142. 8–143. 8–145. a) b) 8–146. a) b) 8–147. 8–149. 8–150. 8–151. 8–153. The wedges do not slip at contact surface AB. The wedges are self-locking. Chapter 9 9–1. y = 2.29 m x = 1.64 m m = 11.8 kg dm = 2y2 + 4 dy dL = 1 2 2y2 + 4 dy FC = 0 NC = 8000 lb F = 1389.2 lb N = 7878.5 lb u = 35.0° M = 2.21 kip # ft M = 2.50 kip # ft T = 1250 lb NB = 2500 lb NA = 1000 lb mB = 1.66 kg W = 6.89 kN W = 1.25 kN W = 15.3 kN T = 6131.25 N NB = 5886.0 N NA = 6376.5 N W = 6.97 kN NA = 5573.86 N T = 2786.93 N s = 0.750 m P = 40 lb = 235 N P = (1200) (9.81) (0.2 + 0.4) 2(15) P = 266 N P = 299 N P = 96.7 N u = 5.74° ms = 0.411 P = 42.2 lb P = 814 N (approx.) P = 814 N (exact) R = 2P2 + (833.85)2 rf = 2.967 mm (rf)B = 3 mm (rf)A = 7.50 mm (rf)B = 0.075 in. (rf)A = 0.2 in. 9–2. 9–3. 9–5. 9–6. 9–7. 9–9. 9–10. 9–11. 9–13. 9–14. 9–15. 9–17. y = 3 5 h x = 3 8 a A = 2 3 ah y ' = y x ' = a 2h1/2 y1/2 dA = a h1/2 y1/2 dy y = 3 10 h x = 3 4 a A = 1 3 ah y = c2 (b - a) 2ab ln b a x = b - a lnb a A = c 2 lnb a y = 1.33 in. y ' = 1 2 x2 dA = x2 dx y = 3 4 2ab x = 3 5 b A = 4 3a1/2 b3/2 y = 0.857 ft x = 2.4 ft A = 2.25 ft2 y = 0.3125 m x = 0.714 m A = 0.4 m2 y ' = x3/2 2 x ' = x dA = x 3/2 dx x = r sin a a y = 1.82 ft x = 0 x = 5 9 L m = 3 2 m0L dm = m0 A1 + x L B dx MO = 3.85 N # m Oy = 7.06 N Ox = 0 x = 0.546 m MA = 32.7 lb # ft Ay = 26.6 lb Ax = 0
- 661. 644 ANSWERS TO SELECTED PROBLEMS 9–43. 9–45. 9–46. 9–47. 9–49. 9–50. 9–51. 9–53. 9–54. 9–55. 9–57. 9–58. 9–59. 9–61. 9–62. 9–63. 9–65. x = 2.4971(10-3 ) 16.347(10-3 ) = 153 mm ©m = 16.4 kg y = 293 mm y = b(W2 - W1) 2b2 - c2 cW x = W1 W b y = 441.2(104 ) 81(104 ) = 544 mm x = 0 y = 2.56 in. x = 4.83 in. x = 4Aro 3 - ri 3 B 3pAro 2 - ri 2 B y = 9.648 6.84 = 1.41 m x = 15.192 6.84 = 2.22 m y = 2.57 in. y = 2.00 in. = 5.125 in. y = 3[2(6)(1)] + 5.5(6)(1) + 9(6)(1) 2(6)(1) + 6(1) + 6(1) y = 12 in. x = 2.64 in. Ax = 0 Ay = 1.32 kN Ey = 342 N y = 9.24 m x = 1.65 m f = 30° - 10.89° = 19.1° u = tan-1 50 400 sin 60° - 88.60 = 10.89° y = 88.6 mm x = -50 mm z = 0.157 in. y = 0.0370 in. x = 0.0740 in. z = 2.14 in. y = 1.07 in. x = -0.590 in. z = 169.44(103 ) 1361.37 = 124 mm y = 60(103 ) 1361.37 = 44.1 mm x = 164.72(103 ) 1361.37 = 121 mm z = 8 15 r m = pkr4 4 9–18. 9–19. 9–21. 9–22. 9–23. 9–25. 9–26. 9–27. 9–29. 9–30. 9–31. 9–33. 9–34. 9–35. 9–37. 9–38. 9–39. 9–41. 9–42. z = a p y = 3 4 h V = pa 2 h 6 y = 23 55 a y ' = y dm = pr0Aa2 - y2 + ay - y3 a B dy y = 4.36 ft z = 2 9 h y = 3.2 m y ' = y dV = p 16 y3 dy y = a 2(10 - 3p) x = 5 9 a m = 3 2 r0 abt y = 1 ft y ' = y dA = Ay 2 - y2 4 B dy y = 0.357 m x = 0.914 m Ay = 73.9 kN Ax = 24.6 kN NB = 55.1 kN y = 0 x = 1.20 m y = n + 1 2(2n + 1) h y = y 2 dA = y dx y = 0.45 m x = 0.45 m y = 1.14 ft x = 1.6 ft A = 2.25 ft2 y ' = 1 2 Ax + x3 9 B x ' = x dA = Ax - x3 9 B dx y = 0.541 in. x = 1.08 in. x = 5a 8 x ' = x dA = 2kAx - x2 2a B dx x = - 0.833a Ay = 1.98 kN Ax = 0 FBC = 2.64 kN
- 662. ANSWERS TO SELECTED PROBLEMS 645 9–66. 9–67. 9–69. 9–70. 9–71. 9–73. 9–74. 9–75. 9–77. 9–78. or 9–79. 9–81. 9–82. 9–83. 9–85. 9–86. 9–87. 9–89. 9–90. 9–91. 9–93. 9–94. 9–95. 9–97. = 536 m3 V = 2pC A4(4) 3p B A1 4p (4)2 B + (2)(8)(4)D 2.26 gallons R = 29.3 kip V c = 20.5 m3 V h = 2p[0.75(6) + 0.6333(0.780) + 0.1(0.240)] V = 0.0376 m3 A = 1.06 m2 V = 50.6 in3 A = 116 in2 = 0.0486 m3 V = 2p[(112.5)(75)(375) + (187.5)(325)(75)] A = 141 in2 A = 1365 m2 = 77.0 m3 + 1.667A2(1.5) 2 B D V = 2pC A4(3) 3p Ap(32 ) 4 B + 0.5(1.5)(1) V = 101 ft3 V = 3485 ft3 A = 2p(184) = 1156 ft2 z = 122 mm h = 48 mm h = 80 mm x = 11.02(106 )p 172(103 )p = 64.1 mm z = 1.67 in. y = 2.79 in. x = 2.19 in. z = 754 mm z = 1.0333p 9.3333p = 111 mm y = 11.0 ft x = 19.0 ft Ay = 5.99 kN By = 4.66 kN y = 3.07 m x = 4.56 m u = 30.2° z = 371 433.63 16 485.84 = 22.5 mm x = 216 000 16 485.84 = 13.1 mm x = L + (n - 1)d 2 y = 3.80 ft x = 5.07 ft z = 1.8221(10-3 ) 16.347(10-3 ) = 111 mm y = -15 mm 9–98. 9–99. 9–101. 9–102. 9–103. 9–105. 9–106. 9–107. 9–109. 9–110. 9–111. 9–113. 9–114. 9–115. 9–117. 9–118. 9–119. 9–121. 9–122. 9–123. 9–125. 9–126. 9–127. 9–129. y = 3.00 m x = 2.74 m FR = 7.62 kN dFR = 6A- 240 x + 1 + 340B dx y = -0.262a y = 1.63 in. x = 0 y = 39.833 27.998 = 1.42 in. x = 76.50 27.998 = 2.73 in. z = 2 3 a x = y = 0 y = 87.5 mm y = 1.33 in. y ' = x2 2 dA = x2 dx FR = 170 kN x = 1.51 m F.S. = 2.66 (Wcon)r = 282.53 kN (Wcon)p = 188.35 kN Fh = 176.58 kN Fv = 39.24 kN mA = 5.89 Mg L = 2.31 m NC = 13.1 kN wC = 58.86 kN wB = 39.24 kN FR = 450 lb FR = 225 lb FR = 41.7 kN dFR = A26.556721 - y2 - 6.9367y21 - y2 B dy h = 2.7071 - 0.7071y FABDC = 1800 lb FCDEF = 750 lb d = 3.65 m d = 2.68 m -176 580(2) + 73 575dA2 3dB = 0 h = 106 mm V = 22.1(103 ) ft3 A = 2p[7.5(2241) + 15(30)] = 3.56(103 ) ft2 14.4 liters A = 43.18 m2 V = 25.5 m3
- 663. 646 ANSWERS TO SELECTED PROBLEMS 10–34. 10–35. 10–37. 10–38. 10–39. 10–41. Consider a large rectangle and a hole. 10–42. 10–43. , 10–45. Consider three segments. 10–46. 10–47. 10–49. Consider three segments. 10–50. 10–51. 10–53. 10–54. 10–55. 10–57. Consider rectangular segments, , , and 10–58. 10–59. 10–61. 10–62. 10–63. 10–65. Ixy = 3.12 m4 y ' = y 2 x ' = x dA = 1 8(x3 + 2x2 + 4x) dx Ixy = 48 in4 Ixy = a2 b2 8 Ixy = 0.667 in4 dA = x dy y = y x = x 2 Ix¿ = 30.2(106 ) mm4 Iy = 153(106 ) mm4 Ix = 115(106 ) mm4 150 mm * 12 mm 100 mm * 12 mm 226 mm * 12 mm Ix = 22.9(106 ) mm4 Ix = 388 in4 Ix¿ = 15.896 + 36.375 = 52.3 in4 y = 61.75 13 = 4.75 in. Ix = 2.51(106 ) mm4 Iy¿ = 1.21(109 ) mm4 Ix¿ = 124(106 ) mm4 Iy = 914(106 ) mm4 Iy = 548(106 ) mm4 Ix = 548(106 ) mm4 Ix¿ = 64.0 in4 y = 2.00 in. Iy = 2.51(106 ) mm4 Ix = 52.7(106 ) mm4 Ix = 2.17(10-3 ) m4 Ix¿ = 722(10)6 mm4 y = 170 mm = 74 in4 Iy = C 1 12(2)(63 )D + 2C 1 12(3)(13 ) + 3(1)(2.5)2 D Iy¿ = 122(106 ) mm4 Ix¿ = 34.4(106 ) mm4 y = 22.5 mm = 10.3(109 ) mm4 + C - p 4(75)4 + (-p(75)2 (450)2 D + C 1 12(200)(300)3 + 200(300)(450)2 Chapter 10 10–1. 10–2. 10–3. 10–5. 10–6. 10–7. 10–9. 10–10. 10–11. 10–13. 10–14. 10–15. 10–17. 10–18. 10–19. 10–21. 10–22. 10–23. 10–25. 10–26. 10–27. 10–29. 10–30. 10–31. 10–33. + 1 2(200)(300)(200)2 D (Iy)triangle = C 1 36(200)(3003 ) Iy = 45.5(106 ) mm4 Ix = 76.6(106 ) mm4 = 54.7 in4 Iy = 1 12(2)(6)3 + 2C 1 12(4)(1)3 + 1(4)(1.5)2 D Ix¿ = 57.9 in4 y = 2.20 in. J0 = pr0 4 4 Iy = pr0 4 8 dA = (rdu) dr Iy = 30.9 in4 Ix = 9.05 in4 Iy = 307 in4 dA = x1/3 dx Iy = 2 15hb3 Ix = 2 7bh3 Iy = 1 12hb3 dA = Ah - h bxB dx Iy = 1.07 in4 Ix = 19.5 in4 Iy = 0.333 in4 dA = (2 - 2x3 ) dx Iy = 10.7 in4 Ix = 307 in4 JO = 0.491 m4 Iy = 0.2857 m4 dA = 2x4 dx Ix = 0.2051 m4 dA = C1 - Ay 2 B1/4 D dy Ix = 0.205 m4 Iy = 4.57 m4 Ix = 2.13 m4 dA = A2 - y2 2 B dy Ix = 0.0606 m4 Iy = 2.67 m4 Ix = 0.533 m4 dA = [2 - (4y)1/3 ] dy
- 664. ANSWERS TO SELECTED PROBLEMS 647 10–66. 10–67. 10–69. 10–70. 10–71. 10–73. Consider three segments. 10–74. 10–75. 10–77. Consider three segments. 10–78. 10–79. 10–81. 10–82. 10–83. 10–85. 10–86. Imax = 31.7 in4 y = 1.68 in. x = 1.68 in. Iuv = 111 in4 Iv = 238 in4 Iu = 109 in4 R = 128.72 in4 Iavg = 173.72 in4 y = 8.25 in. Iuv = -126(106 ) mm4 Iv = 258(106 ) mm4 Iu = 112(106 ) mm4 x = 48.2 mm Iuv = -3.08(106 ) mm4 Iv = 47.0(106 ) mm4 Iu = 43.4(106 ) mm4 y = 82.5 mm Imin = 5.03(106 ) mm4 , (up)2 = -77.7° Imax = 113(106 ) mm4 , (up)1 = 12.3° Ixy = -22.4(106 ) mm4 Iy = 9.907(106 ) mm4 Ix = 107.83(106 ) mm4 Iuv = 111 in4 Iv = 238 in4 Iu = 109 in4 y = 8.25 in. Iuv = 17.5 in4 Iv = 23.6 in4 Iu = 43.9 in4 Ixy = -13.05(106 ) mm4 Iuv = -126(106 ) mm4 Iv = 258(106 ) mm4 Iu = 112(106 ) mm4 x = 48.2 mm Ixy = -110 in4 Ixy = 17.1(106 ) mm4 Ixy = 36.0 in4 Ixy = 35.7 in4 Ixy = 10.7 in4 dA = x1/2 dx, x ' = x, y ' = y 2 Ixy = 3 16b2 h2 Ixy = 0.333 m4 d b 10–87. d b 10–89. 10–90. 10–91. 10–93. 10–94. 10–95. 10–97. 10–98. 10–99. 10–101. 10–102. 10–103. 10–105. 10–106. 10–107. 10–109. 10–110. 10–111. 10–113. Consider four triangles and a rectangle. 10–114. 10–115. 10–117. 10–118. Ix = 0.610 ft4 Iy = 2.13 ft4 dA = 1 4(4 - x2 ) dx y = 0.875 in., Ix¿ = 2.27 in4 Ix = 1 12a4 Iy = 0.187d4 IO = 0.113 kg # m2 IO = 0.276 kg # m2 IA = 222 slug # ft2 IO = 84.94 slug # ft2 Ix = 3.25 g # m2 Iz = 2.25 kg # m2 IG = 4.45 kg # m2 y ' = 1(3) + 2.25(5) 3 + 5 = 1.78 m Iy = 0.144 kg # m2 Iz = 0.150 kg # m2 IO = 53.2 kg # m2 L = 6.39 m 0.5 = 1.5(6) + 0.65[1.3(2)] + 0[L(2)] 6 + 1.3(2) + L(2) Iy = 1.71(103 ) kg # m2 Iz = 63.2 slug # ft2 Iz = 87.7(103 ) kg # m2 dIz = rp 8192z8 dz Ix = 93 10 mb2 Iy = 2 5 mb2 kx = 57.7 mm dIx = rp 2 (2500 x2 ) dx dm = rp(50x) dx Iz = 7 18 ml 2 Ix = 3 10 mr 2 Iz = 3 10 mr0 2 dIz = 1 2 rpar0 - r0 h zb 4 dz dm = rpar0 - r0 h zb 2 dz (up)2 = 77.7° (up)1 = 12.3° Imin = 5.03(106 ) mm4 Imax = 113(106 ) mm4 (u)p)2 = 45° (up)1 = 45° Imin = 8.07 in4
- 665. 648 ANSWERS TO SELECTED PROBLEMS 11–27. 11–29. 11–30. 11–31. 11–33. 11–34. 11–35. 11–37. 11–38. 11–39. 11–41. Thus, the cylinder is in unstable equilibrium at 11–42. 11–43. 11–45. d = h 3 V = W(h - 3d) 4 cos u y = 1 4(h + d) h = 23 r h = 0 u = 0° (Q.E.D.) V = mg(r + a cos u) d2 V du2 = 17.0 7 0 stable u = 20.2° d2 V du2 = -72 6 0 unstable u = 0° d2 V du2 = 135 7 0 stable u = 64.8° mE = 7.10 kg + 202.5 cos2 u - 405 cos u - 9.81 mEb + 202.5 V = -4.415 mE sin u d2 V du2 = -1764 6 0 unstable u = 17.1° d2 V du2 = 1777 7 0 stable u = 70.9° x = 1.23 m u = 36.1° + 24.525a + 4.905b V = 6.25 cos2 u + 7.3575 sin u d2 V dh2 = 70 7 0 stable h = 8.71 in. WD = 275 lb u = 59.0° V = 5886 cos u + 9810 sin u + 39 240 d2 V du2 = -25.6 6 0 unstable u = 36.9° d2 V du2 = 16 7 0 stable u = 90° d2 V dx2 = -12.2 6 0 unstable x = -0.424 ft 10–119. 10–121. Chapter 11 11–1. 11–2. 11–3. 11–5. 11–6. 11–7. 11–9. 11–10. 11–11. 11–13. 11–14. 11–15. 11–17. 11–18. 11–19. 11–21. 11–22. 11–23. 11–25. 11–26. d2 V dx2 = 12.2 7 0 stable x = 0.590 ft F = 259 lb 52 = y2 C + 32 - 2(yC) (3) cos (90° - u) x = 16 in. WG = 2.5 lb k = 10.8 lb/ft Fsp = 4.961 lb xA = 1 sin u yA = 1 cos u yG = 0.5 cos u u = 38.8° F = 200 N k = 166 N/m xC = 0.25 cos u yGt = 0.25 sin u + a yGb = 0.25 sin u + b u = cos-1 A a 2L B 1 3 ml = mAs a B u = 90° u = 13.9° yA = 3 sin u yC = 1 sin u P = 2k tan u (2l cos u - l0) F = 60 N FE = 177 N dyA = 0.5 du yD = 2(0.2 cos u) u = 24.9° FS = 15 lb Fsp = 10.0 lb yC = 3 sin u xB = 6 cos u u = 41.2° u = 0° and u = 73.1° FAD = 3.92 kN yJ = 2(2.4 sin u) + b yD = 2.4 sin u Ixy = 0.1875 m4 dIxy = 1 2 y5/3 dy dA = y1/3 dy Ix¿ = 146(106 ) mm4 Ix = 914(106 ) mm4
- 666. ANSWERS TO SELECTED PROBLEMS 649 11–46. 11–47. 11–49. 11–50. 11–51. 11–53. 11–54. P = Ab - a 2c B mg d2 V du2 = 125.7 7 0 stable u = 37.8° V = 50 sin2 u - 100 sin u - 50 cos u + 50 u = 90° and u = sin-1 A W 2kL B F = 512 N d = 0.586 h V = WC6h2 - 12hd + 3d2 4(3h - d) D cos u y = 6h2 - d2 4(3h - d) h = 1.35 in. u = 0°, d2 V du2 = -12.6 6 0 unstable 11–55. 11–57. 11–58. h = 2kl 2 W d2 V du2 = -45.5 6 0 unstable u = 72.5° d2 V du2 = 35 7 0 stable u = 0° V = 25 sin2 u + 15 cos u d2 V du2 = -1775 6 0 unstable u = 9.47° d2 V du2 = 1524 7 0 stable u = 90°
- 667. Index A Angles, 44–47, 70, 80–81, 389–390, 414–415 Cartesian coordinate direction, 44–47, 80–81 formed between two vectors, 70, 81 friction and, 389–390, 414–415 kinetic friction ( ), 390 screw thread, 414–415 static friction ( ), 389 Applied force (P), 388 Area, 450, 484–487, 511–517, 530–533, 558 axis of symmetry, 530–531 centroid (C) of an, 450 centroidal axis for, 512–513 composite shapes, 485, 522–524, 558 integration and, 450, 512 moments of inertia (I), 512–511, 522–524, 558 parallel–axis theorem for, 512–513, 522, 531, 558 planes, rotation of and, 484–487 procedures for analysis of, 514, 522 product of inertia for, 530–533, 558 radius of gyration for, 513 surface of revolution, 484 volume of revolution, 485 Axial loads, frictional forces and, 429–430 Axis of symmetry, 530–531 Axis systems, 139–143, 183–187, 194, 511–517, 530–536, 550, 558 centroidal axis of, 512–513 distributed loads about, 511–513 inclined, 534–536 mass moments of inertia, 550 moments about, 139–143, 194 moments of inertia (I), 511–517, 534–536, 558 parallel–axis theorem, 512–513 polar moment of inertia, 512 principle moments of inertia, 535 procedure for analysis of, 514 product of inertia and, 530–533, 558 radius of gyration, 513, 550 resultant forces of, 139–143, 183–187, 194 single, 183–187 uniform distributed loads and, 183 B Ball-and-socket joints, 237–238, 240 Base units, 7 Beams, 329–364, 380 bending moments, 330, 280 cantilevered, 345 us uk centroid, 330 couple moments of, 356 distributed loads, 354–356 force equilibrium of, 355 internal forces of, 329–364, 380–382 method of sections for, 329–336 moments and, 355–356, 382 normal force and, 330, 380 procedures for analysis of, 331, 346 resultant forces of, 330, 380 shear and moment diagrams, 345–348, 381 shear force and, 330, 354–356, 380, 382 sign convention for, 331 simply supported, 345 torsional (twisting) moment, 330, 380 Bearings, 237–240, 429–433, 443 axial loads and, 429–430 collar, 429–430, 443 force reactions on, 237–240 frictional forces on, 429–433, 443 journal, 443 lateral loads and, 432 pivot, 429–430 rigid-body supports, 237–240 shaft rotation and, 429–433 Belts (flat), frictional forces on, 421–423, 442 Bending moments, 330, 280 Bending-moment diagram, 345 Bridge loads, trusses, 264 C Cables, 86–88, 365–380, 382 concentrated loads, 365–367 distributed loads of, 368–369 equilibrium and, 86–88 internal forces of, 365–380, 382 weight of as force, 372–375, Cantilevered beams, 345 Cartesian vectors, 33, 43–55, 59–62, 69, 80–81, 122–123, 125 addition of, 46 coordinate direction angle, 44–47, 80–81 coplanar forces, notation for, 33 cross product from, 122–123 dot (scalar) product, 69, 81 line of action, 59–62, 81 magnitude of, 44, 47, 80 moment of a force from, 125 rectangular components, 43, 80 representation, 44 resultant force, 47, 81 right-hand rule for, 43 unit vectors, 43, 80 Center of gravity, 7, 204, 446–509 center of mass and, 449, 505 centroid, and, 446–509 composite bodies, location in, 470–471, 506 free-body diagrams and, 204 location of, 447–448, 505 Newton’s law for, 7 procedure for analysis of, 452, 471 weight (W) and, 204 Center of mass, 449, 460, 474, 505 Centroid, 184, 195, 330, 446–509 area, of a, 450, 484 beam cross-section location, 330 center of gravity and, 446–509 composite bodies, location in, 470–473, 485, 506 distributed loads and, 493, 507 fluid pressure and, 494–500, 507 line, of a, 450–451 location of, 449–459, 505 method of sections and, 330 procedure for analysis of, 452, 471 resultant force, location of, 184, 195, 330 resultant forces and, 493 theorems of Pappus and Guldinus, 484–487, 506 volume, of a, 449, 482 Centroidal axis, 512–513 Coefficient of kinetic friction ( ), 390, 441 Coefficient of rolling resistance (a), 434–435, 443 Coefficient of static friction ( ), 389, 441 Collar bearings, frictional forces on, 429–430, 443 Collinear vectors, 19, 79 Communitative law, 69 Composite bodies, 470–474, 485, 522–524, 550, 558 area of revolution of, 485 center of gravity of, 470–474 mass moment of inertia, 550 moments of inertia (I), 522–524, 550, 558 procedure of analysis for, 471, 522 volume of revolution of, 485 Compressive force members, trusses, 265–267, 280, 323 Concentrated force, 5 Concentrated loads, cables subjected to, 365–367 Concurrent force systems, 170, 264 resultant force of, 170 truss joint connections, 264 ms mk 650
- 668. Disks, 429–430, 443, 545–546, 548, 559 frictional forces on, 429–430, 443 mass moments of inertia of, 545–546, 548, 559 Displacement of virtual work, 564–566, 582, 594 Distributed loads, 183–187, 195, 354–356, 368–369, 493, 507, 511–512 beams subjected to, 354–356 cables subjected to, 368–369 distributed loads and, 493, 507 flat surfaces, 493, 507 force equilibrium, 355 internal forces and, 354–356, 368–369 line of action , 493, 507 location of, 184, 493 magnitude, 183, 493 moments of inertia and, 511–512 reduction of, 183–187, 195 resultant forces, 184, 493 shear force and, 354–355 single-axis, 183–187 uniform, 183, 354 Distributive law, 69 Dot (scalar) product, 69–73, 81 Dry friction, 387–433, 441–443 angles of, 389–390, 414–415 applied force and, 388 bearings, forces on, 429–433, 443 characteristics of, 387–392, 441 coefficients of ( ), 389–390, 441 direction of force and, 394 equilibrium and, 388, 391, 394 impending motion and, 389, 392–393, 414–415 kinetic force, 390–391, 441 motion and, 390–391 problems involving, 392–399 procedure for analysis of, 394 screws, forces on, 414–416, 442 slipping, and, 389–393 static force, 389, 391, 441 theory of, 388 Dynamics, 3 E Elastic potential energy, 580 Equilibrium, 84–115, 198–261, 388, 391, 394, 565–566, 582–588, 595 conditions for, 85, 199–200 coplanar (two-dimensional) systems, 89–93, 113, 200–236, 258–259 criterion for, 582 direction of force and, 394 m equations of, 89, 103, 214–223, 242, 565–566 free-body diagrams, 86–88, 113, 201–210, 237–241, 258–259 friction and, 388, 391, 394 frictionless systems and, 582 neutral, 583 one-degree of freedom systems, 584–588 particles, 84–115 potential-energy and, 582–588, 595 procedures for analysis of, 90, 103, 215, 246, 585 rigid-bodies, 198–261, 582 stable, 583 three-dimensional systems, 103–107, 113, 237–257, 259 three-force coplanar members, 224 tipping effect and, 388 two-force coplanar members, 224 unstable, 583 virtual work and, 565–566, 582–588, 595 Equivalent, force and couple systems, 160–165, 170–177, 195 concurrent force, 170 coplanar force, 170–171 moments, 161 parallel force, 171 perpendicular lines of action, 170–177 procedure for analysis of, 162, 172 reduction of forces, 160–165 resultants of, 160–165, 170–177, 195 wrench (screw), force reduction to, 173, 195 F Fixed supports, 201–203 Flat plates, fluid pressure and, 495, 497 Floor beams, trusses, 264 Fluid mechanics, 3 Fluid pressure, 494–500, 507 centroid (C) and, 494–500, 507 curved plate of constant width, 496 flat plate of constant width, 495 flat plate of variable width, 497 Pascal’s law, 494 Foot-pound, unit of, 564 Force, 4, 5, 8, 16–83, 84–115, 116–197, 198–261, 328–385, 564, 579–581, 594–595. See also Friction;Weight addition of, 20–42 beams subjected to, 329–364, 380 Conservative forces, 579–581, 595 potential energy and, 580–581, 595 spring force, 579–580 virtual work and, 579–581, 595 weight, 579–580 Constraints, 243–251, 259 improper, 244–245 redundant, 243 statical determinacy and, 243–251, 259 Continuous cable, 86 Coordinate direction angles, 44–47, 80–81 Coplanar forces, 32–42, 89–93, 113, 170–171, 200–236, 258–259, 263–264, 266 Cartesian vector notation for, 33 center of gravity (G), 204 equations of equilibrium, 89, 214–223 equilibrium of, 89–93, 113, 200–236, 258–259 free-body diagrams, 86–88, 113, 201–210, 258–259 idealized models, 204–205 internal forces, 204 moments of force and couple systems, 170–171 particle systems, 89–93, 113 procedures for analysis of, 87, 90, 206, 215 resultant forces, 33–34, 170–171 rigid bodies, 200–236, 258–259 scalar notation for, 32 support reactions, 201–203 three-force members, 224 truss analysis, 263–264, 266 two-force members, 224 vector addition of, 32–42 weight (W), 204 Coulomb friction, see Dry friction Couple moments, 148–153, 194, 356, 564 beam segments, 356 equivalent, 149 parallel forces of, 148–153, 194 resultant, 149 rotation, 564 scalar formulation, 148 translation, 564 vector formulation, 148 virtual work of, 564 Cross product, 121–123 Curved plates, fluid pressure and, 496 Cylinders, rolling resistance of, 434–435, 443 D Deformable-body mechanics, 3 Dimensional homogeneity, 10–11 INDEX 651
- 669. cables subjected to, 368–369, 382 components of, 20–21 concentrated, 5 concurrent systems, 170 conservative, 579–581, 595 coplanar systems, 32–42, 89–93, 113, 170–171, 200–236, 258–259 equilibrium and, 84–115, 198–261, 355 equivalent (force and couple) systems, 160–165, 170–177, 195 free-body diagrams (FBD), 86–88, 113, 201–210 internal, 328–385 line of action, 59–62, 81 mechanical concept of, 4 moments of, 116–197 multiple, 21 normal, 330, 380 parallel systems, 171 particle equilibrium and, 84–115 potential energy and, 580–581, 595 procedures for analysis of, 22, 87, 90, 103 reactive, 160 resultant, 20–26, 85, 116–197, 330, 380 rigid-body equilibrium and, 84–115, 198–261 shear, 330, 354–356, 380, 382 spring, 579–580 three-dimensional systems, 103–107, 113, 237–257, 259 units of, 8 vectors and, 16–83 virtual work of a, 564, 579, 594 wrench systems, 173, 195 Frames, 294–322, 325 free-body diagrams for, 294–299 multiforce members, design of, 294–309 procedure for analysis of, 301 structural analysis of, 294–322, 325 Free-body diagrams (FBD), 86–88, 113, 201–210, 237–241, 258–259, 294–299 cables and pulleys, 86–88 center of gravity, 204 coplanar (two-dimensional), 201–210, 258 idealized models, 204–205 internal forces, 204 particle equilibrium, 86–88, 113 procedure for analysis of, 87, 206 rigid-body equilibrium, 201–210, 237–241, 258–259 springs, 86–88 structural analysis and, 294–299 support reactions, 201–203, 237–240 three-dimensional, 237–241, 259 weight, 204 Free vector, 160 Friction, 386–445, 580 axial loads and, 429–430 belts (flat), forces on, 421–423, 442 collar bearings, forces on, 429–430, 443 disks, forces on, 429–430, 443 dry, 387–433, 441 equilibrium and, 388, 391, 394 force of, 387 journal bearings, forces on, 432–433, 443 kinetic force, 390–391, 441 lateral loads and, 432 nonconservative force, as a, 580 pivot bearings, forces on, 429–430 procedure for analysis of, 394 rolling resistance and, 434–435, 443 screws, forces on, 414–416, 442 shaft rotation and, 429–433, 443 static force, 389, 391, 441 virtual work and, 580 wedges and, 412–413, 442 Frictionless systems, 567–582 G Gravitational potential energy, 580 Gravity, see Center of gravity Gusset plate, 264 H Hinge supports, 237, 239 I Idealized models, 204–205 Inclined axis, moments of inertia about, 534–536 Independent coordinates for, 567–568, 594 Inertia, see Moments of inertia Integrals, 450, 511–512 Internal force, 204, 328–385 beams subjected to, 329–364, 380–382 bending moments, 330, 280 cables subjected to, 365–380, 382 couple moments, 356 distributed loads, 354–356, 368–369 force equilibrium, 355 free-body diagrams and, 204 method of sections for, 329–336 moments and, 330, 355–356, 380, 382 normal force and, 330, 380 procedures for analysis of, 331, 346 resultant forces, 330, 380 shear and moment diagrams, 345–348, 381 shear force and, 330, 354–356, 380, 382 sign convention for, 331 torsional (twisting) moment, 330, 380 International System (SI) of units, 8–10 prefixes, 9 rules for use, 10 units of measurement, as, 8 J Joint connections, loads applied to, 264–266 Joule, 564 Journal bearings, frictional forces on, 432–433, 443 K Kinetic friction force, 390–391, 441 L Lateral loads, frictional forces and, 432 Length, 4, 8 Line of action, 59–62, 81, 170–172, 493, 507 distributed loads, 493, 507 force vectors and, 59–62, 81 perpendicular, 170–172 resultant forces, 493, 507 Linear elastic springs, 86 Lines, centroid of, 450–451 Loads, see Concentrated loads; Distributed loads M Machines, 294–322, 325 free-body diagrams for, 294–299 multiforce members, design of, 294–309 procedure for analysis of, 301 structural analysis of, 294–322, 325 Magnitude, 44, 47, 80, 118, 122, 124, 183, 493 Cartesian vectors, 44, 47, 80 cross product, 121 distributed loads, 183, 493 moment of a force, 118, 124 resultant forces, 493 Mass, 4, 8, 449 center of, 449 mechanical concept of, 4 units of, 8 Mass moments of inertia, 545–552, 559 axis systems, 550 composite bodies, 550–552 disk elements, 545–546, 584, 559 radius of gyration for, 550 shell elements, 545–547, 559 652 INDEX
- 670. Mohr’s circle for, 537–539 parallel-axis theorem for, 512–513, 522, 531, 549, 558 polar, 512 principle, 535–536, 559 procedures for analysis of, 514, 522, 538 product of inertia and, 530–533, 558 radius of gyration for, 513, 550 shell elements, 545–547, 559 Motion, 6, 389–393, 414–415, 429–433, 441, 449, 564–566, 582, 594 displacement of virtual work, 564–566, 582, 594 downward, 415, 565 dry friction and, 389–393, 414–415, 441 dynamic response, 449 impending, 389, 392–393, 414–415 Newton’s laws of, 6 rotation of a couple moment, 564 screws and, 414–415 shaft rotation, 429–433 slipping, 389–393, 441 tipping, 388, 393, 441 translation of a couple moment, 564 upward, 414 virtual, 565 Multiplicative scalar law, 69 N Neutral equilibrium, 583 Newton’s laws, 6–7 gravitational attraction, 7 motion, 6 Normal force, 330, 380 Numerical calculations, 10–11, 18–83 dimensional homogeneity, 10–11 rounding off numbers, 11 significant figures, 11 vector operations for, 18–83 O One (single) degree-of-freedom systems, 584–588 P Pappus and Guldinus, theorems of, 484–487, 506 Parallel-axis theorem, 512–513, 522, 531, 549, 558 centroidal axis for, 512–513 composite shapes, 522 moments of inertia, 512–513, 522, 588 products of inertia, 531, 558 Parallel systems, resultant force of, 171 Parallelogram law, 18–19, 79 Particles, 5, 84–115 coplanar force systems, 89–93, 113 equations of equilibrium, 89, 103 equilibrium of, 84–197 force conditions, 84–115 free-body diagrams (FBD), 86–88, 113 idealization of, 5 procedures for analysis of, 87, 90, 103 resultant force, 85 three-dimensional force systems, 103–107, 113 Pascal’s law, 494 Pin supports, 201–203, 239–240, 264 coplanar force systems, 201–203 three-dimensional force systems, 239–240 truss load connections, 264 Pivot bearings, frictional forces on, 429–430 Planar truss, 263–289 Plates, 494–500, 507 centroid of, 494–500, 507 curved of constant width, 496 flat of constant width, 495 flat of variable width, 497 fluid pressure and, 494–500, 507 Polar moments of inertia, 512 Position coordinate, 568, 581–582, 594–595 Position vectors, 56–58, 81 right-hand rule for, 56 x, y, z coordinates, 56–57, 81 Potential energy, 580–588, 595 conservative forces and, 578–581, 595 elastic, 580 equilibrium stability of , 583–588, 595 frictionless systems, 582 function of, 581 gravitational, 580 position coordinate (q), 581–582, 595 procedure for analysis of, 585 single degree-of-freedom systems, 581, 584 stability of systems using, 583–588, 595 virtual work and, 580–582, 595 Pressure, see Fluid pressure Principle moments of inertia, 535–536, 559 Product of inertia, 530–533, 558 area minimum and maximum moments, 530–533, 558 axis of symmetry, 530–531 parallel-axis theorem for, 531 Projection of components, 70, 81 Pulleys, equilibrium and, 86–88 Purlins, 263 Mechanics, 3–7 concentrated force, 5 deformable-body, 3 fluid, 3 Newton’s laws, 6–7 particles, 5 quantities of, 4 rigid-body, 3, 5 weight, 7 Method of joints, 266–279, 323 Method of sections, 280–289, 324, 329–336, 380 beams, 329–336, 380 centroid for, 330 compressive force members, 280 internal forces and, 329–336, 380 procedures for analysis of, 282, 331 tensile force members, 280–281 trusses, 280–289, 324 Mohr’s circle, 537–539 Moments, 116–197, 330, 355–356, 380, 382 axis, about an, 139–143, 194 beams, 330, 355–356, 380, 382 bending, 330, 380 Cartesian vector formulation, 125 changes in ( ), 355–356 couple, 148–153, 194, 356 direction of, 118, 124 equivalent (force and couple) systems, 160–165, 170–177, 195 force system resultants and, 116–197 internal forces and, 330, 355–356, 380, 382 magnitude of, 118, 122, 124 principle of, 128–130 resultant (MR), 118, 125, 149, 170–171, 184 scalar formulation, 117–120, 139, 148, 193 shear and, relationship of, 355–356 torsional (twisting), 330, 380 transmissibility, principle of, 124 Varignon’s theorem, 128–130 vector formulation, 124–127, 140, 148, 193 Moments of inertia, 510–561 area, 512–517, 522–524, 558 axis systems, 511–517, 530–536, 550, 558 composite shapes, 522–524, 550, 558 disk elements, 545–546, 548, 559 distributed loads and, 511–512 inclined axis, area about, 534–536 integrals, 511–512 mass, 545–552, 559 ¢M INDEX 653
- 671. R Radius of gyration, 513, 550 Reactive force, 160 Rectangular components, 43, 80 Resultant force, 20–26, 33–34, 47, 80–81, 85, 116–197, 330, 380, 493, 507 axis, moments about an, 139–143, 194 Cartesian vectors and, 47, 80–80 centroid for location of, 184, 195, 330 components, 20–26 concurrent systems, 170 coplanar systems, 33–34, 170–171 couple moments, 148–153, 194 cross product, 121–123 distributed loads and, 183–187, 195, 493, 507 equilibrium of a particle and, 85 equivalent (force and couple) systems, 160–165, 170–177, 195 internal forces, 330, 380 line of action, 493, 507 magnitude of, 493 moments, 118, 125, 149, 170–171, 184 parallel systems, 171 procedures for analysis of, 162, 172 scalar formulation, 117–120, 139, 148, 193 system moments of, 116–197 transmissibility, principle of, 124, 160 vector formulation, 124–127, 140, 148, 193 wrench (screw) systems, 173, 195 Right-hand rule, 43, 56, 80, 121–122 Cartesian vector coordinates, 43, 56 cross-product direction, 121–122 moment direction, 118 Rigid bodies, 3, 5, 198–261, 567–572, 594 center of gravity, 204 connected, 567–572 constraints, 243–251, 259 coplanar force systems, 200–236, 258–259 equations of equilibrium, 89, 103, 214–223, 242 equilibrium of, 198–261, 565–566, 582–588, 595 free-body diagrams (FBD), 201–210, 237–241, 258–259 idealization of, 5 idealized models, 204–205 independent coordinates, 567–568, 594 internal forces, 204 mechanics, 3 position coordinates, 568, 594 procedures for analysis of, 206, 215, 246, 568 statically indeterminate, 243, 246, 259 support reactions, 201–203, 237–240 three-dimensional systems, 237–257, 259 three-force coplanar members, 224 two-force coplanar members, 224 virtual work for, 567–572, 594 weight, 204 Roller supports, 201–202 Rolling resistance, 434–435, 443 Roof loads, trusses, 263 Rotating shafts and friction, 429–433, 443 Rotation of a couple moment, 564 Rounding off numbers, 11 S Scalar analysis, 117–120, 139, 148, 193, 242 axis, moments about an, 139 couple moments, 148 equations of equilibrium, 242 moments of a force, 117–120, 193 Scalar notation, 32 Scalars and vectors, 17–18, 79 Screws, 414–416, 442 downward impending motion, 415 friction forces on, 414–416, 442 self-locking, 415 thread angle, 414–415 upward impending motion, 414 Shaft rotation, 429–433 Shear and moment diagrams, 345–348, 381 Shear force, 330, 354–356, 380, 382 beams subjected to, 330, 354–355, 380, 382 changes in, 355–356 distributed loads and, 354–355 moments and, 355 Shell elements, mass moments of inertia of, 545–547, 559 Significant figures, 11 Simple truss, 265 Simply supported beams, 345 Single-axis distributed loads, 183–187 Sliding vector, 160 Slipping, 389–393, 441 dry friction and, 389–393, 441 impending motion of, 389, 392–393 motion of, 390–391 Slug, 8 Space truss, 290–293 Specific weight, 470 Spring force, 579–580 conservative force, as a, 579 elastic potential energy, 580 Springs, 86–88 constant, 86 equilibrium and, 86–88 Stability of a system, see Equilibrium Stable equilibrium, 583 Static friction force, 389, 391, 441 Statically indeterminate bodies, 243, 246, 259 Statics, 2–15 development of, 4 mechanics and, 3–7 numerical calculations, 10–11 procedure for analysis, 12 quantities of, 4 units of measurement, 7–10 Stiffness, 86 Stringers, 264 Structural analysis, 262–327 frames, 294–322, 325 free-body diagrams for, 294–299 machines, 294–322, 325 method of joints, 266–279, 323 method of sections, 280–289, 324 procedures for, 267, 282, 290, 301 trusses, 263–293, 323–325 zero-force members, 272–274 Structural members, see Beams Support reactions, 201–203, 237–240 ball-and-socket joint, 237–238, 240 bearing, 237–240 coplanar rigid-body systems, 201–203 fixed, 201–203 hinge, 237, 239 pin, 201–203, 237, 239–240 roller, 201–202 three-dimensional rigid-body systems, 237–240 Surface area of revolution, 484 T Tensile force members, trusses, 265–267, 280–281, 323 Tension and belt friction, 421–422 Tetrahedral truss, 290 Three-dimensional force systems, 103–107, 113, 237–257, 259 constraints, 243–251, 259 equations of equilibrium, 103, 242 free-body diagrams (FBD), 237–241, 259 654 INDEX
- 672. V Varignon’s theorem, 128–130 Vector analysis, 124–127, 140, 148, 193, 242 axis, moments about an, 140 Cartesian, 125 couple moments, 148 equations of equilibrium, 242 magnitude from, 124 moments of a force, 124–127, 193 resultant moment (MR) from, 125 right-hand rule for, 124 transmissibility, principle of, 124 Vectors, 16–83, 121–123, 160 addition, 18–22, 32–37, 46 angles formed between, 70, 81 Cartesian, 43–55, 80–81, 122–123 collinear, 19, 79 coplanar forces, 32–42 cross product, 121–123 division, 18 dot (scalar) product, 69–73, 81 forces and, 20–42, 59–62 free, 160 line of action, 59–62, 81 multiplication, 18 notation for, 32–33 parallelogram law, 18–19, 79 position, 56–58, 81 procedure for analysis of, 22 projection of components, 70, 81 rectangular components, 43, 80 resultant force, 20–26, 33–34 right-hand rule for, 43, 56, 121–122 scalars and, 17–18, 79 sliding, 160 subtraction, 19 Virtual work, 562–597 conservative forces and, 579–581 couple moment, of a, 564 displacement ( ) and, 564–566, 582, 594 equilibrium and, 565–566, 582–588, 595 d force, of a, 564 friction and, 580 frictionless systems, 567–582 independent coordinates for, 567–568, 594 joules (J) as unit of, 564 one degree-of-freedom system, 567, 581, 584, 594 position coordinate for, 568, 581–582, 595 potential energy and, 580–582, 595 principle of, 563–578, 594 procedures for analysis of, 568, 585 rigid-body systems and, 567–572 spring force and, 579 stability of a system, 583–588, 595 weight and, 579 Volume, 449, 485 centroid of a, 449 revolution of a plane area, 485 W Wedges, friction forces and, 412–413, 442 Weight, 7–8, 204, 372–375, 470, 579–580 cables subjected to own, 372–375 center of gravity and, 204 composite bodies, 470 conservative force, as a, 579 free-body diagrams and, 204 gravitational force of, 7 gravitational potential energy, 580 units of measurement, 8 virtual work and, 579–580 Work, see Virtual work Wrench (screw) systems, force reduction to, 173, 195 X x, y, z coordinates, 56–57, 81 Z Zero-force truss members, 272–274 particle equilibrium, 103–107, 113 procedure for analysis of, 90, 246 rigid-body equilibrium, 237–257, 259 statically indeterminate, 243, 246, 259 Three-force coplanar members, 224 Time, 4, 8 Tipping effect, 388, 393, 441 Torque, 117 Torsional (twisting) moment, 330, 380 Translation of a couple moment, 564 Transmissibility, principle of, 124, 160 Triangular truss, 265 Trusses, 263–293, 323–325 compressive force members, 265–267, 280, 323 coplanar loads on, 263–264, 266 design assumptions for, 264–264, 290 joint connections, 264–265 method of joints, 266–279, 323 method of sections, 280–289, 324 planar, 263–289 procedures for analysis of, 267, 282, 290 simple, 265 space, 290–293 tensile force members, 265–267, 280–281, 323 zero-force members, 272–274 Two-dimensional systems, see Coplanar force Two-force coplanar members, 224 U U.S. Customary (FPS) of units, 8 Uniform distribute loads, 183, 354 Unit vectors, 43, 59, 80 Units of measurement, 7–10 conversion of, 9 International System (SI) of units, 8–10 U.S. Customary (FPS) of units, 8 Unstable equilibrium, 583 INDEX 655
- 673. Although each of these planes is rather large, from a distance their motion can be analysed as if each plane were a particle.
- 674. Kinematics of a Particle CHAPTER OBJECTIVES • To introduce the concepts of position, displacement, velocity, and acceleration. • To study particle motion along a straight line and represent this motion graphically. • To investigate particle motion along a curved path using different coordinate systems. • To present an analysis of dependent motion of two particles. • To examine the principles of relative motion of two particles using translating axes. 12 12.1 Introduction Mechanics is a branch of the physical sciences that is concerned with the state of rest or motion of bodies subjected to the action of forces. Engineering mechanics is divided into two areas of study, namely, statics and dynamics. Statics is concerned with the equilibrium of a body that is either at rest or moves with constant velocity. Here we will consider dynamics, which deals with the accelerated motion of a body.The subject of dynamics will be presented in two parts: kinematics, which treats only the geometric aspects of the motion, and kinetics, which is the analysis of the forces causing the motion. To develop these principles, the dynamics of a particle will be discussed first, followed by topics in rigid-body dynamics in two and then three dimensions.
- 675. 4 CHAPTER 12 KINEMATICS OF A PARTICLE 12 Historically, the principles of dynamics developed when it was possible to make an accurate measurement of time. Galileo Galilei (1564–1642) was one of the first major contributors to this field. His work consisted of experiments using pendulums and falling bodies. The most significant contributions in dynamics, however, were made by Isaac Newton (1642–1727), who is noted for his formulation of the three fundamental laws of motion and the law of universal gravitational attraction. Shortly after these laws were postulated, important techniques for their application were developed by Euler, D’Alembert, Lagrange, and others. There are many problems in engineering whose solutions require application of the principles of dynamics. Typically the structural design of any vehicle, such as an automobile or airplane, requires consideration of the motion to which it is subjected. This is also true for many mechanical devices, such as motors, pumps, movable tools, industrial manipulators, and machinery. Furthermore, predictions of the motions of artificial satellites, projectiles, and spacecraft are based on the theory of dynamics. With further advances in technology, there will be an even greater need for knowing how to apply the principles of this subject. Problem Solving. Dynamics is considered to be more involved than statics since both the forces applied to a body and its motion must be taken into account. Also, many applications require using calculus, rather than just algebra and trigonometry. In any case, the most effective way of learning the principles of dynamics is to solve problems. To be successful at this, it is necessary to present the work in a logical and orderly manner as suggested by the following sequence of steps: 1. Read the problem carefully and try to correlate the actual physical situation with the theory you have studied. 2. Draw any necessary diagrams and tabulate the problem data. 3. Establish a coordinate system and apply the relevant principles, generally in mathematical form. 4. Solve the necessary equations algebraically as far as practical; then, use a consistent set of units and complete the solution numerically. Report the answer with no more significant figures than the accuracy of the given data. 5. Study the answer using technical judgment and common sense to determine whether or not it seems reasonable. 6. Once the solution has been completed, review the problem. Try to think of other ways of obtaining the same solution. In applying this general procedure, do the work as neatly as possible. Being neat generally stimulates clear and orderly thinking, and vice versa.
- 676. 12.2 RECTILINEAR KINEMATICS: CONTINUOUS MOTION 5 12 12.2 Rectilinear Kinematics: Continuous Motion We will begin our study of dynamics by discussing the kinematics of a particle that moves along a rectilinear or straight line path. Recall that a particle has a mass but negligible size and shape.Therefore we must limit application to those objects that have dimensions that are of no consequence in the analysis of the motion. In most problems, we will be interested in bodies of finite size, such as rockets, projectiles, or vehicles. Each of these objects can be considered as a particle,as long as the motion is characterized by the motion of its mass center and any rotation of the body is neglected. Rectilinear Kinematics. The kinematics of a particle is characterized by specifying, at any given instant, the particle’s position, velocity, and acceleration. Position. The straight-line path of a particle will be defined using a single coordinate axis s, Fig. 12–1a. The origin O on the path is a fixed point, and from this point the position coordinate s is used to specify the location of the particle at any given instant. The magnitude of s is the distance from O to the particle, usually measured in meters (m) or feet (ft), and the sense of direction is defined by the algebraic sign on s. Although the choice is arbitrary, in this case s is positive since the coordinate axis is positive to the right of the origin. Likewise, it is negative if the particle is located to the left of O. Realize that position is a vector quantity since it has both magnitude and direction. Here, however, it is being represented by the algebraic scalar s since the direction always remains along the coordinate axis. Displacement. The displacement of the particle is defined as the change in its position. For example, if the particle moves from one point to another, Fig. 12–1b, the displacement is In this case is positive since the particle’s final position is to the right of its initial position, i.e., Likewise, if the final position were to the left of its initial position, would be negative. The displacement of a particle is also a vector quantity, and it should be distinguished from the distance the particle travels. Specifically, the distance traveled is a positive scalar that represents the total length of path over which the particle travels. ¢s s¿ 7 s. ¢s ¢s = s¿ - s s s Position (a) O s s Displacement (b) s¿ O s Fig. 12–1
- 677. 6 CHAPTER 12 KINEMATICS OF A PARTICLE 12 Velocity. If the particle moves through a displacement during the time interval the average velocity of the particle during this time interval is If we take smaller and smaller values of the magnitude of becomes smaller and smaller. Consequently, the instantaneous velocity is a vector defined as or (12–1) Since or dt is always positive, the sign used to define the sense of the velocity is the same as that of or ds. For example, if the particle is moving to the right, Fig. 12–1c, the velocity is positive; whereas if it is moving to the left, the velocity is negative. (This is emphasized here by the arrow written at the left of Eq. 12–1.) The magnitude of the velocity is known as the speed, and it is generally expressed in units of Occasionally, the term “average speed” is used. The average speed is always a positive scalar and is defined as the total distance traveled by a particle, divided by the elapsed time i.e., For example, the particle in Fig. 12–1d travels along the path of length in time so its average speed is but its average velocity is vavg = -¢s¢t. 1vsp2avg = sT¢t, ¢t, sT 1vsp2avg = sT ¢t ¢t; sT, ms or fts. ¢s ¢t v = ds dt 1 : + 2 v = lim ¢t:0 1¢s¢t2, ¢s ¢t, vavg = ¢s ¢t ¢t, ¢s s Velocity (c) O s v s s P sT Average velocity and Average speed O P¿ (d) Fig. 12–1 (cont.)
- 678. 12.2 RECTILINEAR KINEMATICS: CONTINUOUS MOTION 7 12 Acceleration. Provided the velocity of the particle is known at two points, the average acceleration of the particle during the time interval is defined as Here represents the difference in the velocity during the time interval i.e., Fig. 12–1e. The instantaneous acceleration at time t is a vector that is found by taking smaller and smaller values of and corresponding smaller and smaller values of so that , or (12–2) Substituting Eq. 12–1 into this result, we can also write Both the average and instantaneous acceleration can be either positive or negative. In particular, when the particle is slowing down, or its speed is decreasing, the particle is said to be decelerating. In this case, in Fig. 12–1f is less than and so will be negative. Consequently, a will also be negative, and therefore it will act to the left, in the opposite sense to Also, note that when the velocity is constant, the acceleration is zero since Units commonly used to express the magnitude of acceleration are or Finally, an important differential relation involving the displacement, velocity, and acceleration along the path may be obtained by eliminating the time differential dt between Eqs. 12–1 and 12–2, which gives (12–3) Although we have now produced three important kinematic equations, realize that the above equation is not independent of Eqs. 12–1 and 12–2. a ds = v dv 1 : + 2 fts2 . ms2 ¢v = v - v = 0. v. ¢v = v¿ - v v, v¿ a = d2 s dt2 1 : + 2 a = dv dt 1 : + 2 a = lim ¢t:0 1¢v¢t2 ¢v, ¢t ¢v = v¿ - v, ¢t, ¢v aavg = ¢v ¢t ¢t s Acceleration (e) O a v v¿ s P Deceleration (f) O P¿ v v¿ a
- 679. 8 CHAPTER 12 KINEMATICS OF A PARTICLE 12 Constant Acceleration, When the acceleration is constant, each of the three kinematic equations and can be integrated to obtain formulas that relate s, and t. Velocity as a Function of Time. Integrate assuming that initially when (12–4) Position as a Function of Time. Integrate assuming that initially when (12–5) Velocity as a Function of Position. Either solve for t in Eq. 12–4 and substitute into Eq. 12–5, or integrate assuming that initially at (12–6) The algebraic signs of and used in the above three equations, are determined from the positive direction of the s axis as indicated by the arrow written at the left of each equation. Remember that these equations are useful only when the acceleration is constant and when A typical example of constant accelerated motion occurs when a body falls freely toward the earth. If air resistance is neglected and the distance of fall is short, then the downward acceleration of the body when it is close to the earth is constant and approximately or The proof of this is given in Example 13.2. 32.2 fts2 . 9.81 ms2 v = v0. s = s0, t = 0, ac, v0, s0, v2 = v0 2 + 2ac1s - s02 Constant Acceleration 1 : + 2 L v v0 v dv = L s s0 ac ds s = s0. v = v0 v dv = ac ds, s = s0 + v0t + 1 2 act2 Constant Acceleration 1 : + 2 L s s0 ds = L t 0 1v0 + act2 dt t = 0. s = s0 v = dsdt = v0 + act, v = v0 + act Constant Acceleration 1 : + 2 L v v0 dv = L t 0 ac dt t = 0. v = v0 ac = dvdt, v, ac, ac ds = v dv v = dsdt, ac = dvdt, a = ac.
- 680. 12.2 RECTILINEAR KINEMATICS: CONTINUOUS MOTION 9 12 Procedure for Analysis Coordinate System. • Establish a position coordinate s along the path and specify its fixed origin and positive direction. • Since motion is along a straight line, the vector quantities position, velocity, and acceleration can be represented as algebraic scalars. For analytical work the sense of s, and a is then defined by their algebraic signs. • The positive sense for each of these scalars can be indicated by an arrow shown alongside each kinematic equation as it is applied. Kinematic Equations. • If a relation is known between any two of the four variables a, s and t, then a third variable can be obtained by using one of the kinematic equations, or since each equation relates all three variables.* • Whenever integration is performed, it is important that the position and velocity be known at a given instant in order to evaluate either the constant of integration if an indefinite integral is used, or the limits of integration if a definite integral is used. • Remember that Eqs. 12–4 through 12–6 have only limited use. These equations apply only when the acceleration is constant and the initial conditions are s = s0 and v = v0 when t = 0. a ds = v dv, v = dsdt a = dvdt, v, v, During the time this rocket undergoes rectilinear motion, its altitude as a function of time can be measured and expressed as Its velocity can then be found using and its acceleration can be determined from a = dvdt. v = dsdt, s = s1t2. s Important Points • Dynamics is concerned with bodies that have accelerated motion. • Kinematics is a study of the geometry of the motion. • Kinetics is a study of the forces that cause the motion. • Rectilinear kinematics refers to straight-line motion. • Speed refers to the magnitude of velocity. • Average speed is the total distance traveled divided by the total time. This is different from the average velocity, which is the displacement divided by the time. • A particle that is slowing down is decelerating. • A particle can have an acceleration and yet have zero velocity. • The relationship is derived from and by eliminating dt. v = dsdt, a = dvdt a ds = v dv *Some standard differentiation and integration formulas are given in Appendix A.
- 681. 10 CHAPTER 12 KINEMATICS OF A PARTICLE 12 The car in Fig. 12–2 moves in a straight line such that for a short time its velocity is defined by where t is in seconds. Determine its position and acceleration when When s = 0. t = 0, t = 3 s. v = 13t2 + 2t2 fts, EXAMPLE 12.1 SOLUTION Coordinate System. The position coordinate extends from the fixed origin O to the car, positive to the right. Position. Since the car’s position can be determined from since this equation relates s, and t. Noting that when we have* When Ans. Acceleration. Since the acceleration is determined from since this equation relates a, and t. When Ans. NOTE: The formulas for constant acceleration cannot be used to solve this problem, because the acceleration is a function of time. a = 6132 + 2 = 20 fts2 : t = 3 s, = 6t + 2 a = dv dt = d dt 13t2 + 2t2 1 : + 2 v, a = dvdt, v = f1t2, s = 1323 + 1322 = 36 ft t = 3 s, s = t3 + t2 s ` 0 s = t3 + t2 ` 0 t L s 0 ds = L t 0 13t2 + 2t2dt v = ds dt = 13t2 + 2t2 1 : + 2 t = 0, s = 0 v, v = dsdt, v = f1t2, s O a, v Fig. 12–2 *The same result can be obtained by evaluating a constant of integration C rather than using definite limits on the integral. For example, integrating yields Using the condition that at then C = 0. s = 0, t = 0, s = t3 + t2 + C. ds = 13t2 + 2t2dt
- 682. 12.2 RECTILINEAR KINEMATICS: CONTINUOUS MOTION 11 12 EXAMPLE 12.2 A small projectile is fired vertically downward into a fluid medium with an initial velocity of Due to the drag resistance of the fluid the projectile experiences a deceleration of where is in Determine the projectile’s velocity and position 4 s after it is fired. SOLUTION Coordinate System. Since the motion is downward, the position coordinate is positive downward, with origin located at O, Fig. 12–3. Velocity. Here and so we must determine the velocity as a function of time using since this equation relates a, and t. (Why not use ) Separating the variables and integrating, with when yields Here the positive root is taken, since the projectile will continue to move downward.When Ans. Position. Knowing we can obtain the projectile’s position from since this equation relates s, and t. Using the initial condition when we have When Ans. s = 4.43 m t = 4 s, s = 1 0.4 e c 1 16022 + 0.8td 12 - 1 60 f m s = 2 0.8 c 1 16022 + 0.8td 12 ` 0 t L s 0 ds = L t 0 c 1 16022 + 0.8td -12 dt v = ds dt = c 1 16022 + 0.8td -12 1+ T2 t = 0, s = 0, v, v = dsdt, v = f1t2, v = 0.559 msT t = 4 s, v = e c 1 16022 + 0.8td -12 f ms 1 0.8 c 1 v2 - 1 16022 d = t 1 -0.4 a 1 -2 b 1 v2 ` 60 v = t - 0 L v 60 ms dv -0.4v3 = L t 0 dt a = dv dt = -0.4v3 1+ T2 t = 0, v0 = 60 ms v = v0 + act? v, a = dvdt, a = f1v2 ms. v a = 1-0.4v3 2 ms2 , 60 ms. s O Fig. 12–3
- 683. 12 CHAPTER 12 KINEMATICS OF A PARTICLE 12 During a test a rocket travels upward at and when it is 40 m from the ground its engine fails. Determine the maximum height reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of due to gravity. Neglect the effect of air resistance. SOLUTION Coordinate System. The origin O for the position coordinate s is taken at ground level with positive upward, Fig. 12–4. Maximum Height. Since the rocket is traveling upward, when At the maximum height the velocity For the entire motion, the acceleration is (negative since it acts in the opposite sense to positive velocity or positive displacement). Since is constant the rocket’s position may be related to its velocity at the two points A and B on the path by using Eq. 12–6, namely, Ans. Velocity. To obtain the velocity of the rocket just before it hits the ground, we can apply Eq. 12–6 between points B and C, Fig. 12–4. Ans. The negative root was chosen since the rocket is moving downward. Similarly, Eq. 12–6 may also be applied between points A and C, i.e., Ans. NOTE: It should be realized that the rocket is subjected to a deceleration from A to B of and then from B to C it is accelerated at this rate. Furthermore, even though the rocket momentarily comes to rest at the acceleration at B is still downward! 9.81 ms2 B 1vB = 02 9.81 ms2 , vC = -80.1 ms = 80.1 ms T = 175 ms22 + 21-9.81 ms2 210 - 40 m2 vC 2 = vA 2 + 2ac1sC - sA2 1+ c2 vC = -80.1 ms = 80.1 ms T = 0 + 21-9.81 ms2 210 - 327 m2 vC 2 = vB 2 + 2ac1sC - sB2 1+ c2 sB = 327 m 0 = 175 ms22 + 21-9.81 ms2 21sB - 40 m2 vB 2 = vA 2 + 2ac1sB - sA2 1+ c2 ac ac = -9.81 ms2 vB = 0. s = sB t = 0. vA = +75ms 9.81 ms2 sB 75 ms, EXAMPLE 12.3 A O vA 75 m/s vB 0 sA 40 m s sB B C Fig. 12–4
- 684. 12.2 RECTILINEAR KINEMATICS: CONTINUOUS MOTION 13 12 EXAMPLE 12.4 A metallic particle is subjected to the influence of a magnetic field as it travels downward through a fluid that extends from plate A to plate B, Fig. 12–5. If the particle is released from rest at the midpoint C, and the acceleration is where s is in meters, determine the velocity of the particle when it reaches plate B, and the time it takes to travel from C to B. SOLUTION Coordinate System. As shown in Fig. 12–5, s is positive downward, measured from plate A. Velocity. Since the velocity as a function of position can be obtained by using Realizing that at we have (1) At Ans. The positive root is chosen since the particle is traveling downward, i.e., in the direction. Time. The time for the particle to travel from C to B can be obtained using and Eq. 1, where when From Appendix A, At Ans. Note: The formulas for constant acceleration cannot be used here because the acceleration changes with position, i.e., a = 4s. t = lnA410.222 - 0.01 + 0.2B + 2.303 2 = 0.658 s s = 0.2 m, lnA4s2 - 0.01 + sB + 2.303 = 2t lnA4s2 - 0.01 + sB ` 0.1 s = 2t ` 0 t L s 0.1 ds 1s2 - 0.01212 = L t 0 2 dt = 21s2 - 0.01212 dt ds = v dt 1+ T2 t = 0. s = 0.1 m v = dsdt +s vB = 0.346 ms = 346 mms T s = 200 mm = 0.2 m, v = 21s2 - 0.01212 ms 1 2 v2 ` 0 v = 4 2 s2 ` 0.1 m s L v 0 v dv = L s 0.1 m 4s ds v dv = a ds 1+ T2 s = 0.1 m, v = 0 v dv = a ds. a = f1s2, s = 200 mm, a = 14s2 ms2 , s = 100 mm, A 200 mm 100 mm B s C Fig. 12–5
- 685. 14 CHAPTER 12 KINEMATICS OF A PARTICLE 12 A particle moves along a horizontal path with a velocity of where t is the time in seconds. If it is initially located at the origin O, determine the distance traveled in 3.5 s, and the particle’s average velocity and average speed during the time interval. SOLUTION Coordinate System. Here positive motion is to the right, measured from the origin O, Fig. 12–6a. Distance Traveled. Since the position as a function of time may be found by integrating with (1) In order to determine the distance traveled in 3.5 s, it is necessary to investigate the path of motion. If we consider a graph of the velocity function, Fig. 12–6b, then it reveals that for the velocity is negative, which means the particle is traveling to the left, and for the velocity is positive, and hence the particle is traveling to the right. Also, note that at The particle’s position when and can now be determined from Eq. 1.This yields The path is shown in Fig. 12–6a. Hence, the distance traveled in 3.5 s is Ans. Velocity. The displacement from to is and so the average velocity is Ans. The average speed is defined in terms of the distance traveled This positive scalar is Ans. Note: In this problem,the acceleration is which is not constant. a = dvdt = 16t - 62 ms2 , 1vsp2avg = sT ¢t = 14.125 m 3.5 s - 0 = 4.04 ms sT. vavg = ¢s ¢t = 6.125 m 3.5 s - 0 = 1.75 ms : ¢s = sƒt=3.5 s - sƒt=0 = 6.125 m - 0 = 6.125 m t = 3.5 s t = 0 sT = 4.0 + 4.0 + 6.125 = 14.125 m = 14.1 m sƒt=0 = 0 sƒt=2 s = -4.0 m sƒt=3.5 s = 6.125 m t = 3.5 s t = 2 s, t = 0, t = 2 s. v = 0 t 7 2 s 0 6 t 6 2 s s = 1t3 - 3t2 2m L s 0 ds = L t 0 (3t2 - 6t) dt = 13t2 - 6t2dt ds = v dt 1 : + 2 s = 0. t = 0, v = dsdt v = f1t2, v = 13t2 - 6t2 ms, EXAMPLE 12.5 O s 4.0 m s 6.125 m t 2 s t 0 s t 3.5 s (a) (0, 0) v (m/s) v 3t2 6t (2 s, 0) t (s) (1 s, 3 m/s) (b) Fig. 12–6
- 686. 12.2 RECTILINEAR KINEMATICS: CONTINUOUS MOTION 15 12 FUNDAMENTAL PROBLEMS F12–1 s F12–2 s F12–3 s F12–4 s F12–5 s s F12–6 s F12–7 s F12–8 F12–5. The position of the particle is given by where is in seconds. Determine the time when the velocity of the particle is zero, and the total distance traveled by the particle when t = 3 s. t s = (2t2 - 8t + 6) m, F12–3. A particle travels along a straight line with a velocity of where is in seconds. Determine the position of the particle when . s = 0 when t = 0 t = 4 s. t v = (4t - 3t2 ) ms, F12–1. Initially, the car travels along a straight road with a speed of If the brakes are applied and the speed of the car is reduced to determine the constant deceleration of the car. 10 ms in 15 s, 35 ms. F12–7. A particle moves along a straight line such that its acceleration is where is in seconds. When the particle is located to the left of the origin, and when it is to the left of the origin. Determine the position of the particle when t = 4 s. 20 m t = 2 s, 2 m t = 0, t a = (4t2 - 2) ms2 , F12–6. A particle travels along a straight line with an acceleration of where s is measured in meters. Determine the velocity of the particle when if at s = 0. v = 5 ms s = 10 m a = (10 - 0.2s) ms2 , F12–4. A particle travels along a straight line with a speed where is in seconds. Determine the acceleration of the particle when t = 2 s. t v = (0.5t3 - 8t) ms, F12–2. A ball is thrown vertically upward with a speed of Determine the time of flight when it returns to its original position. 15 ms. F12–8. A particle travels along a straight line with a velocity of where is in meters. Determine the acceleration of the particle at s = 15 m. s v = (20 - 0.05s2 ) ms,
- 687. 16 CHAPTER 12 KINEMATICS OF A PARTICLE 12 PROBLEMS 12–10. Car A starts from rest at and travels along a straight road with a constant acceleration of until it reaches a speed of . Afterwards it maintains this speed. Also, when , car B located 6000 ft down the road is traveling towards A at a constant speed of . Determine the distance traveled by car A when they pass each other. 60 fts t = 0 80 fts 6 fts2 t = 0 •12–1. A car starts from rest and with constant acceleration achieves a velocity of when it travels a distance of 200 m. Determine the acceleration of the car and the time required. 12–2. A train starts from rest at a station and travels with a constant acceleration of . Determine the velocity of the train when and the distance traveled during this time. 12–3. An elevator descends from rest with an acceleration of until it achieves a velocity of . Determine the time required and the distance traveled. *12–4. A car is traveling at , when the traffic light 50 m ahead turns yellow. Determine the required constant deceleration of the car and the time needed to stop the car at the light. •12–5. A particle is moving along a straight line with the acceleration , where t is in seconds. Determine the velocity and the position of the particle as a function of time.When , and . 12–6. A ball is released from the bottom of an elevator which is traveling upward with a velocity of . If the ball strikes the bottom of the elevator shaft in 3 s, determine the height of the elevator from the bottom of the shaft at the instant the ball is released.Also, find the velocity of the ball when it strikes the bottom of the shaft. 12–7. A car has an initial speed of and a constant deceleration of . Determine the velocity of the car when .What is the displacement of the car during the 4-s time interval? How much time is needed to stop the car? *12–8. If a particle has an initial velocity of to the right, at , determine its position when , if to the left. •12–9. The acceleration of a particle traveling along a straight line is ,where k is a constant.If , when , determine the velocity of the particle as a function of time t. t = 0 v = v0 s = 0 a = kv a = 2 fts2 t = 10 s s0 = 0 v0 = 12 fts t = 4 s 3 ms2 25 ms 6fts s = 15ft v = 0 t = 0 a = (12t – 3t1/2 )fts2 15ms 15fts 5fts2 t = 30s 1ms2 15ms 12–11. A particle travels along a straight line with a velocity , where t is in seconds. When , the particle is located 10 m to the left of the origin. Determine the acceleration when , the displacement from to , and the distance the particle travels during this time period. *12–12. A sphere is fired downwards into a medium with an initial speed of . If it experiences a deceleration of where t is in seconds, determine the distance traveled before it stops. •12–13. A particle travels along a straight line such that in 2 s it moves from an initial position to a position . Then in another 4 s it moves from to . Determine the particle’s average velocity and average speed during the 6-s time interval. 12–14. A particle travels along a straight-line path such that in 4 s it moves from an initial position to a position .Then in another 5 s it moves from to . Determine the particle’s average velocity and average speed during the 9-s time interval. sC = -6 m sB sB = +3 m sA = -8 m sC = +2.5 m sB sB = -1.5 m sA = +0.5 m a = (-6t) ms2 , 27 ms t = 10 s t = 0 t = 4 s t = 1 s v = (12 - 3t2 ) ms A B 6000 ft 60 ft/s Prob. 12–10
- 688. 12.2 RECTILINEAR KINEMATICS: CONTINUOUS MOTION 17 12 •12–21. Two particles A and B start from rest at the origin and move along a straight line such that and , where t is in seconds. Determine the distance between them when and the total distance each has traveled in . 12–22. A particle moving along a straight line is subjected to a deceleration , where is in . If it has a velocity and a position when , determine its velocity and position when . 12–23. A particle is moving along a straight line such that its acceleration is defined as , where is in meters per second. If when and , determine the particle’s position, velocity, and acceleration as functions of time. t = 0 s = 0 v = 20 ms v a = (-2v) ms2 t = 4 s t = 0 s = 10 m v = 8 ms ms v a = (-2v3 ) ms2 t = 4 s t = 4 s aB = (12t2 - 8) fts2 aA = (6t - 3) fts2 s = 0 *12–16. As a train accelerates uniformly it passes successive kilometer marks while traveling at velocities of and then . Determine the train’s velocity when it passes the next kilometer mark and the time it takes to travel the 2-km distance. •12–17. A ball is thrown with an upward velocity of from the top of a 10-m high building. One second later another ball is thrown vertically from the ground with a velocity of . Determine the height from the ground where the two balls pass each other. 12–18. A car starts from rest and moves with a constant acceleration of until it achieves a velocity of . It then travels with constant velocity for 60 seconds. Determine the average speed and the total distance traveled. 12–19. A car is to be hoisted by elevator to the fourth floor of a parking garage, which is 48 ft above the ground. If the elevator can accelerate at decelerate at and reach a maximum speed of determine the shortest time to make the lift, starting from rest and ending at rest. *12–20. A particle is moving along a straight line such that its speed is defined as , where s is in meters. If when , determine the velocity and acceleration as functions of time. t = 0 s = 2 m v = (-4s2 ) ms 8 fts, 0.3 fts2 , 0.6 fts2 , 25 ms 1.5 ms2 10 ms 5 ms 10 ms 2 ms 12–15. Tests reveal that a normal driver takes about before he or she can react to a situation to avoid a collision.It takes about 3 s for a driver having 0.1% alcohol in his system to do the same. If such drivers are traveling on a straight road at 30 mph (44 ) and their cars can decelerate at , determine the shortest stopping distance d for each from the moment they see the pedestrians. Moral: If you must drink, please don’t drive! 2 fts2 fts 0.75 s d v1 44 ft/s Prob. 12–15 *12–24. A particle starts from rest and travels along a straight line with an acceleration , a = (30 - 0.2v) fts2 where is in . Determine the time when the velocity of the particle is . •12–25. When a particle is projected vertically upwards with an initial velocity of , it experiences an acceleration , where g is the acceleration due to gravity, a = -(g + kv2 ) v0 v = 30 fts fts v k is a constant and is the velocity of the particle. Determine the maximum height reached by the particle. 12–26. The acceleration of a particle traveling along a straight line is , where t is in seconds. If , when , determine the velocity and acceleration of the particle at . 12–27. A particle moves along a straight line with an acceleration of , where s is in meters. Determine the particle’s velocity when , if it starts from rest when . Use Simpson’s rule to evaluate the integral. *12–28. If the effects of atmospheric resistance are accounted for, a falling body has an acceleration defined by the equation , where is in and the positive direction is downward. If the body is released from rest at a very high altitude, determine (a) the velocity when , and (b) the body’s terminal or maximum attainable velocity (as ). t : q t = 5 s ms v a = 9.81[1 - v2 (10-4 )] ms2 s = 1 m s = 2 m a = 5(3s13 + s52 ) ms2 s = 4 m t = 0 s = 0 v = 0 a = (0.02et ) ms2 v
- 689. 18 CHAPTER 12 KINEMATICS OF A PARTICLE 12 *12–36. The acceleration of a particle traveling along a straight line is , where s is in meters. If at , determine the velocity of the particle at , and the position of the particle when the velocity is maximum. •12–37. Ball A is thrown vertically upwards with a velocity of . Ball B is thrown upwards from the same point with the same velocity t seconds later. Determine the elapsed time from the instant ball A is thrown to when the balls pass each other, and find the velocity of each ball at this instant. 12–38. As a body is projected to a high altitude above the earth’s surface, the variation of the acceleration of gravity with respect to altitude y must be taken into account. Neglecting air resistance, this acceleration is determined from the formula , where is the constant gravitational acceleration at sea level, R is the radius of the earth, and the positive direction is measured upward. If and , determine the minimum initial velocity (escape velocity) at which a projectile should be shot vertically from the earth’s surface so that it does not fall back to the earth. Hint: This requires that as 12–39. Accounting for the variation of gravitational acceleration a with respect to altitude y (see Prob. 12–38), derive an equation that relates the velocity of a freely falling particle to its altitude. Assume that the particle is released from rest at an altitude from the earth’s surface. With what velocity does the particle strike the earth if it is released from rest at an altitude ? Use the numerical data in Prob. 12–38. *12–40. When a particle falls through the air, its initial acceleration diminishes until it is zero, and thereafter it falls at a constant or terminal velocity If this variation of the acceleration can be expressed as determine the time needed for the velocity to become Initially the particle falls from rest. •12–41. A particle is moving along a straight line such that its position from a fixed point is , where t is in seconds. Determine the total distance traveled by the particle from to .Also, find the average speed of the particle during this time interval. t = 3s t = 1s s = (12 - 15t2 + 5t3 ) m v = vf2. a = 1gv2 f21v2 f - v2 2, vf. a = g y0 = 500 km y0 y : q. v = 0 R = 6356 km g0 = 9.81 ms2 g0 a = -g0[R2 (R + y)2 ] t 6 2v0g v0 s = 2 m s = 0 v = 0 a = (8 - 2s) ms2 •12–29. The position of a particle along a straight line is given by , where t is in seconds. Determine the position of the particle when and the total distance it travels during the 6-s time interval. Hint: Plot the path to determine the total distance traveled. 12–30. The velocity of a particle traveling along a straight line is , where k is constant. If when , determine the position and acceleration of the particle as a function of time. 12–31. The acceleration of a particle as it moves along a straight line is given by where t is in seconds. If and when determine the particle’s velocity and position when Also, determine the total distance the particle travels during this time period. *12–32. Ball A is thrown vertically upward from the top of a 30-m-high-building with an initial velocity of 5 . At the same instant another ball B is thrown upward from the ground with an initial velocity of 20 . Determine the height from the ground and the time at which they pass. •12–33. A motorcycle starts from rest at and travels along a straight road with a constant acceleration of until it reaches a speed of 50 . Afterwards it maintains this speed. Also, when , a car located 6000 ft down the road is traveling toward the motorcycle at a constant speed of 30 . Determine the time and the distance traveled by the motorcycle when they pass each other. 12–34. A particle moves along a straight line with a velocity , where s is in millimeters. Determine the acceleration of the particle at . How long does the particle take to reach this position if when ? 쐍12–35. A particle has an initial speed of If it experiences a deceleration of where t is in seconds, determine its velocity, after it has traveled 10 m. How much time does this take? a = 1-6t2 ms2 , 27 ms. t = 0 s = 500 mm s = 2000 mm v = (200s) mms fts t = 0 fts 6 fts2 t = 0 ms ms t = 6 s. t = 0, v = 2 ms s = 1 m a = 12t - 12 ms2 , t = 0 s = 0 v = v0 - ks t = 6 s s = (1.5t3 - 13.5t2 + 22.5t) ft
- 690. 12.3 RECTILINEAR KINEMATICS: ERRATIC MOTION 19 12 12.3 Rectilinear Kinematics: Erratic Motion When a particle has erratic or changing motion then its position, velocity, and acceleration cannot be described by a single continuous mathematical function along the entire path. Instead, a series of functions will be required to specify the motion at different intervals. For this reason, it is convenient to represent the motion as a graph. If a graph of the motion that relates any two of the variables s, a, t can be drawn, then this graph can be used to construct subsequent graphs relating two other variables since the variables are related by the differential relationships or Several situations occur frequently. The s–t, v–t, and a–t Graphs. To construct the graph given the s–t graph, Fig. 12–7a, the equation should be used, since it relates the variables s and t to .This equation states that For example, by measuring the slope on the s–t graph when the velocity is which is plotted in Fig. 12–7b. The graph can be constructed by plotting this and other values at each instant. The a–t graph can be constructed from the graph in a similar manner, Fig. 12–8, since Examples of various measurements are shown in Fig. 12–8a and plotted in Fig. 12–8b. If the s–t curve for each interval of motion can be expressed by a mathematical function then the equation of the graph for the same interval can be obtained by differentiating this function with respect to time since . Likewise, the equation of the a–t graph for the same interval can be determined by differentiating since . Since differentiation reduces a polynomial of degree n to that of degree n – 1, then if the s–t graph is parabolic (a second-degree curve), the graph will be a sloping line (a first-degree curve), and the a–t graph will be a constant or a horizontal line (a zero-degree curve). v–t a = dvdt v = v(t) v = ds/dt v–t s = s(t), dv dt = a slope of v–t graph = acceleration v–t v–t v1, t = t1, ds dt = v slope of s–t graph = velocity v v = dsdt v–t a ds = v dv. a = dvdt, v = dsdt, v, t O v0 t 0 (a) s ds dt v1 t1 s1 t1 t2 t3 s2 s3 ds dt v2 t2 ds dt v3 t3 ds dt Fig. 12–7 t O (b) v0 v v1 v3 v2 t1 t2 t3 a0 v t t1 t2 t3 v1 v2 v3 v0 a1 a2 O (a) a3 t3 dv dt t2 dv dt t 0 dv dt t1 dv dt t a a0 0 a1 a2 a3 t1 t2 t3 O (b) Fig. 12–8
- 691. 20 CHAPTER 12 KINEMATICS OF A PARTICLE 12 t a a0 t1 v a dt 0 t1 t v v0 t1 v1 v (a) (b) Fig. 12–9 t v v0 t1 t s s0 t1 s1 s (b) (a) s v dt 0 t1 Fig. 12–10 If the a–t graph is given, Fig. 12–9a, the graph may be constructed using written as Hence, to construct the graph, we begin with the particle’s initial velocity and then add to this small increments of area determined from the a–t graph. In this manner successive points, etc., for the graph are determined, Fig. 12–9b. Notice that an algebraic addition of the area increments of the a–t graph is necessary, since areas lying above the t axis correspond to an increase in (“positive” area), whereas those lying below the axis indicate a decrease in (“negative” area). Similarly, if the graph is given, Fig. 12–10a, it is possible to determine the s–t graph using written as In the same manner as stated above, we begin with the particle’s initial position and add (algebraically) to this small area increments determined from the graph, Fig. 12–10b. If segments of the a–t graph can be described by a series of equations, then each of these equations can be integrated to yield equations describing the corresponding segments of the graph. In a similar manner, the s–t graph can be obtained by integrating the equations which describe the segments of the graph.As a result, if the a–t graph is linear (a first-degree curve), integration will yield a graph that is parabolic (a second-degree curve) and an s–t graph that is cubic (third- degree curve). v–t v–t v–t v–t ¢s s0 ¢s = L v dt displacement = area under v–t graph v = dsdt, v–t v v v–t v1 = v0 + ¢v, 1¢v2 v0 v–t ¢v = L a dt change in = area under velocity a–t graph a = dvdt, v–t
- 692. 12.3 RECTILINEAR KINEMATICS: ERRATIC MOTION 21 12 a a0 s1 a ds (v1 2 v0 2 ) 0 s1 (a) 1 — 2 s v v0 s1 v1 (b) s Fig. 12–11 v v0 (a) s dv ds v s a0 (b) s a s a v(dv/ds) Fig. 12–12 The v–s and a–s Graphs. If the a–s graph can be constructed, then points on the graph can be determined by using Integrating this equation between the limits at and at we have, Therefore, if the red area in Fig. 12–11a is determined, and the initial velocity at is known, then Fig. 12–11b. v1 = A21 s1 s0 a ds + v0 2 B12 , s0 = 0 v0 1 21v2 1 - v2 02 = L s1 s0 a ds area under a–s graph s = s1, v = v1 s = s0 v = v0 v dv = a ds. v–s Successive points on the –s graph can be constructed in this manner. If the –s graph is known, the acceleration a at any position s can be determined using written as Thus, at any point (s, ) in Fig. 12–12a, the slope of the –s graph is measured. Then with and known, the value of a can be calculated, Fig. 12–12b. The –s graph can also be constructed from the a–s graph, or vice versa, by approximating the known graph in various intervals with mathematical functions, and then using to obtain the other graph. a ds = v dv v = f(s) or a = g(s), v dvds v v dvds v a = va dv ds b acceleration = velocity times slope of v–s graph a ds = v dv, v v
- 693. 22 CHAPTER 12 KINEMATICS OF A PARTICLE 12 EXAMPLE 12.6 A bicycle moves along a straight road such that its position is described by the graph shown in Fig. 12–13a. Construct the and a–t graphs for 0 … t … 30 s. v–t SOLUTION v–t Graph. Since the graph can be determined by differentiating the equations defining the s–t graph,Fig.12–13a.We have The results are plotted in Fig. 12–13b. We can also obtain specific values of by measuring the slope of the s–t graph at a given instant. For example, at the slope of the s–t graph is determined from the straight line from 10 s to 30 s, i.e., a–t Graph. Since the a–t graph can be determined by differentiating the equations defining the lines of the graph. This yields The results are plotted in Fig. 12–13c. NOTE: Show that when by measuring the slope of the graph. v–t t = 5 s a = 2 fts2 v = 20 fts a = dv dt = 0 10 6 t … 30 s; v = (2t) fts a = dv dt = 2 fts2 0 … t 6 10 s; v–t a = dvdt, v = ¢s ¢t = 500 ft - 100 ft 30 s - 10 s = 20 fts t = 20 s; t = 20 s, v v = ds dt = 20 fts s = (20t - 100) ft 10 s 6 t … 30 s; v = ds dt = (2t) fts s = (t2 ) ft 0 … t 6 10 s; v–t v = dsdt, t (s) s (ft) 500 100 10 30 (a) s t2 s 20t 100 t (s) v (ft/s) 20 10 30 (b) v 2t v 20 t (s) a (ft/s2 ) 2 30 (c) 10 Fig. 12–13
- 694. 12.3 RECTILINEAR KINEMATICS: ERRATIC MOTION 23 12 EXAMPLE 12.7 The car in Fig. 12–14a starts from rest and travels along a straight track such that it accelerates at for 10 s, and then decelerates at . Draw the and s–t graphs and determine the time needed to stop the car. How far has the car traveled? SOLUTION v–t Graph. Since the graph is determined by integrating the straight-line segments of the a–t graph. Using the initial condition when we have When Using this as the initial condition for the next time period, we have v = 101102 = 100 ms. t = 10 s, v = 10t L v 0 dv = L t 0 10 dt, a = (10) ms2 ; 0 … t 6 10 s; t = 0, v = 0 v–t dv = a dt, t¿ v–t 2 ms2 10 ms2 t (s) a (m/s2 ) (a) 10 2 10 A1 A2 t¿ t (s) v (m/s) (b) 100 10 v 10t v 2t 120 t¿ 60 t (s) s (m) (c) 10 60 500 3000 s 5t2 s t2 120t 600 Fig. 12–14 The s–t graph is shown in Fig. 12–14c. NOTE: A direct solution for s is possible when since the triangular area under the graph would yield the displacement from to Hence, t¿ = 60 s. t = 0 ¢s = s - 0 v–t t¿ = 60 s, Ans. ¢s = 1 2160 s21100 ms2 = 3000 m v = (-2t + 120) ms L v 100 ms dv = L t 10 s -2 dt, a = (-2) ms2 ; 10 s 6 t … t¿; When we require This yields, Fig. 12–14b, Ans. A more direct solution for is possible by realizing that the area under the a–t graph is equal to the change in the car’s velocity. We require Fig. 12–14a.Thus Ans. s–t Graph. Since integrating the equations of the graph yields the corresponding equations of the s–t graph. Using the initial condition when we have When Using this initial condition, s = 511022 = 500 m. t = 10 s, s = (5t2 ) m L s 0 ds = L t 0 10t dt, v = (10t) ms; 0 … t … 10 s; t = 0, s = 0 v–t ds = v dt, t¿ = 60 s 0 = 10 ms2 110 s2 + 1-2 ms2 21t¿ - 10 s2 ¢v = 0 = A1 + A2, t¿ t¿ = 60 s v = 0. t = t¿ L s 500 m ds = L t 10 s 1-2t + 1202 dt v = (-2t + 120) ms; 10 s … t … 60 s; When the position is Ans. s = -16022 + 1201602 - 600 = 3000 m t¿ = 60 s, s = (-t2 + 120t - 600) m s - 500 = -t2 + 120t - [-11022 + 1201102]
- 695. 24 CHAPTER 12 KINEMATICS OF A PARTICLE 12 EXAMPLE 12.8 The –s graph describing the motion of a motorcycle is shown in Fig. 12–15a. Construct the a–s graph of the motion and determine the time needed for the motorcycle to reach the position SOLUTION a–s Graph. Since the equations for segments of the –s graph are given, the a–s graph can be determined using The results are plotted in Fig. 12–15b. Time. The time can be obtained using the –s graph and because this equation relates s, and t. For the first segment of motion, when so At Therefore, using these initial conditions for the second segment of motion, Therefore, at Ans. NOTE: The graphical results can be checked in part by calculating slopes. For example, at Also, the results can be checked in part by inspection. The –s graph indicates the initial increase in velocity (acceleration) followed by constant velocity 1a = 02. v a = v1dvds2 = 10150 - 102200 = 2 ms2 . s = 0, t = 400 50 + 4.05 = 12.0 s s = 400ft, t = a s 50 + 4.05b s t - 8.05 = s 50 - 4; L t 8.05 s dt = L s 200 m ds 50 ; dt = ds v = ds 50 v = 50 fts; 200 ft 6 s … 400 ft; t = 5 ln[0.212002 + 10] - 5 ln 10 = 8.05 s. s = 200 ft, t = (5 ln10.2s + 102 - 5 ln 10) s L t 0 dt = L s 0 ds 0.2s + 10 dt = ds v = ds 0.2s + 10 v = (0.2s + 10) fts; 0 … s 6 200 ft; t = 0, s = 0 v, v = dsdt, v a = v dv ds = 1502 d ds 1502 = 0 v = 50 fts 200 ft 6 s … 400 ft; a = v dv ds = 10.2s + 102 d ds 10.2s + 102 = 0.04s + 2 v = (0.2s + 10) fts 0 … s 6 200 ft; a ds = v dv. v s = 400 ft. v (a) v (ft/s) s (ft) 10 50 200 400 v 0.2s 10 v 50 (b) 200 400 s (ft) a (ft/s2 ) 10 2 a 0.04s 2 a 0 Fig. 12–15
- 696. 12.3 RECTILINEAR KINEMATICS: ERRATIC MOTION 25 12 FUNDAMENTAL PROBLEMS t (s) s (m) 6 8 10 108 s 0.5 t3 s 108 F12–9 t (s) v (ft/s) v 4t 80 80 20 F12–10 s (m) (m/s) 10 40 v 0.25 s v F12–11 t (s) s (m) s 30t 75 5 75 225 10 0 s 3t2 F12–12 t (s) t¿ a (m/s2 ) 5 0 20 10 F12–13 15 t (s) v (m/s) v 30 t v 15 t 225 5 150 F12–14 F12–12. The sports car travels along a straight road such that its position is described by the graph. Construct the and graphs for the time interval 0 … t … 10 s. a-t v-t F12–11. A bicycle travels along a straight road where its velocity is described by the graph. Construct the graph for the same time interval. a-s v-s F12–9. The particle travels along a straight track such that its position is described by the graph. Construct the graph for the same time interval. v-t s-t F12–14. The dragster starts from rest and has a velocity described by the graph. Construct the graph during the time interval Also, determine the total distance traveled during this time interval. 0 … t … 15 s. s-t F12–10. A van travels along a straight road with a velocity described by the graph. Construct the and graphs during the same period.Take when t = 0. s = 0 a-t s-t F12–13. The dragster starts from rest and has an acceleration described by the graph. Construct the graph for the time interval where is the time for the car to come to rest. t¿ 0 … t … t¿, v-t
- 697. 26 CHAPTER 12 KINEMATICS OF A PARTICLE 12 PROBLEMS 12–46. A train starts from station A and for the first kilometer, it travels with a uniform acceleration. Then, for the next two kilometers, it travels with a uniform speed. Finally, the train decelerates uniformly for another kilometer before coming to rest at station B. If the time for the whole journey is six minutes, draw the graph and determine the maximum speed of the train. 12–47. The particle travels along a straight line with the velocity described by the graph. Construct the graph. a-s v–t *12–44. A freight train starts from rest and travels with a constant acceleration of . After a time it maintains a constant speed so that when it has traveled 2000 ft. Determine the time and draw the –t graph for the motion. •12–45. If the position of a particle is defined by , where t is in seconds, construct the , , and graphs for . 0 … t … 10 s a-t v-t s-t s = [2 sin (p5)t + 4] m v t¿ t = 160 s t¿ 0.5 fts2 12–42. The speed of a train during the first minute has been recorded as follows: *12–48. The a–s graph for a jeep traveling along a straight road is given for the first 300 m of its motion. Construct the –s graph.At , . v = 0 s = 0 v t (s) 0 20 40 60 ( ) ms v 0 16 21 24 Plot the graph, approximating the curve as straight-line segments between the given points. Determine the total distance traveled. 12–43. A two-stage missile is fired vertically from rest with the acceleration shown. In 15 s the first stage A burns out and the second stage B ignites. Plot the and graphs which describe the two-stage motion of the missile for . 0 … t … 20 s s-t v-t v-t Prob. 12–43 a (m/s2 ) t (s) 15 18 25 20 A B s (m) a (m/s2 ) 2 200 300 Prob. 12–48 s (m) v (m/s) 3 6 4 10 13 v 2s 4 v s 7 Prob. 12–47
- 698. 12.3 RECTILINEAR KINEMATICS: ERRATIC MOTION 27 12 *12–52. A car travels up a hill with the speed shown. Determine the total distance the car travels until it stops ( ). Plot the graph. a-t t = 60 s 12–51. A car starts from rest and travels along a straight road with a velocity described by the graph. Determine the total distance traveled until the car stops. Construct the and graphs. a–t s–t •12–49. A particle travels along a curve defined by the equation where is in seconds. Draw the and graphs for the particle for . 12–50. A truck is traveling along the straight line with a velocity described by the graph. Construct the graph for . 0 … s … 1500 ft a-s 0 … t … 3 s a - t v - t, s - t, t s = (t3 - 3t2 + 2t) m. •12–53. The snowmobile moves along a straight course according to the –t graph. Construct the s–t and a–t graphs for the same 50-s time interval.When , . s = 0 t = 0 v 12–54. A motorcyclist at A is traveling at when he wishes to pass the truck T which is traveling at a constant speed of To do so the motorcyclist accelerates at until reaching a maximum speed of If he then maintains this speed, determine the time needed for him to reach a point located 100 ft in front of the truck. Draw the and graphs for the motorcycle during this time. s-t v-t 85 fts. 6 fts2 60 fts. 60 fts Prob. 12–52 Prob. 12–53 30 t (s) v (m/s) 10 60 Prob. 12–50 s(ft) v (ft/s) 625 1500 75 v 0.6 s3/4 Prob. 12–51 Prob. 12–54 t(s) v(m/s) 30 90 30 v 0.5t 45 v t 60 100 ft 55 ft 40 ft A T (vm)1 60 ft/s (vm)2 85 ft/s vt 60 ft/s t (s) 12 30 50 v (m/s)
- 699. Prob. 12–55 28 CHAPTER 12 KINEMATICS OF A PARTICLE 12 •12–57. The dragster starts from rest and travels along a straight track with an acceleration-deceleration described by the graph. Construct the graph for and determine the distance traveled before the dragster again comes to rest. s¿ 0 … s … s¿, v-s *12–56. The position of a cyclist traveling along a straight road is described by the graph. Construct the and graphs. a–t v–t 12–55. An airplane traveling at lands on a straight runway and has a deceleration described by the graph. Determine the time and the distance traveled for it to reach a speed of . Construct the and graphs for this time interval, . 0 … t … t¿ s-t v–t 5 ms t¿ 70 ms 12–58. A sports car travels along a straight road with an acceleration-deceleration described by the graph. If the car starts from rest, determine the distance the car travels until it stops. Construct the graph for . 0 … s … s¿ v-s s¿ t(s) a(m/s2 ) 5 t¿ 4 10 Prob. 12–56 t (s) s (m) s 0.625 t2 27.5t 162.5 10 20 50 137.5 s 0.05 t3 Prob. 12–58 s(ft) a(ft/s2 ) s¿ 4 6 1000 Prob. 12–57 s(m) a(m/s2 ) 200 s¿ 25 5 15 a 0.1s 5
- 700. 12.3 RECTILINEAR KINEMATICS: ERRATIC MOTION 29 12 •12–61. The graph of a car while traveling along a road is shown. Draw the and graphs for the motion. a-t s-t v-t *12–60. A motorcyclist starting from rest travels along a straight road and for 10 s has an acceleration as shown. Draw the graph that describes the motion and find the distance traveled in 10 s. v-t 12–59. A missile starting from rest travels along a straight track and for 10 s has an acceleration as shown. Draw the graph that describes the motion and find the distance traveled in 10 s. v-t 12–62. The boat travels in a straight line with the acceleration described by the graph. If it starts from rest, construct the graph and determine the boat’s maximum speed.What distance does it travel before it stops? s¿ v-s a-s Prob. 12–59 30 5 10 40 t(s) a (m/s2 ) a 2t 20 a 6t Prob. 12–60 6 10 t (s) 6 1 — 6 a (m/s2 ) a t2 Prob. 12–61 20 20 30 5 t (s) v (m/s) Prob. 12–62 s(m) s¿ a(m/s2 ) 3 150 4 6 a 0.02s 6
- 701. 30 CHAPTER 12 KINEMATICS OF A PARTICLE 12 •12–65. The acceleration of the speed boat starting from rest is described by the graph. Construct the graph. v-s *12–64. The jet bike is moving along a straight road with the speed described by the graph. Construct the graph. a-s v-s 12–63. The rocket has an acceleration described by the graph. If it starts from rest, construct the and graphs for the motion for the time interval . 0 … t … 14s s-t v-t 12–66. The boat travels along a straight line with the speed described by the graph. Construct the and graphs. Also, determine the time required for the boat to travel a distance if . s = 0 when t = 0 s = 400 m a-s s–t Prob. 12–63 t(s) a(m/s2 ) 38 18 9 14 a2 36t a 4t 18 Prob. 12–65 a(ft/s2 ) 10 2 200 500 a 0.04s 2 s(ft) Prob. 12–64 v(m/s) v 5s1/2 75 15 225 525 v 0.2s 120 s(m) Prob. 12–66 v(m/s) 100 400 20 80 s(m) v2 4s v 0.2s
- 702. 12.3 RECTILINEAR KINEMATICS: ERRATIC MOTION 31 12 •12–69. The airplane travels along a straight runway with an acceleration described by the graph. If it starts from rest and requires a velocity of to take off, determine the minimum length of runway required and the time for take off. Construct the and graphs. s–t v-t t¿ 90 ms *12–68. The airplane lands at on a straight runway and has a deceleration described by the graph. Determine the distance traveled before its speed is decreased to . Draw the graph. s-t 25 fts s¿ 250 fts 12–67. The s–t graph for a train has been determined experimentally. From the data, construct the and a–t graphs for the motion. v-t 12–70. The graph of the bullet train is shown. If the train starts from rest, determine the elapsed time before it again comes to rest. What is the total distance traveled during this time interval? Construct the and graphs. s–t v–t t¿ a–t Prob. 12–67 t (s) 600 360 30 40 s (m) s 24t 360 s 0.4t2 Prob. 12–69 t(s) a(m/s2 ) t¿ 10 8 a 0.8t Prob. 12–68 a(ft/s2 ) 1750 7.5 15 s¿ s(ft) Prob. 12–70 t(s) a(m/s2 ) 3 30 75 a ( )t 5 1 15 a 0.1t t¿
- 703. Fig. 12–16 32 CHAPTER 12 KINEMATICS OF A PARTICLE 12 12.4 General Curvilinear Motion Curvilinear motion occurs when a particle moves along a curved path. Since this path is often described in three dimensions, vector analysis will be used to formulate the particle’s position, velocity, and acceleration.* In this section the general aspects of curvilinear motion are discussed, and in subsequent sections we will consider three types of coordinate systems often used to analyze this motion. Position. Consider a particle located at a point on a space curve defined by the path function s(t), Fig. 12–16a.The position of the particle, measured from a fixed point O, will be designated by the position vector Notice that both the magnitude and direction of this vector will change as the particle moves along the curve. Displacement. Suppose that during a small time interval the particle moves a distance along the curve to a new position, defined by Fig. 12–16b. The displacement represents the change in the particle’s position and is determined by vector subtraction; i.e., Velocity. During the time the average velocity of the particle is The instantaneous velocity is determined from this equation by letting and consequently the direction of approaches the tangent to the curve. Hence, or (12–7) Since dr will be tangent to the curve, the direction of v is also tangent to the curve, Fig. 12–16c. The magnitude of v, which is called the speed, is obtained by realizing that the length of the straight line segment in Fig. 12–16b approaches the arc length as we have or (12–8) Thus, the speed can be obtained by differentiating the path function s with respect to time. v = ds dt v = lim ¢t:0 1¢r¢t2 = lim ¢t:0 1¢s¢t2, ¢t : 0, ¢s ¢r v = dr dt v = lim ¢t:0 1¢r¢t2 ¢r ¢t : 0, vavg = ¢r ¢t ¢t, ¢r = r¿ - r. ¢r r¿ = r + ¢r, ¢s ¢t r = r1t2. *A summary of some of the important concepts of vector analysis is given in Appendix B. s r O Position (a) Path s Displacement (b) r r¿ s r s O Velocity (c) r v s O
- 704. 12.4 GENERAL CURVILINEAR MOTION 33 12 Acceleration. If the particle has a velocity v at time t and a velocity at Fig. 12–16d, then the average acceleration of the particle during the time interval is where To study this time rate of change, the two velocity vectors in Fig. 12–16d are plotted in Fig. 12–16e such that their tails are located at the fixed point and their arrowheads touch points on a curve. This curve is called a hodograph, and when constructed, it describes the locus of points for the arrowhead of the velocity vector in the same manner as the path s describes the locus of points for the arrowhead of the position vector, Fig. 12–16a. To obtain the instantaneous acceleration, let in the above equation. In the limit will approach the tangent to the hodograph, and so or (12–9) Substituting Eq. 12–7 into this result, we can also write By definition of the derivative, a acts tangent to the hodograph, Fig. 12–16f, and, in general it is not tangent to the path of motion, Fig. 12–16g. To clarify this point, realize that and consequently a must account for the change made in both the magnitude and direction of the velocity v as the particle moves from one point to the next along the path, Fig. 12–16d. However, in order for the particle to follow any curved path, the directional change always “swings” the velocity vector toward the “inside” or “concave side” of the path, and therefore a cannot remain tangent to the path. In summary, v is always tangent to the path and a is always tangent to the hodograph. ¢v a = d2 r dt2 a = dv dt a = lim ¢t:0 1¢v¢t2, ¢v ¢t : 0 O¿ ¢v = v¿ - v. aavg = ¢v ¢t ¢t t + ¢t, v¿ = v + ¢v v v¿ (d) v v¿ (e) v O¿ v a (f) O¿ Hodograph Acceleration (g) a path Fig. 12–16 (cont.)
- 705. 34 CHAPTER 12 KINEMATICS OF A PARTICLE 12 12.5 Curvilinear Motion: Rectangular Components Occasionally the motion of a particle can best be described along a path that can be expressed in terms of its x, y, z coordinates. Position. If the particle is at point (x, y, z) on the curved path s shown in Fig. 12–17a, then its location is defined by the position vector (12–10) When the particle moves, the x, y, z components of r will be functions of time; i.e., so that At any instant the magnitude of r is defined from Eq. C–3 in Appendix C as And the direction of r is specified by the unit vector Velocity. The first time derivative of r yields the velocity of the particle. Hence, When taking this derivative, it is necessary to account for changes in both the magnitude and direction of each of the vector’s components. For example, the derivative of the i component of r is The second term on the right side is zero, provided the x, y, z reference frame is fixed, and therefore the direction (and the magnitude) of i does not change with time. Differentiation of the j and k components may be carried out in a similar manner, which yields the final result, (12–11) where (12–12) vx = x # vy = y # vz = z # v = dr dt = vxi + vyj + vzk d dt 1xi2 = dx dt i + x di dt v = dr dt = d dt 1xi2 + d dt 1yj2 + d dt 1zk2 ur = rr. r = 4x2 + y2 + z2 r = r1t2. z = z1t2, y = y1t2, x = x1t2, r = xi + yj + zk y x z r xi yj zk z y x s k i j Position (a) y x z s Velocity (b) v vxi vyj vzk Fig. 12–17
- 706. 12.5 CURVILINEAR MOTION: RECTANGULAR COMPONENTS 35 12 The “dot” notation represents the first time derivatives of respectively. The velocity has a magnitude that is found from and a direction that is specified by the unit vector As discussed in Sec. 12–4, this direction is always tangent to the path, as shown in Fig. 12–17b. Acceleration. The acceleration of the particle is obtained by taking the first time derivative of Eq. 12–11 (or the second time derivative of Eq. 12–10).We have (12–13) where (12–14) Here represent, respectively, the first time derivatives of or the second time derivatives of the functions The acceleration has a magnitude and a direction specified by the unit vector Since a represents the time rate of change in both the magnitude and direction of the velocity, in general a will not be tangent to the path, Fig. 12–17c. ua = aa. a = 4ax 2 + ay 2 + az 2 z = z1t2. y = y1t2, x = x1t2, vz = vz1t2, vy = vy1t2, vx = vx1t2, az ay, ax, ax = v # x = x $ ay = v # y = y $ az = v # z = z $ a = dv dt = axi + ayj + azk uv = vv. v = 4vx 2 + vy 2 + vz 2 z = z1t2, y = y1t2, x = x1t2, z # y # , x # , y x z s a axi ayj azk Acceleration (c)
- 707. 36 CHAPTER 12 KINEMATICS OF A PARTICLE 12 Procedure for Analysis Coordinate System. • A rectangular coordinate system can be used to solve problems for which the motion can conveniently be expressed in terms of its x, y, z components. Kinematic Quantities. • Since rectilinear motion occurs along each coordinate axis, the motion along each axis is found using and or in cases where the motion is not expressed as a function of time, the equation can be used. • In two dimensions, the equation of the path can be used to relate the x and y components of velocity and acceleration by applying the chain rule of calculus. A review of this concept is given in Appendix C. • Once the x, y, z components of v and a have been determined, the magnitudes of these vectors are found from the Pythagorean theorem, Eq. B–3, and their coordinate direction angles from the components of their unit vectors, Eqs. B–4 and B–5. y = f(x) a ds = v dv a = dvdt; v = dsdt Important Points • Curvilinear motion can cause changes in both the magnitude and direction of the position, velocity, and acceleration vectors. • The velocity vector is always directed tangent to the path. • In general, the acceleration vector is not tangent to the path, but rather, it is tangent to the hodograph. • If the motion is described using rectangular coordinates, then the components along each of the axes do not change direction, only their magnitude and sense (algebraic sign) will change. • By considering the component motions, the change in magnitude and direction of the particle’s position and velocity are automatically taken into account.
- 708. 12.5 CURVILINEAR MOTION: RECTANGULAR COMPONENTS 37 12 EXAMPLE 12.9 At any instant the horizontal position of the weather balloon in Fig. 12–18a is defined by where t is in seconds. If the equation of the path is determine the magnitude and direction of the velocity and the acceleration when SOLUTION Velocity. The velocity component in the x direction is To find the relationship between the velocity components we will use the chain rule of calculus. (See Appendix A for a full explanation.) vx = x # = d dt 18t2 = 8 fts : t = 2 s. y = x2 10, x = 18t2 ft, vy = y # = d dt 1x2 102 = 2xx # 10 = 2116218210 = 25.6 fts c When the magnitude of velocity is therefore Ans. The direction is tangent to the path, Fig. 12–18b, where Ans. Acceleration. The relationship between the acceleration components is determined using the chain rule. (See Appendix C.) We have Thus, Ans. The direction of a, as shown in Fig. 12–18c, is Ans. NOTE: It is also possible to obtain and by first expressing and then taking successive time derivatives. y = f1t2 = 18t22 10 = 6.4t2 ay vy ua = tan-1 12.8 0 = 90° a = 4(022 + (12.822 = 12.8 ft s2 = 21822 10 + 2116210210 = 12.8 fts2 c ay = v # y = d dt 12xx # 102 = 21x # 2x # 10 + 2x1x $ 210 ax = v # x = d dt 182 = 0 uv = tan-1 vy vx = tan-1 25.6 8 = 72.6° v = 4(8 fts22 + (25.6 fts22 = 26.8 fts t = 2 s, y A B x 16 ft (a) y x2 10 (b) B v 26.8 ft/s uv 72.6 (c) a 12.8 ft/s2 B ua 90 Fig. 12–18
- 709. 38 CHAPTER 12 KINEMATICS OF A PARTICLE 12 EXAMPLE 12.10 For a short time, the path of the plane in Fig. 12–19a is described by If the plane is rising with a constant velocity of 10 , determine the magnitudes of the velocity and acceleration of the plane when it is at SOLUTION When , then or . Also, since , then ; Velocity. Using the chain rule (see Appendix C) to find the relationship between the velocity components, we have (1) Thus The magnitude of the velocity is therefore Ans. Acceleration. Using the chain rule, the time derivative of Eq. (1) gives the relation between the acceleration components. When The magnitude of the plane’s acceleration is therefore Ans. These results are shown in Fig. 12–19b. = 0.791 ms2 a = 4ax 2 + ay 2 = 4(-0.791 ms2 22 + (0 ms2 22 0 = 0.002((15.81 ms)2 + 316.2 m(ax)) ax = -0.791 ms2 x = 316.2 m, vx = 15.81 ms, v # y = ay = 0, ay = v # y = 0.002x # vx + 0.002xv # x = 0.002(vx 2 + xax) = 4(15.81 ms)2 + (10 ms)2 = 18.7 ms v = 4vx 2 + vy 2 10 ms = 0.002(316.2 m)(vx) vx = 15.81 ms vy = y # = d dt (0.001x2 ) = (0.002x)x # = 0.002xvx t = 10 s 100 m = (10 ms) t y = vyt vy = 10 ms x = 316.2 m 100 = 0.001x2 y = 100 m y = 100 m. ms y = (0.001x2 ) m. y x x y (b) 100 m vy v a vx Fig. 12–19 x y (a) y 0.001x2 100 m
- 710. 12.6 MOTION OF A PROJECTILE 39 12 12.6 Motion of a Projectile The free-flight motion of a projectile is often studied in terms of its rectangular components. To illustrate the kinematic analysis, consider a projectile launched at point ( ), with an initial velocity of having components and Fig. 12–20.When air resistance is