Solving Forces and Angles to Load Fuel Drum

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The operator can use the plank to load the 200L fuel drum because tilting the plank reduces the gravitational force acting perpendicular to its surface, allowing it to support more weight without breaking. The critical weight that caused the original plank to break was 166 kg when loaded horizontally, but at an angle, the effective weight is decreased. To find the minimum angle at which the plank can still safely load the drum, the equation (197 kg) * cos(A) = 166 kg can be used. This calculation shows that as the angle increases, the load capacity of the plank also increases. Therefore, the plank can be utilized effectively as a ramp for the fuel drum.
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hi, i need help on a question

its like... there is a operator supported a plank horizontally by its two ends, but when loaded bricks on its centre and found that the weight of 166kg of bricks was just enough to break the plank. But an identical plank is now being used as a ramp to load a full 200L fuel drum onto a truck. The drum and its contents have a mass of 197kg.

a) Carefully explain why the operator could still use this plank to load the fuel drum.
b) what is the minimum angle to the horizontal at which the plank can still used to load the drum?

i don't know where to start... as i don't know how to answer part a) :mad:
 
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min angle would be = cos-1(166/197)
 
and the reason the greater weight could be carried is b/c when the plank is at angle the internal moment arm is less
 
k3l said:
hi, i need help on a question

its like... there is a operator supported a plank horizontally by its two ends, but when loaded bricks on its centre and found that the weight of 166kg of bricks was just enough to break the plank. But an identical plank is now being used as a ramp to load a full 200L fuel drum onto a truck. The drum and its contents have a mass of 197kg.

a) Carefully explain why the operator could still use this plank to load the fuel drum.
b) what is the minimum angle to the horizontal at which the plank can still used to load the drum?

i don't know where to start... as i don't know how to answer part a) :mad:

The original plank broke at 166 kg because the gravitational force NORMAL to its surface exceeded the plank's capacity. When the plank is tilted, the component of gravitational force NORMAL to the surface is reduced by a factor of cos(A), where "A" is the plank's tilt angle with respect to horizontal ground. Thus, the minimum "A" required for 197 kg can be determined from:
(197)*cos(A) = (166) <--- (derive this yourself by resolving grav force into components normal and parallel to plank surface)
~
 
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how do i attach a document
 
plank

this should help
 

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Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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