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Question

A plank with a mass M=6.00 kg rests on top of two identical, solid, cylindrical rollers that have R=5.00 cm and m=2.00 kg. The plank is pulled by a constant force F of magnitude 6.00 N applied to the end of the plank and perpendicular to the axes of the cylinders ( which are parallel). The cylinders roll without slipping on a flat surface. There is also no slipping between the cylinders and the plank.
What friction forces are acting at this moment?
1865823_09d96e41a1cd463b9da3a1408f00d072.png

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Solution

Force exterted on system are shown in diagram.Here,
ft is the frictional force exerted by each roller backward on the plank. fb is the rolling resistance exerted backward by the ground on each roller. Suppose the rollers are equally far from the ends of the plank.

For the plank force-acceleration equation is:

x=max6.00N2ft=(6.00kg)ap

The center of each roller moves forward only half as far as the plank. Each roller has acceleration ap2 and angular acceleration

ap/2(5.00cm)=ap(0.100m)
Then for each,

Fx=max+ftfb=(2.00kg)ap2

τ=Iαft(5.00cm)+fb(5.00cm)=12(2.00kg)(5.00cm)2ap10.0cm

So ft+fb=(12kg)ap

Add to eliminate fb

2ft=(1.50kg)ap

And 6.00N(1.50kg)ap=(6.00kg)ap

ap=(6.00N)(7.50kg)=0.800m/s2

For each roller, a=ap2=0.400m/s2

on Substituting , 2ft=(1.50kg)0.800m/s2

ft=0.600N

0.600N+fb=12kg(0.800m/s2)

fb=0.200N

The negative sign means that the horizontal force of ground on each roller is 0.200N forward rather than backward as we assumed.

1894877_1865823_ans_db18cf75cc0b41c7a0341f7d794e4dc1.png

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