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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 (figure). The plank is pulled by a constant horizontal 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. All surfaces are rough. 


The initial acceleration of the plank at the moment when the rollers are equidistant from the ends of the plank is
  m/s2.


Solution

Let ft be the force of friction between plank and the cylinder and fb be the force of friction between cylinder and the ground. 

For the plank, Fx=max
6.0N2ft=(6.00kg)ap .       .........(i)
Now, as there is no slipping between the plank and the cylinders we can say 
ap=ac+Rα...........(ii)
Also there is no slipping between ground and the cylinders so,
ac=Rα............(iii)
By using (ii) and (iii) we can say
each roller has acceleration ap2 and angular acceleration
α=ap/25.00cm=ap0.100mThen for each cylinders; Fx=max
ftfb=(2.00kg)ap2....(iv)
Now applying torque equation τ=Iα
ft(5cm)+fb(5cm)=12(2kg)(5cm)2ap10cmSo ft+fb=(12kg)ap ..........(v)
Substituting the value for 2ft into equation (i) it gives
6.00 N(1.50kg)ap=(6.0kg)ap
ap=6.00N7.50kg=0.80 m/s2
 

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