A man pushes a cylinder of mass `m_1` with the help of a plank of mass `m_2` as shown in figure. There in no slipping at any contact. The horizontal component of the force applied by the man is F.
(a) the acceleration fo the plank and the center of mass of the cylinder, and

(b) the magnitudes and direction of frictional force at contact points.

Explanation

Analysis of the force `F` in the horizontal direction on the plank as shwon in figure. Therefore the point of contact of the plank with the cylinder will try to move towards the righ. Therefore, the friction force `f` will act towers the left on the plank.

To each and every action, there is an equal and opposite reaction. Therefore, a frictional force of will act on the top of he cylinder towareds the right. Direction of `f`:
For finding the direction of `f'` first assume `f'` is not acting. A force `f` is acting on the cylinder. This force is trying to move the point of contact towards right by an acceleration `a_(CM)=f/m_(1)` acting towards the right.
At the same time, the force `f` is trying to rotate the cylinder about its centre of mass `m` the clockwise direction.
`fxxR=Ixxalphaimplies alpha=(fxxR)/I=(fxxR)/(1/2m_(1)R^(2))=(2f)/(m_(1)R)`
Therefore, acceleration of the point of contact
`a_(P)=a_(CM)-alphaR=f/(m_(1))-(2f)/(m_(1)R)xxR=f/(m_(1))`
i.e,, towards the left.
Therefore the point of contact of the cylinder with the ground moves towards the left. therefore, fiction foce acts towards the right on the cylinder.
Here `k^(2)ltRx` as `k^(2)=(R^(2))/2`
and `Rx=R.R=R^(2)`
Hence, friction should act in forward direction.
Applying Newton's law on the plank we get
`F-f=m_(2)a_(2)`.....i
Also `a_(2)=2a_(1)`....ii
Because `a_(2)` is the acceleration of the top most point of the cylinder and there is no slipping. Applying Newton's law on the cylinder.
`m_(1)a_(1)=f+f'`.........iii
The torque equation for the cylinder is
`fxxR-f'xxR=Ialpha=1/2m_(1)R^(2)xx(a/R)`
`:'I=m_(1)R^(2)` and `Ralpha=a_(1)`
`:. 1=m_(1)R^(2)` and `Ralpha=a_(1)`
`:. (f-f')R=1/2m_(1)Ra_(1)`
`:. f+f'=1/2m_(1)Ra_(1)`..iv
solving eqn iii and iv we get `f=3/4m_(1)a_(1)` ............v
and `f'=1/4m_(1)a_(1)`..........vi
From eqn i and iii `F-f=2m_(2)a_(1)`
`implies F=3/4m_(1)a_(1)=2m_(2)a_(1)`
`:. a_(1)=(4F)/(3m_(1)+8m_(2))`
`:. a_(2)=(8F)/(3m_(1)+8m_(2))`
From eqn v and vi we get
`f=3/4m_(1)xx(4F)/(3m_(1)+8m_(2))=(3Fm_(1))/(3m_(1)+8m_(2))`
and `f'=1/4m_(1)xxa_(1)-(3Fm)/(3m_(1)+8m_(2))`