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

We can choose any arbitrary direction of frictional forces at different contacts.
In the final answer, the negative value will show for opposite direction
let `f_(1)=` friction between plank and cylinder
`f_(2)=` friction between cylinder and ground
`a_(1)=` acceleration of plank
`a_(2)=` acceleration of centre of mass of cylinder
and `alpha=` angular acceleration of cylinder about is COM.

Decreation of `f_(1)` and `f_(2)` are as shown here.
Since, there is not slipping anywhere
`thereforea_(1)=2a_(2)` ..(i)
(acceleration of plank `=` acceleration fo top point of cylinder)
`a_(1)=(F-f_(1))/(m_(2))` ...(ii)
`a_(2)=(f_(1)+f_(2))/(m_(1))`
`alpha((f_(1)-f_(2))R)/(I)`
`(I=` moment of inertia of cylinder about COM)

`thereforealpha=((f_(1)-f_(2))R)/((1)/(2)m_(1)R^(2))`
`alpha=(2(f_(1)-f_(2))/(m_(1)R)` ..(iv)
`a_(2)=Ralpha(2(f_(1)-f_(2))/(m_(1))` ....(v)
(acceleration of bottommost point of cylinder `=0`)
(a). solving eqs (i), (ii), (iii) and (V) we get
`a_(1)=(8F)/(3m_(1)+8m_(2))`
and `a_(2)=(4F)/(3m_(1)+8m_(2))`
(b). `f_(1)=(m_(1)F)/(3m_(1)+8m_(2))`
Since all quantities are positive, they are correctly shown in figures.