A man pushes a cylinder of mass `m_(1)` with the help of a plank of mass `m_(2)` as shown. There is no shipping at any contact. The horizontal component of the force applied by the man is F. Find:
(a). the acceleration of the plank and the centre of mass the cylinder and
(b). The magnitudes and direction of frictional forces at contact points.

Explanation

(i) `(8F)/(3m_(1) + 8m_(2))`, (ii) `(4F)/(3m_(1) + 8m_(2))`, (iii) `(3Fm_(1))/(3m_(1) + 8m_(2))`, (iv) `(Fm_(1))/(3m_(1) + 8m_(2))`
The man applies a force F in the horizontal direction on the plank as shown in the figure. Therefore, the point of contact of the plank with the cylinder will try to move towards the right. Therefore, the friction force f will act towards the left on the plank.
To each and every action, there is an equal and opposite reaction. Therefore, a frictional force f will act on the top of the cylinder towards the right.
Direction of friction by surface on cylinder:
For finding the direction on the cylinder, the force f 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 in the clockwise direction.
`f xx R =I xx alpha rArr alpha =(f xx R)/I =(f xx R)/(1/2m_(1)R^(2)) =(2f)/(m_(1)R)`
Therefore, acceleration of the point of contact, `a_(p) =a_(CM) -alpha R =f/(m_(1)) -(2f)/(m_(1)R) xx R =-f/m_(1)`
i.e., towards the left.
Therefore, the point of contact of the cylinder with the ground move towards the left. Therefore, friction force acts towards the right on the cylinder.
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 topmost point of the cylinder and there is no slipping.
Apply Newton's law on the cylinder.
`f+f^(') =m_(1)a_(1)`........(iii)
The torque equation for the cylinder is,
`f xx R -f xx R =I alpha =1/2m_(1)R^(2) xx (a_(1)/R) (therefore I=m_(1)R^(2)` and `R alpha =a_(1)`)
`therefore f-f' =1/2 m_(1)a_(1)` ...........(iv)
Solving Eqs. (iii) and (iv), we get `f=3/4 m_(1)a_(1)` and `f'=1/4m_(1)a_(1)`
From Eqs. (i) and (ii), `F-f=2m_(2)a_(1) rArr F=-3/4 m_(1)a_(1) = 2m_(2)a_(1)`
`therefore a_(1) =(4F)/(3m_(1)+8m_(2))` and `a_(2) = (8F)/(3m_(1) + 8m_(2))`
`f=3/4 m_(1) xx (4F)/(3m_(1) + 8m_(2))` and `f^(') =1/4 m_(1) xx a_(1) =(Fm_(1))/(3m_(1) + 8m_(2))`.