Consider a cylinder of mass `M` and radius `R` lying on a rough horizontal plane. It has a plank lying on its top as shown in the figure. A force `F` is applied on the plank such that the plank moves and causes the cylinder to roll. The plank always remains horizontal. There is no slipping at any point to contact.

`(a)` what are the directions of the fricition forces acting at `A` and `B` on the plank and the cylinder ?
`(b)` Calculate the acceleration of the cylinder.
`(c )` Find the value of frictional force at `A` & `B`.

Explanation

F.B.D of plank

FBD of cylinder

Since there is no slipping , `a=alphaR`
`:.a_(1)=a+alphaR=2a`
Applying Newton's Second Law for, plank
`sumF=Ma_(CM)impliesF-f_(1)=m(2a)`.........`(1)`
For cylinder
`sumF=Ma_(CM)impliesf_(1)-f_(2)=Ma`..........`(2)`
`sum tau=I_(CM)alphaimpliesf_(1)R+f_(2)R=(MR^(2))/(2)*(a)/(R )impliesf_(1)+f_(2)=(Ma)/(2)`..........`(3)`
Adding all the above three equation
`2F-2f_(1)=4ma`
`ul{:(f_(1)-f_(2)=Ma),(f_(1)+f_(2)=(Ma)/(2)):}`
`2F=a(4m+(3M)/(2))`
`:.a=(4F)/(8m+3M))`
From equation `(1)`,
`f_(1)=F-2ma`
`=F-(8Fm)/(8m+3M)=(3FM)/(8m+3M)`
Similarly from `(2)`
`f_(2)=(FM)/(3M+8m)`
(-ve sign shows that `f_(2)` will act opposite to the direction assumed. It will act righward on the cylinder). On the plank at point `A` friction will act leftward and on cylinder at point `B`, it will act rightward.