Consider a cylinder of mass `M=1 kg` and radius `R=1 m`lying on a rough horizontal plane. It has a plank lying on its stop as shown in the figure.

A force `F=55N` is applied on the plank moves and causes the cylinder to roll. The plank always remains horizontal. there is no slipping at any point of contact.
The value of frictional force at `A` is

Options

A. `7.5N`
B. `5.0 N`
C. `2.5N`
D. `1.5N`

Answer

A

Explanation

Drawing the `F.B.D.` of the plank and the cylinder

Equations of motion are
`F cos theta-f_(1)=ma.....(1)`
`F sin theta+N_(1)=mg.....(2)`
`f_(1)+f_(2)=MA......(3)`
`f_(1)R-f_(2)R=Ialpha......(4)`
`A=Ralpha.....(5)`
`a=(4Fcos theta)/(3M+8m)=(4xx55xx1/2)/([(3xx1)+(8xx1)])=10 m//s^(2)`
`f_(1)=(3MFcos theta)/(3M+8m)=(3xx1+55xx1/2)/(3xx1+8xx1)=7.5 N`
and `f_(2)=(MF cos theta)/(3M+8m)=(1+55xx1/2)/(3xx1+8xx1)=2.5 N`