Question

Two forces of magnitude $F$ are acting on a uniform disc kept on a horizontal rough surface as shown in the figure. Friction force by the horizontal surface on the disc is $xF$. Find the value of $x$.

Answer 1

The Torque equation at the contact point with ground point,
$FR+\left(F\right)2R=3FR=I\alpha$-------(1) where $I=\frac{1}{2}m{R}^2+m{R}^2=\frac{3}{2}m{R}^2$ is the MI about the contact point with the ground i.e about an axis $\perp$ to the plane of the disc at the boundary.
$\Rightarrow \alpha =\frac{3FR}{\frac{3}{2}M{R}^2}$
let the frictional force direction be in the direction of forces.
Now apply torque equation at centre of disc
$\frac{1}{2}M{R}^2\alpha =FR-nFR=(1-n)FR$
$\Rightarrow \alpha =\frac{(1-n)FR}{\frac{1}{2}M{R}^2}$---------(2)
by equating 1 and 2.we get ,
$n=0$.