Question

A solid cylinder of mass $m$ is kept in balance on a fixed incline of angle $\alpha ={37}^{\circ }$ with the help of a thread fastened to its jacket. The cylinder does not slip. What will be the value of force $F$ that is required to keep the cylinder in balance when the thread is held vertically?

• A. $\frac{mg}{8}$
• B. $\frac{2mg}{7}$
• C. $\frac{3mg}{5}$
• D. $\frac{3mg}{8}$

Answer 1

The correct option is D $\frac{3mg}{8}$

For complete equilibrium of the solid cylinder,
$F_{\text{net}}=0,\&{\tau }_{\text{net}}=0$


Applying torque balance about point $C$:
${\tau }_{\text{net}}=0$
$\Rightarrow {\tau }_N+{\tau }_{mg}+{\tau }_f+{\tau }_F=0$
(Taking anticlockwise sense of rotation as $+ve$)
$\Rightarrow 0+0+fr-Fr=0$
$\Rightarrow Fr=fr$
$\therefore F=f...\left(i\right)$

Applying translational equilibrium condition,
Component of $F$ along the inclined plane is $F\sin \alpha$ and $\perp$ to inclined plane is $F\cos \alpha$
$N+F\cos \alpha =mg\cos \alpha ...\left(ii\right)$
$f+F\sin \alpha =mg\sin \alpha ...\left(iii\right)$

From Eq.$\left(i\right)\&\left(iii\right)$,
$F\sin \alpha +F=mg\sin \alpha [\because f=F]$
Putting $\alpha ={37}^{\circ }$
$\Rightarrow F\left(\frac{3}{5}\right)+F=mg\left(\frac{3}{5}\right)$
$\Rightarrow \frac{8F}{5}=\frac{3mg}{5}$
$\therefore F=\frac{3mg}{8}$