Question

A solid cylinder of mass $m$ is wrapped with an inextensible light string and, is placed on a rough inclined plane as shown in the figure. The frictional force acting between the cylinder and the inclined plane is :

$($Coefficient of static friction, ${\mu }_s=0.4)$

• A. $\frac{7}{2}mg$
  
• B. $0$
• C. $\frac{1}{5}mg$
  
• D. $5mg$

Answer 1

The correct option is C $\frac{1}{5}mg$


From linear equilibrium,
$T+f=mg\sin {60}^{\circ }...\left(1\right)$

From angular equilibrium,
$TR=fR\Rightarrow T=f...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,

$f=\frac{mg\sin {60}^{\circ }}{2}=\frac{\sqrt{3}}{4}mg=0.43mg\left(\text{Required}\right)$

Limiting friction,
$f={\mu }_{s}N={\mu }_{s}mg\cos {60}^{\circ }=0.4\times mg\times \frac{1}{2}=0.2mg$

Since, limiting friction $<$ required friction, therefore, cylinder will not be in equilibrium.

$\therefore f$ will be kinetic.

As coefficient of kinetic friction is not mentioned, we have to assume that, coefficient of kinetic friction is equal to coefficient of static friction.

Therefore, friction force,
$f={\mu }_{k}N={\mu }_{s}mg\cos {60}^{\circ }\Rightarrow f=0.4\times mg\times \frac{1}{2}=\frac{1}{5}mg$