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 :

[The coefficient of static friction, ${\mu }_s$, is 0.4]

• A. $72mg$
• B. $0$
• C. $mg/5$
• D. $5mg$

Answer 1

The correct option is C

$mg/5$

Step 1: Given data

Mass of the cylinder $=m$

The coefficient of static friction, ${\mu }_s=0.4$

Step 2: To find the frictional force acting between the cylinder and the inclined plane

Consider the solid cylinder is in equilibrium

$$\begin{gathered} T+f=mg\sin {60}^0 \\ TR-fR=0 \end{gathered}$$

Let solving the above equation we get

$$\begin{gathered} T=f_{required} \\ =\frac{mg\sin \theta }{2} \end{gathered}$$

limiting friction < required friction

$\mu mg\cos {60}^0<\frac{mg\sin {60}^0}{2}$

The cylinder will not always equilibrium, Therefore

$$\begin{gathered} f=kinetic \\ f={\mu }_{k}N \\ f={\mu }_{k}mg\cos {60}^0 \\ f=\frac{mg}{5} \end{gathered}$$

Hence the frictional force $f=\frac{mg}{5}$

Therefore the correct Option is C.