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]
Answer: (C) $\frac{mg}{5}$
First check whether the cylinder can stay at rest. With tension $T$ (along the incline, at distance $R$ from the centre) and friction $f$ at the contact point, torque balance gives $T = f$, and force balance along the incline gives $T + f = mg\sin60^\circ$, so the friction needed is
$$f = \frac{mg\sin60^\circ}{2} = \frac{\sqrt3}{4}mg \approx 0.43\,mg$$
The maximum static friction is $\mu_sN = 0.4\times mg\cos60^\circ = 0.2\,mg$, which is smaller. So the cylinder cannot stay at rest and slides at the contact point. The friction then takes its limiting value:
$$f = \mu N = 0.4\times\frac{mg}{2} = \frac{mg}{5}$$
Solution by Sreeraj P, M.Sc Physics