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 $\left.0.4\right]$

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

Answer 1

Answer: (C)

Let's take solid cylinder is in equilibrium
$T+f=m g \sin 60\,\,\,\,\,\dots(i)$
$TR - fR =0\,\,\,\,\,\dots(ii)$
Solving we get
$T = f _{\text {seq }}=\frac{ mg \sin \theta}{2}$
But limiting friction $<$ required friction
$\mu mgcos 60^{\circ}<\,\frac{ mg \sin 60^{\circ}}{2}$
$\therefore $ Hence cylinder will not remain in equilibrium
Hence $f =$ kinetic
$=\mu_{ k } N$
$=\mu_{ k } mg cos 60^{\circ}$
$=\frac{ mg }{5}$