Question 117 of 280

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$\dfrac{7}{2}mg$B$5mg$C$\dfrac{mg}{5}$D$0$

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Answer: (C) $\dfrac{mg}{5}$

Let's take solid cylinder is in equilibrium
$T+f=mg\sin 60$
$TR-fR=0$
Solving we get
$T={f}_{req}=\dfrac{mg\sin \theta }{2}$
But limiting friction $\lt$ required friction
$\mu mg\cos 60^{\circ}\lt \dfrac{mg\sin 60^{\circ}}{2}$
$∴$ Hence cylinder will not remain in equilibrium
Hence $f=$ kinetic
$={\mu }_{k}N$
$={\mu }_{k}mg\cos 60^{\circ}$
$=\dfrac{mg}{5}$