Question

A thin hoop of weight $500N$ and radius $1m$ rests on a rough inclined plane as shown in the figure. The minimum coefficient of friction needed for this configuration is

• A. $\frac{1}{3\sqrt{3}}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{1}{2}$
• D. $\frac{1}{2\sqrt{3}}$

Answer 1

The correct option is D $\frac{1}{2\sqrt{3}}$

The minimum co-efficient of friction corresponds to friction acting. The forces are

Balancing forces along the incline,
$mg\sin \theta -T-f_l=0$ .....$\left(i\right)$
Balancing torque about 0
$TR-f_{l}R=0\Rightarrow T=f_l....\left(ii\right)$

$\left(ii\right)$ in $\left(i\right)$ gives,

$f_l=\frac{mg\sin {30}^{\circ }}{2}$
But, $f_l=\mu mg\cos {30}^{\circ }$

Now,

$\frac{\sqrt{3}\mu mg}{2}=\frac{mg}{4}\mu =\frac{1}{2\sqrt{3}}$