Question

A thin hoop of weight 500 N and radius 1 m rests on a rough inclined plane as shown in the figure. The minimum coefficient of friction needed for this configuration to be in equilibrium 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 $\frac{1}{2\sqrt{3}}$

$Given$ :- Weight of thin hoop = 500 $N$

Radius of hoop ( $R$ ) = 1 $m$

$ToFind$ :- Minimum coefficient of friction in equilibrium ( ${\mu }_{min}$ )

$Solution$ :- According to diagram , $N=MgCos{30}^{\circ }$

$\therefore$ $f=\mu mgCos{30}^{\circ }$ $(\because f=\mu N)$

Now , taking equilibrium along the plane , we get

$MgSin{30}^{\circ }$ - $\mu mgCos{30}^{\circ }$ = $\tau$ $-----$ (i)

We know , In rotational equilibrium about centre of mass

$\tau \times R$ = $f\times R$ $⟹$ $\tau$ = $f$ $⟹$ $\tau$ = $\mu mgCos{30}^{\circ }$

Applying in eq. (i) we get ,

$MgSin{30}^{\circ }$ = 2$\mu mgCos{30}^{\circ }$

$\therefore$ ${\mu }_{min}$ = $\frac{1}{4}$ × $\frac{2}{\sqrt{3}}$ $⟹$ $\underset{–}{\frac{1}{2\sqrt{3}}}$

Hence , $\underset{–}{OptionD\left(\frac{1}{2\sqrt{3}}\right)iscorrect.}$