Question

A body of $6kg$ rests in limiting equilibrium on an inclined plane whose slope is $30^{\circ }$. If the plane is raised to slope of $60^{\circ }$ , then force (in kg -wt) )along the plane required to support it is

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

Answer 1

Answer: (B)

Let $P$ be the force required to support the body and $\mu$ be the coefficient of friction.
Case I
When plane make inclination of $30^{\circ }$

In this case, $R=6g\cos 30^{\circ }$
$\mu R=6g\sin 30^{\circ }$ [limiting equilibrium]
$\therefore \mu =\tan 30^{\circ }=\frac{1}{\sqrt{3}}$
Case II
When plane raised to the slope of $60^{\circ }$

In this case,
$S=6g\cos 60^{\circ },P+\mu S=6g\sin 60^{\circ }$
$\therefore P+\frac{1}{\sqrt{3}}\left(6g\cos 60^{\circ }\right)=6g\sin 60^{\circ }$
$\Rightarrow P=6g\left(\frac{\sqrt{3}}{2}-\frac{1}{2\sqrt{3}}\right)=2\sqrt{3}g$
Hence, $P=2\sqrt{3}$ kg - wt