Question

A body of $6 \,kg$ 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=6 g \cos 30^{\circ}$
$\mu R=6 g \sin 30^{\circ}$ [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 =6 g \cos 60^{\circ}, P+\mu S=6 g \sin 60^{\circ} $
$ \therefore \, P+\frac{1}{\sqrt{3}}\left(6 g \cos 60^{\circ}\right)=6 g \sin 60^{\circ} $
$ \Rightarrow \, P=6 g\left(\frac{\sqrt{3}}{2}-\frac{1}{2 \sqrt{3}}\right)=2 \sqrt{3} g$
Hence, $P=2 \sqrt{3}$ kg - wt