Question

A body of $6kg$ rests in limiting equilibrium on an inclined plane whose slope is ${30}^o$. If the plane is raised to slope of ${60}^o$, 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

The correct option is B $2\sqrt{3}$

Let $P$ be the force required to support the body and $\mu$ be the coefficient of friction.
Case $I$
When plane make inclination of ${30}^o$
In this case, $R=6g\cos {30}^o$
$\mu R=6g\sin {30}^o$ [limiting equilibrium]
$\therefore \mu =\tan {30}^o=\frac{1}{\sqrt{3}}$
Case $II$
When plane raised to the slope of ${60}^o$
In this case,
$S=6g\cos {60}^o,P+\mu S=6g\sin {60}^o$
$\therefore P+\frac{1}{\sqrt{3}}\left(6g\cos {60}^o\right)=6g\sin {60}^o$
$\Rightarrow P=6g(\frac{\sqrt{3}}{2}-\frac{1}{2\sqrt{3}})=2\sqrt{3}g$
Hence, $P=2\sqrt{3}kg-wt$