Question

A cubical block rests on a plane of $\mu =\sqrt{3}$. The angle through which the plane should be inclined to the horizontal so that the block just slides down will be

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$

The given problem can be depicted in the pictorial view as shown.


Since, the block just slides down, it means the force acting down the plane will be nearly equal to the limiting static friction force acting between the block and the inclined surface.
i.e $mg\sin \theta =f_s=\mu N$.........(i)
and $N=mg\cos \theta$

So, from eq (i), we have,
$mg\sin \theta =\mu mg\cos \theta$
$\Rightarrow \tan \theta =\mu$
$\Rightarrow \theta ={\tan }^{-1}\mu ={\tan }^{-1}\sqrt{3}={60}^{\circ }$