Question

A cubical block rests on a plane of $\mu =\sqrt{3}$. The angle through which the plane 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 ${60}^{\circ }$

$I-$ Method

The block will have force $mg\sin \theta$ along the incline plane and the frictional force will have $f_s=\mu N$
where $N=mg\cos \theta$
$f_s=mg\sin \theta$

For equilibrium

$\mu mg\cos \theta =mg\sin \theta$
$\tan \theta =\mu =\sqrt{3}$
$\theta ={60}^{\circ }$

$II-$ Method
This is the case of angle of repose. Angle of repose is defined as the minimum angle of the inclined plane and we have the formulae as $\tan \theta =\mu$ by putting the value of $\mu =\sqrt{3}$ we get $\theta ={60}^{\circ }$