Q. 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
Laws of Motion
Solution:
$f_{s}=m g \sin \theta$
$\mu m g \cos \theta=m g \sin \theta$
$\tan \theta=\mu=\sqrt{3}$
$\theta=60^{\circ}$
