Angle of repose is defined as the angle of an inclined plane, at which a body placed on it just begins to slide. Place a body of mass $m$ on an inclined plane. Increase the angle of inclined plane till the body just begins to slide. If $\theta$ is the angle of inclination at which the body just begins to slide, then $\theta$ is the angle of repose.

The weight $mg$ of the body acts vertically downwards. Because of the angle of inclination, this weight can be resolved into two components; $mg\cos\theta$ and $mg\sin\theta$ as shown in fig. If the body just begins to slide, then 1. $mg\sin\theta$ is equal to the frictional force \[mg\sin\theta=F_c\text{ __(a)}\] 2. $mg\cos\theta$ is equal to the normal reaction \[mg\cos\theta=R\text{ __(b)}\]

Dividing $\text{(a)}$ by $\text{(b)}$, \[\frac{mg\sin\theta}{mg\cos\theta}=\frac{F_c}{R}\] \[\tan\theta=\frac{F_c}{R}\] \[\tan\theta=\mu\text{ ___(2)}\]

Thus, coefficient of limiting friction is equal to the tangent of the angle of repose.

Relation between Angle of Friction and Angle of Repose

From $\text{(1)}$ and $\text{(2)}$, \[\tan\alpha=\mu=\tan\theta\] \[\tan\alpha=\tan\theta\] \[\alpha=\theta\] Thus, angle of friction is equal to the angle of repose.