Prove that angle of friction is equal to angle of repose. (asked in MU Dec 2025)

Answer: On an inclined plane at the angle of repose \(\alpha\), \(\tan\alpha=F/N=\mu=\tan\phi\), so \(\alpha=\phi\).

Definitions. Angle of friction \(\phi\): the angle between the normal reaction N and the resultant of N and the limiting friction F, so \(\tan\phi=\dfrac{F}{N}=\mu\). Angle of repose \(\alpha\): the maximum angle of an inclined plane at which a body placed on it just remains at rest (is on the point of sliding) under its own weight.

Proof. A block of weight W rests on a rough plane inclined at \(\alpha\), on the point of sliding down. Forces: W (vertical), N (normal to the plane), limiting friction F (up the plane).

Resolving parallel to the plane: \(F=W\sin\alpha\) ...(1)

Resolving perpendicular to the plane: \(N=W\cos\alpha\) ...(2)

Dividing (1) by (2): \(\dfrac{F}{N}=\tan\alpha\).

But for limiting friction \(\dfrac{F}{N}=\mu=\tan\phi\). Therefore \(\tan\alpha=\tan\phi\), i.e.

\[\alpha=\phi\]

The angle of repose equals the angle of friction.