Prove that angle of friction is equal to angle of repose.

Angle of friction (\(\phi\)): the angle between the resultant of the limiting friction F and the normal reaction N, and the normal reaction: \(\tan\phi = \dfrac{F}{N} = \mu\).

Angle of repose (\(\alpha\)): the maximum inclination of a plane at which a body placed on it just remains at rest (is on the point of sliding) under its own weight.

            N
            ^   /
            |  / F (up the plane)
          [block]
         /  |
        /   W (vertical)
       /____alpha______

Proof: consider a block of weight W on a plane inclined at \(\alpha\), in limiting equilibrium (about to slide down). Forces: W vertically down, N normal to the plane, limiting friction \(F = \mu N\) up the plane.

Perpendicular to plane: \(N = W\cos\alpha\). Along the plane: \(F = W\sin\alpha\).

Dividing: \(\dfrac{F}{N} = \tan\alpha\). But in limiting condition \(\dfrac{F}{N} = \mu = \tan\phi\).

\(\therefore \tan\alpha = \tan\phi \Rightarrow\) \(\alpha = \phi\): angle of repose = angle of friction.