Angle of Repose

Consider a block placed on a plane inclined at angle $\theta$ to the horizontal. For small $\theta$, the block stays put; as $\theta$ is increased, at some critical angle the block just begins to slide down — this critical angle is called the angle of repose.

The weight $mg$ resolves into a component along the incline ($mg\sin\theta$, tending to slide the block down) and a component perpendicular to it ($mg\cos\theta$, balanced by the normal force $N=mg\cos\theta$). At the point of sliding, the static friction has reached its maximum, and this maximum exactly balances the down-slope component of gravity:

Equating:

$\mu_s=\tan\theta.$ …