Angle of Repose

What the angle of repose is

Place a body on an inclined surface and gradually increase the angle of inclination. At some critical angle, the body just begins to slide. This critical angle is the angle of repose $\theta_r$:

$\tan\theta_r = \mu_s$

Below $\theta_r$: body stays at rest. At or above $\theta_r$: body slides.

Derivation

Consider a body of mass $m$ on an inclined plane at angle $\theta$ to the horizontal.

Forces acting:

• Weight $mg$ downward
• Normal reaction $N$ perpendicular to surface
• Static friction $f_s$ up the slope (opposing tendency to slide down)

Resolving forces:

Perpendicular to surface:

$N = mg\cos\theta$

Along the surface (down the slope positive):

$mg\sin\theta - f_s = 0 \quad \text{(body in equilibrium)}$

$f_s = mg\sin\theta$

For the body to remain stationary, static friction must not exceed its maximum:

$f_s \leq \mu_s N$

$mg\sin\theta \leq \mu_s \cdot mg\cos\theta$

$\tan\theta \leq \mu_s$

The maximum angle before sliding begins — the angle of repose:

$\tan\theta_r = \mu_s$

$\boxed{\theta_r = \tan^{-1}(\mu_s)}$

What happens at exactly $\theta_r$

At $\theta = \theta_r$, the body is on the verge of sliding. Static friction has reached its maximum value $\mu_s N$.

If $\theta$ increases even slightly above $\theta_r$, the body slides. Once sliding begins, kinetic friction $\mu_k N < \mu_s N$ takes over, and since $\mu_k < \mu_s$, the friction force drops — the body accelerates down.

Practical significance

The angle of repose is why:

• Sand dunes have a characteristic slope (angle of repose of sand ≈ 30–35°)
• Gravel piles form a cone with a specific angle
• Engineers design slopes and embankments not to exceed $\theta_r$ for the material

Relation to the angle of friction

Both the angle of repose and the angle of friction equal $\tan^{-1}(\mu_s)$. They are geometrically different descriptions of the same physical quantity $\mu_s$.

• Angle of friction: angle the resultant contact force makes with the normal
• Angle of repose: maximum slope angle before sliding

Both give $\tan^{-1}(\mu_s)$.