Acceleration on Rough Inclined Plane

The situation

A body slides down a rough inclined plane of angle $\theta$ with coefficient of kinetic friction $\mu_k = \mu$. What is its acceleration?

Derivation

Forces on the body:

• Weight $mg$ vertically downward
• Normal reaction $N$ perpendicular to surface
• Kinetic friction $f_k$ up the slope (opposing downward sliding)

Perpendicular to incline:

$N = mg\cos\theta$

Kinetic friction:

$f_k = \mu_k N = \mu mg\cos\theta$

Along the incline (taking down the slope as positive):

$mg\sin\theta - f_k = ma$

$mg\sin\theta - \mu mg\cos\theta = ma$

$mg(\sin\theta - \mu\cos\theta) = ma$

$\boxed{a = g(\sin\theta - \mu\cos\theta)}$

Condition for sliding to occur

For the body to slide at all, $a > 0$:

$g(\sin\theta - \mu\cos\theta) > 0$

$\sin\theta > \mu\cos\theta$

$\tan\theta > \mu$

$\theta > \tan^{-1}(\mu) = \theta_r$

The angle must exceed the angle of repose. This is consistent with the definition of angle of repose.

Going up vs coming down

Sliding down: friction acts up the slope

$a_{down} = g(\sin\theta - \mu\cos\theta)$

Sliding up (body given initial velocity up the slope): friction acts down the slope (opposing upward motion)

$a_{up} = g(\sin\theta + \mu\cos\theta)$

The deceleration going up is greater than the acceleration going down — the body decelerates faster going up than it accelerates coming down.

Effect of $\mu$ on acceleration