Put together Inclined Planes and Static Friction, and a ramp becomes a friction meter. Tilt a board slowly with a block on it: at first static friction holds it, then at one particular angle it lets go.
Key idea
A block starts to slip when $\tan\theta = \mu_s$. Once sliding down, kinetic friction points up the slope: $a = g(\sin\theta - \mu_k\cos\theta)$. Sliding up, both gravity's part and friction point down the slope: the deceleration is $g(\sin\theta + \mu_k\cos\theta)$.
Why it works
Along the slope, gravity pulls with $mg\sin\theta$; static friction can resist with at most $\mu_s F_N = \mu_s mg\cos\theta$. The block holds while $mg\sin\theta \le \mu_s mg\cos\theta$, that is, while $\tan\theta \le \mu_s$. The mass cancels, so a heavy block and a light one of the same material slip at the same angle. Friction always opposes sliding (or the sliding that would happen), so its direction flips when the motion does.
Worked examples
Worked example: The slipping angle
A block sits on a board with $\mu_s = 0.6$. At what angle does it slip?
- At the limit: $mg\sin\theta = \mu_s mg\cos\theta$, so $\tan\theta = 0.6$.
- $\theta = \tan^{-1}(0.6) \approx 31.0^\circ$.
Answer: about $31.0^\circ$
Worked example: Sliding down
A block slides down a $35^\circ$ ramp with $\mu_k = 0.2$. Find its acceleration.
- Friction points up the slope, against the sliding.
- $$a = 9.8(\sin 35^\circ - 0.2\cos 35^\circ) = 9.8(0.574 - 0.164) \approx 4.02\ \text{m/s}^2$$ down the slope.
Answer: about $4.02\ \text{m/s}^2$
Worked example: Shoved up the slope
The same block is shoved up the same ramp. While it slides up, what is its deceleration?
- Now gravity's part and friction both point down the slope.
- $$a = 9.8(0.574 + 0.164) \approx 7.23\ \text{m/s}^2,$$ so it stops quickly, then slides back down at $4.02\ \text{m/s}^2$.
Answer: about $7.23\ \text{m/s}^2$
Common mistake
Pointing friction down the slope for a block that is sliding down. Friction resists the sliding, so it points up the slope. Check: for a block sliding down, its acceleration must be smaller than the frictionless $g\sin\theta$; if your answer is bigger, friction has the wrong sign.
Try it in the Science Lab: Forces at Work.