Coefficient of friction from the sliding threshold

Problem

A block is on an inclined plane of adjustable angle $\theta$. Slowly increasing $\theta$, the block starts to slide when $\theta = 22^\circ$. What is the coefficient of static friction?

Solution

Free-body diagram. Three forces act on the block: the weight $\vec P$ (vertical, downward), the normal reaction $\vec N$ (perpendicular to the plane) and the static friction $\vec f_s$ (along the plane, upward, opposing the incipient sliding).

Choice of axes. Take $x$ along the plane (positive upward) and $y$ perpendicular to the plane. Resolve the weight: driving component $P_x = mg\sin\theta$, pressing component $P_y = mg\cos\theta$.

Threshold condition. As long as the block is still, static friction exactly balances the driving component. The block is on the verge of sliding when friction reaches its maximum value $f_{s,\max} = \mu_s N$. At that instant, equilibrium along the plane gives: $mg\sin\theta = \mu_s N$

Along the perpendicular (no motion in that direction): $N = mg\cos\theta$

Combination. Substituting $N$ into the first equation and dividing term by term, mass and $g$ cancel: $\mu_s = \frac{mg\sin\theta}{mg\cos\theta} = \tan\theta$

Numerical result. The threshold angle (or angle of friction) directly measures the coefficient: $\ev{\mu_s = \tan 22^\circ \approx 0{,}40}$

Note: the mass does not appear. Any block of the same material would slide at the same angle — it is the simplest way to measure $\mu_s$ in the lab.

Links

Topics: Dinamica · Attrito Concepts: Attrito statico · Forza normale · Forza peso Skills: Scomposizione sul piano inclinato Objects: Piano inclinato · Blocco