Slipping threshold

Problem

A box rests on an inclined plane at $\theta = 20^\circ$. What is the minimum static friction coefficient so that the box does not slip?

Solution

We decompose the weight along the directions parallel and perpendicular to the plane. The component that tends to make the box slide down is $mg\sin\theta$; the component that presses it against the plane determines the normal force: $N = mg\cos\theta$ For the box to remain still, static friction must balance the parallel component of the weight. But static friction cannot exceed its maximum value $\mu_s N$: $mg\sin\theta \le \mu_s\, N = \mu_s\, mg\cos\theta$ Simplifying $mg$ and dividing by $\cos\theta$: $\mu_s \ge \frac{\sin\theta}{\cos\theta} = \tan\theta$ Substituting $\theta = 20^\circ$: $\mu_s \ge \tan 20^\circ \approx 0{,}364$ The minimum value is thus $\mu_s = \tan\theta$: note that it is independent of the mass of the box. $\ev{\mu_s \ge \tan 20^\circ \approx 0{,}364}$

Linked items

Topics: Dinamica · Attrito Concepts: Attrito statico Skills: Scomposizione sul piano inclinato · Applicazione delle leggi di Newton Objects: Piano inclinato · Blocco