Friction — Question 4

Question 4

A horizontal load can reverse the direction of friction on a slope

Take $g=10$ m/s$^2$. A 2 kg block rests on a fixed $30^\circ$ incline rising to the right. A horizontal force P pushes rightward. Static friction is sufficient for the three cases below. Treat the block as a particle.

1. Express the required signed friction along the incline, taking uphill as positive.

2. Find its magnitude and direction for $P=0$, 8 and 16 N.

3. Find the P for which friction vanishes and the normal force there.

Question 4 - Solution

A horizontal load can reverse the direction of friction on a slope

(a) Balance tangential components.

Take uphill positive. Tangential equilibrium gives $$f+P\cos30^\circ-mg\sin30^\circ=0.$$ $$\boxed{f=mg\sin30^\circ-P\cos30^\circ=10-\frac{\sqrt3}{2}P}\quad\text{(SI units)}.$$

(b) Let the sign determine the direction.

Evaluate the signed force for each prescribed push: $$f(0)=+10\ \mathrm N,\qquad f(8)=10-4\sqrt3=+3.072\ \mathrm N,$$ $$f(16)=10-8\sqrt3=-3.856\ \mathrm N.$$ The first two forces point uphill; the last has magnitude 3.856 N downhill.

(c) Check both components.

Set the tangential demand to zero: $$\boxed{P_0=\frac{10}{\cos30^\circ}=\frac{20}{\sqrt3}=11.547\ \mathrm N}.$$ The horizontal push also increases the normal load: $$\boxed{N=mg\cos30^\circ+P_0\sin30^\circ=\frac{40}{\sqrt3}=23.094\ \mathrm N}.$$