Properties of Friction — Question 2

Question 2

A wall clamp has a lower and an upper useful load

Take $g=10$ m/s$^2$. Use ideal dry friction, $|f_s|\le\mu_sN$ and $|f_k|=\mu_kN$. A 2 kg block is pressed against a vertical wall by a force P directed $30^\circ$ above horizontal toward the wall. The static coefficient is 0.5. Treat the block as a particle.

1. Find N and the signed upward friction required for rest.

2. Find the full interval of nonnegative P allowing rest.

3. Explain the two endpoint slip tendencies and locate the zero-friction load.

Question 2 - Solution

A wall clamp has a lower and an upper useful load

(a) Resolve relative to the wall.

Horizontal equilibrium fixes the wall normal; vertical equilibrium fixes friction: $$\boxed{N=P\cos30^\circ=\frac{\sqrt3}{2}P},\qquad\boxed{f=mg-P\sin30^\circ=20-\frac P2}.$$ Here $f>0$ means upward friction, with all force values in newtons.

(b) Use both sides of the inequality.

Rest requires both signs of the static bound: $$\left|20-\frac P2\right|\leq\mu_sN=\frac{\sqrt3}{4}P.$$ $$20\leq\left(\tfrac12+\tfrac{\sqrt3}{4}\right)P,\qquad20\geq\left(\tfrac12-\tfrac{\sqrt3}{4}\right)P.$$ $$\boxed{\frac{80}{2+\sqrt3}\leq P\leq\frac{80}{2-\sqrt3}}\ \mathrm N \quad\Longrightarrow\quad\boxed{21.436\leq P\leq298.564}\ \mathrm N.$$

(c) Interpret the signed demand.

At the lower endpoint, $f=+\mu_sN$ prevents downward slip. At the upper endpoint, $f=-\mu_sN$ prevents upward slip. Between them, $$f=0\Longrightarrow20-P/2=0\Longrightarrow\boxed{P=40\ \mathrm N}.$$ Large $P$ can therefore destroy equilibrium by demanding too much downward friction.