Properties of Friction — Question 6

Question 6

A hanging mass has an entire equilibrium interval

Take $g=10$ m/s$^2$. Use ideal dry friction, $|f_s|\le\mu_sN$ and $|f_k|=\mu_kN$. A 4 kg block on a $30^\circ$ incline is connected by a light taut cord over an ideal fixed pulley to hanging mass M. The incline has $\mu_s=0.25$. All bodies start at rest.

1. Find normal force and maximum static friction on the incline block.

2. Find all M for which equilibrium is possible.

3. For $M=2$ kg, determine tension and friction; identify the slip tendency beyond each interval endpoint.

Question 6 - Solution

A hanging mass has an entire equilibrium interval

(a) Separate normal and tangential forces.

Only the weight has a normal component: $$\boxed{N=mg\cos30^\circ=20\sqrt3\ \mathrm N},\qquad\boxed{f_{s,\max}=.25N=5\sqrt3\ \mathrm N}.$$

(b) Bound the force needed for equilibrium.

Hanging equilibrium gives $T=Mg=10M$ in SI units. With uphill positive on the incline, $$T+f-mg\sin30^\circ=0\Longrightarrow f=20-10M.$$ $$|20-10M|\leq5\sqrt3\Longrightarrow \boxed{2-\frac{\sqrt3}{2}\leq M\leq2+\frac{\sqrt3}{2}}\ \mathrm{kg}.$$ The numerical interval is $[1.134,2.866]$ kg.

(c) Interpret the interval.

For the specified hanging mass, $$\boxed{T=2(10)=20\ \mathrm N},\qquad\boxed{f=20-20=0}.$$ Below the allowed interval, insufficient tension leaves a downhill tendency for the incline block. Above it, the hanging mass tends downward and pulls the incline block uphill.