Properties of Friction — Question 5

Question 5

Static friction has one vector budget

Take $g=10$ m/s$^2$. Use ideal dry friction, $|f_s|\le\mu_sN$ and $|f_k|=\mu_kN$. A 2 kg puck is at rest on a level rough table with $\mu_s=0.5$. Two horizontal forces act: 6 N east and P north, where $P\ge0$. No other horizontal force acts.

1. Find the required friction vector.

2. Find the largest P for equilibrium.

3. Why is checking each component against 10 N separately insufficient? Test $P=9$ N.

Question 5 - Solution

Static friction has one vector budget

(a) Cancel the applied vector.

Choose east and north as positive coordinate directions. Horizontal equilibrium gives $$\vec f_s+(6\hat{\mathbf i}+P\hat{\mathbf j})=0 \Longrightarrow\boxed{\vec f_s=-6\hat{\mathbf i}-P\hat{\mathbf j}}\ \mathrm N.$$

(b) Apply a magnitude bound.

The available friction is a vector-magnitude bound: $$N=mg=20\ \mathrm N,\qquad\sqrt{6^2+P^2}\leq\mu_sN=10\ \mathrm N.$$ $$P^2\leq100-36=64\Longrightarrow\boxed{P_{\max}=8\ \mathrm N}.$$

(c) Use the circular feasible region.

At $P=9$ N, each component separately is smaller than 10 N, yet $$|\vec f_s|=\sqrt{6^2+9^2}=\sqrt{117}=10.817\ \mathrm N>10\ \mathrm N.$$ The feasible region is the disk $f_x^2+f_y^2\leq100$, not the larger square $|f_x|,|f_y|\leq10$. Equilibrium fails.