Properties of Friction — Question 1

Question 1

The easiest pull is not horizontal

Take $g=10$ m/s$^2$. Use ideal dry friction, $|f_s|\le\mu_sN$ and $|f_k|=\mu_kN$. A 5 kg crate on a level floor is pulled with magnitude P at angle $\theta$ above horizontal, $0\le\theta<90^\circ$. The static coefficient is 0.5. Treat the crate as a particle and exclude tipping.

1. Derive the pull needed for impending rightward slip as a function of angle.

2. Find the angle and magnitude that minimize this pull.

3. Check that the optimum retains positive normal force.

Question 1 - Solution

The easiest pull is not horizontal

(a) Include vertical unloading.

Vertical equilibrium and impending rightward slip require $$N=mg-P\sin\theta,\qquad P\cos\theta=\mu_sN.$$ Eliminating $N$ gives $$\boxed{P(\theta)=\frac{\mu_smg}{\cos\theta+\mu_s\sin\theta}=\frac{25}{\cos\theta+.5\sin\theta}\ \mathrm N}.$$

(b) Maximize the denominator.

Maximize the denominator using $$\cos\theta+.5\sin\theta=\sqrt{1.25}\cos(\theta-\tan^{-1}.5)\leq\sqrt{1.25}.$$ Therefore $$\boxed{\theta_{\mathrm{opt}}=\tan^{-1}(.5)=26.565^\circ},\qquad\boxed{P_{\min}=\frac{25}{\sqrt{1.25}}=10\sqrt5=22.361\ \mathrm N}.$$

(c) Verify the constraint.

At this angle, $\sin\theta=1/\sqrt5$ and $\cos\theta=2/\sqrt5$: $$\boxed{N=50-(10\sqrt5)(1/\sqrt5)=40\ \mathrm N}>0.$$ The horizontal check is $P\cos\theta=20$ N $=\mu_sN$. Thus the optimized solution remains in contact with the floor.