Equilibrium — Question 2

Question 2

Translational equilibrium permits uniform motion

A nonrotating crate slides right at $4\,\mathrm{m/s}$ on a level surface. A $15\,\mathrm N$ pull exactly balances $15\,\mathrm N$ kinetic friction, and the other forces balance vertically. All moments about its mass center cancel.

1. Classify the motion as static equilibrium, uniform motion with balanced forces, or accelerated motion.

2. Over a $2\,\mathrm m$ displacement, find the work of the pull, friction and their sum.

Question 2 - Solution

Translational equilibrium permits uniform motion

(a) Derivation and calculation

The given forces and moments satisfy $$\sum F_x=15-15=0,\quad\sum F_y=0,\quad\sum\tau_C=0.$$ Thus $$a_C=0,\qquad v_C(t)=4\ \mathrm{m/s},\qquad\omega(t)=0.$$ This is uniform translation with force and torque balance, but $$\boxed{\text{not static equilibrium in the ground frame}}.$$

(b) Result and interpretation

For rightward displacement $s=2$ m, $$\boxed{W_{\mathrm{pull}}=Fs=15(2)=30\ \mathrm J},$$ $$\boxed{W_f=fs\cos\pi=-30\ \mathrm J},\qquad \boxed{W_{\mathrm{net}}=0=\Delta K}.$$ The pull supplies energy while friction dissipates it; constant kinetic energy does not imply zero work by each force.

Physical check

The displacement takes $s/v=.50$ s, with no change in speed.