Torque Revisited — Question 7

Question 7

An equal and opposite pair can do rotational work

A rigid bar of length $2a$ lies at angle $\theta$ above $+x$. A constant upward force $F$ acts at its positive end and an equal downward force at its negative end. The midpoint is free to translate.

1. Find the net force and torque about the midpoint.

2. Find work by the pair as the bar turns from $0$ to $\pi/2$, and explain why center translation contributes none.

Question 7 - Solution

An equal and opposite pair can do rotational work

(a) Derivation and calculation

The applied pair has zero resultant: $$\vec F_++\vec F_-=F\hat y-F\hat y=\boxed{\vec0}.$$ At $\vec r_\pm=\pm a(\cos\theta\hat x+\sin\theta\hat y)$, $$\boxed{\tau_z=aF\cos\theta+aF\cos\theta=2aF\cos\theta}.$$

(b) Result and interpretation

The rotational work is $$\boxed{W=\int_0^{\pi/2}\tau_z\,d\theta =2aF[\sin\theta]_0^{\pi/2}=2aF}.$$ Center translation contributes no power: $$\mathcal P_{\mathrm{trans}}=(\vec F_++\vec F_-)\cdot\vec V_C=0.$$ Because the resultant vanishes, shifting the moment origin cannot change the couple torque.

Physical check

A zero net force does not imply zero rotational work when a nonzero couple acts.