Torque Revisited — Question 8

Question 8

Some force systems have torque that no origin shift removes

A force system has resultant $\vec F=5\hat z\,\mathrm N$ and moment $\vec\tau_O=(2,3,4)\,\mathrm{N\,m}$. For an origin shifted by $\vec a$, use $\vec\tau_A=\vec\tau_O-\vec a\times\vec F$.

1. Can any origin make the total torque zero?

2. Find an origin shift that removes the transverse torque components.

Question 8 - Solution

Some force systems have torque that no origin shift removes

(a) Derivation and calculation

For any origin shift, $$\vec\tau_A=\vec\tau_O-\vec a\times\vec F.$$ The subtracted term is perpendicular to $\vec F=5\hat z$, so $$\boxed{\tau_{A,z}=\tau_{O,z}=4\ \mathrm{N\,m}}.$$ No origin makes the full torque zero. Equivalently, the invariant $\vec\tau_A\cdot\vec F=20\ \mathrm{N^2\,m}\ne0$ proves impossibility.

(b) Result and interpretation

Write $\vec a=(a_x,a_y,a_z)$. Then $$\vec a\times\vec F=(5a_y,-5a_x,0).$$ Choose $5a_y=2$ and $-5a_x=3$: $$\boxed{a_x=-.600\ \mathrm m,\qquad a_y=.400\ \mathrm m,\qquad a_z\ \text{arbitrary}}.$$ The residual torque is $\boxed{\vec\tau_A=4\hat z\ \mathrm{N\,m}}$.

Physical check

Shifts along the resultant force leave the torque unchanged, explaining the arbitrary $a_z$.