Torque — Question 10

Question 10

Changing the moment origin exposes a nonzero resultant

A planar force system has resultant $\vec F_{\mathrm{net}}=(4,6)\,\mathrm N$ and torque $\tau_{O,z}=10\,\mathrm{N\,m}$ about the origin. A new origin is at $\vec a=(1,2)\,\mathrm m$.

1. Find torque about the new origin.

2. Find the locus of origins about which the torque is zero.

Question 10 - Solution

Changing the moment origin exposes a nonzero resultant

(a) Derivation and calculation

Translate all position vectors by $-\vec a$: $$\vec\tau_A=\sum(\vec r_i-\vec a)\times\vec F_i =\vec\tau_O-\vec a\times\vec F_{\mathrm{net}}.$$ $$\boxed{\tau_{A,z}=10-[1(6)-2(4)]=12\ \mathrm{N\,m}}.$$

(b) Result and interpretation

For a general origin $(a_x,a_y)$, set the new moment to zero: $$10-(6a_x-4a_y)=0\Rightarrow\boxed{6a_x-4a_y=10}.$$ Its normal $(6,-4)$ is perpendicular to the resultant $(4,6)$ because $6(4)-4(6)=0$. The moment depends on origin when the resultant is nonzero.

Physical check

Shifting any origin along the resultant direction leaves its moment unchanged.