Torque — Question 6

Question 6

Only the torque component along a fixed axis drives that axis

A force $\vec F=(0,5,-1)\,\mathrm N$ acts at $\vec r=(1,2,3)\,\mathrm m$ on a body mounted on a fixed $z$ shaft.

1. Find the full torque vector and its $z$ component.

2. Explain why using the torque magnitude in a scalar equation for rotation about $z$ would be wrong.

Question 6 - Solution

Only the torque component along a fixed axis drives that axis

(a) Derivation and calculation

Expand the cross product component by component: $$\vec\tau=\bigl(yF_z-zF_y,\ zF_x-xF_z,\ xF_y-yF_x\bigr).$$ $$\boxed{\vec\tau=(2(-1)-3(5),\ 3(0)-1(-1),\ 1(5)-2(0)) =(-17,1,5)\ \mathrm{N\,m}}.$$

(b) Result and interpretation

The fixed-$z$ rotational equation uses the projection: $$\boxed{\tau_z=\vec\tau\cdot\hat z=5\ \mathrm{N\,m}},\qquad |\vec\tau|=\sqrt{(-17)^2+1^2+5^2}=\sqrt{315}\ \mathrm{N\,m}.$$ Transverse components are not moments driving the permitted $z$ rotation. Shaft supports constrain the other rotations and can supply reaction moments.

Physical check

Projection retains the sign and axis information that a nonnegative magnitude would discard.