Torque — Question 1

Question 1

Torque uses the perpendicular component of a force

A force $\vec F=(5,-2)\,\mathrm N$ acts at $\vec r=(3,4)\,\mathrm m$ relative to a pivot at the origin.

1. Find torque about the pivot and its rotational sense in the $xy$ plane.

2. Find the perpendicular distance from the pivot to the force's line of action.

Question 1 - Solution

Torque uses the perpendicular component of a force

(a) Derivation and calculation

For an in-plane force, the axial moment is the signed determinant: $$\tau_z=(\vec r\times\vec F)_z=xF_y-yF_x.$$ $$\boxed{\tau_z=3(-2)-4(5)=-26\ \mathrm{N\,m}}.$$ The negative $z$ direction corresponds to clockwise turning viewed from $+z$.

(b) Result and interpretation

The line-of-action distance satisfies $|\tau|=F d_\perp$: $$F=\sqrt{5^2+(-2)^2}=\sqrt{29}\ \mathrm N,\qquad \boxed{d_\perp=\frac{26}{\sqrt{29}}=4.828\ \mathrm m}.$$ The full radius is $|\vec r|=5$ m; only its component perpendicular to the force enters the moment.

Physical check

The lever arm is smaller than the radius and satisfies $F d_\perp=26$ N m.