Moments of forces at an angle

By the end of this lesson

By the end you can find the moment of a force acting at any angle, using either the perpendicular distance or the perpendicular component.

Key to the symbols

The main symbols used in this lesson:

Symbol	Meaning
$\alpha$	an angle

The rope trick

Tie a rope to a door handle and pull. Pull at right angles to the door and it swings open.

Pull almost along the door, towards the hinge, and nothing happens however hard you heave. Same force: what changed is the perpendicular distance from the hinge to the force's line of action.

Easy cases first

A $20\ \text{N}$ force acts on the end of a $0.50\ \text{m}$ lever:

Angle between force and lever	Perpendicular distance	Moment
90°	$0.50\ \text{m}$	$10\ \text{N m}$
30°	$0.50\sin 30° = 0.25\ \text{m}$	$5.0\ \text{N m}$
0° (along the lever)	$0$	$0$

At 30° you get exactly half the right-angle moment, because $\sin 30° = 0.5$ .

Two methods, same answer

For a force $F$ at angle $\theta$ to a lever of length $L$ :

Perpendicular distance. Extend the line of action and drop a perpendicular onto it from the pivot. Its length is $d = L\sin\theta$ , so the moment is $F \times L\sin\theta$ .
Perpendicular component. Resolve $F$ into $F\sin\theta$ (at right angles to the lever) and $F\cos\theta$ (along it). $F\cos\theta$ points straight through the pivot, so it has no moment. The moment is $F\sin\theta \times L$ .

Pick whichever picture you find easier.

Worked example: A spanner at an angle

A force of $60\ \text{N}$ acts on the end of a $0.30\ \text{m}$ spanner at $50°$ to the spanner. Find its moment about the nut.

Have a go first?

Quick check

Multiple choiceQuestion 1 · Recall

CalculateQuestion 2 · Apply

Answer the 2 questions above to continue.