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Mechanics · Moments

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Moments of forces

The moment (turning effect) of a force as force × perpendicular distance from the pivot, measured clockwise or anticlockwise, in newton-metres — including forces acting at an angle to a rod.

Mechanics · Moments

Moments of forces

The moment (turning effect) of a force as force × perpendicular distance from the pivot, measured clockwise or anticlockwise, in newton-metres — including forces acting at an angle to a rod.

Why it works

A force doesn't just push a body along — applied away from a pivot, it makes it turn. The size of that turning effect is the moment of the force. It depends on two things: how big the force is, and how far its line of action is from the pivot. A small push far out (like a long spanner) turns more than a big push close in.

The moment of a force about a point is moment=F×d,\text{moment} = F \times d, where FF is the force and dd is the perpendicular distance from the pivot to the force's line of action. The unit is the newton-metre (N m). Each moment has a sense — clockwise or anticlockwise — and you must keep track of which, usually by calling one direction positive.pivotFSo a 1010 N force acting 33 m from a pivot has a moment of 10×3=3010 \times 3 = 30 N m. The catch is the word perpendicular: dd is the distance at right angles to the force, not just the distance to where it acts.

Forces at an angle. If a force acts at an angle θ\theta to a rod, only the part of it perpendicular to the rod does any turning. So for a force FF acting at the end of a rod of length LL, at angle θ\theta to the rod, the moment about the other end is Fsin⁡θ×L,F\sin\theta \times L, using the perpendicular component Fsin⁡θF\sin\theta (equivalently, FF times the perpendicular distance Lsin⁡θL\sin\theta — the same thing). A force acting along the rod, or straight through the pivot, has zero perpendicular distance and so no moment.