Moments, Couples and Equilibrium

Until now we have treated objects as point masses — no shape, no rotation. Real objects have extent, and forces applied to them can cause rotation as well as translation. To handle this we introduce the concept of a moment (the rotational effect of a force) and the conditions for full mechanical equilibrium. This lesson covers OCR Module 3.2.3 (Equilibrium) and gives you the toolkit for beams, bridges, seesaws, cranes, and the balance practical (PAG 4).

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Moment of a Force

The moment of a force about a point is a measure of its tendency to cause rotation about that point. For a force F acting at perpendicular distance x from the pivot:

Moment = F × x

• SI unit: newton metre (N m). This is not the same as a joule, despite having the same base units — moment is a vector-like quantity describing rotation, while a joule is scalar energy.
• The perpendicular distance is measured from the pivot to the line of action of the force.
• A moment has a sense of rotation — clockwise or anticlockwise — which must be stated.

Exam Tip: OCR questions sometimes give the angle between the force and the lever arm rather than the perpendicular distance. In that case, use: Moment = F × d × sin θ where d is the length of the lever arm and θ is the angle between F and d.

Worked Example — Spanner

A spanner of length 20 cm is used to turn a nut. The force applied is 60 N, perpendicular to the spanner. Find the moment.

Moment = 60 × 0.20 = 12 N m (clockwise or anticlockwise depending on the direction of pushing)

Worked Example — Spanner at an Angle

The same spanner, but the 60 N force is applied at 70° to the spanner. Find the moment.

Moment = F × d × sin θ = 60 × 0.20 × sin 70° = 60 × 0.20 × 0.940 = 11.3 N m

Notice how a non-perpendicular force gives a smaller moment, and the moment is zero if the force is along the spanner (sin 0° = 0). This is why you instinctively pull perpendicular to a lever for maximum turning effect.

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Principle of Moments

A rigid body is in rotational equilibrium if, and only if:

Sum of clockwise moments about any point = sum of anticlockwise moments about the same point

This is the principle of moments. It is equivalent to saying the net torque about any chosen pivot is zero.

Worked Example — Seesaw

A 30 kg child sits 1.5 m from the pivot of a uniform seesaw. Where must a 45 kg child sit to balance it?

Assume the pivot is at the centre of a uniform seesaw (so the seesaw's own weight contributes zero moment about the pivot).

Clockwise moment = 45 × 9.81 × x Anticlockwise moment = 30 × 9.81 × 1.5

Setting them equal:

45 × 9.81 × x = 30 × 9.81 × 1.5 45 x = 45 x = 1.0 m from the pivot

Notice the weights cancel — only mass ratios matter for balance on a pivot. This is why children of different sizes on a seesaw naturally settle at different distances.

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Full Conditions for Equilibrium

A rigid body is in complete equilibrium when both:

1. Translational equilibrium: ΣF = 0 (resultant force is zero — body does not accelerate in any direction)
2. Rotational equilibrium: ΣM = 0 (resultant moment is zero — body does not angularly accelerate)

Both conditions must be checked. An object can have zero net force but still rotate (a couple — see below), or it can have zero net moment but still accelerate translationally.

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Couples and Torque

A couple consists of two equal and opposite forces whose lines of action do not coincide, separated by a perpendicular distance d. The two forces produce no net translation (they cancel) but do produce a rotation. The turning effect of a couple is called its torque:

Torque = F × d

where d is the perpendicular distance between the lines of action of the two forces.

Familiar Examples

• Turning a car steering wheel — you push with one hand and pull with the other, producing a couple.
• A screwdriver turning a screw.
• A spinning gymnast in the air.
• Tightening a bolt with a torque wrench.

Note that "torque" is often used synonymously with "moment of a couple". Both are measured in newton metres.

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Worked Example — Beam Problem

A uniform horizontal beam of length 4.0 m and mass 20 kg rests on two supports: one at the left end (A) and one 3.0 m along (B). A 30 kg mass sits 1.0 m from A. Find the reaction forces at A and B.