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Forces and Moments overview

Topic 3 of 4

Moments, couples and centre of gravity

A force's turning effect depends on its line of action as well as its magnitude. State the point about which the moment is being calculated.

Moment of a force = Fd

Here d is the perpendicular distance from the chosen point to the force's line of action. It is the shortest distance to that line, which may be extended beyond the body. Moment has unit N m.

For a planar diagram, choose a sign convention such as anticlockwise positive in the stated view. A force whose line of action passes through the pivot has zero moment about it, even when the force itself is not zero.

Worked angled force

Two ways to find the same moment

A horizontal arm extends 0.30 m to the right of a pivot. A 20 N force at its end acts 30° above the arm.

  1. Use the full force: perpendicular distance d = 0.30 sin 30° = 0.15 m. Moment = 20 × 0.15 = 3.0 N m anticlockwise.
  2. Alternatively, resolve the force: its upward component is 20 sin 30° = 10 N, acting 0.30 m from the pivot. Its moment is 10 × 0.30 = 3.0 N m anticlockwise. The horizontal component acts along a line through the pivot, so its moment is zero.

Use the perpendicular distance to the line of action

A twenty-newton angled force gives an anticlockwise moment of three newton metresA horizontal arm extends 0.30 metres right of its pivot. At its end, a twenty-newton force acts thirty degrees above the arm. The force's line of action is continued backwards as a dashed line below the pivot. At a common distance scale of six hundred drawing units per metre, the perpendicular from the pivot meets this extension at physical coordinates 0.075 metres right and 0.1299 metres below the pivot. Its length is 0.15 metres and a right-angle mark is shown. The moment is twenty times 0.15, or three newton metres anticlockwise. The 0.30-metre arm itself is not perpendicular to this force.Pivot0.30 m30°20 Nd = 0.15 mLine of action

The green distance is perpendicular to the dashed line of action. Moment = 20 N × 0.15 m = 3.0 N m anticlockwise. Resolving the force is an alternative way to obtain the same result.

The perpendicular distance to the 20 N force's line of action is 0.15 m. The 0.30 m arm can instead be paired with the 10 N perpendicular component. The two calculations describe the same turning effect.

Use one method or the other. Adding the full-force moment to its component moment would count the same force twice. Multiplying 20 N by 0.30 m would wrongly treat the force as perpendicular to the arm.

A couple has zero resultant force but a turning effect

A couple consists of two equal, opposite, parallel forces on the same body, acting along different lines. The forces cancel as vectors, but their moments give a tendency to rotate. The torque of the couple is:

τ = Fd

F is the magnitude of one force, and d is the perpendicular separation of the two lines of action. Torque may be written τ or T; in another context T can mean tension, so state the quantity and unit.

Worked couple

Equal forces can turn in the same sense

A 12 N force acts upward on the right and a 12 N force downward on the left. Their parallel lines are 0.18 m apart horizontally.

Resultant force = 0.
Couple torque = 12 × 0.18 = 2.16 N m anticlockwise.

About the midpoint, each force contributes 12 × 0.09 = 1.08 N m anticlockwise. The two moments add. About the left force's line, that force has zero moment and the right force contributes all 2.16 N m. The total couple torque is the same about any chosen point.

A couple has zero resultant force and a nonzero torque

Two equal opposite forces form an anticlockwise coupleOn the same body a twelve-newton force acts down on the left and a twelve-newton force up on the right. Equal arrow lengths represent equal force magnitudes. The vertical lines of action are horizontally separated by 0.18 metres. The resultant force is zero, while both forces contribute anticlockwise turning. The torque is twelve times 0.18, or 2.16 newton metres; the separation is not the distance of one force from the body's centre.12 N12 NAnticlockwisePerpendicular separation: 0.18 m

Torque = 12 N × 0.18 m = 2.16 N m anticlockwise. Both forces act on this one body; a third-law pair acts on different bodies.

Both forces act on the same body. Their moments have the same anticlockwise sense, although their resultant force is zero. The 0.18 m distance is the full separation between their lines of action.

A third-law pair acts on different bodies, so it is not a couple on one body's free-body diagram. Also, two equal opposite forces acting on the same line have no couple torque: their line separation is zero.

Moment and torque have the same dimensions as energy, but they describe a different physical quantity. Report torque in N m, rather than joules.

Optional check A body experiences a 12 N upward force on the right and a 12 N downward force on the left, on parallel lines 0.18 m apart. With no other forces or moments, is it in equilibrium?
A body experiences a 12 N upward force on the right and a 12 N downward force on the left, on parallel lines 0.18 m apart. With no other forces or moments, is it in equilibrium?

Replace distributed weight by one equivalent force

Every part of an extended body has weight. For force and moment calculations, their combined effect can be represented by a single weight acting at the body's centre of gravity.

A uniform straight beam in a uniform gravitational field has its centre of gravity at its midpoint. A nonuniform or loaded body need not. In some shapes the centre of gravity lies outside the material, so do not automatically place the weight at a convenient contact point.

For scale, a few newtons acting through a perpendicular arm of a few tenths of a metre give a moment of order 1 N m. For example, 5 N × 0.2 m = 1 N m. State the assumptions and convert centimetres to metres before interpreting a calculated torque.