Physics 9702 · Paper 2

The Moment of a Force

A force can do two things: push an object somewhere, and turn it.
The turning effect of a force is called its . This lesson measures it.

The definition

A hand pushing the far end of a long spanner fitted onto a nut. — The moment of a force = force × perpendicular distance from the pivot. A push at the far end of the spanner acts a large distance from the nut, so it gives a big turning effect.

\text{moment} = F \times d

Symbols

$F$ = force (N)
$d$ = perpendicular distance of the line of action of the force from the point (m)

The written definition carries marks word by word: force × perpendicular distance of the line of action of the force from the point (9702/22/M/J/25 Q2(a)).
The unit is the newton metre (N m). It has no special name.
A bigger force, or a bigger distance from the pivot, gives a bigger turning effect. To turn a stiff nut, you push at the far end of the spanner, at right angles to it.
In base units a moment is kg m² s⁻²: the unit check skill from chapter 1 (9702/11/O/N/24 Q1).

Worked example

Smallest case: force at right angles

A force of 6.0 N acts at right angles to a spanner, 0.50 m from the centre of the nut. Find its moment.

$\text{moment} = F \times d = 6.0 \times 0.50 = 3.0 \text{ N m}$ .

Answer

\text{moment} = 3.0 \text{ N m}

Worked example

One change: the force is a weight

A plank of mass 100 kg rests on a pivot. Its centre of gravity is 4.0 m from the pivot. Find the moment of its weight about the pivot. (9702/11/M/J/22 Q11)

The weight acts at the centre of gravity: $W = mg = 100 \times 9.81 = 981 \text{ N}$ .
$\text{moment} = 981 \times 4.0 = 3900 \approx 4 \times 10^{3} \text{ N m}$ .

Answer

\text{moment} \approx 4 \times 10^{3} \text{ N m}

When the force is at an angle

If the force is not at right angles to the arm, the turning effect is smaller. The word perpendicular in the definition handles this.

Left: force at 90° to the lever, moment = F × x. Right: force at 30°, so the perpendicular distance from the pivot to the force line is only d sin 30°, drawn to scale.

There are two equal ways to compute it, and both give $F\,d\sin\theta$ for a force at angle $\theta$ to an arm of length $d$ .
Way 1: keep the full force, but measure the perpendicular distance from the pivot to the force's line of action: $d\sin\theta$ .
Way 2: keep the full distance $d$ , but take only the part of the force at right angles to the arm: $F\sin\theta$ .
A force whose line of action passes through the pivot has distance zero, so its moment is zero. It cannot turn the object at all.

Common mistake

Using the distance along the rod to where the force is applied, instead of the perpendicular distance to the line of the force. When the force is at an angle, these are different lengths, and the multiple-choice options are built from the wrong one (9702/12/M/J/23 Q11).

Your turn— tap to reveal the worked answer (9702/12/M/J/23 Q11)

A rod is pivoted at O. A force F acts on the rod at point W, at an angle to it. OY is the perpendicular from O to the line of the force. Which product gives the moment about O? (9702/12/M/J/23 Q11)

Answer: $F \times OY$ . Not $F \times OW$ : OW is the distance along the rod, not the perpendicular distance.

Ways the exam changes it: a hanging sign at an angle to the wall (find the perpendicular distance with sin or cos) (9702/12/M/J/25 Q15); the moment of a rod's weight as the rod is tilted (largest when horizontal, zero when vertical) (9702/12/F/M/24 Q13).

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The Moment of a Force

A force can do two things: push an object somewhere, and turn it.
The turning effect of a force is called its . This lesson measures it.

The definition

\text{moment} = F \times d

Symbols

$F$ = force (N)
$d$ = perpendicular distance of the line of action of the force from the point (m)

The written definition carries marks word by word: force × perpendicular distance of the line of action of the force from the point (9702/22/M/J/25 Q2(a)).
The unit is the newton metre (N m). It has no special name.
A bigger force, or a bigger distance from the pivot, gives a bigger turning effect. To turn a stiff nut, you push at the far end of the spanner, at right angles to it.
In base units a moment is kg m² s⁻²: the unit check skill from chapter 1 (9702/11/O/N/24 Q1).

Worked example

Smallest case: force at right angles

A force of 6.0 N acts at right angles to a spanner, 0.50 m from the centre of the nut. Find its moment.

$\text{moment} = F \times d = 6.0 \times 0.50 = 3.0 \text{ N m}$ .

Answer

\text{moment} = 3.0 \text{ N m}

Worked example

One change: the force is a weight

A plank of mass 100 kg rests on a pivot. Its centre of gravity is 4.0 m from the pivot. Find the moment of its weight about the pivot. (9702/11/M/J/22 Q11)

The weight acts at the centre of gravity: $W = mg = 100 \times 9.81 = 981 \text{ N}$ .
$\text{moment} = 981 \times 4.0 = 3900 \approx 4 \times 10^{3} \text{ N m}$ .

Answer

\text{moment} \approx 4 \times 10^{3} \text{ N m}

When the force is at an angle

If the force is not at right angles to the arm, the turning effect is smaller. The word perpendicular in the definition handles this.

Left: force at 90° to the lever, moment = F × x. Right: force at 30°, so the perpendicular distance from the pivot to the force line is only d sin 30°, drawn to scale.

There are two equal ways to compute it, and both give $F\,d\sin\theta$ for a force at angle $\theta$ to an arm of length $d$ .
Way 1: keep the full force, but measure the perpendicular distance from the pivot to the force's line of action: $d\sin\theta$ .
Way 2: keep the full distance $d$ , but take only the part of the force at right angles to the arm: $F\sin\theta$ .
A force whose line of action passes through the pivot has distance zero, so its moment is zero. It cannot turn the object at all.

Common mistake

Your turn— tap to reveal the worked answer (9702/12/M/J/23 Q11)

A rod is pivoted at O. A force F acts on the rod at point W, at an angle to it. OY is the perpendicular from O to the line of the force. Which product gives the moment about O? (9702/12/M/J/23 Q11)

Answer: $F \times OY$ . Not $F \times OW$ : OW is the distance along the rod, not the perpendicular distance.

Ways the exam changes it: a hanging sign at an angle to the wall (find the perpendicular distance with sin or cos) (9702/12/M/J/25 Q15); the moment of a rod's weight as the rod is tilted (largest when horizontal, zero when vertical) (9702/12/F/M/24 Q13).