Moment and torque

CIE A-Level Physics· 4.2(b), 4.2(c)· 15 min read

1. Moment of a Force: Definition and Calculation★★☆☆☆⏱ 4 min

📘 Definition

Moment of a force

M

The turning effect of a force about a fixed pivot, calculated as $M = F \times d_{⊥}$ , where $F$ is force magnitude and $d_{⊥}$ is the perpendicular distance from the pivot to the line of action of $F$ . Units are newton-metres (Nm).

Example:

A 10 N force 2 m from a pivot gives a moment of 20 Nm.

If the force is not perpendicular to the distance from the pivot, we use the angle between the force and the position vector from the pivot to get the correct moment:

M = Fd \sin\theta

📐 Worked Example

A 5.0 m uniform beam pivoted at one end has a force of 12 N applied at the free end, at 30° to the beam. Calculate the moment of the force about the pivot.

1
Identify given values: $F = 12$ N, $d = 5.0$ m, $θ = 3 0^{\circ}$
2
Use the moment formula for angled forces:
3
$M = Fd \sin\theta$
4
Substitute values and use $sin 3 0^{\circ} = 0.5$ :
5
$M = (12)(5.0)(0.5) = 30 \text{ Nm}$
6
The moment of the force about the pivot is 30 Nm, turning clockwise.

2. Couples and Torque★★★☆☆⏱ 5 min

📘 Definition

Couple

A pair of equal-magnitude, opposite-direction, parallel forces separated by a perpendicular distance. The resultant force of a couple is zero, so it produces only rotation, no linear movement.

📘 Definition

Torque of a couple

\tau

The total turning effect of a couple, calculated as $τ = F \times d$ , where $F$ is the magnitude of one force and $d$ is the perpendicular distance between the lines of action of the two forces. Units are Nm, same as moment.

📐 Worked Example

A steering wheel of diameter 40 cm has two hands applying opposite tangential forces of 15 N each to opposite edges. Calculate the applied torque.

1
Convert diameter to metres: $d = 40$ cm $= 0.40$ m
2
The perpendicular distance between the two tangential forces equals the diameter of the wheel.
3
Use the torque formula for a couple:
4
$\tau = Fd = 15 \times 0.40 = 6.0 \text{ Nm}$
5
The total torque applied to the steering wheel is 6.0 Nm.

3. Principle of Moments for Rotational Equilibrium★★★☆☆⏱ 6 min

📘 Definition

Principle of Moments

For a rigid body to be in rotational equilibrium, the sum of all clockwise moments about any pivot equals the sum of all anticlockwise moments about that pivot, written as: $\sum M_{c l oc k w i se} = \sum M_{an t i c l oc k w i se}$

For a body to be in full (translational + rotational) equilibrium, two conditions must hold: 1) resultant force on the body is zero, 2) sum of moments about any pivot is zero.

📐 Worked Example

A uniform 4.0 m beam of mass 20 kg is pivoted 1.0 m from the left end. What mass must be placed at the left end to keep the beam balanced?

1
The weight of a uniform beam acts at its centre of mass, which is 2.0 m from the left end, 1.0 m right of the pivot, creating a clockwise moment.
2
Let the unknown mass be $m$ . Its weight is $m g$ , acting 1.0 m left of the pivot, creating an anticlockwise moment.
3
Apply the principle of moments:
4
$m g \times 1.0 = (20 g) \times 1.0$
5
Acceleration due to gravity $g$ cancels from both sides, giving $m = 20$ kg.
6
A 20 kg mass at the left end keeps the beam in equilibrium.

Exam tip:

Always remember the centre of mass of a uniform beam is at its midpoint — this is a common point examiners test.

4. Solving Complex Equilibrium Problems★★★★☆⏱ 6 min

For problems with multiple unknown forces, we use both equilibrium conditions: sum of forces in $x$ and $y$ directions equal zero, and sum of moments equal zero. Choosing the pivot at the point of an unknown force eliminates that force from the calculation, simplifying the problem.

📐 Worked Example

A 3.0 m uniform ladder of mass 15 kg leans against a smooth wall, foot on rough ground, at 60° to the horizontal. Find the horizontal reaction force from the wall.

1
Smooth wall means only horizontal reaction $R$ at the top. Ground exerts vertical normal $N$ and horizontal friction $F$ at the foot. Choose pivot at the foot to eliminate $N$ and $F$ .
2
Weight $W = 15 g$ acts at 1.5 m from the foot. Perpendicular distance to pivot = $1.5 cos 6 0^{\circ}$ (clockwise moment). Perpendicular distance for $R$ = $3.0 sin 6 0^{\circ}$ (anticlockwise moment).
3
Apply principle of moments:
4
$R \times 3.0 \sin 60^\circ = 15g \times 1.5 \cos 60^\circ$
5
Substitute $sin 6 0^{\circ} = \frac{3}{2}$ , $cos 6 0^{\circ} = 0.5$ , $g = 9.8$ m/s²:
6
$R = \frac{15 \times 9.8 \times 1.5 \times 0.5 \times 2}{3 \times \sqrt{3}} \approx 42 \text{ N}$
7
The horizontal reaction force from the wall is approximately 42 N.

5. Common Pitfalls

Wrong move:

Multiplying force by straight line distance instead of perpendicular distance for angled forces

Why:

Parallel components of force do not contribute to turning, so this overestimates or underestimates moment

Correct move:

Always use $M = F d sin θ$ to account for the angle between force and distance from the pivot

Wrong move:

Using radius instead of diameter for torque when forces act on opposite edges of a wheel

Why:

Torque uses perpendicular distance between the two forces of the couple, which is diameter here not radius

Correct move:

Always measure the perpendicular distance between the lines of action of the two forces directly

Wrong move:

Taking the weight of a uniform beam at the pivot or end instead of the midpoint

Why:

This gives an incorrect moment for the beam's own weight, which is almost always included in exam problems

Correct move:

Always mark the centre of mass at the midpoint of a uniform body before starting calculations

Wrong move:

Claiming torque of a couple depends on the position of the pivot

Why:

This confuses torque of a couple with moment of a single force; a couple has zero resultant force so its turning effect is constant

Correct move:

Remember: torque of a couple is independent of pivot position, unlike moment of a single force

Wrong move:

Mixing up the direction (clockwise/anticlockwise) of moments when applying the principle of moments

Why:

Moments in opposite directions cancel, so wrong direction leads to incorrect final values

Correct move:

Always label each moment as clockwise or anticlockwise before grouping and summing

6. Quick Reference Cheatsheet

Concept	Formula	Key Notes
Moment of a force	$M = F d_{⊥} = F d sin θ$	$d$ = distance from pivot, $θ$ = angle between $F$ and $d$
Torque of a couple	$τ = F d$	$d$ = perpendicular distance between forces, independent of pivot
Principle of moments	$\sum M_{c l oc k w i se} = \sum M_{an t i c l oc k w i se}$	Holds for any pivot for rotational equilibrium
Full equilibrium	$\sum F_{x} = 0$ , $\sum F_{y} = 0$ $\sum M = 0$	Two conditions required for complete equilibrium

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 12
Calculate torque of a steering wheel
2023 · 22
Equilibrium of a balanced beam
2021 · 13
Moment of a force at an angle

Going deeper

SyllabusCIE 9702 A-Level Physics Official Syllabus

What's Next

Moment and torque are foundational concepts for further topics in CIE A-Level Mechanics, including centre of mass calculations, static equilibrium of extended bodies, and A-Level rotational motion. Mastery of this sub-topic is essential for solving calculation-based questions in both Paper 1 multiple choice and Paper 2 structured questions, with frequent appearances in every exam series. The principle of moments also forms the basis for understanding more complex dynamic systems later in the course.