Moments and Torque

So far we have treated objects as particles — points with mass but no size. In reality, forces can act at different points on an object, and this matters. A force applied at the edge of a door swings it open easily; the same force applied near the hinge barely moves it. The turning effect of a force depends not just on its magnitude, but also on where and how it is applied. This is the concept of the moment of a force.

Moment of a Force

The moment of a force about a point (or pivot) is defined as:

Moment = Force x Perpendicular distance from the line of action of the force to the pivot

Moment = F d

where F is the force in newtons and d is the perpendicular distance in metres. The unit of moment is the newton-metre (N m).

The line of action of a force is an imaginary line extending along the direction of the force. The perpendicular distance is measured at right angles from this line to the pivot.

If the force acts at an angle theta to the line joining the point of application to the pivot, then the perpendicular distance is r sin(theta), where r is the distance from the pivot to the point of application. So:

Moment = F r sin(theta)

Worked Example 1

A spanner is 0.30 m long. A force of 40 N is applied at the end, perpendicular to the spanner. What is the moment about the bolt?

Moment = F d = 40 x 0.30 = 12 N m

Worked Example 2

The same 40 N force is applied at 60° to the spanner instead of perpendicular. What is the moment?

Moment = F r sin(theta) = 40 x 0.30 x sin(60°) = 40 x 0.30 x 0.866 = 10.4 N m

The moment is reduced because the force is no longer perpendicular.

Worked Example 3

A gate is 1.2 m wide. A child pushes with 30 N at the far edge, perpendicular to the gate. An adult pushes with 30 N at 0.40 m from the hinge, also perpendicular. Compare the moments.

Child: Moment = 30 x 1.2 = 36 N m Adult: Moment = 30 x 0.40 = 12 N m

The child produces 3 times the turning effect despite using the same force, because the distance from the pivot is 3 times greater.

Direction of Moments: Clockwise and Anticlockwise

Moments can cause rotation in two directions:

Clockwise moments tend to rotate the object clockwise.
Anticlockwise moments tend to rotate the object anticlockwise.

By convention, we often take anticlockwise as positive and clockwise as negative, though any consistent convention is acceptable.

The Principle of Moments

For an object in rotational equilibrium (not rotating, or rotating at constant angular velocity):

The sum of clockwise moments about any point equals the sum of anticlockwise moments about the same point.

This is the principle of moments, and it is the rotational equivalent of Newton’s first law for forces.

flowchart TD
    A["Object in rotational equilibrium"]
    A --> B["Choose a pivot point"]
    B --> C["Calculate all clockwise\nmoments about the pivot"]
    B --> D["Calculate all anticlockwise\nmoments about the pivot"]
    C --> E["Sum CW = Sum ACW"]
    D --> E
    E --> F["Solve for unknown\nforce or distance"]

Worked Example 4

A uniform beam of length 4.0 m and weight 200 N is supported at its centre. A 300 N weight is placed 1.5 m from the left end. Where must a 500 N weight be placed to balance the beam?

Taking moments about the centre (2.0 m from each end):

The 300 N weight is 2.0 - 1.5 = 0.5 m to the left of the centre. Anticlockwise moment from 300 N weight = 300 x 0.5 = 150 N m

Let the 500 N weight be at distance x to the right of the centre. Clockwise moment = 500 x

For equilibrium: 500 x = 150 x = 150 / 500 = 0.30 m to the right of the centre

Worked Example 5: Beam with Two Supports

A uniform plank of weight 500 N and length 6.0 m rests on two supports: A at the left end and B at 4.0 m from A. A person of weight 700 N stands at the right end (6.0 m from A). Find the reactions at A and B.

Taking moments about A: R_B x 4.0 = 500 x 3.0 + 700 x 6.0 R_B x 4.0 = 1500 + 4200 = 5700 R_B = 5700 / 4.0 = 1425 N

Resolving vertically: R_A + R_B = 500 + 700 = 1200 R_A = 1200 - 1425 = -225 N

The negative sign means the reaction at A is downward — support A must be clamped down, not just resting. The plank would topple about B without it.

Couples

A couple is a pair of equal and opposite parallel forces that do not act along the same line. A couple produces pure rotation with no translational movement.

The moment of a couple (also called the torque) is:

Torque = F x d

where F is the magnitude of one of the forces and d is the perpendicular distance between the two forces.

Examples of couples: turning a steering wheel (hands push in opposite directions), turning a tap, the forces on a compass needle in a magnetic field.

A couple has no resultant force (the forces cancel), but it does have a resultant moment. This means it causes rotation without translation.

Important property: The moment of a couple is the same about any point. This makes it different from the moment of a single force, which depends on the chosen pivot.

Centre of Gravity

The centre of gravity of an object is the point at which the entire weight of the object can be considered to act. For a uniform object, the centre of gravity is at the geometric centre.

For a uniform beam, the centre of gravity is at the midpoint. This means when calculating moments, the weight of the beam acts at its centre.

Common Centres of Gravity

Object	Centre of Gravity
Uniform rod	Midpoint
Uniform rectangular sheet	Intersection of diagonals
Uniform circular disc	Centre of circle
Uniform triangular sheet	1/3 of way from base to apex
Uniform sphere	Centre of sphere
L-shaped object	Must be calculated (often outside the object)

Finding the Centre of Gravity Experimentally

For irregular objects, the centre of gravity can be found experimentally by suspending the object from different points. When suspended freely, the object hangs so that the centre of gravity is directly below the suspension point. Drawing a vertical line from each suspension point, the intersection of these lines is the centre of gravity.

Toppling and Stability

An object will topple when its centre of gravity is no longer above its base of support. More precisely, if the vertical line through the centre of gravity falls outside the base, the weight creates a moment that tips the object over.

Stability depends on:

The height of the centre of gravity — lower is more stable.
The width of the base — wider base is more stable.
The mass of the object — heavier objects require larger forces to tilt.

A racing car has a low centre of gravity and wide wheelbase for stability. A tall, narrow bookshelf is less stable than a wide, short one.

Worked Example 6

A uniform box is 0.60 m wide and 1.2 m tall, resting on a slope. At what angle does the box topple?

The box topples when the centre of gravity (at the geometric centre) is directly above the edge of the base. The centre of gravity is 0.30 m from the downhill edge and 0.60 m above the base.

tan(theta) = 0.30 / 0.60 = 0.50 theta = arctan(0.50) = 26.6°

At angles greater than 26.6°, the box topples.

Worked Example 7: Toppling vs Sliding

The same box (0.60 m wide, 1.2 m tall, mass 20 kg) is on a slope with mu = 0.30. Does it topple or slide first as the angle increases?

Toppling angle: arctan(0.30/0.60) = arctan(0.50) = 26.6° Sliding angle: arctan(mu) = arctan(0.30) = 16.7°

Since 16.7° < 26.6°, the box slides before it topples. The sliding angle is reached first.

If mu were 0.60 instead: sliding angle = arctan(0.60) = 31.0°. Since 26.6° < 31.0°, the box would topple before it slides.

Conditions for Complete Equilibrium

For a rigid body to be in complete equilibrium:

Resultant force = 0 (no translational acceleration) — Newton’s first law.
Resultant moment about any point = 0 (no rotational acceleration) — principle of moments.

Both conditions must be satisfied simultaneously.

Common Mistakes and Misconceptions

Forgetting the weight of the beam/plank: In moments problems, uniform beams have their own weight acting at the centre. Students often only include the external loads.
Using the wrong distance: The moment uses the perpendicular distance from the pivot to the line of action of the force, not the distance along the beam to where the force is applied (unless the force is vertical and the beam is horizontal).
Taking moments about the wrong point: You can take moments about any point, but choosing a point where an unknown force acts eliminates that force from the equation, simplifying the calculation.
Confusing torque of a couple with moment of a single force: A couple’s torque uses the distance between the two forces, not the distance from a pivot.
Forgetting that moment direction matters: Always state whether each moment is clockwise or anticlockwise before summing.

Exam Tips for Edexcel Physics

When taking moments, always state the point about which you are taking moments and your sign convention.
Choose your pivot wisely. Placing it at a point where an unknown force acts eliminates that unknown from the moment equation.
For beams with multiple loads, systematically work through each force and its distance from the pivot. A table can help organise this.
In "toppling" questions, clearly identify the pivot point (the edge of the base) and show that the moment due to weight has changed direction.
Remember: the Edexcel specification uses "moment" for a single force and "torque" for a couple. Use these terms correctly.

Key Points

Moment = F x perpendicular distance from the pivot. Units: N m.
The principle of moments: for equilibrium, sum of clockwise moments = sum of anticlockwise moments.
A couple is two equal, opposite, parallel forces — it produces rotation without translation. Its torque is the same about any point.
Centre of gravity is where the weight effectively acts; for uniform objects, it is at the geometric centre.
An object topples when the line of action of its weight falls outside the base.
Complete equilibrium requires both zero resultant force and zero resultant moment.
Choose a pivot where an unknown force acts to simplify calculations.

A-Level Deep Dive: Moments and Torque

Spec mapping

Edexcel 9PH0 specification, Topic 2 — Mechanics, sub-topic 2.4 (Statics) requires candidates to use the principle of moments, define the moment of a force as force times perpendicular distance from the line of action to the pivot, identify a couple as a pair of equal and opposite forces producing rotation, and apply the conditions for equilibrium of coplanar forces (refer to the official Pearson Edexcel specification document for exact wording). The techniques recur in Topic 4 (Materials) for bending of loaded beams, Topic 6 (Further mechanics) where rotational analogues such as $\tau = I\alpha$ reuse the moment concept, and Topic 7 (Electric and magnetic fields) where the torque on a current loop $\tau = BIAN\sin\theta$ underlies the moving-coil meter and the electric motor. The Edexcel data, formulae and relationships booklet does not list the principle of moments — it must be quoted in working when invoked.

Worked example with full mark scheme

Question (8 marks):

A uniform horizontal beam of length 4.00 m and weight 120 N rests on two supports, A at the left-hand end and B which is 0.500 m from the right-hand end. A load of 80.0 N hangs from a point 1.00 m from A and a second load of 200 N hangs from the right-hand end of the beam.

(a) Calculate the magnitude of the upward force exerted by support B on the beam. (4)

(b) Hence calculate the upward force exerted by support A. (2)

(c) State the assumption you made about the beam when locating its weight, and explain how the value of the force at B would change if the beam were not uniform but had its centre of gravity 0.30 m closer to B. (2)

Solution with mark scheme:

(a) Step 1 — choose the pivot. Take moments about A. This eliminates the unknown reaction at A from the moment equation.

M1 — clear statement that moments are taken about A, with sign convention (e.g. "anticlockwise positive").

Step 2 — identify the perpendicular distances. The weight of the beam acts at its centre, 2.00 m from A (because the beam is uniform). The 80.0 N load is 1.00 m from A. The 200 N load is 4.00 m from A. Support B is $4.00 - 0.500 = 3.50$ m from A.

Step 3 — apply $\Sigma M_A = 0$ for equilibrium. Taking clockwise moments (caused by the downward forces) equal to anticlockwise moments (caused by the upward force at B):

$N_B \times 3.50 = 120 \times 2.00 + 80.0 \times 1.00 + 200 \times 4.00$

M1 — correct equation, each term being force times perpendicular distance from A.

$N_B \times 3.50 = 240 + 80.0 + 800 = 1120 \text{ N m}$

A1 — correct evaluation of the right-hand side.

$N_B = \dfrac{1120}{3.50} = 320 \text{ N}$

A1 — final answer to 3 sf with units.

(b) Vertical equilibrium: the sum of the upward forces equals the sum of the downward forces.

$N_A + N_B = 120 + 80.0 + 200 = 400 \text{ N}$

M1 — correct vertical-equilibrium statement.

$N_A = 400 - 320 = 80.0 \text{ N}$

A1 — $N_A = 80.0$ N (3 sf).

B1 — if the centre of gravity moved 0.30 m closer to B, the weight would act 2.30 m from A, increasing the clockwise moment about A by $120 \times 0.30 = 36$ N m, so $N_B$ would increase by $36 / 3.50 \approx 10.3$ N (and $N_A$ would correspondingly decrease).

Total: 8 marks (M3 A3 B2).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A uniform plank of length 3.00 m and mass 12.0 kg is balanced horizontally on a single pivot 1.20 m from its left-hand end. A child of mass 25.0 kg stands on the plank. Take $g = 9.81$ m s⁻².

(a) Draw a free-body diagram of the plank, labelling all forces and their points of application. (1)

(b) Calculate the distance, measured from the pivot, at which the child must stand on the left-hand side for the plank to remain in equilibrium. (3)

(c) State and explain whether your answer to (b) would change if the plank were not uniform but had its centre of gravity displaced toward the right-hand end. (2)

Mark scheme decomposition by AO: