Moments

Understanding the turning effect of forces, couples, and equilibrium conditions

SPEC 3.4.1.2 • MECHANICS AND MATERIALS

What is a Moment?

A moment is the turning effect of a force about a point or axis. When you push a door handle, turn a steering wheel, or use a spanner, you're applying moments.

Definition

The moment of a force about a point is defined as:

M = F \times d

where d is the perpendicular distance from the point to the line of action of the force.

Practical Note

Although the definition uses perpendicular distance, in most practical cases it is easier to find the perpendicular component of the force instead.

Units

Moment is measured in newton metres (N m)

Note: This is NOT the same as joules, even though both are N × m. Moments and energy are fundamentally different quantities.

Perpendicular Distance

The key to calculating moments correctly is understanding perpendicular distance.

Common Error

Always use the perpendicular distance from the pivot to the line of action of the force — not just any distance along the object!

Worked Example01

A spanner of length 0.25 m is used to tighten a nut. A force of 80 N is applied perpendicular to the spanner at its end. Calculate the moment about the nut.

1

Identify the values: F = 80 N, d = 0.25 m (perpendicular)

2

Apply M = Fd since force is already perpendicular

3

M = 80 × 0.25 = 20 N m

Answer: 20 N m

Force at an Angle

When a force acts at an angle θ to the lever arm, we must find the perpendicular component.

Formula

When force acts at angle θ to the lever arm:

M = F d sin θ

Or equivalently: resolve the force and use the perpendicular component.

Practical Note

Although the definition uses perpendicular distance, in most practical cases it is easier to find the perpendicular component of the force instead.

Worked Example02

A force of 50 N acts at the end of a 0.4 m lever at an angle of 60° to the perpendicular. Calculate the moment about the pivot.

1

Identify: F = 50 N, d = 0.4 m, angle to perpendicular = 60°

2

The angle to the lever arm is 90° − 60° = 30°, so we can use M = Fd sin 30°

3

Alternatively: perpendicular component = F cos 60° (since angle is measured from perpendicular)

4

M = Fd cos 60° = 50 × 0.4 × 0.5 = 10 N m

Answer: 10 N m

Clockwise vs Anticlockwise

Moments have a direction — they cause rotation either clockwise or anticlockwise about the pivot.

Sign Convention

In calculations, we often take:

Anticlockwise moments as positive (+)

Clockwise moments as negative (−)

However, either convention works — just be consistent!

Couples and Torque

A couple is a special arrangement of forces that produces pure rotation.

Definition

A couple is a pair of coplanar forces (forces in the same plane) that:

are equal in magnitude

act in opposite directions

do not pass through the same point (lines of action do not coincide)

Terminology

Since a couple involves two forces, the turning effect is called the torque. The terms "moment of a couple" and "torque" are used interchangeably.

Key Property

A couple produces rotation only — no linear acceleration. The net force is zero, but the net moment is not!

Torque of a Couple

Formula

The torque (moment of a couple) is:

Torque = F \times d

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

Forces at an Angle

If the forces act at an angle θ to the line joining their points of application:

Torque = F \times d \times sin θ

where d is the separation between the points of application.

Key Properties

A couple produces rotation only — no linear acceleration (net force is zero)

The torque is the same about any point — no need to specify a pivot

03Worked Example

Two forces, each of 30 N, act on a rod pivoted at its centre. The ends of the forces on the rod are 60 cm apart. The forces act at an angle of 20° to the rod as shown. Calculate the torque produced.

1

Identify: F = 30 N (each force), d = 60 cm = 0.60 m, θ = 20°

2

The forces act at an angle to the rod, so we need the perpendicular distance between lines of action

3

Torque = F × d × sin θ = 30 × 0.60 × sin 20°

4

Torque = 30 × 0.60 × 0.342 = 6.2 N m

Answer: 6.2 N m

Worked Example04

A steering wheel has diameter 0.36 m. A driver applies forces of 15 N with each hand at opposite ends of a diameter, perpendicular to the wheel. Calculate the torque applied to the steering wheel.

1

Identify: F = 15 N (each hand), d = 0.36 m (diameter = perpendicular separation)

2

The forces are equal, opposite, coplanar, and perpendicular (θ = 90°)

3

Torque = F × d = 15 × 0.36 = 5.4 N m

Answer: 5.4 N m

Principle of Moments

For a body to be in rotational equilibrium, there must be no resultant turning effect.

Principle of Moments

For a body in equilibrium:

\sum Clockwise moments = \sum Anticlockwise moments

Or equivalently:

\sum M = 0

about any point.

Complete Equilibrium

A body is in equilibrium when acted on by two or more coplanar forces if the resultant of these forces and their moments is zero.

This means two conditions must be satisfied:

ΣF = 0 (no resultant force — translational equilibrium)

ΣM = 0 (no resultant moment — rotational equilibrium)

Common Mistake

Forgetting that the resultant moment must be zero when stating the condition for equilibrium. Both force AND moment conditions are required!

Choosing the Pivot Point

When solving moment problems, you can take moments about any point. Clever choice simplifies calculations!

Strategy

Choose your pivot point where an unknown force acts. That force then has zero moment (d = 0), eliminating it from your equation!

Example

To find R_B: take moments about point A. Then R_A × 0 = 0, so R_A disappears from the equation, leaving only R_B and the known force W.

Centre of Mass and Centre of Gravity

Definitions

The centre of mass is the point at which an object's mass acts.

The centre of gravity is the point at which the weight of an object may be considered to act.

Important Note

If the gravitational field is uniform over the size of the object, the centre of mass and centre of gravity coincide. For all A-Level problems, you can treat them as the same point.

Uniform Regular Solids

If an object is described as uniform, its centre of mass is exactly at its geometric centre:

ShapeCentre of Mass Location

Uniform rod/beamMidpoint (half the length)

Rectangular plateCentre (intersection of diagonals)

Sphere / Cylinder / CubeGeometric centre

Worked Example05

A uniform beam of length 4.0 m and weight 200 N is supported at both ends. A 500 N load is placed 1.0 m from the left end. Calculate the reaction forces at each support.

1

Let R₁ = left support, R₂ = right support. Beam weight acts at centre (2.0 m from left).

2

Take moments about left end (R₁ has zero moment):

3

Clockwise: (500 × 1.0) + (200 × 2.0) = 500 + 400 = 900 N m

4

Anticlockwise: R₂ × 4.0

5

Equilibrium: R₂ × 4.0 = 900 → R₂ = 225 N

6

Force equilibrium: R₁ + R₂ = 500 + 200 = 700 N → R₁ = 475 N

Answer: R₁ = 475 N, R₂ = 225 N

Worked Example06

A seesaw is 3.0 m long with the pivot at its centre. A child of weight 400 N sits 1.2 m from the pivot. Where must a child of weight 300 N sit to balance the seesaw?

1

For balance: clockwise moment = anticlockwise moment

2

Child 1: W₁ = 400 N at d₁ = 1.2 m

3

Child 2: W₂ = 300 N at d₂ = ? (other side)

4

400 × 1.2 = 300 × d₂

5

d₂ = 480 ÷ 300 = 1.6 m from pivot

Answer: 1.6 m from the pivot (on opposite side)

Worked Example07

A non-uniform rod AB of length 4.0 m and weight 80 N is supported horizontally by two vertical strings attached at A and B. When a 60 N weight is hung from a point 1.0 m from A, the tension in string A equals the tension in string B. Find the distance of the centre of mass from A.

1

Let the centre of mass be at distance x from A

2

Force equilibrium: T_A + T_B = 80 + 60 = 140 N

3

Since T_A = T_B, we have T_A = T_B = 70 N

4

Take moments about A: Clockwise = (80 × x) + (60 × 1.0)

5

Anticlockwise = T_B × 4.0 = 70 × 4.0 = 280 N m

6

80x + 60 = 280 → 80x = 220 → x = 2.75 m

Answer: Centre of mass is 2.75 m from A

Worked Example08

A horizontal beam of negligible mass is 2.0 m long. It is hinged at one end (A) and held horizontal by a cable attached at B, 0.5 m from A. The cable makes an angle of 60° with the beam. A 100 N weight hangs from the far end. Find the tension in the cable.

1

Take moments about A (hinge force has zero moment)

2

Clockwise moment from weight: 100 × 2.0 = 200 N m

3

Anticlockwise moment from tension: T × 0.5 × sin 60°

4

(The perpendicular component of T creates the moment)

5

T × 0.5 × sin 60° = 200

6

T = 200 ÷ (0.5 × 0.866) = 200 ÷ 0.433 = 462 N

Answer: T = 462 N (or 460 N to 2 s.f.)

Real-World Applications

Moments and equilibrium are fundamental to engineering and everyday physics.

Engineering

Bridges: Support forces calculated using moments

Cranes: Counterweights balance loads

Levers: Mechanical advantage from moments

Biology

Forearm: Bicep creates moment about elbow

Jaw: Muscles create moments for chewing

Standing: Balance requires moment equilibrium

Mechanical Advantage

A lever allows a small force at large distance to balance a large force at small distance. This is the principle behind tools like crowbars, wheelbarrows, and scissors.

Practice

Question 1: A door handle is 0.80 m from the hinge. A force of 12 N is applied perpendicular to the door at the handle. Calculate the moment about the hinge.

1

M = F × d (force is perpendicular)

2

M = 12 × 0.80 = 9.6 N m

Answer: 9.6 N m

Practice

Question 2: A uniform metre rule is pivoted at the 30 cm mark. A 2.0 N weight is hung at the 10 cm mark. Where must a 1.5 N weight be hung to balance the rule? (Ignore the weight of the ruler)

1

Distance of 2.0 N from pivot: 30 − 10 = 20 cm = 0.20 m

2

Anticlockwise moment: 2.0 × 0.20 = 0.40 N m

3

For balance: 1.5 × d = 0.40 → d = 0.267 m = 26.7 cm

4

Position: 30 + 26.7 = 56.7 cm mark (on the other side of pivot)

Answer: At the 56.7 cm mark (or 57 cm to 2 s.f.)

Practice

Question 3: Two forces of magnitude 8 N act on a rectangular object as a couple. The perpendicular distance between the lines of action is 0.15 m. Calculate the moment of the couple. Explain why this moment is the same about any point.

1

Moment of couple = F × d = 8 × 0.15 = 1.2 N m

2

Explanation: If we take moments about any point P, one force creates a clockwise moment F × x, and the other creates an anticlockwise moment F × (d + x) or vice versa.

3

The net moment = F(d + x) − Fx = Fd, which is constant regardless of where P is located.

Answer: 1.2 N m; the moment of a couple is independent of the pivot point because the two equal, opposite forces create moments that differ by Fd about any point.

Summary

Key Formulas

QuantityFormulaUnits

Moment of a forceM = FdN m

Moment (force at angle)M = Fd sin θN m

Torque (moment of couple)Torque = F × dN m

Torque (couple at angle)Torque = Fd sin θN m

Key Concepts

Moment: Turning effect of a force; uses perpendicular distance to pivot

Couple: Two equal, opposite, coplanar forces with non-coincident lines of action

Torque: The moment of a couple; same about any point

Principle of moments: ΣM_clockwise = ΣM_anticlockwise for equilibrium

Equilibrium: Requires both ΣF = 0 AND ΣM = 0

Centre of mass/gravity: Point where weight acts; at geometric centre for uniform objects