Introduction to Moments

Understanding the turning effect of forces and calculating moments about a point

SPEC 9.1 • MOMENTS

What is a Moment?

Definition

A moment is the turning effect of a force about a point (pivot).

When you push a door, it rotates about its hinges. The further from the hinges you push, and the harder you push, the greater the turning effect.

Key Concept

The moment depends on two things:

The magnitude of the force

The perpendicular distance from the line of action of the force to the pivot

Calculating Moments

Formula

M = F \times d

Moment = Force × Perpendicular distance

SymbolMeaningUnit

M

Moment about the pivotNewton-metres (Nm)

F

Force appliedNewtons (N)

d

Perpendicular distance from pivot to line of actionMetres (m)

Important

The distance must be perpendicular to the line of action of the force. This is the key requirement that students often miss.

Sign Convention

When calculating moments, we must distinguish between forces that cause rotation in different directions.

Anticlockwise

↺

Positive moment

Clockwise

↻

Negative moment

Tip

The choice of positive/negative is a convention. Some textbooks use the opposite. The key is to be consistent throughout a problem. In A-Level exams, anticlockwise is typically taken as positive.

01Worked Example

A horizontal lever arm is 1.5 m long. A force of 40 N is applied vertically downwards at the end of the lever. Calculate the moment about the pivot at the other end.

1

Identify the values:

F = 40 N

,

d = 1.5 m

2

The force is perpendicular to the lever (vertical force on horizontal lever), so we can use the formula directly.

3

Apply the formula:

M = F \times d = 40 \times 1.5 = 60 Nm

4

The force causes clockwise rotation, so:

M = - 60 Nm

(or state "60 Nm clockwise")

Answer60 Nm clockwise (or −60 Nm)

Forces at Angles

When a force acts at an angle, we need the perpendicular distance from the pivot to the line of action of the force.

Definition

The perpendicular distance is the shortest distance from the pivot point to the line along which the force acts (extended if necessary).

Method 1: F sin θ

M = F sin θ \times d

Use when θ is the angle between force and rod. Multiply the force by sin θ.

Method 2: d sin θ

M = F \times d sin θ

Use when you know the distance along the rod. Multiply the distance by sin θ.

Which to Use?

Both methods give the same answer! The component (sin or cos) depends on how the angle is measured:

If θ is measured from the rod: use sin θ

If θ is measured from the perpendicular: use cos θ

Draw a diagram to find the perpendicular distance or perpendicular component!

02Worked Example

A force of 25 N acts at the end of a 2 m rod, at an angle of 30° above the horizontal. Calculate the moment about the pivot at the other end of the rod (which is horizontal).

1

Identify the values:

F = 25 N

,

d = 2 m

,

θ = 30°

(angle above horizontal)

2

The rod is horizontal, so we need the vertical component of the force.

3

Perpendicular component:

F sin θ = 25 sin 30° = 25 \times 0.5 = 12.5 N

4

Calculate moment:

M = F sin θ \times d = 12.5 \times 2 = 25 Nm

Answer25 Nm (anticlockwise)

Multiple Forces

When several forces act on a body, we find the resultant moment by adding the individual moments, taking account of their signs.

Resultant Moment

Σ M = M_{1} + M_{2} + M_{3} + \dots

Method

Choose a pivot point (often given, or choose strategically)

Calculate the moment of each force about that point

Assign positive/negative based on direction of rotation

Add all moments together

Alternative Method

You can also calculate clockwise and anticlockwise moments separately, then find the difference:

Resultant = Sum of CW moments − Sum of ACW moments

03Worked Example

A horizontal rod has length 4 m. Forces act on the rod as shown: 5 N upwards at 2 m from P, 4 N downwards at 3 m from P, and 3 N upwards at the left end (4 m from P). Find the resultant moment about P.

1

Taking moments about P (clockwise positive):

2

5 N at 2 m from P: Upward on left → CW →

M_{1} = + 5 \times 2 = + 10 Nm

3

4 N at 3 m from P: Downward → ACW →

M_{2} = - 4 \times 3 = - 12 Nm

4

3 N at 4 m from P: Upward → CW →

M_{3} = + 3 \times 4 = + 12 Nm

5

Resultant:

Σ M = 10 - 12 + 12 = 10 Nm

Answer10 Nm clockwise

04Worked Example

Forces act on a light rod as shown. A 5 N force acts at 80° to the rod at 1 m from P. A 6 N force acts vertically downwards at 4 m from P. A 4 N force acts at 50° to the rod at the end (6 m from P), angled to the left. Calculate the resultant moment about P.

1

5 N at 80° (1 m from P): Perpendicular component =

5 sin 80° = 4.92

N. ACW →

M_{1} = + 5 sin 80° \times 1 = + 4.92 Nm

2

6 N downwards (4 m from P): Already perpendicular. CW →

M_{2} = - 6 \times 4 = - 24 Nm

3

4 N at 50° (6 m from P): Perpendicular component =

4 sin 50° = 3.06

N. ACW →

M_{3} = + 4 sin 50° \times 6 = + 18.39 Nm

4

Resultant:

Σ M = 4.92 - 24 + 18.39 = - 0.69 Nm

Answer0.69 Nm clockwise (3 s.f.)

Practice

Question 1: A force of 50 N acts at the end of a rod of length 0.8 m. The force makes an angle of 60° above the horizontal. The rod is horizontal. Calculate the moment of this force about the pivot at the other end.

1

F = 50 N

,

d = 0.8 m

,

θ = 60°

2

Perpendicular component:

F sin 60° = 50 \times \frac{3}{2} = 253 N

3

M = F sin θ \times d = 253 \times 0.8 = 203 \approx 34.6 Nm

Answer

203

Nm ≈ 34.6 Nm

Practice

Question 2: A horizontal rod has length 4 m with pivot P at the right end. Forces act as shown: 2 N upwards at 2 m from P, 3 N upwards at 1 m from P, and 4 N downwards at 1 m from P. Calculate the resultant moment about P.

1

2 N at 2 m from P: Upward on left → CW →

M_{1} = + 2 \times 2 = + 4 Nm

2

3 N at 1 m from P: Upward → CW →

M_{2} = + 3 \times 1 = + 3 Nm

3

4 N at 1 m from P: Downward → ACW →

M_{3} = - 4 \times 1 = - 4 Nm

4

Resultant:

Σ M = 4 + 3 - 4 = + 3 Nm

Answer3 Nm clockwise

Practice

Question 3: A horizontal plank of length 5 m is pivoted at its centre. A 40 N force acts vertically downwards at the left end. What force must be applied vertically downwards at 0.5 m from the right end to produce a resultant moment of 30 Nm clockwise about the pivot?

1

40 N at left end: 2.5 m from pivot. Causes ACW rotation →

M_{1} = - 40 \times 2.5 = - 100 Nm

2

F at 0.5 m from right end = 2 m from pivot. Causes CW rotation →

M_{2} = + F \times 2

3

Resultant moment = 30 Nm clockwise = +30 Nm

4

- 100 + 2 F = + 30

5

2 F = 130

, so

F = 65 N

Answer65 N

Summary

Key Formulas

FormulaWhen to Use

M = F d

Force perpendicular to lever arm

M = F sin θ \times d

Force at angle θ to horizontal (horizontal rod)

M = F \times d sin θ

Finding perpendicular distance via d sin θ

Σ M = M_{1} + M_{2} + \dots

Multiple forces (with signs)

Exam Skills

Draw a diagram to identify perpendicular distances

Check whether to use sin or cos based on angle definition

Use consistent sign convention (ACW = +ve typically)

Choose your pivot strategically to simplify calculations

State your units clearly (Nm)