3.6.1 Moment of a force (A-level only) - Revision Guide

What you'll learn

What the moment of a force means and how it measures a turning effect.
How to identify the correct perpendicular distance from an axis to a force’s line of action.
How to calculate moments using $M = F d$ or $M=Frsin⁡θM=Fr\sin\theta$ .
How to assign signs to clockwise and anticlockwise moments.

Forces and turning effects

A force is a push or pull, measured in newtons (N). As well as changing an object’s motion, a force can make a rigid body turn.

For example, pushing a door near its handle produces a much greater turning effect than pushing close to its hinges. The force may be the same, but its distance from the axis of rotation is different.

An axis is a line about which a body can rotate. In a two-dimensional mechanics diagram, the axis passes through a point in the plane of the body and is perpendicular to that plane. It is often represented by a pivot labelled $O$ .

Definition

Moment of a force

The moment of a force about an axis is the turning effect of the force about that axis.

A moment may act clockwise or anticlockwise, depending on which way the force would tend to rotate the body.

The line of action

The line of action of a force is the straight line extending through the force arrow in both directions.

To calculate a moment about an axis, you need the shortest distance from the axis to this line. The shortest distance between a point and a line is always measured perpendicularly.

The diagram shows the key quantities. The force $F$ acts at $A$ , while $d$ is the perpendicular distance from $O$ to the force’s line of action.

A force acting at an angle to a rigid bar, with its line of action and perpendicular distance from the axis labelled

Key Idea

Use the perpendicular distance

The distance used in a moment calculation is the perpendicular distance from the axis to the force’s line of action, not necessarily the distance from the axis to the point where the force is applied.

Calculating a moment

If a force of magnitude $F$ has a line of action at perpendicular distance $d$ from the axis, the magnitude of its moment is

M = F d .

The SI unit of moment is the newton metre, written N m.

Notice that moment is measured in N m, not joules (J). Although both units involve newtons and metres, moment and energy are different physical quantities.

Example

Using a perpendicular distance

A horizontal force of 15 N acts on a body. Its line of action is 0.40 m above an axis through $O$ . Calculate the magnitude of its moment about the axis.

The given distance is perpendicular to the force’s horizontal line of action, so use $md=0.40\text{ m}$ .
Apply the moment formula:
$M=Fd=15×0.40.M=Fd=15\times 0.40.$
Therefore, the magnitude of the moment is 6 N m.

Common Mistake

Using the wrong length

Do not automatically multiply the force by the distance to its point of application. That length is valid only when it is perpendicular to the force’s line of action.

A force acting at an angle

Suppose a force $F$ acts at a point whose distance from the axis is $r$ . If the angle between the position line and the force is $θ\theta$ , the perpendicular distance is

d=rsin⁡θ.d=r\sin\theta.

Therefore,

M=F(rsin⁡θ)=Frsin⁡θ.M=F(r\sin\theta)=Fr\sin\theta.

This is an alternative form of the moment formula:

M=Frsin⁡θ.\boxed{M=Fr\sin\theta}.

Here:

$F$ is the magnitude of the force;
$r$ is the distance from the axis to the point where the force acts;
$θ\theta$ is the angle between the position line and the force.

Example

Finding the moment of an angled force

A force of 20 N is applied at the end of a rod, 0.60 m from an axis through $O$ . The force makes an angle of $35∘35^\circ$ with the rod. Find the magnitude of its moment about the axis.

The rod gives the distance $mr=0.60\text{ m}$ , but it is not perpendicular to the force. Use $M=Frsin⁡θM=Fr\sin\theta$ .
Substitute the force, distance and included angle:
$M=20×0.60×sin⁡35∘.M=20\times 0.60\times\sin 35^\circ.$
This gives
$m.M\approx 6.88\text{ N m}.$

Tip

Check the angle

In $M=Frsin⁡θM=Fr\sin\theta$ , $θ\theta$ must be the angle between the force and the line joining the axis to the point of application. If the question gives the complementary angle, use it carefully.

Resolving the force instead

There is another way to reach the same result. Resolve the force into components:

the component parallel to the position line is $Fcos⁡θF\cos\theta$ ;
the component perpendicular to the position line is $Fsin⁡θF\sin\theta$ .

The parallel component acts along a line through the axis, so it produces no moment. Only the perpendicular component turns the body:

M=(Fsin⁡θ)r.M=(F\sin\theta)r.

This is the same as $Frsin⁡θFr\sin\theta$ .

Key Idea

Only the perpendicular effect turns

You may either multiply the whole force by its perpendicular distance from the axis, or multiply the perpendicular component of the force by the distance to its point of application.

When the moment is zero

If the line of action of a force passes through the axis, its perpendicular distance from the axis is zero. Hence,

M=F×0=0.M=F\times 0=0.

The force may still cause linear acceleration, but it has no turning effect about that particular axis.

This also follows from $M=Frsin⁡θM=Fr\sin\theta$ . When the force acts directly towards or away from the axis, $θ=0∘\theta=0^\circ$ or $180∘180^\circ$ , so $sin⁡θ=0\sin\theta=0$ .

Common Mistake

The chosen axis matters

A force can have zero moment about one axis but a non-zero moment about another. Always calculate distances from the specific axis named in the question.

Clockwise and anticlockwise moments

When several forces act, their moments may oppose one another. You therefore need a sign convention.

A common choice is:

anticlockwise moments are positive;
clockwise moments are negative.

You may choose the opposite convention, but you must use it consistently throughout your calculation.

Example

Combining moments about an axis

Two forces act on a horizontal bar about an axis through $O$ :

a downward force of 8 N acts 0.50 m to the right of $O$ ;
an upward force of 5 N acts 0.30 m to the right of $O$ .

Find the resultant moment about $O$ , taking anticlockwise as positive.

The 8 N downward force tends to turn the bar clockwise, so its moment is
$m.-8\times 0.50=-4\text{ N m}.$
The 5 N upward force tends to turn the bar anticlockwise, so its moment is
$m.5\times 0.30=1.5\text{ N m}.$
Add the signed moments:
$m.M_{\text{resultant}}=-4+1.5=-2.5\text{ N m}.$
The negative sign means the resultant moment is 2.5 N m clockwise.

Common Mistake

Adding all moment magnitudes

Opposing moments must not simply be added. Decide whether each force acts clockwise or anticlockwise, then combine the moments using a consistent sign convention.

A reliable calculation method

For each force:

Identify the axis about which moments are required.
Extend the force mentally to identify its line of action.
Find the perpendicular distance from the axis to that line.
Calculate the moment using $M = F d$ .
Determine whether the turning effect is clockwise or anticlockwise.
If there are several forces, add their signed moments.

A useful modelling assumption is that the body is rigid, meaning that it does not deform as the forces act. This lets you treat all distances within the body as fixed.

Exam technique

In the exam

Mark the named axis and draw or imagine each force’s complete line of action.
Use a perpendicular distance with $M = F d$ ; if only a sloping length is available, use $M=Frsin⁡θM=Fr\sin\theta$ or resolve the force.
State a sign convention before combining several moments, and interpret the sign of your final answer.
Include the unit N m and keep full calculator accuracy until the final rounding step.

Self review

Check yourself

Why does a force whose line of action passes through the axis have zero moment about that axis?
A force $F$ acts a distance $r$ from an axis at angle $θ\theta$ to the position line. How would you calculate its moment?
How would you decide whether a moment should be recorded as positive or negative?

What you'll learn

What the moment of a force means and how it measures a turning effect.
How to identify the correct perpendicular distance from an axis to a force’s line of action.
How to calculate moments using $M = F d$ or $M=Frsin⁡θM=Fr\sin\theta$ .
How to assign signs to clockwise and anticlockwise moments.

Forces and turning effects

A force is a push or pull, measured in newtons (N). As well as changing an object’s motion, a force can make a rigid body turn.

Definition

Moment of a force

The moment of a force about an axis is the turning effect of the force about that axis.

A moment may act clockwise or anticlockwise, depending on which way the force would tend to rotate the body.

The line of action

The line of action of a force is the straight line extending through the force arrow in both directions.

To calculate a moment about an axis, you need the shortest distance from the axis to this line. The shortest distance between a point and a line is always measured perpendicularly.

The diagram shows the key quantities. The force $F$ acts at $A$ , while $d$ is the perpendicular distance from $O$ to the force’s line of action.

A force acting at an angle to a rigid bar, with its line of action and perpendicular distance from the axis labelled

Key Idea

Use the perpendicular distance

Calculating a moment

If a force of magnitude $F$ has a line of action at perpendicular distance $d$ from the axis, the magnitude of its moment is

M = F d .

The SI unit of moment is the newton metre, written N m.

Notice that moment is measured in N m, not joules (J). Although both units involve newtons and metres, moment and energy are different physical quantities.

Example

Using a perpendicular distance

A horizontal force of 15 N acts on a body. Its line of action is 0.40 m above an axis through $O$ . Calculate the magnitude of its moment about the axis.

The given distance is perpendicular to the force’s horizontal line of action, so use $md=0.40\text{ m}$ .
Apply the moment formula:
$M=Fd=15×0.40.M=Fd=15\times 0.40.$
Therefore, the magnitude of the moment is 6 N m.

Common Mistake

Using the wrong length

Do not automatically multiply the force by the distance to its point of application. That length is valid only when it is perpendicular to the force’s line of action.

A force acting at an angle

Suppose a force $F$ acts at a point whose distance from the axis is $r$ . If the angle between the position line and the force is $θ\theta$ , the perpendicular distance is

d=rsin⁡θ.d=r\sin\theta.

Therefore,

M=F(rsin⁡θ)=Frsin⁡θ.M=F(r\sin\theta)=Fr\sin\theta.

This is an alternative form of the moment formula:

M=Frsin⁡θ.\boxed{M=Fr\sin\theta}.

Here:

$F$ is the magnitude of the force;
$r$ is the distance from the axis to the point where the force acts;
$θ\theta$ is the angle between the position line and the force.

Example

Finding the moment of an angled force

A force of 20 N is applied at the end of a rod, 0.60 m from an axis through $O$ . The force makes an angle of $35∘35^\circ$ with the rod. Find the magnitude of its moment about the axis.

The rod gives the distance $mr=0.60\text{ m}$ , but it is not perpendicular to the force. Use $M=Frsin⁡θM=Fr\sin\theta$ .
Substitute the force, distance and included angle:
$M=20×0.60×sin⁡35∘.M=20\times 0.60\times\sin 35^\circ.$
This gives
$m.M\approx 6.88\text{ N m}.$

Tip

Check the angle

Resolving the force instead

There is another way to reach the same result. Resolve the force into components:

the component parallel to the position line is $Fcos⁡θF\cos\theta$ ;
the component perpendicular to the position line is $Fsin⁡θF\sin\theta$ .

The parallel component acts along a line through the axis, so it produces no moment. Only the perpendicular component turns the body:

M=(Fsin⁡θ)r.M=(F\sin\theta)r.

This is the same as $Frsin⁡θFr\sin\theta$ .

Key Idea

Only the perpendicular effect turns

You may either multiply the whole force by its perpendicular distance from the axis, or multiply the perpendicular component of the force by the distance to its point of application.

When the moment is zero

If the line of action of a force passes through the axis, its perpendicular distance from the axis is zero. Hence,

M=F×0=0.M=F\times 0=0.

The force may still cause linear acceleration, but it has no turning effect about that particular axis.

This also follows from $M=Frsin⁡θM=Fr\sin\theta$ . When the force acts directly towards or away from the axis, $θ=0∘\theta=0^\circ$ or $180∘180^\circ$ , so $sin⁡θ=0\sin\theta=0$ .

Common Mistake

The chosen axis matters

A force can have zero moment about one axis but a non-zero moment about another. Always calculate distances from the specific axis named in the question.

Clockwise and anticlockwise moments

When several forces act, their moments may oppose one another. You therefore need a sign convention.

A common choice is:

anticlockwise moments are positive;
clockwise moments are negative.

You may choose the opposite convention, but you must use it consistently throughout your calculation.

Example

Combining moments about an axis

Two forces act on a horizontal bar about an axis through $O$ :

a downward force of 8 N acts 0.50 m to the right of $O$ ;
an upward force of 5 N acts 0.30 m to the right of $O$ .

Find the resultant moment about $O$ , taking anticlockwise as positive.

The 8 N downward force tends to turn the bar clockwise, so its moment is
$m.-8\times 0.50=-4\text{ N m}.$
The 5 N upward force tends to turn the bar anticlockwise, so its moment is
$m.5\times 0.30=1.5\text{ N m}.$
Add the signed moments:
$m.M_{\text{resultant}}=-4+1.5=-2.5\text{ N m}.$
The negative sign means the resultant moment is 2.5 N m clockwise.

Common Mistake

Adding all moment magnitudes

Opposing moments must not simply be added. Decide whether each force acts clockwise or anticlockwise, then combine the moments using a consistent sign convention.

A reliable calculation method

For each force:

Identify the axis about which moments are required.
Extend the force mentally to identify its line of action.
Find the perpendicular distance from the axis to that line.
Calculate the moment using $M = F d$ .
Determine whether the turning effect is clockwise or anticlockwise.
If there are several forces, add their signed moments.

A useful modelling assumption is that the body is rigid, meaning that it does not deform as the forces act. This lets you treat all distances within the body as fixed.

Exam technique

In the exam

Mark the named axis and draw or imagine each force’s complete line of action.
Use a perpendicular distance with $M = F d$ ; if only a sloping length is available, use $M=Frsin⁡θM=Fr\sin\theta$ or resolve the force.
State a sign convention before combining several moments, and interpret the sign of your final answer.
Include the unit N m and keep full calculator accuracy until the final rounding step.

Self review

Check yourself

Why does a force whose line of action passes through the axis have zero moment about that axis?
A force $F$ acts a distance $r$ from an axis at angle $θ\theta$ to the position line. How would you calculate its moment?
How would you decide whether a moment should be recorded as positive or negative?

3.4.1 Moment of a force (A-level only)

What you'll learn

Forces and turning effects

The line of action

Calculating a moment

A force acting at an angle

Resolving the force instead

When the moment is zero

Clockwise and anticlockwise moments

A reliable calculation method

Review 3.6.1 Moment of a force (A-level only) by teaching Genie

3.4.1 Moment of a force (A-level only)

What you'll learn

Forces and turning effects

The line of action

Calculating a moment

A force acting at an angle

Resolving the force instead

When the moment is zero

Clockwise and anticlockwise moments

A reliable calculation method

Review 3.6.1 Moment of a force (A-level only) by teaching Genie

3.4.1 Moment of a force (A-level only)

What you'll learn

Forces and turning effects

The line of action

Calculating a moment

A force acting at an angle

Resolving the force instead

When the moment is zero

Clockwise and anticlockwise moments

A reliable calculation method

Review 3.6.1 Moment of a force (A-level only) by teaching Genie

3.6.1 Moment of a force (A-level only) Revision Guide

3.4.1 Moment of a force (A-level only)

What you'll learn

Forces and turning effects

The line of action

Calculating a moment

A force acting at an angle

Resolving the force instead

When the moment is zero

Clockwise and anticlockwise moments

A reliable calculation method

Review 3.6.1 Moment of a force (A-level only) by teaching Genie

3.6.1 Moment of a force (A-level only) Revision Guide