Moments and Couples Revision Notes for AQA A-Level Further Maths

Moments and Couples

What is a moment?

A moment measures the turning effect that a force produces about a fixed point. When you push a door to open it, you create a moment about the hinges. The size of this turning effect depends on two things: how hard you push (the force) and how far from the hinge you push (the perpendicular distance).

The fixed point about which rotation occurs is called a pivot. A moment is sometimes also called a torque.

infoNote

Understanding the concept: Think of opening a door - pushing near the edge (far from the hinges) makes it much easier to open than pushing near the hinges. This everyday experience demonstrates how the perpendicular distance affects the moment.

Definition: The moment of a force $F$ about a point $A$ is given by:

$\text{Moment} = F \times x$

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

The perpendicular distance means the shortest distance measured at right angles (90°) from the pivot point to the line along which the force acts.

Direction of moments

Moments can cause rotation in two directions:

Anticlockwise moments are taken as positive (+)
Clockwise moments are taken as negative (−)

chatImportant

Sign Convention: This sign convention is crucial when calculating resultant moments in equilibrium problems. Always state your convention clearly at the start of your solution, and be consistent throughout.

Units of moments

Moments are measured in Newton metres (Nm). Although this is the same unit as work (Joules), moments and work represent different physical quantities. While work can be measured in either Nm or J, moments are always expressed as Nm.

infoNote

Why not Joules? Even though Nm and Joules have the same dimensions, they represent fundamentally different quantities. Work involves movement along the direction of force, while moments involve turning effects. Using Nm for moments helps maintain this important distinction.

Resultant moments

When several forces act on an object simultaneously, each force produces its own turning effect. The total turning effect is found by adding all the individual moments together.

Key principle: The resultant moment of several forces equals the sum of the moments of the individual forces, taking into account their directions (clockwise or anticlockwise).

To find the resultant moment:

Calculate the moment of each force about the chosen point
Assign positive signs to anticlockwise moments and negative signs to clockwise moments
Add all the moments algebraically

Rotational equilibrium

An object that can rotate about a fixed point is in rotational equilibrium when the total moment acting on it is zero. This means the clockwise moments exactly balance the anticlockwise moments.

chatImportant

Equilibrium condition: If a system of forces is in equilibrium, the resultant moment about any point is zero.

Mathematically: $\sum M = 0$ or Clockwise moments = Anticlockwise moments

This principle is powerful because when a system is in equilibrium, you can calculate moments about any convenient point to find unknown forces or distances.

Worked example 1: Rectangular lamina with three forces

lightbulbExample

Worked Example: Calculating Resultant Moment

Problem: A light rectangular lamina $L$ can rotate about point $A$ . Find the resultant moment of three forces about $A$ : 6 N acting upwards at distance 5 m, 7 N acting horizontally at the top, and 8 N acting downwards at distance 3 m from $A$ .

Solution:

Taking anticlockwise as positive:

$\text{Total moment} = (6 \times 5) + (7 \times 0) - (8 \times 3)$

$= 30 + 0 - 24 = 6 \text{ Nm}$

The resultant moment is 6 Nm anticlockwise.

Note: The 7 N force passes through point $A$ , so it produces zero moment about $A$ (perpendicular distance = 0).

Couples

Sometimes two forces work together to create a turning effect without causing any overall translation (movement in a straight line). This special arrangement is called a couple.

Definition: A couple comprises two equal and opposite forces which do not act in the same straight line.

infoNote

Examples of couples in everyday life:

Unscrewing a bottle top (your hands apply two opposite forces)
Turning a steering wheel
Using a screwdriver
Turning a doorknob

In each case, two equal and opposite forces create a pure turning effect without moving the object in any direction.

Moment of a couple

The moment of a couple has a unique property: it is the same regardless of which point you calculate it about.

Formula: The moment of a couple equals:

$M = F \times d$

where:

$F$ is the magnitude of each force
$d$ is the perpendicular distance between the two forces

chatImportant

This independence from the reference point makes couples particularly useful in mechanics problems. You can choose any convenient point for calculations and get the same answer.

Worked example 2: Cantilever beam equilibrium

lightbulbExample

Worked Example: Equilibrium of a Cantilever Beam

Problem: A cantilever is formed by clamping a uniform horizontal beam $AB$ at $A$ . The beam is 4 m long, weighs 20 N, and has a load of 15 N hanging vertically at $B$ . Find the force and couple that must act at $A$ to keep the beam in equilibrium.

Solution:

Let the force at $A$ have components $X$ and $Y$ , and let the couple at $A$ have moment $M$ .

Step 1: Resolve horizontally

$X = 0$

No horizontal forces act, so $X = 0$ .

Step 2: Resolve vertically

$Y = 20 + 15 = 35 \text{ N}$

The upward force must balance the two downward forces.

Step 3: Take moments about A

Taking anticlockwise as positive:

$20 \times 2 + 15 \times 4 = M$

$40 + 60 = M$

$M = 100 \text{ Nm}$

Answer: The force at $A$ is 35 N upwards and the couple at $A$ is 100 Nm anticlockwise.

Physical interpretation: The components of the force at $A$ prevent translation (movement), while the couple prevents rotation. In equilibrium, you need both a balance of forces and a balance of moments.

Calculating moments of forces at angles

When a force acts at an angle (not perpendicular to the beam or rod), you can use one of two methods to calculate its moment.

Method 1: Resolve the force into components

Resolve the force into two perpendicular components, then calculate the moment of each component separately.

For a force $F$ at angle $\theta$ :

Component along the line: $F \cos \theta$
Component perpendicular to the line: $F \sin \theta$

$\text{Moment} = F\cos\theta \times 0 + F\sin\theta \times l = Fl\sin\theta$

Method 2: Use perpendicular distance from line of action

Draw the line of action of the force and find the perpendicular distance from the pivot to this line.

$\text{Moment} = F \times x$

where $x = l\sin\theta$ is the perpendicular distance from the pivot to the line of action.

infoNote

Key insight: Both methods give the same result: $M = Fl\sin\theta$

Choose whichever method seems more natural for the problem at hand. Method 1 is often easier when the geometry is simple, while Method 2 can be more elegant when you can easily visualize the perpendicular distance.

Worked example 3: Force at an angle to a rod

lightbulbExample

Worked Example: Moment of an Angled Force

Problem: A force $F$ of 10 N acts at end $B$ of a light rod $AB$ of length $l$ , making an angle $\theta$ with the rod. If $l = 3$ m and $\theta = 30°$ , find the moment of $F$ about $A$ .

Solution using Method 1:

Resolve $F$ into components parallel and perpendicular to $AB$ :

$\text{Moment of } F \text{ about } A = F\cos\theta \times 0 + F\sin\theta \times l$

$= 0 + 10\sin 30° \times 3$

$= 10 \times 0.5 \times 3 = 15 \text{ Nm}$

Solution using Method 2:

Find perpendicular distance from $A$ to line of action of $F$ :

$\text{Moment of } F \text{ about } A = F \times x$

$= F \times l\sin\theta$

$= 10 \times 3\sin 30° = 15 \text{ Nm}$

Both methods confirm the moment is 15 Nm anticlockwise.

Worked example 4: Square lamina in equilibrium

lightbulbExample

Worked Example: Equilibrium of a Square Lamina

Problem: A light square lamina $ABCD$ of side 10 cm can rotate about point $A$ . Three forces act on it: $F$ acting horizontally at $A$ , 2 N acting downwards at $C$ , and $8\sqrt{2}$ N acting at 45° at $B$ . Find the magnitude of force $F$ for the lamina to be in equilibrium.

Solution:

First, find the diagonal length: In $\triangle AOB$ , $x^2 + x^2 = 10^2$ , so $x = \frac{10}{\sqrt{2}} = 5\sqrt{2}$ cm.

Taking moments about point $A$ for equilibrium:

Anticlockwise moments = Clockwise moments

$8\sqrt{2} \times 5\sqrt{2} = (2 \times 10) + (F \times 5)$

$80 = 20 + 5F$

$60 = 5F$

$F = 12 \text{ N}$

Alternative approach: Set total moment about $A$ equal to zero:

$(8\sqrt{2} \times 5\sqrt{2}) - (2 \times 10) - (F \times 5) = 0$

$80 - 20 - 5F = 0$

$F = 12 \text{ N}$

Therefore, $F =$ 12 N.

Exam tip: The reaction at point $A$ doesn't appear in the moment equation because moments are taken about $A$ . This is a strategic choice that simplifies the calculation. You still need to check force balance separately.

Problem-solving strategy

When solving problems involving moments and couples, follow this systematic approach:

Step 1: Draw a clear diagram showing all forces and distances

Step 2: Choose a convenient point to take moments about (often the pivot or a point where unknown forces act, as these forces will have zero moment about that point)

Step 3: Establish a sign convention (typically anticlockwise = positive)

Step 4: Calculate the moment of each force using $M = F \times x$

For forces at angles, choose either component method or perpendicular distance method
Remember that forces passing through the chosen point have zero moment

Step 5: Apply equilibrium condition: Sum of anticlockwise moments = Sum of clockwise moments

Step 6: Solve for unknown quantities

Step 7: Check your answer makes physical sense

chatImportant

Strategic thinking: The choice of which point to take moments about can make the difference between a simple calculation and a complex one. Always look for points where unknown forces act - this eliminates them from your moment equation.

Exam tips and common pitfalls

chatImportant

Choosing the reference point wisely:

When taking moments, choose a point where unknown forces act. This eliminates those forces from your moment equation, making calculations simpler.

chatImportant

Perpendicular distance is crucial:

Always use the perpendicular (shortest) distance from the point to the line of action of the force. This is the most common error in moment calculations. If in doubt, draw the line of action of the force and measure the perpendicular distance from your reference point to this line.

infoNote

Sign conventions matter:

Be consistent with your sign convention throughout a problem. State clearly whether you're taking clockwise or anticlockwise as positive at the start of your solution.

infoNote

Couples are special:

Remember that the moment of a couple is independent of the point chosen. Use this property to simplify problems - you can calculate the moment about any convenient point.

chatImportant

Units check:

Moments are always in Nm, not Joules, even though the dimensions are the same. This distinction is important for clarity and avoiding confusion.

infoNote

Forces through the pivot:

Any force passing through the pivot point produces zero moment about that point. This is a powerful simplification - choose your pivot wisely to eliminate unknown forces from your calculations.

Remember!

bookmarkSummary

Key Points to Remember:

Moment = Force × Perpendicular distance: The turning effect depends on both how hard you push and where you push from
Equilibrium means zero total moment: When clockwise moments equal anticlockwise moments, there's no net rotation
Couples are independent: The moment of a couple ( $F \times d$ ) has the same value regardless of which point you calculate it about
Choose your point strategically: Taking moments about a point where unknown forces act simplifies calculations significantly
Two methods for angled forces: You can either resolve into components or use perpendicular distance – both give $M = Fl\sin\theta$

Moments and Couples (AQA A-Level Further Maths): Revision Notes