What is the equilibrium of forces in the context of Physics?

In Physics, particularly within the Cambridge International A Levels curriculum, the equilibrium of forces refers to a state where all the forces acting on an object are balanced. This means that the net force is zero, resulting in the object being in a state of rest or moving with constant velocity. This concept is crucial for understanding how forces interact and is fundamental in solving problems related to statics and dynamics.

How can you determine if an object is in equilibrium?

To determine if an object is in equilibrium, you need to ensure that the vector sum of all forces acting on the object is zero. This involves analyzing both the horizontal and vertical components of the forces. Additionally, for rotational equilibrium, the sum of all torques (moments) about any axis must also be zero. These conditions ensure that the object is not accelerating and is either at rest or moving with constant velocity.

What are the conditions for equilibrium in a system of forces?

The conditions for equilibrium in a system of forces include two main criteria: translational equilibrium and rotational equilibrium. Translational equilibrium requires that the sum of all horizontal forces and the sum of all vertical forces are zero. Rotational equilibrium requires that the sum of all torques about any point is zero. These conditions ensure that the object does not experience any linear or angular acceleration.

Why is understanding equilibrium of forces important in Physics?

Understanding the equilibrium of forces is crucial in Physics because it helps explain how objects behave under various force conditions. It is essential for solving problems related to structures, such as bridges and buildings, ensuring they remain stable. Additionally, it aids in understanding mechanical systems and is a foundational concept in engineering and design, making it a vital topic in the Cambridge International A Levels Physics curriculum.

Can you provide an example of equilibrium of forces in everyday life?

An example of equilibrium of forces in everyday life is a book resting on a table. The book is in equilibrium because the gravitational force pulling it downward is balanced by the normal force exerted by the table pushing it upward. These forces are equal in magnitude and opposite in direction, resulting in no net force acting on the book, thus keeping it stationary. This simple scenario illustrates the principle of equilibrium of forces.

Why is "Resolving forces but ignoring moments" a common mistake in Equilibrium of forces?

Why it happens: Rigid body confused with particle. How to avoid it: Rigid body equilibrium requires BOTH net force = 0 AND net moment = 0. Choose pivot strategically. Source: 9702 Examiner Reports 2022-2024

Why is "Confusing centre of mass with geometric centre for stability" a common mistake in Equilibrium of forces?

Why it happens: For uniform bodies they coincide. How to avoid it: Centre of MASS determines stability. For uniform body = geometric centre. For non-uniform, may differ. Source: 9702 Examiner Reports 2022-2024

Cambridge International A Levels Physics (9702)

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Short Notes - Equilibrium of forces

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Detailed Study Notes

Detailed notes on Forces, Density and Pressure for Cambridge International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Equilibrium of Forces Study Notes — Cambridge International A Level Physics 9702 (2025-2027 syllabus)

Conditions for translational and rotational equilibrium. Three-force problems. Stability.

At a glance

Translational equilibrium: $\sum F = 0$ .
Rotational equilibrium: $\sum M = 0$ .
Both required for rigid body.
Stability: stable / unstable / neutral.
3-force concurrent equilibrium: closed triangle.

What you’ll learn

Mapped to the Cambridge International A Level 9702 syllabus (2025-2027).

4.2.1 — Apply equilibrium conditions.
4.2.2 — Solve 3-force concurrent problems.
4.2.3 — Classify stability.

Equilibrium conditions

Force AND moment.

Two conditions for rigid body equilibrium:

Translational. $\sum F = 0$ (resolved in perpendicular directions: $\sum F_x = 0$ and $\sum F_y = 0$ ).
Rotational. $\sum M = 0$ about ANY point.

For a particle, only condition 1 applies (no rotation possible).

Method.

Sketch free-body diagram.
Resolve forces into perpendicular axes.
Apply $\sum F_x = 0$ , $\sum F_y = 0$ .
Take moments about chosen pivot, $\sum M = 0$ .
Solve system.

Cambridge tip. Choose pivot through an unknown force to eliminate it from moment equation — simplifies algebra.

Two conditions for rigid body.
Free-body diagram first.
Strategic pivot choice.

Three-force concurrent equilibrium

Closed triangle of forces.

Three concurrent forces in equilibrium form a CLOSED TRIANGLE when drawn head-to-tail.

Approach 1: components. Resolve into perpendicular directions; apply $\sum F = 0$ in each.

Approach 2: triangle of forces. Draw triangle scale-and-direction; apply trigonometry / sine rule.

Lami's theorem. For three concurrent forces in equilibrium: $\frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma}$ where $\alpha$ is opposite $F_1$ , etc.

Example. Object 20 N hanging from two strings at $30°$ and $45°$ to horizontal.

Resolve: $T_1\cos 30° = T_2\cos 45°$ ; $T_1\sin 30° + T_2\sin 45° = 20$ .
Solve: $T_1 \approx 14.6$ N; $T_2 \approx 17.9$ N.

Cambridge tip. Both methods OK; pick whichever gives cleaner algebra.

Three concurrent forces are in equilibrium when their vector sum is zero; placed head-to-tail they form a closed triangle.

Closed triangle.
Components or trig.
Lami's theorem optional.

See the full worked example for equilibrium of forces →

Stability

Centre of mass behaviour.

Stable equilibrium. Body returns to original position after small displacement.

Centre of mass at MINIMUM height locally.
Example: ball at bottom of bowl.

Unstable equilibrium. Body moves further from equilibrium.

Centre of mass at MAXIMUM height locally.
Example: ball balanced on dome.

Neutral equilibrium. Body remains in displaced position.

Centre of mass at CONSTANT height.
Example: ball on horizontal surface.

Toppling. A body topples when its vertical line through centre of mass falls OUTSIDE its base of support.

Cambridge tip. Sketch + label centre of mass for top marks.

Stable: min height.
Unstable: max height.
Neutral: constant height.

See the full worked example for equilibrium of forces →

Quick recap

Equilibrium: $\sum F = 0$ AND $\sum M = 0$ .
Three forces: closed triangle / Lami.
Stability classified by centre of mass behaviour.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Two equilibrium conditions.
Three stability classes.
Lami's theorem (optional).

How it’s examined

Equilibrium heavily tested. Most-tested: 3-force problem (8 marks), ladder/beam (10 marks), stability identification (5 marks).

Sources: Cambridge International A Level Physics 9702 syllabus (2025-2027); 9702 Examiner Reports 2022-2024; 9702/22 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

Take this whole topic with you

Step-by-step worked examples — Equilibrium of forces

Step-by-step solutions to past-paper-style questions on equilibrium of forces, written exactly the way a tutor would explain them at the board.

1Three forces in equilibrium (8 marks)
Extended• Adapted from 9702/22 May/Jun 2024• equilibrium
Question
An object of weight 20 N hangs from two strings, one at $30°$ and the other at $45°$ to the horizontal ceiling. Find the tensions $T_1$ (at $30°$ ) and $T_2$ (at $45°$ ). (8 marks)
Step-by-step solution
1. Step 1
  Resolve horizontally. $T_1 \cos 30° = T_2 \cos 45°$ .
2. Step 2
  Resolve vertically. $T_1 \sin 30° + T_2 \sin 45° = 20$ .
3. Step 3
  From horizontal: $T_1 = T_2 \dfrac{\cos 45°}{\cos 30°} = T_2 \times 0.8165$ .
4. Step 4
  Substitute into vertical.
  $0.8165 T_2 \times 0.5 + 0.7071 T_2 = 20 \Rightarrow 1.116 T_2 = 20 \Rightarrow T_2 \approx 17.9 \text{ N}$
5. Step 5
  $T_1 = 0.8165 \times 17.9 \approx 14.6$ N.
Answer
$T_1 \approx 14.6$ N; $T_2 \approx 17.9$ N.
2Stability (5 marks)
Extended• stability
Question
Define stable, unstable and neutral equilibrium. (5 marks)
Step-by-step solution
1. Step 1
  Stable equilibrium. When slightly displaced, body returns to original position (centre of mass rises when displaced, then falls back).
2. Step 2
  Unstable. Slight displacement causes body to move further from equilibrium (centre of mass falls when displaced).
3. Step 3
  Neutral. Body remains in new position after displacement (centre of mass at same height).
Answer
Stable: returns. Unstable: moves further. Neutral: stays.
3Ladder against wall (10 marks)
Extended• ladder
Question
A uniform ladder of weight 200 N rests at $60°$ to the horizontal against a smooth vertical wall. Find the horizontal reaction $N_w$ from the wall and the friction $F$ at the ground. (10 marks)
Step-by-step solution
1. Step 1
  Forces. Weight $W = 200$ N (centre); wall normal $N_w$ (horizontal); ground normal $N_g$ (vertical); friction $F$ (horizontal).
2. Step 2
  Vertical equilibrium. $N_g = W = 200$ N.
3. Step 3
  Horizontal equilibrium. $F = N_w$ .
4. Step 4
  Take moments about the foot of the ladder (length $L$ ). Weight acts at $L/2$ from foot. Wall reaction at $L\sin 60°$ vertical height (= $L\cos 60°$ horizontal arm).
5. Step 5
  Anticlockwise (wall force). $N_w \times L\sin 60°$ . Clockwise (weight). $W \times (L/2)\cos 60°$ .
6. Step 6
  Equate.
  $N_w \sin 60° = \dfrac{W}{2}\cos 60° \Rightarrow N_w = \dfrac{200 \cos 60°}{2 \sin 60°} \approx 57.7 \text{ N}$
7. Step 7
  $F = N_w \approx 57.7$ N.
Answer
$N_w \approx 57.7$ N; $F \approx 57.7$ N.

Key Formulae — Equilibrium of forces

The formulae you need to memorise for equilibrium of forces on the Cambridge International A Level 9702 paper, with every variable defined in plain English and a note on when to use it.

Equilibrium conditions
$\sum F_x = 0, \quad \sum F_y = 0, \quad \sum M = 0$
When to use
Rigid body in equilibrium. Both translational AND rotational.

Key Definitions and Keywords — Equilibrium of forces

Definitions to memorise and the exact keywords mark schemes credit for equilibrium of forces answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9702 sitting.

Equilibrium
Examiner keyword
Net force AND net moment both zero. Body at rest or moving uniformly with constant rotation.
Stable equilibrium
Examiner keyword
Body returns to original position after small displacement. Centre of mass at minimum height.
Unstable equilibrium
Examiner keyword
Body moves further from equilibrium after small displacement. Centre of mass at maximum height.
Neutral equilibrium
Body remains at new position after displacement. Centre of mass at constant height.

Common Mistakes and Misconceptions — Equilibrium of forces

The traps other students keep falling into on equilibrium of forces questions — taken from recent Cambridge International A Level 9702 examiner reports and mark schemes — and how to avoid them.

✕Resolving forces but ignoring moments
9702 Examiner Reports 2022-2024
Why it happens
Rigid body confused with particle.
How to avoid it
Rigid body equilibrium requires BOTH net force = 0 AND net moment = 0. Choose pivot strategically.
✕Confusing centre of mass with geometric centre for stability
9702 Examiner Reports 2022-2024
Why it happens
For uniform bodies they coincide.
How to avoid it
Centre of MASS determines stability. For uniform body = geometric centre. For non-uniform, may differ.

Equilibrium of forces — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Cambridge International A Levels Physics (9702)

Equilibrium of forces

0% Complete

3 tasks remaining

Short Notes - Equilibrium of forces

Page 1/0

Detailed Study Notes

Detailed notes on Forces, Density and Pressure for Cambridge International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Equilibrium of Forces Study Notes — Cambridge International A Level Physics 9702 (2025-2027 syllabus)

Conditions for translational and rotational equilibrium. Three-force problems. Stability.

At a glance

Translational equilibrium: $\sum F = 0$ .
Rotational equilibrium: $\sum M = 0$ .
Both required for rigid body.
Stability: stable / unstable / neutral.
3-force concurrent equilibrium: closed triangle.

What you’ll learn

Mapped to the Cambridge International A Level 9702 syllabus (2025-2027).

4.2.1 — Apply equilibrium conditions.
4.2.2 — Solve 3-force concurrent problems.
4.2.3 — Classify stability.

Equilibrium conditions

Force AND moment.

Two conditions for rigid body equilibrium:

Translational. $\sum F = 0$ (resolved in perpendicular directions: $\sum F_x = 0$ and $\sum F_y = 0$ ).
Rotational. $\sum M = 0$ about ANY point.

For a particle, only condition 1 applies (no rotation possible).

Method.

Sketch free-body diagram.
Resolve forces into perpendicular axes.
Apply $\sum F_x = 0$ , $\sum F_y = 0$ .
Take moments about chosen pivot, $\sum M = 0$ .
Solve system.

Cambridge tip. Choose pivot through an unknown force to eliminate it from moment equation — simplifies algebra.

Two conditions for rigid body.
Free-body diagram first.
Strategic pivot choice.

Three-force concurrent equilibrium

Closed triangle of forces.

Three concurrent forces in equilibrium form a CLOSED TRIANGLE when drawn head-to-tail.

Approach 1: components. Resolve into perpendicular directions; apply $\sum F = 0$ in each.

Approach 2: triangle of forces. Draw triangle scale-and-direction; apply trigonometry / sine rule.

Lami's theorem. For three concurrent forces in equilibrium: $\frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma}$ where $\alpha$ is opposite $F_1$ , etc.

Example. Object 20 N hanging from two strings at $30°$ and $45°$ to horizontal.

Resolve: $T_1\cos 30° = T_2\cos 45°$ ; $T_1\sin 30° + T_2\sin 45° = 20$ .
Solve: $T_1 \approx 14.6$ N; $T_2 \approx 17.9$ N.

Cambridge tip. Both methods OK; pick whichever gives cleaner algebra.

Three concurrent forces are in equilibrium when their vector sum is zero; placed head-to-tail they form a closed triangle.

Closed triangle.
Components or trig.
Lami's theorem optional.

See the full worked example for equilibrium of forces →

Stability

Centre of mass behaviour.

Stable equilibrium. Body returns to original position after small displacement.

Centre of mass at MINIMUM height locally.
Example: ball at bottom of bowl.

Unstable equilibrium. Body moves further from equilibrium.

Centre of mass at MAXIMUM height locally.
Example: ball balanced on dome.

Neutral equilibrium. Body remains in displaced position.

Centre of mass at CONSTANT height.
Example: ball on horizontal surface.

Toppling. A body topples when its vertical line through centre of mass falls OUTSIDE its base of support.

Cambridge tip. Sketch + label centre of mass for top marks.

Stable: min height.
Unstable: max height.
Neutral: constant height.

See the full worked example for equilibrium of forces →

Quick recap

Equilibrium: $\sum F = 0$ AND $\sum M = 0$ .
Three forces: closed triangle / Lami.
Stability classified by centre of mass behaviour.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Two equilibrium conditions.
Three stability classes.
Lami's theorem (optional).

How it’s examined

Equilibrium heavily tested. Most-tested: 3-force problem (8 marks), ladder/beam (10 marks), stability identification (5 marks).

Sources: Cambridge International A Level Physics 9702 syllabus (2025-2027); 9702 Examiner Reports 2022-2024; 9702/22 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

Take this whole topic with you

Step-by-step worked examples — Equilibrium of forces

Step-by-step solutions to past-paper-style questions on equilibrium of forces, written exactly the way a tutor would explain them at the board.

1Three forces in equilibrium (8 marks)
Extended• Adapted from 9702/22 May/Jun 2024• equilibrium
Question
An object of weight 20 N hangs from two strings, one at $30°$ and the other at $45°$ to the horizontal ceiling. Find the tensions $T_1$ (at $30°$ ) and $T_2$ (at $45°$ ). (8 marks)
Step-by-step solution
1. Step 1
  Resolve horizontally. $T_1 \cos 30° = T_2 \cos 45°$ .
2. Step 2
  Resolve vertically. $T_1 \sin 30° + T_2 \sin 45° = 20$ .
3. Step 3
  From horizontal: $T_1 = T_2 \dfrac{\cos 45°}{\cos 30°} = T_2 \times 0.8165$ .
4. Step 4
  Substitute into vertical.
  $0.8165 T_2 \times 0.5 + 0.7071 T_2 = 20 \Rightarrow 1.116 T_2 = 20 \Rightarrow T_2 \approx 17.9 \text{ N}$
5. Step 5
  $T_1 = 0.8165 \times 17.9 \approx 14.6$ N.
Answer
$T_1 \approx 14.6$ N; $T_2 \approx 17.9$ N.
2Stability (5 marks)
Extended• stability
Question
Define stable, unstable and neutral equilibrium. (5 marks)
Step-by-step solution
1. Step 1
  Stable equilibrium. When slightly displaced, body returns to original position (centre of mass rises when displaced, then falls back).
2. Step 2
  Unstable. Slight displacement causes body to move further from equilibrium (centre of mass falls when displaced).
3. Step 3
  Neutral. Body remains in new position after displacement (centre of mass at same height).
Answer
Stable: returns. Unstable: moves further. Neutral: stays.
3Ladder against wall (10 marks)
Extended• ladder
Question
A uniform ladder of weight 200 N rests at $60°$ to the horizontal against a smooth vertical wall. Find the horizontal reaction $N_w$ from the wall and the friction $F$ at the ground. (10 marks)
Step-by-step solution
1. Step 1
  Forces. Weight $W = 200$ N (centre); wall normal $N_w$ (horizontal); ground normal $N_g$ (vertical); friction $F$ (horizontal).
2. Step 2
  Vertical equilibrium. $N_g = W = 200$ N.
3. Step 3
  Horizontal equilibrium. $F = N_w$ .
4. Step 4
  Take moments about the foot of the ladder (length $L$ ). Weight acts at $L/2$ from foot. Wall reaction at $L\sin 60°$ vertical height (= $L\cos 60°$ horizontal arm).
5. Step 5
  Anticlockwise (wall force). $N_w \times L\sin 60°$ . Clockwise (weight). $W \times (L/2)\cos 60°$ .
6. Step 6
  Equate.
  $N_w \sin 60° = \dfrac{W}{2}\cos 60° \Rightarrow N_w = \dfrac{200 \cos 60°}{2 \sin 60°} \approx 57.7 \text{ N}$
7. Step 7
  $F = N_w \approx 57.7$ N.
Answer
$N_w \approx 57.7$ N; $F \approx 57.7$ N.

Key Formulae — Equilibrium of forces

The formulae you need to memorise for equilibrium of forces on the Cambridge International A Level 9702 paper, with every variable defined in plain English and a note on when to use it.

Equilibrium conditions
$\sum F_x = 0, \quad \sum F_y = 0, \quad \sum M = 0$
When to use
Rigid body in equilibrium. Both translational AND rotational.

Key Definitions and Keywords — Equilibrium of forces

Equilibrium
Examiner keyword
Net force AND net moment both zero. Body at rest or moving uniformly with constant rotation.
Stable equilibrium
Examiner keyword
Body returns to original position after small displacement. Centre of mass at minimum height.
Unstable equilibrium
Examiner keyword
Body moves further from equilibrium after small displacement. Centre of mass at maximum height.
Neutral equilibrium
Body remains at new position after displacement. Centre of mass at constant height.

Common Mistakes and Misconceptions — Equilibrium of forces

The traps other students keep falling into on equilibrium of forces questions — taken from recent Cambridge International A Level 9702 examiner reports and mark schemes — and how to avoid them.

✕Resolving forces but ignoring moments
9702 Examiner Reports 2022-2024
Why it happens
Rigid body confused with particle.
How to avoid it
Rigid body equilibrium requires BOTH net force = 0 AND net moment = 0. Choose pivot strategically.
✕Confusing centre of mass with geometric centre for stability
9702 Examiner Reports 2022-2024
Why it happens
For uniform bodies they coincide.
How to avoid it
Centre of MASS determines stability. For uniform body = geometric centre. For non-uniform, may differ.

Equilibrium of forces — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Equilibrium of forces

Short Notes - Equilibrium of forces

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Three forces in equilibrium (8 marks)

2Stability (5 marks)

3Ladder against wall (10 marks)

Equilibrium conditions

Equilibrium

Stable equilibrium

Unstable equilibrium

Neutral equilibrium

✕Resolving forces but ignoring moments

✕Confusing centre of mass with geometric centre for stability

Equilibrium of forces — frequently asked questions

Equilibrium of forces

Short Notes - Equilibrium of forces

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Three forces in equilibrium (8 marks)

2Stability (5 marks)

3Ladder against wall (10 marks)

Equilibrium conditions

Equilibrium

Stable equilibrium

Unstable equilibrium

Neutral equilibrium

✕Resolving forces but ignoring moments

✕Confusing centre of mass with geometric centre for stability

Equilibrium of forces — frequently asked questions