# Force equilibrium

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u4-force-equilibrium/

This sub-topic covers the two core conditions for force equilibrium of rigid bodies, explains how to resolve forces and apply the principle of moments to solve common static problems including ladders, beams, and systems of coupled forces.

**Prerequisites:** [Force resolution in two dimensions](https://www.owlsprep.com/study/cie-9702-u4-force-resolution/); [Basic concept of moment of a force](https://www.owlsprep.com/study/cie-9702-u4-moment-of-force/)

## Learning objectives

- Distinguish between translational and rotational equilibrium of rigid bodies
- Apply the principle of moments to solve static equilibrium problems
- Resolve forces in two dimensions for systems in full equilibrium
- Calculate torque for systems of coupled forces

## Conditions for Full Equilibrium

For any rigid body to be completely at rest (no linear or angular acceleration), it must satisfy two independent conditions. Both must hold for full equilibrium to exist.

**Full Equilibrium** — A rigid body is in full equilibrium if and only if both conditions are satisfied: (1) Net resultant force in any direction is zero (translational equilibrium), (2) Net resultant moment about any point is zero (rotational equilibrium).

*Example:* A stationary ladder leaning against a wall is in full equilibrium, while a spinning top moving at constant velocity is only in translational, not rotational, equilibrium.

**Worked example:** A body has three forces acting on it: 10 N right along the x-axis, 6 N left along the x-axis, and 8 N up along the y-axis. Is the body in translational equilibrium? If not, what is the resultant force?

1. Sum forces along the x-axis first:

   $$\sum F_x = 10 - 6 = 4 \text{ N}$$
2. Sum forces along the y-axis:

   $$\sum F_y = 8 \text{ N}$$
3. For translational equilibrium, net force must be zero. Calculate the magnitude of the resultant force:

   $$F_{resultant} = \sqrt{(4)^2 + (8)^2} = \sqrt{80} = 4\sqrt{5} \approx 8.94 \text{ N}$$
4. Conclusion: The body is not in translational equilibrium, because the net force is non-zero.

> **Exam tip:** Always check BOTH conditions of equilibrium, even if the problem only asks for one force. Many students forget to check moment equilibrium and end up with wrong answers.

## Applying the Principle of Moments

The principle of moments is a direct result of the rotational equilibrium condition. It states that for any body in rotational equilibrium, the sum of clockwise moments about any pivot equals the sum of anticlockwise moments about the same pivot.

> **tip**
>
> To simplify calculations, choose your pivot at the point where an unknown force acts. This eliminates that force from the moment equation, as its distance from the pivot is zero.

**Worked example:** A uniform 4.0 m long beam of mass 20 kg is pivoted 1.0 m from its left end. A 30 kg mass is hung from the left end. What mass must be hung from the right end to keep the beam horizontal and in equilibrium?

1. Weight of the uniform beam acts at its center, 2.0 m from the left end = 1.0 m to the right of the pivot. The right end is 3.0 m right of the pivot. Gravitational acceleration g cancels out so we can omit it early.
2. Calculate total anticlockwise moments (left of the pivot):

   $$M_{anticlockwise} = (30g)(1.0) = 30g$$
3. Calculate total clockwise moments (right of the pivot, including beam weight):

   $$M_{clockwise} = (20g)(1.0) + M g (3.0) = 20g + 3Mg$$
4. Equate moments per the principle of moments:

   $$30g = 20g + 3Mg$$
5. Cancel g and rearrange for M:

   $$10 = 3M \implies M = \frac{10}{3} \approx 3.3 \text{ kg}$$
6. Final answer: A mass of approximately 3.3 kg is required.

## Solving Ladder and Hinged Beam Problems

Most CIE exam questions on equilibrium involve either a ladder leaning against a wall or a beam hinged at a point. These problems require applying both equilibrium conditions: force balance and moment balance.

1. For a ladder: the ground exerts both a normal reaction force and a frictional force (to balance horizontal force from the wall), while the wall is usually smooth (no friction) unless stated otherwise.
2. For a hinged beam: the hinge exerts an unknown reaction force with horizontal and vertical components, which you can solve for using force balance after finding unknowns from moments.
3. Always include the weight of the body unless it is explicitly described as *light* (massless).

**Worked example:** A uniform ladder of length L and mass m rests against a smooth vertical wall, with its base on rough horizontal ground a distance d from the wall. What is the frictional force at the ground in terms of m, g, L and d?

1. Label all forces: Normal reaction N (horizontal, from smooth wall), normal reaction R (vertical, from ground), friction f (horizontal at ground, balances N), weight mg acts at L/2 (center of the ladder).
2. Take moments about the base of the ladder to eliminate f and R. Height of the top of the ladder: $h = \sqrt{L^2 - d^2}$
3. Equate anticlockwise moment from N to clockwise moment from mg:

   $$N\sqrt{L^2-d^2} = mg \frac{d}{2}$$
4. Solve for N:

   $$N = \frac{mg d}{2 \sqrt{L^2-d^2}}$$
5. Horizontal force balance: net force is zero, so $f = N$

   $$f = \frac{mg d}{2 \sqrt{L^2 - d^2}}$$

> **Exam tip:** Always read the question carefully to check if the wall is smooth or rough. A rough wall adds an extra vertical friction force and unknown to your calculation.

## Couples and Torque Calculations

**Couple** — A couple is a system of two equal magnitude, opposite direction, parallel forces separated by a perpendicular distance. A couple produces no net force, only net torque (turning effect).

*Notation:* \tau

*Example:* The forces applied to turn a car steering wheel form a couple.

The torque (total moment) of a couple is constant about any pivot point, equal to the magnitude of one force multiplied by the perpendicular distance between the lines of action of the two forces.

**Worked example:** A steering wheel of diameter 40 cm has two 15 N forces applied to opposite edges, forming a couple. What is the torque of the couple?

1. Convert diameter to meters: $d = 40 \text{ cm} = 0.40 \text{ m}$. The perpendicular distance between the two forces equals the diameter.
2. Use the torque formula for a couple $\tau = F \times d$

   $$\tau = 15 \text{ N} \times 0.40 \text{ m} = 6.0 \text{ Nm}$$
3. Final answer: Torque of the couple is 6.0 Nm, regardless of which pivot point you choose.

## Common pitfalls

- **Wrong:** Ignoring the weight of the beam/ladder when the question doesn't explicitly mention it
  - Why it fails: CIE almost always expects you to include the weight of the rigid body unless it is explicitly called 'light' or 'massless'
  - Correct: Always check for the word 'light'—if it is not present, add the weight acting at the center of mass of the uniform body
- **Wrong:** Using straight-line distance from the pivot instead of perpendicular distance to calculate moments
  - Why it fails: Moment is defined as force multiplied by perpendicular distance to the line of action of the force, not distance to the point of application
  - Correct: Always resolve the force into components perpendicular and parallel to the body, or calculate the perpendicular distance from the pivot to the force line, to get the correct moment value
- **Wrong:** Only applying one equilibrium condition (either force balance or moment balance)
  - Why it fails: Full equilibrium requires both conditions, so missing one leaves unknowns or gives incorrect force values
  - Correct: Write down both $\sum F_x = 0$, $\sum F_y = 0$ and $\sum M = 0$ about your chosen pivot to get enough equations to solve for all unknowns
- **Wrong:** Assuming all walls exert friction on ladders
  - Why it fails: Unless stated otherwise, CIE exam questions assume vertical walls are smooth (no friction), so only exert a horizontal normal force
  - Correct: Only add friction at the wall if the question explicitly says the wall is rough

## Cheatsheet

| Concept | Formula/Rule | Key Note |
| --- | --- | --- |
| Full Equilibrium | $\sum F_x = 0, \sum F_y = 0, \sum M = 0$ | All three conditions must hold |
| Principle of Moments | Sum clockwise M = Sum anticlockwise M | Works for any pivot point |
| Torque of a Couple | $\tau = F \times d$ | d = perpendicular distance between forces |
| Uniform body weight | Acts at midpoint of the body | Include unless body is 'light' |
| Smooth wall | Only horizontal normal force | No vertical friction force |

## What's next

Force equilibrium is a core foundation for many subsequent topics in CIE A-Level Physics, including fluid statics, circular motion, and rigid body rotation. Mastery of equilibrium problem-solving techniques — especially choosing optimal pivot points and correctly resolving forces into components — will save you time and marks in both Paper 1 multiple choice and Paper 2 structured questions. This sub-topic is frequently tested in both AS and A-Level papers, so regular practice of ladder and beam problems is highly recommended to build speed and accuracy.

- [Moment and torque](https://www.owlsprep.com/study/cie-9702-u4-moment-and-torque/)
- [Density](https://www.owlsprep.com/study/cie-9702-u4-density/)
- [Fluid pressure](https://www.owlsprep.com/study/cie-9702-u4-fluid-pressure/)

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