Force equilibrium

CIE A-Level Physics· Unit 4: Forces, density and pressure· 15 min read

1. Conditions for Full Equilibrium★★☆☆☆⏱ 4 min

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.

📘 Definition

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.

2. Applying the Principle of Moments★★★☆☆⏱ 5 min

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.

📐 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.

3. Solving Ladder and Hinged Beam Problems★★★★☆⏱ 6 min

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.

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.
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.
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: $h = L^{2} - d^{2}$
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.

4. Couples and Torque Calculations★★★☆☆⏱ 3 min

📘 Definition

Couple

\tau

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).

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: $d = 40 cm = 0.40 m$ . The perpendicular distance between the two forces equals the diameter.
2
Use the torque formula for a couple $τ = F \times d$
$\tau = 15 \text{ N} \times 0.40 \text{ m} = 6.0 \text{ Nm}$
3
Final answer: Torque of the couple is 6.0 Nm, regardless of which pivot point you choose.

5. Common Pitfalls

Wrong move:

Ignoring the weight of the beam/ladder when the question doesn't explicitly mention it

Why:

CIE almost always expects you to include the weight of the rigid body unless it is explicitly called 'light' or 'massless'

Correct move:

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 move:

Using straight-line distance from the pivot instead of perpendicular distance to calculate moments

Why:

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 move:

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 move:

Only applying one equilibrium condition (either force balance or moment balance)

Why:

Full equilibrium requires both conditions, so missing one leaves unknowns or gives incorrect force values

Correct move:

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 move:

Assuming all walls exert friction on ladders

Why:

Unless stated otherwise, CIE exam questions assume vertical walls are smooth (no friction), so only exert a horizontal normal force

Correct move:

Only add friction at the wall if the question explicitly says the wall is rough

6. Quick Reference 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	$τ = F \times d$	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

7. Frequently Asked

Do I need to include the weight of the beam/ladder if it is not mentioned?

Unless the question explicitly describes the body as light (massless), you must always include its weight acting at the center of mass, usually the midpoint of a uniform body.

Can I choose any pivot point for moment calculations?

Yes! For equilibrium, net moment is zero about any point. Choosing a pivot at the location of an unknown force eliminates that force from your equation, simplifying calculations.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 11
Moment calculation for a leaning ladder
2023 · 22
Equilibrium of a hinged uniform beam
2021 · 13
Torque calculation for a couple

Going deeper

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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.