GCSE Maths · Ratio & proportion

Pressure, force and area GCSE Questions and Worked Answers

Pressure is force per unit contact area: P = F/A. Rearrange to F = PA or A = F/P when another quantity is missing. Match the area unit to the pressure unit: pascals mean newtons per square metre.

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Start with the meaning

What you need to know about pressure, force and area

Imagine pressing the same block onto soft sand in two positions. On its broad face it spreads its push over a large patch; on its narrow face the push is concentrated into a smaller patch. The force is the push, measured in newtons (N). Pressure measures how much force acts on each unit of the contact area.

See the idea first

Divide the push among units of area

If a 120 N force acts evenly over 60 cm², each square centimetre carries 120 ÷ 60 = 2 N. We write the pressure as 2 N/cm². Let P mean pressure, F force and A contact area: P = F/A. This is average pressure over the area. One pascal, Pa, is 1 N/m², not 1 N/cm².

The same 120 N force spread over different contact areas
ForceContact areaForce per cm²
120 N60 cm²2 N/cm²
120 N30 cm²4 N/cm²
120 N10 cm²12 N/cm²
From problem to method

Typical problems you need to be able to solve

These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.

Find pressure

What the problem asks: A force of 200 N acts over 0.5 m². Find the pressure.

How to solve it: Divide force by area: 200 ÷ 0.5 = 400 N/m² = 400 Pa.

Find force

What the problem asks: Pressure is 6 N/cm² over 15 cm². Find the force.

How to solve it: Each cm² carries 6 N, so 15 cm² carries 6 × 15 = 90 N. Thus F = PA.

Find contact area

What the problem asks: A 240 N force creates pressure 8 N/cm². Find the area.

How to solve it: At 8 N per cm², 240 N requires 240 ÷ 8 = 30 cm². Thus A = F/P.

A reliable routine

Solve a pressure, force or contact-area problem

Use this when a perpendicular force is spread over a stated contact area and average pressure is required. Division gives force per unit area; multiplication or division reverses the relationship to find a missing quantity.

  1. Identify P, F and the actual contact area A.
  2. Make the area units agree with the pressure units.
  3. Use P = F/A, F = PA or A = F/P for the requested quantity.
  4. Check units and whether changing the contact area should raise or lower pressure.

Check: For a block resting on one face, use that face's area, not total surface area. Mass in kilograms is not force in newtons; any required conversion must be given.

Fully worked

Pressure, force and area GCSE worked examples

Each solution explains the clue, why the method fits and how to check the result.

Example 1

Pressure in pascals

2 marks
Question

A 360 N force acts at right angles to a surface over a contact area of 0.12 m². Find the average pressure.

P=FA=3600.12=3000 PaP=\frac FA=\frac{360}{0.12}=3000\text{ Pa}

The area is in m², so the result is N/m², or Pa.

Example 2

Force

2 marks
Question

Pressure is 12 N/cm² over a contact area of 25 cm². Find the force.

F=PA=12×25=300 NF=PA=12\times25=300\text{ N}

Every cm² contributes 12 N.

Example 3

Area

2 marks
Question

A force of 450 N produces pressure 15 N/cm². Find the contact area.

A=FP=45015=30 cm2A=\frac FP=\frac{450}{15}=30\text{ cm}^2

The pressure denominator tells us the resulting area unit.

Example 4

Rectangular contact

3 marks
Question

A block exerts 180 N on a face measuring 6 cm by 5 cm. Find the pressure in N/cm².

Contact area:

A=6×5=30 cm2A=6\times5=30\text{ cm}^2

Pressure:

P=180/30=6 N/cm2P=180/30=6\text{ N/cm}^2

Do not add the areas of its other faces.

Example 5

Convert before calculating

4 marks
Question

A 500 N force acts over 200 cm². Find the pressure in Pa.

200 cm2=200/10000=0.02 m2200\text{ cm}^2=200/10000=0.02\text{ m}^2 P=500/0.02=25000 PaP=500/0.02=25000\text{ Pa}

Using 200 directly would instead give 2.5 N/cm², not 2.5 Pa.

Example 6

Choose the least pressure

4 marks
Question

A cuboid measures 2 cm × 4 cm × 6 cm and exerts a 48 N force. Find the least possible pressure when it rests flat on a face.

Possible contact areas are 2 × 4 = 8, 2 × 6 = 12 and 4 × 6 = 24 cm². The largest area gives the least pressure at fixed force.

P=48/24=2 N/cm2P=48/24=2\text{ N/cm}^2
10 original questions · total 25 marks

Pressure, force and area GCSE exam-style questions

Try each question before opening its fully worked answer. The difficulty rises through the set.

Before you startAllow about 30 minutes · Show your working. Give exact answers unless a question asks you to round, and include units where needed. · answers start collapsed
1

Find pressure

2 marks

A force of 90 N acts over 3 m². Find the pressure.

Show worked answer
P=90/3=30 PaP=90/3=30\text{ Pa}
2

Small area

2 marks

A force of 240 N acts over 0.08 m². Find the pressure.

Show worked answer
P=240/0.08=3000 PaP=240/0.08=3000\text{ Pa}
3

Find force

2 marks

Pressure is 500 Pa over 0.4 m². Find the force.

Show worked answer
F=500×0.4=200 NF=500\times0.4=200\text{ N}
4

Find area

2 marks

A force of 600 N creates 2000 Pa. Find the contact area.

Show worked answer
A=600/2000=0.3 m2A=600/2000=0.3\text{ m}^2
5

A face

3 marks

A 300 N force acts on a rectangular contact area 10 cm by 6 cm. Find pressure in N/cm².

Show worked answer
A=10×6=60 cm2A=10\times6=60\text{ cm}^2 P=300/60=5 N/cm2P=300/60=5\text{ N/cm}^2
6

Two feet

3 marks

A person exerts 600 N, shared across two feet each with contact area 150 cm². Find average pressure over both feet.

Show worked answer

Total area = 2 × 150 = 300 cm².

P=600/300=2 N/cm2P=600/300=2\text{ N/cm}^2

Use total force with total area.

7

Convert area

3 marks

Find pressure in Pa when 80 N acts over 40 cm².

Show worked answer
40 cm2=0.004 m240\text{ cm}^2=0.004\text{ m}^2 P=80/0.004=20000 PaP=80/0.004=20000\text{ Pa}
8

Tripled area

2 marks

At fixed force, contact area triples. How does pressure change?

Show worked answer
Pnew=F/(3A)=P/3P_{\text{new}}=F/(3A)=P/3

Pressure becomes one third of the original.

9

Greatest pressure

3 marks

A block exerts 72 N and can rest on faces of area 8, 12 or 24 cm². Find the greatest pressure.

Show worked answer

Choose the smallest area:

P=72/8=9 N/cm2P=72/8=9\text{ N/cm}^2

A smaller contact concentrates the same force.

10

Supplied mass conversion

3 marks

In this model, each kilogram of mass has a weight of 10 N. A 12 kg object rests on 0.03 m². Find the pressure.

Show worked answer

Convert mass to force using the supplied model:

F=12×10=120 NF=12\times10=120\text{ N} P=120/0.03=4000 PaP=120/0.03=4000\text{ Pa}
Examiner-style feedback

Common pressure, force and area mistakes

Using all faces

Pressure uses contact area, not the block's total surface area.

Mixing area units

Pa requires m²; N/cm² requires cm².

Using kilograms as newtons

Mass and force are different. Use a supplied weight conversion when needed.

30-second recap

Choose, carry out, check

Translate the wording before you reach for a rule.

  1. Pressure is force per contact area.
  2. Match compound units.
  3. Rearrange for the missing quantity.
  4. At fixed force, smaller area means greater pressure.
Quick answers

Pressure, force and area FAQ

What is a pascal?

One newton per square metre: 1 Pa = 1 N/m².

Why are snowshoes useful in this model?

They increase contact area, reducing average pressure for the same downward force.

Build connected skills

What to revise next

Content standards

Curriculum and rights review

AQA R11 density and pressure are additional Foundation content, also available at Higher. These simplified tasks use force perpendicular to a contact surface. Questions, suggested marks, solutions and diagrams are original practice material, not official past-paper questions.

Official specification references