Pressure: revision notes and practice

What you will learn

• Calculate solid-surface pressure using force and contact area.
• Rearrange the pressure equation and convert square units correctly.
• Explain pressure changes and plan fair comparisons of contact faces.

Key rules

Calculate pressure

p = F/A, with F in newtons, A in square metres and p in pascals. One Pa = one N/m².

Rearrange

F = pA and A = F/p. Identify the actual contact area rather than the whole outside surface.

Convert areas

1 m² = 10 000 cm². Convert each length before multiplying, or divide an area in cm² by 10 000.

Worked examples

Example 1: A 120 N force acts on 0.04 m². Find the pressure.

1. Use p = F/A.
2. 120 ÷ 0.04 = 3000.

3000 Pa

Example 2: A surface experiences 2000 Pa over 0.03 m². Find the force.

1. Rearrange to F = pA.
2. 2000 × 0.03 = 60.

60 N

A practice question

A perpendicular force of 126 N acts on a solid surface with contact area 0.07 m². Calculate the pressure in pascals.

1. Pressure describes force per unit area

Pressure measures how concentrated a force is on a surface. For a perpendicular force spread over a stated contact area, divide force by area. A force of 100 newtons on 0.05 square metres produces 2000 pascals. If the same force acts over 0.10 square metres, the pressure is 1000 pascals. The total force stayed the same while its concentration changed.

Use the force perpendicular to the surface in this model. A force pushing sideways is not the normal contact force needed in the simple pressure equation. The app’s pressure questions concern solid contact surfaces; fluid pressure at depth is a separate topic with different conditions. Always identify the actual surface that supports the load.

p = F/A; p in Pa, F in N, A in m²; 1 Pa = 1 N/m²

2. Identify and measure contact area

A rectangular block may touch a table through just one face. The contact area is the area of that face, not the sum of all six faces. Multiply the two perpendicular lengths of the face. A face measuring 0.20 metres by 0.30 metres has area 0.060 square metres. Rotating the block can change the supporting face without changing the block’s weight.

For an irregular footprint, the question may give a measured area or a model for estimating it. State the model when relevant. Surface roughness and uneven loading mean real pressures can vary from point to point; the simple equation gives an average pressure across the stated contact area. Do not assume that every part of a large object necessarily touches the support.

3. Convert square units carefully

A metre contains one hundred centimetres, but a square metre contains ten thousand square centimetres. A square 1 metre long and 1 metre wide is 100 centimetres by 100 centimetres, so its area is 10 000 square centimetres. Divide an area in square centimetres by 10 000 to obtain square metres.

Alternatively, convert each length first, then multiply. A 20 centimetre by 15 centimetre face becomes 0.20 metres by 0.15 metres, giving 0.030 square metres. Multiplying the original lengths gives 300 square centimetres, and 300/10 000 gives the same answer. Dividing 300 by 100 would be a length conversion applied incorrectly to an area.

4. Rearrange without losing the physical meaning

To find force, multiply pressure by area: F = pA. A pressure of 4000 pascals across 0.025 square metres corresponds to 100 newtons. To find the area required for a target pressure, use A = F/p. A 200-newton force limited to 5000 pascals needs an area of at least 0.040 square metres in this uniform-pressure model.

Check the units as well as the calculation. Pascals multiplied by square metres produce newtons. A larger area at the same pressure needs a larger force. A larger permissible pressure for the same force allows a smaller contact area. These checks help detect the common mistake of dividing when the rearranged equation requires multiplication.

F = pA; A = F/p

5. Compare changes with one variable held fixed

When force is fixed, pressure is inversely proportional to area. Doubling contact area halves pressure; reducing area to one third makes pressure three times as large. When area is fixed, doubling force doubles pressure. These conclusions follow from p = F/A, but only if the other quantity remains unchanged.

If both quantities change, use their separate scale factors. Doubling force while making the area four times larger gives a pressure half as large as before. Do not say that a bigger object always gives lower pressure: a bigger object may also have a greater weight. State the controlled quantity before explaining the predicted pattern.

6. Connect calculations to everyday examples

Broad supports distribute a load over a larger area. Snowshoes can reduce the average pressure on snow compared with ordinary footwear supporting the same person. A sharp point concentrates a force on a small area, producing a larger local pressure. These comparisons rely on the same force being applied, not on area alone.

The relevant force for a resting object is often its weight. If mass is supplied instead of force, calculate weight using W = mg and the given gravitational field strength before applying p = F/A. Mass in kilograms is not itself a force in newtons. In simple horizontal resting situations, the perpendicular load equals the weight when no other vertical forces act.

7. Design a fair and sensible investigation

A supervised comparison can place the same rectangular block on different faces. Measure each face’s two dimensions, calculate its area, and keep the block and added load unchanged. The force is then constant while area varies. Compare the calculated pressure values with an appropriate observed effect, such as an indentation in a suitable test material.

Indentation is only an indirect indication of pressure: it also depends on the material and how long the load is applied. Keep those conditions consistent, repeat measurements and record uncertainty in dimensions. Follow the teacher’s handling procedure and keep hands clear of loads. Explain each control’s purpose instead of listing vague instructions such as make it fair.

8. Present and assess the result

Show the contact-area calculation before substituting into the pressure equation when dimensions are supplied. Write square metres for the area and pascals for the final pressure, and retain calculator precision until the end. A small area can produce a pressure number much larger than the force number; this is reasonable because dividing by a number smaller than one increases the numerical result.

For reasoning questions, name both quantities in the equation. A larger area gives lower pressure only if the perpendicular force is unchanged. If someone claims that rotating a block changes its weight, explain that its mass and gravitational field strength remain the same; the changed face alters contact area and therefore pressure. Separate a measured observation from a calculation made using a simplified model.

Check your understanding

How to earn the marks

• Use the contact face, not total outside area.
• Convert lengths before multiplying, or convert areas by the squared factor.
• Convert mass to weight if force is not given.
• State what stays constant in any proportional comparison.

A common mistake

Do not convert cm² to m² by dividing by 100; an area uses the square of the length conversion.

Course context and official specifications

This guide covers selected skills. Some extensions are Higher content; choose your official tier separately from practice difficulty in the app. These are original questions and explanations.

• AQA GCSE Physics — 4.5.5.1.1

  Force-per-area calculations are available at both tiers. Liquid-column calculations in 4.5.5.1.2 are Higher-only and are not included here.

• Edexcel GCSE Physics — 15.10P–15.11P

  Force-per-area calculations are available at both tiers. The current bank omits fluid-depth, upthrust and gas-law calculations.

• OCR Gateway GCSE Physics A — PM2.3v; P1.3

  The selected p = F/A calculations are available at both tiers. Liquid-column pressure is Higher content and is outside this bank.

• OCR Twenty First Century GCSE Physics B — P6.4.1–P6.4.2

  The selected p = F/A relationship is available at both tiers. The bank omits fluid-depth, upthrust and gas-law calculations; P6.4.8 liquid-depth calculations are Higher-only.

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Try five questions, read the feedback, and revisit one thing you found difficult. The app tracks a 20-answer practice goal separately from accuracy.

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