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IGCSE · Cambridge (CIE) · 0652

Define and calculate force per unit area

Use the normal contact force and actual area to calculate average pressure in pascals.

Section P1.5.4 · Pressure

Pressure · Cambridge IGCSE Physical Science 0652 · P1.5.4

Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.

Before you begin

  • Measure force in newtons.
  • Distinguish an area from a length or a volume.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

Which unit measures an area?

What you will learn

  • Define pressure as force per unit area and recall p = F/A.
  • Identify the force acting normally on the chosen contact area.
An original flat-contact schematic assigns a normal 120 N force to the actual contact area 0.0060 m², giving average pressure 20 000 Pa. The force arrow, block shape and patch drawing are schematic rather than measured dimensions or calibrated forces. Average pressure need not be uniform locally; an unrelated sideways force or another support's share does not belong in this contact's numerator.Use the named normal contact force120 N pushing downwardActual contact area: 0.0060 m²p = F/A = 120/0.0060Average pressure: 20 000 PaForce / area · not force alone
An original flat-contact schematic assigns a normal 120 N force to the actual contact area 0.0060 m², giving average pressure 20 000 Pa. The force arrow, block shape and patch drawing are schematic rather than measured dimensions or calibrated forces. Average pressure need not be uniform locally; an unrelated sideways force or another support's share does not belong in this contact's numerator.

Pressure spreads a force over an area

Pressure is force per unit area: p = F/A. For a flat contact in this model, F is the force pushing normally, or perpendicularly, on the stated area A. Use newtons and square metres to obtain pascals, where 1 Pa = 1 N/m². A normal 120 N force spread over 0.0060 m² gives 120/0.0060 = 20 000 Pa, or 20 kPa.

This quotient gives average pressure across the stated contact. Local pressure need not be uniform if a real surface is uneven. A force alone is not a pressure, and a width alone is not an area. Do not put a sideways force into a normal-pressure calculation or assume that the whole visible object touches the surface.

Choose the contact and force together

A standing person's total downward contact force can equal their weight in a simple steady model with no other vertical support. If two feet share the contact, use the combined area with the combined force, or each foot's share with its own area. Giving each foot the whole weight and then adding those pressures would not describe the same account.

When another support carries some weight, the contact of interest need not carry the full weight. Identify the named contact force from the supplied balance before using its area. A larger mass is relevant only after converting it into the appropriate force using supplied gravitational field strength and support conditions.

Sources for this lesson

Pause and explain

Can a sideways force alone be used as the normal force on a flat horizontal contact?

Put the idea to work

Worked example

A flat contact carries a normal 120 N force over 0.0060 m². Find average pressure.

Show the worked solution
  1. Identify F = 120 N and A = 0.0060 m².
  2. Use p = F/A = 120/0.0060.
  3. The result is 20 000 N/m² = 20 000 Pa = 20 kPa.

Answer 20 000 Pa or 20 kPa.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Calculate average pressure for 120 N acting normally over 0.0060 m².

Give me a hint

Divide the specified normal force by its contact area.

Compare my reasoning
  1. Force is 120 N.
  2. Area is 0.0060 m².
  3. p = 120/0.0060 = 20 000 Pa.

20 000 Pa.

Look for these in your work

  • Use force per area.
  • State the pressure unit.
2 · Independent

Find pressure for 0.40 kN over 0.020 m²; then explain whether a total 600 N on two feet with combined contact area 0.030 m² gives the same pressure.

Give me a hint

Convert the force prefix and keep each total force with its total area.

Compare my reasoning
  1. 0.40 kN = 400 N, so 400/0.020 = 20 000 Pa.
  2. The two-foot total gives 600/0.030 = 20 000 Pa.
  3. The supplied totals give equal average pressures.

20 000 Pa in both cases.

Look for these in your work

  • Convert kilonewtons.
  • Match total force and total area.
3 · Transfer

A 600 N load is partly supported elsewhere, leaving 200 N on a contact of area 0.010 m². A calculation uses the whole 600 N. Correct it.

Give me a hint

Use the force actually assigned to the contact of interest.

Compare my reasoning
  1. The named contact carries 200 N.
  2. Pressure is 200/0.010 = 20 000 Pa.
  3. 600/0.010 would incorrectly assign the other support's share to this contact.

20 000 Pa; use 200 N, not the whole 600 N.

Look for these in your work

  • Identify the stated support share.
  • Keep the pressure calculation tied to its contact.

Common mistakes

  • Use area, not width or volume.
  • Use the force on the named contact.
  • Average pressure need not describe every local point.
Recall without your notes

What can you explain now?

What does 1 Pa mean?

Compare with the explanation

1 N per square metre.

A pascal is the pressure associated with one newton per square metre.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Define pressure and the pascal, calculate an average, and identify the actual normal contact force and area.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Physical Science

Course scope and assessment

This page maps the 2025–2026 Cambridge IGCSE Physical Science 0652 syllabus: 38 Chemistry sections and 46 Physics sections. Every section links to original 0620/0625 teaching or one of five new 0652 lessons, with guided, independent and transfer practice and explained answers. Cambridge limits entry to Administrative Zone 3 and identifies November 2026 as the final exam series. The section crosswalk is complete as a first coverage map; a numbered-objective and Core/Supplement audit, independent subject-teacher review and cumulative assessment are still needed. Written practice does not award a Cambridge grade. Online activities cannot demonstrate supervised laboratory competence or complete Paper 5/6.

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