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IGCSE · Cambridge (CIE) · 0625

Convert areas and rearrange pressure calculations

Square length conversions correctly and find missing force or contact area.

Pressure pathway · 1.8

Core and Extended; Physics 0625, 2026–2028, section 1.8. Define and use p = F/A, explain everyday force/area effects, and describe liquid depth and density qualitatively. Numerical liquid-pressure changes belong to the clearly labelled Supplement lessons. Written work is self-checked.

Before you begin

  • Use p = F/A.
  • Convert centimetres and millimetres as lengths.

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

Quick readiness check

How many square metres is 1 cm²?

What you will learn

  • Convert cm² and mm² to m² before obtaining pascals.
  • Use F = pA and A = F/p with physically consistent data.
An original controlled comparison applies the same normal 120 N force to contact areas 0.0060 and 0.012 m², giving average pressures 20 and 10 kPa. Square sides are 110 times the square root of area divided by 0.0060 in drawing units, so drawn square area doubles with the physical contact area. Pressure bars use one drawing unit per 200 Pa, giving widths 100 and 50. The shapes represent contact patches viewed face-on, not block masses, and the force labels do not encode motion or tool performance.Same normal force: 120 NContact patches · face-on view0.0060 m²20 kPa0.0120 m²10 kPaTwice the area · half the pressureBoth force and area matter
An original controlled comparison applies the same normal 120 N force to contact areas 0.0060 and 0.012 m², giving average pressures 20 and 10 kPa. Square sides are 110 times the square root of area divided by 0.0060 in drawing units, so drawn square area doubles with the physical contact area. Pressure bars use one drawing unit per 200 Pa, giving widths 100 and 50. The shapes represent contact patches viewed face-on, not block masses, and the force labels do not encode motion or tool performance.

An area conversion squares the length factor

One centimetre is 0.01 m, so 1 cm² is 0.0001 m². One millimetre is 0.001 m, so 1 mm² is 0.000001 m². A 3 cm by 2 cm rectangular contact has area 6 cm² = 0.0006 m². A normal 30 N over that contact gives 50 000 Pa. Converting cm² by only dividing by 100 would be a length conversion, not the required area conversion.

If a contact's dimensions are both supplied, multiply their consistently converted lengths. For a 20 mm by 10 mm patch, area is 200 mm² = 0.0002 m². Do not divide by an object's total external surface area when only one face or a small patch touches the support.

Rearrange without changing the meaning

From p = F/A, force is F = pA and contact area is A = F/p. For pressure 20 kPa over 0.0060 m², F = 20 000 × 0.0060 = 120 N. To keep average pressure at 10 kPa with force 120 N, the required area is 120/10 000 = 0.012 m² or 120 cm².

Use a positive non-zero area for a finite-contact average. A zero normal force on a finite area gives zero pressure in this simple model. If both supplied F and p are zero, F/p is 0/0 and does not determine a unique area. A non-zero force with zero requested pressure cannot be achieved on a finite area by this model.

Pause and explain

Do F = 0 and p = 0 establish a unique contact area by A = F/p?

Put the idea to work

Worked example

A normal 30 N acts on a 3 cm by 2 cm rectangular patch. Find pressure, then find the area needed for 120 N at 10 kPa.

Show the worked solution
  1. The patch is 6 cm² = 0.0006 m².
  2. Pressure is 30/0.0006 = 50 000 Pa.
  3. Required area is 120/10 000 = 0.012 m² = 120 cm².

Answer 50 000 Pa; 0.012 m² or 120 cm².

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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

Find pressure for normal 30 N on a 3 cm by 2 cm rectangular contact.

Give me a hint

Multiply the lengths and convert the resulting squared unit.

Compare my reasoning
  1. Area = 6 cm².
  2. Area = 0.0006 m².
  3. Pressure = 30/0.0006 = 50 000 Pa.

50 000 Pa.

Look for these in your work

  • Convert the whole area correctly.
  • Use the normal force.
2 · Independent

Find force at 20 kPa on 0.0060 m², and the area for 120 N at 10 kPa.

Give me a hint

Use F = pA and A = F/p after converting kPa to Pa.

Compare my reasoning
  1. 20 kPa = 20 000 Pa, giving F = 120 N.
  2. 10 kPa = 10 000 Pa, giving A = 0.012 m².
  3. 0.012 m² is 120 cm².

120 N; 0.012 m² or 120 cm².

Look for these in your work

  • Choose the right rearrangement.
  • Convert pressure and area units.
3 · Transfer

A 20 mm by 10 mm patch carries normal 10 N. A student uses 200 m² as its area. Correct the area and pressure.

Give me a hint

The numeric product retains mm² until that squared unit is converted.

Compare my reasoning
  1. Area = 200 mm² = 0.0002 m².
  2. Pressure = 10/0.0002 = 50 000 Pa.
  3. Using 200 m² would describe a completely different contact.

0.0002 m²; 50 000 Pa.

Look for these in your work

  • Preserve the original squared unit.
  • Calculate with m² for Pa.

Common mistakes

  • Square the conversion factor.
  • Use the area that actually touches.
  • Zero divided by zero does not determine an area.
Recall without your notes

What can you explain now?

How many square metres is 200 mm²?

Compare with the explanation

0.0002 m².

Each square millimetre is one millionth of a square metre.

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Make it stick

Convert contact areas, rearrange pressure for force or area and explain a zero-data limit.

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 Physics

Course scope and assessment

491 sequenced Physics lessons provide 1473 guided, independent and application problems. 58 of 58 syllabus sections have full linked teaching sequences; 0 sections remain pending. Three earlier starter overviews remain available but are not counted as full sequences. The map covers all six subject areas for 2026–2028 and keeps review and assessment limits visible. This is available content, not full programme completion, an exam grade or certified mastery. Written practice is self-checked. Supervised practical experience, cumulative assessment and human subject-teacher review remain pending. Paper method exercises do not demonstrate practical competence or complete either practical-assessment route.

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