Convert areas and rearrange pressure calculations
Square length conversions correctly and find missing force or contact area.
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
- 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 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.
Sources for this lesson
Pause and explain
Do F = 0 and p = 0 establish a unique contact area by A = F/p?
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
- The patch is 6 cm² = 0.0006 m².
- Pressure is 30/0.0006 = 50 000 Pa.
- Required area is 120/10 000 = 0.012 m² = 120 cm².
Answer 50 000 Pa; 0.012 m² or 120 cm².
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →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.
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
- Area = 6 cm².
- Area = 0.0006 m².
- 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.
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
- 20 kPa = 20 000 Pa, giving F = 120 N.
- 10 kPa = 10 000 Pa, giving A = 0.012 m².
- 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.
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
- Area = 200 mm² = 0.0002 m².
- Pressure = 10/0.0002 = 50 000 Pa.
- 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.
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.