Explain everyday changes in contact pressure
Compare force and area separately in snowshoes, sharp tools and differently supported loads.
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
- Recall p = F/A.
- Identify total contact force and actual contact area.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
At fixed force, what does a larger contact area do?
What you will learn
- Explain how larger force or smaller area changes pressure.
- Use everyday comparisons with explicit controlled quantities.
Hold one quantity fixed
At fixed normal force, doubling the area halves average pressure; at fixed area, doubling the force doubles pressure. Spreading the same supported weight over snowshoes can reduce pressure on snow. A sharp blade or point concentrates a given force on a smaller contact, increasing pressure. These explanations need a stated similar force rather than assuming every user pushes equally.
Pressure alone does not prove how far a real tool cuts or a surface deforms: material strength, shape, motion and other conditions matter. The syllabus comparison asks why force and area affect pressure, not whether a particular product has certified performance. Do not turn 'smaller area gives larger pressure for the same force' into 'every smaller object always creates larger pressure'.
When both force and area change, compare their ratio
A contact has 120 N over 0.0060 m², giving 20 kPa. Keeping 120 N but using 0.012 m² halves this to 10 kPa. If both force and area double instead, 240/0.012 remains 20 kPa. More force does not necessarily mean more pressure when the contact area also changes.
For a steady symmetrical two-foot stance with the stated total force unchanged, lifting one foot and retaining the same individual footprint halves total area and doubles average pressure. The remaining foot then carries the whole force in this model. If another support or altered contact patch is supplied, account for it rather than automatically applying that doubled-pressure conclusion.
Sources for this lesson
Pause and explain
Does doubling force guarantee doubled pressure if area also doubles?
Worked example
Compare 120 N on 0.0060 m² with 120 N on 0.012 m², then with 240 N on 0.012 m².
Show the worked solution
- The first average is 20 kPa.
- Doubling area alone gives 10 kPa.
- Doubling force and area together returns the ratio to 20 kPa.
Answer 20 kPa; 10 kPa; 20 kPa.
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.
Compare the three force/area pairs 120 N/0.0060 m², 120 N/0.012 m² and 240 N/0.012 m².
Give me a hint
Calculate each ratio and name what changes.
Compare my reasoning
- First pressure is 20 kPa.
- Second pressure is 10 kPa because area doubles alone.
- Third pressure is 20 kPa because force and area both double.
20 kPa; 10 kPa; 20 kPa.
Look for these in your work
- Control the quantities correctly.
- Use the ratio when both vary.
A steady 600 N person has two equal 150 cm² foot contacts. Find average pressure with both feet down, then on one unchanged footprint with no other support.
Give me a hint
Match the whole supported force with the total area in each stance.
Compare my reasoning
- Two-foot area is 300 cm² = 0.030 m², giving 20 kPa.
- One-foot area is 0.015 m² and carries the whole 600 N.
- One-foot pressure is 40 kPa.
20 kPa on two feet; 40 kPa on one footprint.
Look for these in your work
- Use the combined area initially.
- Apply the stated force balance.
A wider snowshoe carries twice the normal force over twice the area. May its wider shape alone prove lower pressure or less sinking? Explain.
Give me a hint
Compare the force/area ratio before making a real-material claim.
Compare my reasoning
- Both factors double, so their ratio stays the same.
- Average pressure is not reduced in this supplied comparison.
- Actual sinking also depends on snow and other conditions not supplied.
Same average pressure; less sinking is not established by width alone.
Look for these in your work
- Keep both changed quantities.
- Separate a pressure result from an unsupported performance claim.
Common mistakes
- State which quantity is held fixed.
- Compare F/A when both change.
- Pressure is not proof of cutting depth or material failure.
What can you explain now?
What happens to pressure if both force and area triple?
Compare with the explanation
It stays the same.
The numerator and denominator have the same factor, so their ratio is unchanged.
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Make it stick
Explain tools and snowshoes with controlled force/area comparisons, then handle both quantities changing together.
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.