Pressure, force and area
Key idea: Pressure is the perpendicular force divided by the contact area. The same force produces greater pressure when spread over a smaller area.
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Guided lesson
Guided steps
- Compare pressure with the total force on a surface.
- Calculate the three pressures from the forces into the floor.
- Find the force from the pressure and area.
- Follow the weight calculation and square-unit conversion.
- Find the car’s weight, total contact area and pressure.
- Compare the brick’s largest and smallest contact faces.
- Convert the pin’s areas and compare the two pressures.
- Find both pressures and their ratio.
- Follow mass to weight to pressure in the weaker field.
- Find density, weight and pressure; classify the two forces.
- Compare pressures when both force and area change.
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ContinueA sharp knife cuts easily and a blunt one does not, though you push just as hard. Snowshoes stop you sinking into snow that your boots would sink straight into. In both cases the force is the same; what changes is the area it is spread over, and with it the pressure.
Pressure: force per unit area
Press a drawing pin into a cork board with your thumb. Your thumb and the board feel the same push, yet the pin goes into the board and not into your thumb. The difference is the area the push is spread over: the flat head spreads it over your thumb, and the sharp point concentrates it on a tiny patch of the board.
Pressure, p, is the force acting perpendicular to a surface per unit area.
For these contact calculations, F/A is the average pressure across the stated area; actual pressure need not be the same at every point.
The SI unit of pressure is the pascal (Pa): 1 Pa = 1 N/m², a force of 1 N spread over 1 m². That is a small pressure: a sheet of paper lying flat on a desk presses on it with about 1 Pa. Larger pressures are often given in kilopascals: 1 kPa = 1000 Pa.
- For the same force, a smaller area gives a greater pressure.
- For the same area, a greater force gives a greater pressure.
The relationship rearranges to F = pA to find a force, and A = F ÷ p to find an area.
Use the force at right angles to the surface
Only the force that pushes straight into the surface, at right angles to it, causes pressure on that surface. Before you calculate, decide which surface you mean and which force presses on it.
| Situation | Surface | Force that presses on it |
|---|---|---|
| a box resting on a level floor | floor | the box’s weight |
| a box on a level floor, with a person pushing down on its top | floor | the box’s weight plus the downward push |
| a box dragged along a level floor by a horizontal rope | floor | still only the box’s weight: the pull is along the floor |
| a hand pushing horizontally on a wall | wall | the horizontal push; the person’s weight presses on the floor |
| a thumb pushing a drawing pin straight into a wall | wall | the push, passed on by the pin |
Check your understanding 1
A 400 N crate rests on a level floor on a face of area 0.50 m². Assume it remains in full contact on that face without tipping or vertical acceleration. Find the average pressure on the floor. Then find it again when a worker pushes down on top with 100 N, and when the worker instead pushes the crate sideways with 100 N.
Show the steps Optional working
Show answer
Resting: p = 400 ÷ 0.50 = 800 Pa. Pushed down: the force on the floor is 400 + 100 = 500 N, so p = 500 ÷ 0.50 = 1000 Pa. Pushed sideways: the push is along the floor, not into it, so the pressure stays 800 Pa.
Check your understanding 2
A pressure of 2.5 × 10⁵ Pa acts on an area of 0.040 m². Calculate the force.
Show answer
F = pA = 2.5 × 10⁵ × 0.040 = 1.0 × 10⁴ N.
Choose the force and the contact area before calculating
- Identify the force perpendicular to the surface. For an object resting on a level surface, this is its weight.
- If you are given a mass, calculate the weight with W = mg.
- Identify the face that touches the surface. Its area is the contact area; the other faces, and the total surface area, play no part.
- Convert the contact area to square metres.
- Use p = F ÷ A and give the answer in pascals.
Area conversions are squared: 1 cm = 10⁻² m, so 1 cm² = 10⁻⁴ m², not 10⁻² m². Likewise 1 mm² = 10⁻⁶ m².
Worked example 1
Pressure caused by weight
Problem
A 60 kg student stands on two shoes with a total contact area of 400 cm². Take g = 10 N/kg. Find the pressure on the floor.
Show full solution
Find the force
Method
Use the student’s weight.Reason
The weight is the force pressing on the floor.Working
Answer
F = W = mg = 60 × 10 = 600 N
Convert the area
Method
Change cm² to m².Reason
Pressure in pascals needs the area in m²; 1 cm² is 10⁻⁴ m².Working
Answer
A = 400 × 10⁻⁴ = 0.040 m²
Find the pressure
Method
Divide force by area.Reason
Pressure is force per unit area.Working
Answer
p = 600 ÷ 0.040 = 1.5 × 10⁴ Pa
Check
Working
Answer
If the student lifts one foot, the area halves and the pressure doubles to 3.0 × 10⁴ Pa.
Check your understanding 3
A 1200 kg car rests on four tyres. Each tyre touches the road over 150 cm². Calculate the pressure on the road.
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Show the steps Optional working
Show answer
W = 1200 × 10 = 12 000 N. Total area = 4 × 150 = 600 cm² = 0.060 m². p = 12 000 ÷ 0.060 = 2.0 × 10⁵ Pa.
Worked example 2
Choose the face in contact
Problem
A brick of weight 20 N measures 20 cm by 10 cm by 5.0 cm. Find the pressure it exerts on the ground when it rests on its largest face and when it rests on its smallest face.
Show full solution
Find the force
Method
Use the weight.Reason
The force on the ground is the weight, whichever face is down.Working
Answer
F = 20 N
Largest face
Method
Use the area of the face in contact.Reason
Pressure depends on the contact area only.Working
Answer
20 × 10 = 200 cm² = 0.020 m², so p = 20 ÷ 0.020 = 1.0 × 10³ Pa
Smallest face
Method
Repeat with the smallest face.Reason
The same force on a smaller area gives a larger pressure.Working
Answer
10 × 5.0 = 50 cm² = 0.0050 m², so p = 20 ÷ 0.0050 = 4.0 × 10³ Pa
Compare
Working
Answer
A quarter of the contact area gives four times the pressure. Using the brick’s total surface area, 700 cm², would give one pressure for every way the brick lies, which cannot be right.
Designing for high or low pressure
Many everyday objects work by changing the contact area, because the force is often fixed by a person’s push or an object’s weight.
| Object | Contact area | Effect |
|---|---|---|
| knife blade, needle, nail point | very small | a large pressure, so it cuts or pierces with a modest force |
| football boot studs | small | a large pressure pushes the studs into the ground for grip |
| snowshoes, skis | large | a small pressure, so the wearer does not sink into snow |
| tractor tyres, tank and digger tracks | large | a small pressure, so the vehicle does not sink into soft ground |
| wide bag straps | large | a small pressure on the shoulders, so the bag hurts less |
| building foundations | large | a small pressure on the ground, so the building does not sink |
A blunt knife needs a greater force to cut than a sharp one: its edge is wider, so the same push gives a smaller pressure.
Worked example 3
A drawing pin
Problem
A thumb pushes a drawing pin into a board with a force of 20 N. The head of the pin has an area of 1.0 cm² and the point has an area of 0.10 mm². Compare the pressure on the thumb with the pressure on the board.
Show full solution
Find the force
Method
Use the thumb’s push.Reason
The pin passes the same 20 N push from the thumb to the board.Working
Answer
F = 20 N on both
Pressure on the thumb
Method
Convert the head area to m², then divide.Reason
Pressure is force per unit area; 1 cm² is 10⁻⁴ m².Working
Answer
20 ÷ (1.0 × 10⁻⁴) = 2.0 × 10⁵ Pa
Pressure on the board
Method
Convert the point area to m², then divide.Reason
1 mm² is 10⁻⁶ m².Working
Answer
0.10 mm² = 1.0 × 10⁻⁷ m², so p = 20 ÷ (1.0 × 10⁻⁷) = 2.0 × 10⁸ Pa
Compare
Working
Answer
The point area is 1000 times smaller, so the pressure on the board is 1000 times greater. The board gives way; the thumb does not.
Check your understanding 4
A 54 kg hiker (g = 10 N/kg) stands in snow. Her boots have a total contact area of 0.050 m²; with snowshoes, the area is 0.30 m². Compare the two pressures. Start with the force, which is the same in both cases.
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Show the steps Optional working
Show the steps Optional working
Show answer
The force is her weight: 54 × 10 = 540 N. In boots, p = 540 ÷ 0.050 = 1.08 × 10⁴ Pa, about 1.1 × 10⁴ Pa. In snowshoes, p = 540 ÷ 0.30 = 1.8 × 10³ Pa. The snowshoes have six times the area, so the pressure is six times smaller and she sinks much less.
Bring the topic together
A pressure problem often needs every part of this topic: mass gives weight in a gravitational field, and weight spread over an area gives pressure. Density links the mass to the size of the object.
Worked example 4
An equipment case in a weaker field
Problem
A 60 kg equipment case rests on a base of area 0.080 m² where g = 10 N/kg. Find the pressure on the floor, and say what changes in a place with a weaker gravitational field.
Show full solution
Find the weight
Method
Use W = mg.Reason
The weight is the force on the floor.Working
Answer
W = 60 × 10 = 600 N
Find the pressure
Method
Divide force by area.Reason
Pressure is force per unit area.Working
Answer
p = 600 ÷ 0.080 = 7.5 × 10³ Pa
Move to a weaker field
Working
Answer
The case has a smaller weight and exerts a smaller pressure on the same base, but its mass, volume and density do not change.
Check your understanding 5
A 45 kg crate has a volume of 0.018 m³ and rests on a face of area 0.030 m² where g = 4.0 N/kg. Find its density, its weight and the pressure it exerts. Then classify its weight and the normal force from the ground as contact or non-contact.
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Show answer
ρ = 45 ÷ 0.018 = 2.5 × 10³ kg/m³. W = 45 × 4.0 = 180 N. p = 180 ÷ 0.030 = 6.0 × 10³ Pa. Weight is a gravitational force, so it is non-contact; the normal force is a contact force.
Check your understanding 6 Two changes need one force-per-area comparison
A storage platform supports a 100 N load over 0.020 m². A new arrangement halves the load and also halves the contact area. Another arrangement keeps the original load but doubles the area. Compare each new average pressure with the original. Why is “a smaller area always means greater pressure” incomplete?
Show answer
Original pressure = 100 ÷ 0.020 = 5000 Pa. Halving both factors gives 50 ÷ 0.010 = 5000 Pa: the ratio is unchanged. Keeping 100 N and doubling the area gives 100 ÷ 0.040 = 2500 Pa, half the original. A smaller area raises pressure only when the perpendicular force stays the same; when both change, compare force divided by contact area.
Common mistakes
- Putting the mass into p = F/A. Use a force in newtons: for a resting object, its weight W = mg.
- Using the whole surface area of the object. Only the face touching the surface counts.
- Writing 1 cm² = 10⁻² m². Areas convert with the square: 1 cm² = 10⁻⁴ m².
- Treating pressure and force as the same thing. The same force can give very different pressures, depending on the area.
- Saying a heavier person always presses harder on the ground. A heavier person on snowshoes can exert less pressure than a lighter one in boots.
- Saying a sideways push on a box increases its pressure on the floor. Only forces at right angles to the surface count.
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Syllabus and review details
- SEC G3 Combined Science Physics component 2027 · 2027
Physics section, PDF pages 8–24
Last reviewed: