Cambridge IGCSE Physics 0625
1.8 Pressure
Detailed Core and Extended notes covering force per unit area, everyday pressure applications and pressure changes beneath liquid surfaces.
2026-2028 syllabus
Syllabus checklist
Core
- Define pressure as force per unit area; recall and use the equation p = F ÷ A.
- Describe how pressure varies with force and area in the context of everyday examples.
- Describe qualitatively how pressure beneath the surface of a liquid changes with depth and density of the liquid.
Supplement
- Recall and use the equation for the change in pressure beneath the surface of a liquid: Δp = ρgΔh.
Key scientific language
Definitions
Pressure
Force acting per unit area. Pressure describes how concentrated a force is on a surface.
Pascal
The SI unit of pressure. One pascal is one newton per square metre: 1 Pa = 1 N/m2.
Normal force
The component of force acting perpendicular to a surface. This is the force used in the pressure equation.
Liquid pressure
Pressure produced by the weight of liquid above a point beneath its surface.
Depth
The vertical distance below a liquid's surface. It is not the distance measured along a sloping container.
Density
Mass per unit volume. A denser liquid produces a greater pressure increase at the same depth.
Pressure change
The difference in pressure between two depths. It is represented by Δp.
Core
Pressure on solid surfaces
Force spread over an area
Pressure tells us how concentrated a force is. The same force can produce different pressures depending on the area over which it acts.
If a force acts over a small area, the force is concentrated and the pressure is large. If the same force is spread over a larger area, the pressure is smaller.
The force used in the pressure equation must act perpendicular to the surface. For an object resting on a horizontal floor, this force is usually the object's weight, provided no other vertical forces affect it.
Pressure on a surface
- p is pressure in pascals (Pa).
- F is the perpendicular force in newtons (N).
- A is the contact area in square metres (m2).
Because 1 Pa = 1 N/m2, a pressure of 2000 Pa means that a force of 2000 N acts on each square metre of area.
Increasing force
For a constant area, increasing the perpendicular force increases pressure in direct proportion.
Doubling the force while keeping the area unchanged doubles the pressure.
Increasing area
For a constant force, increasing the contact area decreases pressure.
Doubling the area while keeping the force unchanged halves the pressure.
Rearranging the pressure equation
The same relationship can be rearranged when force or area is unknown.
Calculating force
- Use this form when pressure and area are known.
- The force is calculated in newtons.
Calculating area
- Use this form when force and pressure are known.
- The area is calculated in m2.
Area conversions
Since 1 m = 100 cm, one square metre contains 100 × 100 square centimetres.
Example: 200 cm2 = 200 ÷ 10 000 = 0.020 m2.
Core applications
Force, area and everyday pressure
Everyday pressure examples can be explained by identifying what happens to the perpendicular force, the contact area, or both.
| Example | Area or force change | Effect on pressure |
|---|---|---|
| Standing on one foot | The same body weight acts through a smaller total area. | Pressure on the floor increases. |
| Standing on tiptoe | Body weight acts through a very small contact area. | Pressure increases further. |
| Snowshoes | A large area spreads the person's weight over more snow. | Pressure decreases, reducing sinking. |
| Wide tractor tyres | A larger area spreads the vehicle's weight over soft ground. | Pressure decreases. |
| Sharp knife | A narrow cutting edge produces a very small contact area. | A large pressure is produced for the applied force. |
| Drawing pin | The small point concentrates the pushing force. | High pressure allows the point to enter the surface. |
| Building foundations | A wide base spreads the building's weight over a larger area. | Pressure on the ground decreases. |
Worked example: a box on the floor
A box presses on the floor with a perpendicular force of 600 N. Its base area is 0.20 m2.
p = F ÷ A
p = 600 ÷ 0.20
p = 3000 Pa
If the box is turned onto a face with half the area, its weight remains unchanged but its pressure doubles.
Core
Pressure beneath a liquid surface
Why pressure increases with depth
A point deeper in a liquid has a taller column of liquid above it. This column contains more liquid and therefore has greater weight.
The greater force per unit area produces a greater pressure. Pressure therefore increases as vertical depth beneath the liquid surface increases.
This is why a swimmer feels greater pressure on the ears when swimming deeper and why the lower parts of a dam must withstand greater liquid pressure than the upper parts.
Why liquid density matters
A denser liquid has more mass in each unit volume. For the same depth, a column of denser liquid contains more mass and has greater weight than a column of less dense liquid.
The denser liquid therefore produces a greater pressure increase at the same vertical depth.
At the same depth, saltwater produces a slightly greater pressure increase than freshwater because saltwater has a greater density.
Supplement only
Calculating liquid pressure changes
Liquid pressure equation
The following equation calculates the difference in pressure between two vertical depths in a liquid.
If one point is at the liquid surface, the equation gives the pressure increase caused by the liquid below that surface.
Change in pressure beneath a liquid surface
- Δp is the pressure change in pascals (Pa).
- ρ is the liquid density in kg/m3.
- g is gravitational field strength in N/kg.
- Δh is the vertical depth change in metres (m).
- 1
Identify the two depths and calculate the vertical depth change, Δh.
- 2
Convert density to kg/m3 and depth to metres when necessary.
- 3
Substitute the values into Δp = ρgΔh.
- 4
Calculate the pressure change and give the answer in pascals.
Rearranging the liquid pressure equation
Calculating density
Calculating vertical depth change
Example 1: pressure increase in water
Calculate the pressure increase 5.0 m below the surface of water. Use ρ = 1000 kg/m3 and g = 9.8 N/kg.
Δp = ρgΔh
Δp = 1000 × 9.8 × 5.0
Δp = 49 000 Pa
This is the pressure increase above the pressure at the water's surface.
Example 2: container of liquid
A rectangular container has base dimensions 0.12 m by 0.16 m and contains 4.8 kg of liquid. The liquid depth is 0.32 m. Use g = 9.8 N/kg.
1. Calculate the weight of the liquid
W = mg
W = 4.8 × 9.8
W = 47.04 N ≈ 47 N
2. Calculate the pressure increase at the base
base area = 0.12 × 0.16 = 0.0192 m2
p = F ÷ A
p = 47.04 ÷ 0.0192
Δp = 2450 Pa
3. Calculate the liquid density
ρ = Δp ÷ (gΔh)
ρ = 2450 ÷ (9.8 × 0.32)
ρ = 781 kg/m3 ≈ 780 kg/m3
4. Distinguish pressure increase from total pressure
The calculated liquid pressure is the increase above the pressure at the surface.
If the surface is open to the atmosphere, total pressure at the base equals atmospheric pressure plus the pressure increase caused by the liquid.
Cambridge-style practice
Practice questions
Define pressure.
Pressure is force per unit area.
A force of 500 N acts normally on an area of 0.25 m2. Calculate the pressure.
p = F ÷ A
p = 500 ÷ 0.25
p = 2000 Pa
A pressure of 12 000 Pa acts on an area of 0.030 m2. Calculate the force.
F = pA
F = 12 000 × 0.030
F = 360 N
A person's weight is unchanged when they stand on one foot instead of two. Explain what happens to the pressure on the floor.
The force remains the same.
The contact area decreases.
Pressure increases because the same force acts over a smaller area.
Explain why liquid pressure is greater at a greater depth.
There is a taller column of liquid above the point.
The column contains more liquid and has greater weight.
This produces a greater force per unit area and therefore greater pressure.
Two liquids are at the same depth. Liquid A is denser than liquid B. Compare their pressure increases.
Liquid A produces the greater pressure increase.
A denser liquid has more mass and weight in the same volume.
Calculate the pressure increase 12 m below the surface of water. Use ρ = 1000 kg/m3 and g = 9.8 N/kg.
Δp = ρgΔh
Δp = 1000 × 9.8 × 12
Δp = 117 600 Pa
Oil has density 800 kg/m3. Calculate the depth that produces a pressure increase of 15 680 Pa. Use g = 9.8 N/kg.
Δh = Δp ÷ (ρg)
Δh = 15 680 ÷ (800 × 9.8)
Δh = 2.0 m
A pressure increase of 29 400 Pa occurs at a depth of 3.0 m. Use g = 9.8 N/kg to calculate the liquid density.
ρ = Δp ÷ (gΔh)
ρ = 29 400 ÷ (9.8 × 3.0)
ρ = 1000 kg/m3
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