Learning Objectives
4 objectivesBy the end of this note, you should be able to:
- Define pressure as force per unit area and use p = F/A.
- Describe how pressure varies with force and area in everyday examples.
- Describe qualitatively how liquid pressure changes with depth and density.
- Recall and use the equation Δp = ρgΔh.
CORE VS EXTENDED GUIDE
- Core students study only the unlabelled sections.
- Extended students must study everything, including Extended Extended points.
- Extended = Core + Supplement.
Defining Pressure
Pressure is a measure of how concentrated a force is over an area.
Pressure: the force per unit area.
When the same force acts over a smaller area, the pressure increases. When the same force spreads over a larger area, the pressure decreases.
A drawing pin pushed into a board illustrates this clearly: the sharp point has a tiny area, so the pressure there is very large, which is why it pierces the board. The flat head has a much larger area, so the pressure on your thumb remains small enough to avoid injury.
Pressure is a scalar quantity — it acts equally in all directions at a point and has no specific direction.
Reading the equation
In physics equations, each letter represents a measurable quantity. A fraction means "divided by," so p = F/A reads as "pressure equals force divided by area." Whenever a quantity sits in the denominator (bottom of the fraction), increasing that quantity decreases the result.
Key Equations
Pressure:
$$p=\frac{F}{A}$$
Variables:
- $p$ = pressure, in pascals (Pa)
- $F$ = force acting perpendicular to the surface, in newtons (N)
- $A$ = area over which the force acts, in metres squared (m²)
SI unit: Pa (1 Pa = 1 N m⁻²)
Rearrangements
Starting from $p=\frac{F}{A}$:
Rearranging for force:
$$F=p\times A$$
Rearranging for area:
$$A=\frac{F}{p}$$
Proportionality
- Pressure is directly proportional to force when area is constant. Doubling the force doubles the pressure.
- Pressure is inversely proportional to area when force is constant. Doubling the area halves the pressure.
Worked Example
A crate weighing 600 N stands on the floor. The base of the crate has an area of 0.20 m². Calculate the pressure the crate exerts on the floor.
Equation used
$$p=\frac{F}{A}$$
$$F=600\text{ N}$$
$$A=0.20{\text{ m}}^{2}$$
Substitution
$$p=\frac{600}{0.20}$$
$$p=3000\text{ Pa}$$
$$p=3000\text{ Pa}$$
MisconceptionStudents sometimes write the unit of pressure as N/m or forget to square the metre. The SI unit is the pascal (Pa), equivalent to N m⁻². Always check that area is in m², not cm².
Exam TipIf area is given in cm², convert to m² by dividing by 10⁴ before substituting.

Pressure in Everyday Contexts
The relationship between force, area, and pressure explains many everyday observations.
When designers want high pressure, they reduce the area on which a force acts. When they want low pressure, they increase the area. The table below summarises common examples.
| Situation | Design feature | Effect on pressure |
|---|---|---|
| Knife blade | Very thin cutting edge (small area) | High pressure cuts through food easily |
| Snowshoes / skis | Wide base (large area) | Low pressure prevents sinking into snow |
| Tractor tyres | Broad, wide tyres (large area) | Low pressure prevents sinking into soft soil |
| Drawing pin point | Tiny sharp tip (small area) | High pressure pierces the board |
| Stiletto heel | Very small heel area | High pressure — can damage soft floors |
| Camel's feet | Wide, flat feet (large area) | Low pressure prevents sinking into sand |
In every case, the force involved (usually the weight of the person or object) stays the same. The change in pressure results entirely from the change in area.
Examiner InsightCIE frequently asks students to explain an everyday situation using the pressure equation. Always state three things: (1) what the force is, (2) whether the area is large or small, and (3) the effect on pressure using the word "inversely proportional" or the equation p = F/A.
Exam TipName the force (e.g. "weight") — do not just write "force."
Pressure Beneath a Liquid Surface
Liquid pressure increases with depth below the surface because a greater depth of liquid sits above the point, so a greater weight of liquid pushes down.
Three key facts describe how pressure in a liquid behaves:
1. Pressure increases with depth. The deeper the point beneath the surface, the greater the weight of liquid above it, so the greater the pressure.
2. Pressure increases with the density of the liquid. A denser liquid has more mass in the same volume, so the weight of liquid above a given depth is greater, which produces a greater pressure. Mercury (density ≈ 13 500 kg m⁻³) exerts a much higher pressure at the same depth than water (density ≈ 1000 kg m⁻³).
3. Pressure acts equally in all directions at a given depth. At any point in a liquid, the pressure pushes up, down, and sideways with the same magnitude.
A simple demonstration shows these ideas: a tall container of water with small holes at different heights leaks water furthest from the lowest hole, because the pressure is greatest at the greatest depth.

MisconceptionStudents sometimes state that pressure in a liquid depends on the total volume of liquid. Pressure at a point depends only on the depth below the surface and the density of the liquid, not on the shape or total volume of the container. Two containers with different widths but the same liquid and the same depth produce the same pressure at their bases.
Exam TipWrite "depth" not "amount of liquid."
Extended Calculating Pressure Change in a Liquid
Extended The pressure change beneath the surface of a liquid can be calculated using the equation Δp = ρgΔh.
Reading the equation
The symbol Δ (Greek capital delta) means "change in." So Δp means "change in pressure" and Δh means "change in depth." The equation tells us the additional pressure caused by a column of liquid of height Δh, not the total pressure (which would also include atmospheric pressure above the liquid surface).
Key Equations
Pressure change beneath a liquid surface:
$$\Delta p=\rho g\Delta h$$
Variables:
- $\Delta p$ = change in pressure, in pascals (Pa)
- $\rho $ = density of the liquid, in kilograms per metre cubed (kg m⁻³)
- $g$ = gravitational field strength, in newtons per kilogram (N kg⁻¹). Use $g=9.8$ N kg⁻¹ (or 10 N kg⁻¹ if the question states so).
- $\Delta h$ = change in depth below the surface, in metres (m)
SI unit: Pa
Rearrangements
Starting from $\Delta p=\rho g\Delta h$:
Rearranging for depth change:
$$\Delta h=\frac{\Delta p}{\rho g}$$
Rearranging for density:
$$\rho =\frac{\Delta p}{g\Delta h}$$
Proportionality
- Δp is directly proportional to Δh when ρ and g are constant. Doubling the depth doubles the pressure change.
- Δp is directly proportional to ρ when Δh and g are constant. A liquid twice as dense produces twice the pressure change at the same depth.
Worked Example
A submarine dives to a depth of 250 m below the surface of the sea. The density of seawater is 1030 kg m⁻³ and g = 9.8 N kg⁻¹. Calculate the pressure change due to the seawater at this depth.
Equation used
$$\Delta p=\rho g\Delta h$$
$$\rho =1030{\text{ kg m}}^{-3}$$
$$g=9.8{\text{ N kg}}^{-1}$$
$$\Delta h=250\text{ m}$$
Substitution
$$\Delta p=1030\times 9.8\times 250$$
$$\Delta p=2 523 500\text{ Pa}$$
$$\Delta p\approx 2 520 000\text{ Pa}\text{ (3 s.f.)}$$
This is approximately 2520 kPa, which is about 25 times atmospheric pressure.
Worked Example with Unit Conversion
A fish swims at a depth of 800 cm in a freshwater lake. The density of fresh water is 1000 kg m⁻³ and g = 9.8 N kg⁻¹. Calculate the pressure change at this depth.
Converting depth to SI units
$$\Delta h=\frac{800}{100}$$
$$\Delta h=8.0\text{ m}$$
Equation used
$$\Delta p=\rho g\Delta h$$
$$\rho =1000{\text{ kg m}}^{-3}$$
$$g=9.8{\text{ N kg}}^{-1}$$
$$\Delta h=8.0\text{ m}$$
Substitution
$$\Delta p=1000\times 9.8\times 8.0$$
$$\Delta p=78 400\text{ Pa}$$
$$\Delta p=78 400\text{ Pa}$$
Examiner InsightCIE questions sometimes ask for the total pressure on an object at a certain depth. The total pressure equals atmospheric pressure plus the pressure due to the liquid: ${p}_{\text{total}}={p}_{\text{atm}}+\rho g\Delta h$. Read the question carefully to determine whether it asks for the pressure change (Δp only) or the total pressure.
Exam TipIf the question says "pressure at a depth," check whether atmospheric pressure is given — if so, add it.
QUICK RECAP
Key Points
- Pressure is defined as force per unit area, measured in pascals (Pa).
- p = F/A; pressure is directly proportional to force and inversely proportional to area.
- Smaller area with the same force results in greater pressure.
- Knife blades, pins, and needles use small area to produce high pressure.
- Snowshoes, wide tyres, and flat foundations use large area to reduce pressure.
- Liquid pressure increases with depth below the surface.
- Liquid pressure increases with the density of the liquid.
- Pressure at a point in a liquid acts equally in all directions.
- Liquid pressure does not depend on the shape or volume of the container.
- Extended Δp = ρgΔh gives the pressure change due to a liquid column.
- Extended Doubling the depth doubles the pressure change.
- Extended Total pressure = atmospheric pressure + ρgΔh.
CAN I…? PROGRESS CHECK
Self-Assessment
- Define pressure using the exact CIE wording?
- Calculate pressure, force, or area using p = F/A?
- Explain everyday examples of pressure using force and area?
- State how liquid pressure changes with depth and density?
- State that liquid pressure acts equally in all directions at a given depth?
- Extended Calculate pressure change in a liquid using Δp = ρgΔh?
- Extended Rearrange Δp = ρgΔh to find depth or density?
- Extended Distinguish between pressure change due to liquid and total pressure?