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How is pressure calculated, and how does pressure in a liquid change with depth?

Pressure as force per unit area, P = F / A, pressure in liquids increasing with depth and density, and the equation P = rho g h.

A CCEA GCSE Physics answer on pressure as force per unit area, the equation P = F / A, how pressure in a liquid increases with depth and density, and using P = rho g h.

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  1. What this dot point is asking
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  3. Examples in context
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What this dot point is asking

CCEA wants you to define pressure as force per unit area, use the equation P = F / A, explain why pressure in a liquid increases with depth and density, and use the equation P = rho g h for the pressure due to a column of liquid.

The answer

Pressure on a surface

The same force over a smaller area gives a larger pressure. This is why a sharp knife (small area) cuts easily and why snowshoes (large area) stop you sinking into snow.

Pressure in liquids

A liquid exerts pressure on any surface in contact with it, and the pressure acts equally in all directions at a given depth.

The liquid pressure equation

This explains why a dam is built thicker at the bottom, where the water pressure is greatest, and why deep-sea divers experience very high pressures.

Worked example: pressure on a dam

Examples in context

Example 1. Drawing pins and skis
A drawing pin has a tiny point area, so even a small push gives a huge pressure that pierces a board. Skis spread weight over a large area, lowering the pressure so a skier does not sink into snow.
Example 2. A water tower
Putting a water tank high above houses creates a large depth hh, so the equation P=ρghP = \rho g h gives a high pressure that drives water through the pipes to the taps below.
Example 3. A submarine hull
As a submarine dives deeper, hh increases and so does the water pressure on the hull. The hull must be built strong enough to withstand the very large pressures at depth, which is why submarines have thick, curved metal hulls.

Liquid pressure acting equally in all directions also explains hydraulics: a force applied to a small piston creates a pressure that is transmitted through the liquid and acts on a larger piston, producing a much bigger force. This is how car brakes and hydraulic lifts multiply forces.

Try this

Q1. State the equation for pressure on a surface and the unit of pressure. [2 marks]

  • Cue. P=F/AP = F / A; the unit is the pascal (Pa), equal to N per square metre.

Q2. A force of 200 N200\ \text{N} acts on an area of 0.50 m20.50\ \text{m}^2. Find the pressure. [2 marks]

  • Cue. P=200/0.50=400 PaP = 200 / 0.50 = 400\ \text{Pa}.

Q3. State two factors that affect the pressure at a point in a liquid. [2 marks]

  • Cue. The depth and the density of the liquid (and gg).

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA style3 marksA box weighs 360 N and its base measures 0.40 m by 0.30 m. Calculate the pressure it exerts on the floor.
Show worked answer →

Area of the base:

A=0.40×0.30=0.12 m2.A = 0.40 \times 0.30 = 0.12\ \text{m}^2.

Pressure:

P=FA=3600.12=3000 Pa.P = \dfrac{F}{A} = \dfrac{360}{0.12} = 3000\ \text{Pa}.

Markers reward the area, the pressure equation, and the value 3000 Pa (N per square metre).

CCEA style3 marksCalculate the pressure due to the water at the bottom of a swimming pool 2.5 m deep. The density of water is 1000 kg per cubic metre and g is 9.8 N/kg.
Show worked answer →

Use the liquid pressure equation:

P=ρgh=1000×9.8×2.5=24500 Pa.P = \rho g h = 1000 \times 9.8 \times 2.5 = 24500\ \text{Pa}.

So the pressure due to the water is about 24500 Pa (about 25 kPa).

Markers reward P=ρghP = \rho g h, correct substitution, and the value 24500 Pa.

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