Pressure in a fluid; atmospheric pressure (physics only)

SPEC 4.5.5 • TOPIC 5 FORCES — PAPER 2

Pressure in Fluids & the Atmosphere

Two equations. One you must recall. One on the equation sheet — but only for Higher Tier. Plus the conversion that loses marks every year.

⚡ PHYSICS ONLY — NOT IN COMBINED SCIENCE

What You'll Learn

Pressure shows up everywhere: snow boots, knife edges, scuba divers, weather. The maths is two equations, and the trap is unit conversion — but for area this time, not distance.

Lesson Goals

Define pressure, the pascal, and atmospheric pressure in mark-scheme language

Apply p = F/A (recall) with correct unit conversion (cm² → m²)

Apply p = h ρ g (HT, equation sheet) to pressure in liquids

Explain upthrust in terms of the pressure difference between top and bottom of a submerged object

Describe why atmospheric pressure decreases with altitude

2

Equations (1 recall, 1 HT)

⭐

High-priority spec point

÷10,000

cm² → m² (not ÷100)

Key Definitions

Mark-scheme wording. The words "perpendicular" / "normal" and "column of fluid" recur in mark schemes — use them.

Definition

Pressure: the force acting normal (at right angles) to a surface, per unit area. Measured in pascals (Pa). 1 Pa = 1 N/m².

Definition

Fluid: a substance that can flow — either a liquid or a gas. (Spec wording — both count as fluids.)

Definition

Atmospheric pressure: the pressure created by air molecules colliding with a surface. At sea level, about 1.0 × 10⁵ Pa.

Definition

Upthrust: the resultant upward force on an object submerged in a fluid, caused by the pressure on its bottom surface being greater than the pressure on its top surface.

Equation 1 — Pressure on a Surface

This is the recall equation — not on the Physics Equations Sheet. You must know it from memory. It applies to any pressure: solid on solid, fluid on a surface.

Pressure = Force ÷ Areap = F / Ap — pressure, PaF — force, N (normal)A — area, m²

Rearranged forms

F = p × A | A = F ÷ p

Critical: F must be NORMAL to the surface

F is the component of force acting at right angles to the surface — not parallel, not at an angle. If a force is applied at an angle, only the perpendicular component contributes to pressure.

Force, Area & Pressure — Intuition

Pressure depends on both the force AND the area it acts over. Same force ÷ smaller area = larger pressure. That's why a sharp knife cuts and a blunt one doesn't.

Real-world examples

Snowshoes / skis — large area, low pressure, don't sink into snow. Knife edge / drawing pin — small area, high pressure, cuts/pierces easily. Tractor tyres — large area, low pressure, won't sink into mud.

Force must be perpendicular to the surface

p = F/A only works when F is at right angles to the surface. If a force is applied at an angle, only its perpendicular component contributes to pressure — a force pulling along a surface (parallel to it) produces zero pressure on that surface. This is why mark schemes specifically require the word "normal" or "perpendicular".

ESUA Worked Example

A wooden block has a weight of 50 N. It rests on a flat floor on its base, which has an area of 200 cm².

Calculate the pressure the block exerts on the floor. 3 MARKS

E

Equation: p = F ÷ A

S

Convert area: 200 cm² = 200 ÷ 10,000 = 0.020 m²

Substitute: p = 50 ÷ 0.020

U

Unit: pascals (Pa)

A

Answer: p = 2,500 Pa (or 2.5 × 10³ Pa)

Pressure In A Liquid — Increases With Depth

At any point in a liquid, the pressure is caused by the weight of the column of liquid above that point. Deeper down → taller column above → more weight → more pressure.

Spec wording you must use

Pressure in a liquid increases with the height of the column of liquid above the point, and with the density of the liquid. Pressure at any point acts in all directions — sideways, upward, downward.

Equation 2 — Pressure In a Column of Liquid HT Equation Sheet

The Higher Tier equation. You don't need to recall it — it's on the equation sheet — but you must know what each symbol means and apply it correctly.

Pressure = height × density × gp = h ρ gp — Pah — mρ — kg/m³g — N/kg

Why three variables on the right

Pressure in a liquid increases with h (taller column → more weight pressing down), with ρ (denser liquid → more mass per unit volume → heavier column), and with g (stronger gravity → more weight). All three together give the pressure at depth h.

Don't forget g

The most common error on this equation is calculating p = h × ρ and forgetting g. Three variables on the RHS. Every time.

For pressure DIFFERENCES at two depths

The spec also requires you to "calculate the differences in pressure at different depths in a liquid". The shortcut: Δp = ρ × g × Δh, where Δh is the difference in depth between the two points. Same equation, applied to the height difference rather than the absolute depth.

ESUA Worked Example HT

A submarine sits at a depth of 50 m in seawater.

Density of seawater = 1030 kg/m³. g = 9.8 N/kg.

Calculate the pressure on the submarine due to the seawater above it. 4 MARKS

E

Equation (from sheet): p = h × ρ × g

S

Substitute: p = 50 × 1030 × 9.8

U

Unit: pascals (Pa)

A

Answer: p = 504,700 Pa ≈ 5.0 × 10⁵ Pa (2 s.f.)

Upthrust — Why Things Float

When an object is submerged in a fluid, the bottom of the object is deeper than the top. So the pressure at the bottom is greater than the pressure at the top. This pressure difference creates a resultant upward force called upthrust.

Spec-mandated explanation

"A partially or totally submerged object experiences a greater pressure on the bottom surface than on the top surface. This creates a resultant force upwards. This force is called the upthrust." — explain it this way and you get the marks.

Floating vs Sinking

An object floats if the upthrust is large enough to balance its weight. The steady-state outcome depends on how the object's density compares to the fluid's density.

Density comparisonSteady-state outcome

Object density < fluid densityFloats at the surface, partially submerged. The object settles until the weight of fluid it displaces equals its own weight — at which point upthrust = weight.

Object density = fluid densityNeutrally buoyant — fully submerged but does not sink or rise. Upthrust = weight at any depth.

Object density > fluid densitySinks . Even when fully submerged, upthrust is less than weight, so the object accelerates downwards.

Equilibrium condition (all cases)

In every case, an object settles at upthrust = weight. The density rule just tells you where that balance occurs: at the surface (floats), at any depth (neutrally buoyant), or never (sinks).

Quick check — why a steel ship floats

Steel is denser than water, but a ship is mostly air inside. Its overall (average) density — steel hull + enclosed air — is lower than water. Replace the air with water and the average density rises above water's, so the ship sinks.

Atmospheric Pressure

The atmosphere is a thin layer of air around the Earth. Air molecules are in constant motion — they collide with surfaces, and the force per unit area of those collisions is atmospheric pressure.

Spec mechanism

Atmospheric pressure is created by air molecules colliding with a surface. The number of air molecules above a surface — and so the weight of air pressing on it — depends on altitude.

10⁵

Pa at sea level (≈ 1.0 × 10⁵)

~80

km of atmosphere above us

↓

Decreases with altitude

Why we don't feel it

You're under 10 tonnes of air right now — but it doesn't crush you, because the same pressure acts in all directions, including from inside your body outwards. The pressures balance.

Why Atmospheric Pressure Decreases With Altitude

The further up you go, the fewer molecules are above you, and the less they weigh. Mark schemes want this chain of reasoning, in order.

3-mark explanation chain

As altitude increases, the atmosphere becomes less dense — there are fewer air molecules per unit volume (1)

So there are fewer air molecules above the surface and the weight of air above decreases (1)

So the number of collisions per second with the surface decreases → atmospheric pressure decreases (1)

Exam-Style Multi-Part Question HT

A cylindrical tank stands on a flat floor. The tank's base has area 0.020 m². A liquid fills the tank to a depth of 2.0 m. The total downward force on the floor due to the weight of the liquid is 392 N.

g = 9.8 N/kg

(a) Write down the equation that links pressure, force and area. 1 MARK

(b) Calculate the pressure on the floor due to the liquid. 3 MARKS

(c) Use your answer to part (b) to calculate the density of the liquid. 3 MARKS

(a)

p = F ÷ A (1 mark)

(b)

Substitute: p = 392 ÷ 0.020 (M1)

Calculate: p = 19,600 Pa (M1)

Unit: Pa or N/m² (A1)

(c)

Rearrange p = h × ρ × g for ρ: ρ = p ÷ (h × g) (M1)

Substitute: ρ = 19,600 ÷ (2.0 × 9.8) (M1)

Answer: ρ = 1,000 kg/m³ — the liquid is water (A1)

Examiner Alerts

Trap 1 — cm² → m² is ÷ 10,000, not ÷ 100

Students convert cm to m by ÷ 100 — that's correct for cm. But cm² → m² requires ÷ (100 × 100) = ÷ 10,000. Get this wrong and your pressure is off by a factor of 100. Worked example: 200 cm² = 200 ÷ 10,000 = 0.020 m², NOT 2 m² and NOT 20 m².

Trap 2 — Don't forget g in p = h ρ g

The HT equation has three variables on the right. Students often forget g and calculate p = h × ρ, giving an answer ~10× too small. The equation sheet shows all three — copy it carefully every time.

Trap 3 — F must be the NORMAL force

In p = F/A, F is the component of force at right angles to the surface. Vague answers ("the force") lose marks where the question gives a force at an angle.

Weak vs Strong Answers

Same question — "Calculate the pressure when 50 N acts on an area of 200 cm²".

✗ Weak

"p = 50 ÷ 200 = 0.25 Pa"Area substituted in cm². No conversion. Magnitude wrong by 10,000.

✓ Strong

"200 cm² = 0.020 m²; p = 50 ÷ 0.020 = 2,500 Pa"Conversion at substitution. All marks: equation, substitution, unit.

Same question — "A weather balloon rises through the atmosphere. Explain why atmospheric pressure outside the balloon decreases as it rises. [3 marks]"

✗ Weak

"There's less air higher up."Captures the idea but no chain of reasoning. Usually scores 1/3.

✓ Strong

"At higher altitude the atmosphere is less dense, so fewer molecules are above the balloon. The weight of air above is smaller, so fewer molecules collide with the balloon per second — pressure is lower."Three linked points (less dense → less weight → fewer collisions) → 3/3.

Checkpoint

You can now

Define pressure, the pascal, fluid (liquid or gas), atmospheric pressure, and upthrust

Apply p = F/A with confident cm² → m² conversion (÷ 10,000)

Apply p = h ρ g (HT, equation sheet) — including all three RHS variables

Explain upthrust in terms of greater pressure on the bottom than on the top of a submerged object

Predict whether an object floats or sinks using upthrust vs weight (or density comparison)

Explain why atmospheric pressure decreases with altitude in 3 mark-scheme steps

Two equations, one habit

For both pressure equations, the habit is the same: (1) write the equation, (2) convert all units at substitution, (3) state the answer with the unit "Pa". Most students who lose marks on a pressure question are skipping step 2.

Next lesson

5.08 — Describing motion along a line: distance–time graphs. Reading gradients to find speed, recognising acceleration/deceleration as curves, and the difference between distance and displacement.