Density, Upthrust and Pressure

This lesson covers AQA Spec 3.4.2.1 (Density) and related concepts including Archimedes' principle, atmospheric pressure, pressure in fluids, and hydraulic systems.

Spec mapping (AQA 7408 §3.4.2.1): This lesson develops density ρ = m/V as a fundamental material property, pressure p = F/A in solids and gases, the hydrostatic pressure variation in a fluid column p = ρgh, Archimedes' principle (upthrust equal to weight of displaced fluid), atmospheric pressure as the weight of an air column, and Pascal's principle as the basis for hydraulic systems. The treatment includes experimental methods for measuring density of regular solids, irregular solids and liquids. (Refer to the official AQA specification document for exact wording.)

Synoptic links: Density and pressure connect to a wide range of A-Level physics. (i) Mechanics of equilibrium (Lesson 6, §3.4.1.1) uses upthrust as another force to balance — floating bodies are static equilibrium with weight balanced by buoyancy. (ii) Gas laws (Year 13, §3.6.2.2) generalise pressure to gases at the molecular level: pV = nRT links macroscopic pressure to microscopic kinetic theory, and atmospheric pressure becomes the macroscopic consequence of molecular bombardment. (iii) Materials and stress (Lesson 7, §3.4.2.2) uses pressure as the compressive analogue of tensile stress — bulk modulus K = −V(dp/dV) is the volumetric stiffness counterpart of Young's modulus. (iv) Engineering applications — submarine design, balloon flight, hydraulic machinery, dam construction — all depend on the principles developed here. The hydrostatic equation p = ρgh is the foundation of all fluid statics.

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1 Density

Key Definition: Density is the mass per unit volume of a substance.

ρ = m / V

SI unit: kg m⁻³

Material	Density (kg m⁻³)
Air (at STP)	1.2
Water	1000
Ice	917
Aluminium	2700
Iron/Steel	7800
Copper	8900
Lead	11 340
Gold	19 300
Mercury	13 600

Worked Example 1 — Finding Density

A block of metal has dimensions 5.0 cm × 4.0 cm × 3.0 cm and a mass of 0.47 kg. Find its density and identify the metal.

Solution:

V = 0.050 × 0.040 × 0.030 = 6.0 × 10⁻⁵ m³

ρ = m/V = 0.47 / (6.0 × 10⁻⁵) = 7833 kg m⁻³ ≈ 7800 kg m⁻³

This is closest to iron/steel.

Worked Example 2 — Mass of Air in a Room

A room measures 5.0 m × 4.0 m × 3.0 m. The density of air is 1.2 kg m⁻³. Find the mass of air in the room.

Solution:

V = 5.0 × 4.0 × 3.0 = 60 m³

m = ρV = 1.2 × 60 = 72 kg

Exam Tip: Always convert volumes to m³ before using the density formula. 1 cm³ = 1 × 10⁻⁶ m³. 1 litre = 1 × 10⁻³ m³.

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2 Pressure

Key Definition: Pressure is the force per unit area acting perpendicular to a surface.

p = F / A

SI unit: pascal (Pa). 1 Pa = 1 N m⁻².

Worked Example 3

A person of mass 70 kg stands on the ground. Each shoe has a contact area of 200 cm². Find the pressure on the ground.

Solution:

Total area = 2 × 200 cm² = 400 cm² = 400 × 10⁻⁴ m² = 0.040 m²

Weight = mg = 70 × 9.81 = 686.7 N

p = F/A = 686.7 / 0.040 = 17 200 Pa ≈ 17.2 kPa

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3 Pressure in a Fluid Column

The pressure at depth h in a fluid of density ρ:

p = ρgh

This is the pressure due to the fluid column only (gauge pressure). The absolute pressure at depth h is:

p_absolute = p_atm + ρgh

where p_atm ≈ 101 325 Pa ≈ 101 kPa.

Derivation

Consider a column of fluid of height h, cross-sectional area A, and density ρ.

Weight of fluid column: W = mg = ρVg = ρAhg

Pressure at the base: p = F/A = W/A = ρAhg/A = ρgh

Worked Example 4 — Pressure at Depth

Find the total pressure at a depth of 25 m in a lake. Density of water = 1000 kg m⁻³, atmospheric pressure = 101 kPa.

Solution:

Pressure due to water: p_water = ρgh = 1000 × 9.81 × 25 = 245 250 Pa = 245.3 kPa

Total pressure = p_atm + p_water = 101 + 245.3 = 346 kPa

This is about 3.4 atmospheres.

Worked Example 5 — Mercury Barometer

A mercury barometer reads a column height of 760 mm. Density of mercury = 13 600 kg m⁻³. Calculate atmospheric pressure.

Solution:

p_atm = ρgh = 13 600 × 9.81 × 0.760 = 101 400 Pa ≈ 101 kPa ✓

Exam Tip: The pressure at a given depth depends only on the depth, the fluid density, and g — it does NOT depend on the shape of the container. This is sometimes called the "hydrostatic paradox."

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4 Archimedes' Principle

Archimedes' Principle: When an object is wholly or partially immersed in a fluid, it experiences an upward force (upthrust) equal to the weight of the fluid displaced.

Upthrust (buoyancy force) = ρ_fluid × V_displaced × g

where V_displaced is the volume of the object that is submerged.

Floating Condition

An object floats when the upthrust equals its weight:

ρ_fluid × V_displaced × g = m_object × g

For a fully submerged object: V_displaced = V_object

If ρ_object < ρ_fluid, the object floats (only partially submerged). If ρ_object > ρ_fluid, the object sinks. If ρ_object = ρ_fluid, the object is neutrally buoyant.

Worked Example 6 — Floating Object

An ice cube of mass 0.050 kg floats in water (ρ = 1000 kg m⁻³). What volume of ice is submerged?

Solution:

At equilibrium: upthrust = weight

ρ_water × V_submerged × g = mg

V_submerged = m / ρ_water = 0.050 / 1000 = 5.0 × 10⁻⁵ m³ = 50 cm³

Total volume of ice = m / ρ_ice = 0.050 / 917 = 5.45 × 10⁻⁵ m³ = 54.5 cm³

Fraction submerged = 50 / 54.5 = 0.917 = 91.7% (consistent with ρ_ice / ρ_water = 917/1000)

Worked Example 7 — Apparent Weight

A block of aluminium (ρ = 2700 kg m⁻³) has a volume of 0.0020 m³. It is weighed in air and then fully submerged in water. Find (a) its weight in air and (b) its apparent weight in water.

Solution:

(a) m = ρV = 2700 × 0.0020 = 5.4 kg

Weight in air = mg = 5.4 × 9.81 = 53.0 N

(b) Upthrust = ρ_water × V × g = 1000 × 0.0020 × 9.81 = 19.6 N

Apparent weight = true weight − upthrust = 53.0 − 19.6 = 33.4 N

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5 Atmospheric Pressure

Atmospheric pressure is caused by the weight of the column of air above a surface. At sea level:

p_atm ≈ 101 325 Pa ≈ 101 kPa ≈ 1 atmosphere

Atmospheric pressure decreases with altitude because there is less air above.

The atmospheric column model: Imagine a column of air of cross-section 1 m² extending from sea level to the top of the atmosphere. The weight of this column:

F = p × A = 101 325 × 1 = 101 325 N

Mass of air column: m = F/g = 101 325/9.81 ≈ 10 330 kg

This means each square metre of the Earth's surface supports about 10 tonnes of air above it.

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6 Hydraulic Systems

Hydraulic systems use the principle that pressure is transmitted equally through an enclosed fluid (Pascal's principle).

For a hydraulic press with two pistons of areas A₁ and A₂:

p₁ = p₂ → F₁/A₁ = F₂/A₂

Therefore: F₂ = F₁ × (A₂/A₁)

This gives a force multiplier with a ratio equal to the area ratio.

However, energy is conserved. If piston 1 moves distance d₁ and piston 2 moves d₂:

F₁d₁ = F₂d₂ (work done by piston 1 = work done by piston 2)

So: d₂ = d₁ × (A₁/A₂) — the larger piston moves a smaller distance.

Worked Example 8

A hydraulic jack has pistons of diameters 2.0 cm and 20 cm. A force of 50 N is applied to the small piston. Find (a) the force on the large piston and (b) how far the large piston rises when the small piston is pushed down 0.40 m.

Solution:

A₁ = π(0.010)² = 3.14 × 10⁻⁴ m²

A₂ = π(0.10)² = 3.14 × 10⁻² m²

Area ratio: A₂/A₁ = 0.0314/0.000314 = 100

(a) F₂ = F₁ × (A₂/A₁) = 50 × 100 = 5000 N

(b) F₁d₁ = F₂d₂ → d₂ = F₁d₁/F₂ = 50 × 0.40 / 5000 = 0.0040 m = 4.0 mm

Or equivalently: d₂ = d₁ × (A₁/A₂) = 0.40 × (1/100) = 0.004 m ✓

Common Misconception: A hydraulic system is a force multiplier, NOT an energy multiplier. The work done on the input piston equals the work done by the output piston (ignoring losses). You gain force but lose distance.

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7 Measuring Density Experimentally

Regular Solid

1. Measure mass with a balance.
2. Measure dimensions with a ruler/vernier caliper/micrometer.
3. Calculate volume from dimensions.
4. ρ = m/V.

Irregular Solid

1. Measure mass with a balance.
2. Use displacement (Eureka) can or measuring cylinder to find volume by water displacement.
3. ρ = m/V.

Liquid

1. Measure mass of an empty measuring cylinder.
2. Add a known volume of liquid.
3. Measure the total mass.
4. Mass of liquid = total − empty.
5. ρ = m/V.

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8 Worked Example 9 — Hot Air Balloon Lift

A spherical hot-air balloon has a diameter of 12 m. The air inside is heated to 90 °C (363 K) while the outside air is at 15 °C (288 K). Both are at atmospheric pressure 101 kPa. The molar mass of air is 29 g mol⁻¹, and the molar gas constant is 8.31 J K⁻¹ mol⁻¹.

(a) Calculate the density of the air at 15 °C and at 90 °C.

(b) Calculate the upthrust on the balloon.

(c) Calculate the weight of the heated air inside.

(d) Hence find the maximum mass (envelope + basket + payload) that the balloon can lift.

Solution:

(a) Using pV = nRT and ρ = m/V = nM/V = pM/(RT):

ρ_cold = (101 000 × 0.029) / (8.31 × 288) = 2929/2393 = 1.224 kg m⁻³.

ρ_hot = (101 000 × 0.029) / (8.31 × 363) = 2929/3017 = 0.971 kg m⁻³.

The hot air is significantly less dense — only 79% of cold-air density.

(b) Volume of balloon: V = (4/3) π r³ = (4/3) π × 6³ = (4/3) × π × 216 = 904.8 m³.

Upthrust = weight of displaced cold air = ρ_cold × V × g = 1.224 × 904.8 × 9.81 = 10 868 N (≈ 10.9 kN).

(c) Weight of heated air inside the envelope = ρ_hot × V × g = 0.971 × 904.8 × 9.81 = 8623 N (≈ 8.6 kN).

(d) Maximum payload (envelope, basket, occupants): The net upward force available is upthrust − weight of heated air = 10 868 − 8623 = 2245 N. Converting to mass: m = F/g = 2245/9.81 = 229 kg.

This is enough to lift a typical hot-air balloon's envelope and basket (~150 kg) plus one or two adult passengers. Real balloons use larger volumes (typically 2000-3000 m³) to carry full passenger loads. The principle — heating reduces density, increasing upthrust relative to weight — is the basis of all lighter-than-air aviation.

Worked Example 10 — Pressure on a Submarine Window

A submarine has a porthole window of diameter 25 cm. The submarine descends to a depth of 200 m in seawater of density 1030 kg m⁻³. The internal cabin pressure is maintained at atmospheric (101 kPa).