Density, Pressure and Archimedes' Principle

Spec mapping: OCR H556 Module 3.2 — Forces in action (density $\rho = m/V$ ; pressure in a fluid $p = h\rho g$ ; upthrust on an object in a fluid and Archimedes' principle). Refer to the official OCR H556 specification document for exact wording.

Density and pressure are deceptively simple quantities with profound consequences. They explain why steel ships can float, why aeroplane wings generate lift, why your ears pop when you dive into a pool, why a single drawing pin supports your full weight without breaking the skin, and why a single sneeze can launch a droplet at $30$ m s $^{-1}$ . This lesson covers density, hydrostatic pressure (pressure in a fluid at rest), and the magnificent insight of Archimedes — paraphrased rather than quoted from the apocryphal Syracusan-bath story of c. 250 BCE — that the upthrust on a submerged body equals the weight of the fluid it displaces.

This lesson is the gateway to a wider range of fluid mechanics that you will meet at undergraduate level: hydrodynamics, aerodynamics, oceanography, atmospheric science. At A-Level we restrict ourselves to hydrostatic (no-flow) fluid pressure plus Archimedes' upthrust, but the conceptual scaffolding here generalises directly.

Key Definition: Density is the mass per unit volume of a material: $\rho = m/V$ , with SI unit kg m $^{-3}$ . Pressure is the magnitude of normal force per unit area: $p = F/A$ , with SI unit pascal (Pa $=$ N m $^{-2}$ ). Both are scalars; both are intensive quantities (independent of sample size, for uniform materials).

Density

Density is the mass per unit volume:

$\rho = \frac{m}{V}$

Material	$\rho$ (kg m $^{-3}$ )
Hydrogen gas (STP)	0.090
Helium gas (STP)	0.179
Air (sea level, 15 °C)	1.23
Cork	240
Pine wood	510
Oak wood	750
Ice (0 °C)	917
Water (4 °C, pure)	1000
Seawater	1025
Human body (average)	1010
Aluminium	2700
Iron	7870
Steel	7850
Copper	8960
Lead	11 340
Mercury (liquid)	13 534
Gold	19 320
Platinum	21 450
Osmium (densest natural element)	22 590
Neutron-star matter	$\sim 4 \times 10^{17}$

Water's density of $1000$ kg m $^{-3}$ is a useful benchmark — by definition, in fact, the metric system was originally calibrated so that 1 kg = 1 litre of water. Materials denser than water sink in it; materials less dense float. Ice (917 kg m $^{-3}$ ) floats; its expansion on freezing is the reason why pipes burst in winter — water is one of very few substances that expand on freezing rather than contracting.

Worked Example 1 — Copper Block

A cuboid copper block measures $5.0$ cm $\times$ $4.0$ cm $\times$ $2.0$ cm. Find its mass.

Volume $V = 0.050 \times 0.040 \times 0.020 = 4.0 \times 10^{-5}$ m $^3$ .
Density of copper $\rho = 8960$ kg m $^{-3}$ .
Mass $m = \rho V = 8960 \times 4.0 \times 10^{-5} = 0.358$ kg (358 g).

Common Exam Mistake: Using cm $^3$ as volume but kg m $^{-3}$ as density — mixed units. Always convert to SI (m $^3$ , kg, kg m $^{-3}$ ) before substituting. The conversion $1$ cm $^3 = 10^{-6}$ m $^3$ is the most-needed.

Worked Example 2 — Density of an Alloy

A brass cylinder (alloy of copper and zinc) has mass $0.85$ kg and volume $1.0 \times 10^{-4}$ m $^3$ . Find its density and estimate the mass fraction of copper, given $\rho_{\text{Cu}} = 8960$ and $\rho_{\text{Zn}} = 7140$ kg m $^{-3}$ .

Density: $\rho = m/V = 0.85 / 10^{-4} = 8500$ kg m $^{-3}$ .
Approximate by linear mixing (volume-weighted, which approximately equals mass-weighted for similar densities): if mass fraction $x$ is copper, then $\rho \approx x \rho_{\text{Cu}} + (1 - x) \rho_{\text{Zn}}$ . Solving: $8500 = 8960 x + 7140 (1 - x) = 7140 + 1820 x$ , so $x = (8500 - 7140) / 1820 = 0.747$ . About 75% copper, 25% zinc by mass.

This is the linear-mixing rule for ideal binary alloys; real alloys deviate slightly due to atomic-packing effects.

Worked Example 3 — Average Density of a Hollow Object

A hollow steel sphere has outer radius $0.20$ m, inner radius $0.18$ m, and the steel has density $7850$ kg m $^{-3}$ . Find (a) its mass and (b) its average density.

Volume of steel shell: $V_{\text{shell}} = \tfrac{4}{3}\pi(R^3 - r^3) = \tfrac{4}{3}\pi(0.008 - 0.005832) = \tfrac{4}{3}\pi \times 0.002168 = 9.07 \times 10^{-3}$ m $^3$ .
Mass: $m = \rho V_{\text{shell}} = 7850 \times 9.07 \times 10^{-3} = 71.2$ kg.
Outer volume (sphere): $V_{\text{outer}} = \tfrac{4}{3}\pi R^3 = \tfrac{4}{3}\pi \times 0.008 = 3.35 \times 10^{-2}$ m $^3$ .
Average density: $\rho_{\text{avg}} = m / V_{\text{outer}} = 71.2 / 3.35 \times 10^{-2} = 2125$ kg m $^{-3}$ .

Although steel itself is denser than water, the hollow-sphere assembly has average density (2125 kg m $^{-3}$ ) still greater than water and so would sink. A hollow ship-hull achieves average density below water by virtue of a much greater enclosed-air-to-shell-steel ratio. This is the key to why ships float despite steel being 7.85× denser than water.

Pressure

Pressure is the magnitude of normal force per unit area:

$p = \frac{F}{A}$

SI unit: pascal (Pa $=$ N m $^{-2}$ ). Several other units are in common use:

Unit	Value in Pa
Pa	1
kPa	$10^3$
MPa	$10^6$
bar	$10^5$
atmosphere (atm)	$1.013 \times 10^5$
millimetre of mercury (mmHg)	133.3
inch of water	249.1
pound per square inch (psi)	6895

At sea level the atmospheric pressure is $\approx 101$ kPa $\approx 1$ atm. This is the weight of a column of air of cross-section $1$ m $^2$ extending from sea level to the top of the atmosphere — about 10 tonnes per square metre. We do not feel this immense pressure because it acts on us equally from all directions, including from inside our bodies (lungs, blood vessels, sinuses).

Pressure is Inversely Related to Area

For a fixed applied force, a small contact area produces a large pressure. Examples:

A sharp knife has a cutting edge $\sim 10^{-6}$ m thick over perhaps $0.10$ m long, giving a contact area $\sim 10^{-7}$ m $^2$ . A few newtons of force produces pressures of tens of MPa — easily enough to exceed the yield stress of soft materials (bread, fruit, flesh).
A drawing pin has a tip of $\sim 10^{-7}$ m $^2$ area. Push it with $5$ N and the pressure on the wood is $5 \times 10^7$ Pa — through the pin and into the wood.
A stiletto heel concentrates a person's weight ( $\sim 600$ N) on perhaps $1$ cm $^2 = 10^{-4}$ m $^2$ , giving $6$ MPa — sufficient to dent timber flooring. By contrast, an elephant's foot (area $\sim 0.05$ m $^2$ , weight $\sim 30000$ N) exerts only $600$ kPa, less than a stiletto.

Pressure in a Fluid at Depth

Imagine a column of fluid of cross-sectional area $A$ , height $h$ , density $\rho$ , with its top at the surface. The volume of the column is $V = Ah$ , the mass is $m = \rho V = \rho A h$ , and the weight is $W = mg = \rho A h g$ . This weight presses down on the base, giving a pressure:

$p = \frac{F}{A} = \frac{\rho A h g}{A} = \rho g h$

So fluid pressure increases linearly with depth:

$\boxed{p = \rho g h}$

where $p$ is the gauge pressure (pressure due to the fluid alone, above atmospheric). The total absolute pressure at depth $h$ below the surface is:

$p_{\text{abs}} = p_{\text{atm}} + \rho g h$

This derivation assumes (i) the fluid is incompressible (true for liquids; reasonable for gases over moderate height changes), (ii) the fluid is at rest (no flow, no Bernoulli effects), and (iii) the density $\rho$ is uniform with depth.

Worked Example 4 — Pressure at the Bottom of a Swimming Pool

A swimming pool is $2.5$ m deep. Find (a) the gauge pressure at the bottom and (b) the absolute pressure.

(a) Gauge: $p = \rho g h = 1000 \times 9.81 \times 2.5 = 24525$ Pa $\approx 24.5$ kPa.

(b) Absolute: $p_{\text{abs}} = 101325 + 24525 = 125850$ Pa $\approx 126$ kPa.

The pressure at the bottom of a $2.5$ m pool is about 25% greater than atmospheric. Deep-sea divers face much more: at $100$ m depth in seawater, gauge pressure is $\rho g h = 1025 \times 9.81 \times 100 = 1.005 \times 10^6$ Pa $\approx 10$ atm. At the bottom of the Mariana Trench ( $\sim 11000$ m depth), pressure exceeds $110$ MPa $\approx 1100$ atm — equivalent to the weight of an elephant balanced on a postage stamp.

Worked Example 5 — Pressure on Submarine Hull

A submarine's pressure hull is rated for safe operation to $300$ m depth in seawater. Find (a) the gauge pressure at this depth, (b) the inward net force on a hatch of area $0.50$ m $^2$ if the cabin is held at atmospheric pressure.

(a) $p = 1025 \times 9.81 \times 300 = 3.02 \times 10^6$ Pa $\approx 3$ MPa $\approx 30$ atm.

(b) Net inward force on the hatch: $F = p \times A = 3.02 \times 10^6 \times 0.50 = 1.51 \times 10^6$ N $\approx 1.5$ MN. That is the equivalent of 150 tonnes pressing inward on a single square-metre-ish hatch. Submarine hulls require enormous structural engineering — typically thick HY-100 steel or titanium alloy.

Worked Example 6 — The Hydraulic Press

A small piston of area $A_1 = 1.0 \times 10^{-4}$ m $^2$ is pushed with force $F_1 = 50$ N. The fluid is incompressible and connects to a larger piston of area $A_2 = 2.5 \times 10^{-3}$ m $^2$ . Find the force exerted by the larger piston.

By Pascal's principle, pressure transmits undiminished through a fluid in static equilibrium. So:

$p = \frac{F_1}{A_1} = \frac{50}{10^{-4}} = 5.0 \times 10^5 \text{ Pa}$

Force on the large piston:

$F_2 = p \times A_2 = 5.0 \times 10^5 \times 2.5 \times 10^{-3} = 1250 \text{ N}$

The amplification factor is $A_2 / A_1 = 25$ . This is the operating principle of car jacks, hydraulic excavators, hydraulic disc brakes, and dental chairs. Energy is conserved, however — the small piston must move 25 times further than the large piston for the same work input: $W_1 = F_1 d_1 = F_2 d_2 = W_2$ , so $d_1 / d_2 = F_2 / F_1 = 25$ . Hydraulic amplification trades force for distance.

Upthrust and Archimedes' Principle

Objects submerged in a fluid experience an upward force called upthrust (or buoyancy). Archimedes' principle (paraphrased from the apocryphal On Floating Bodies, c. 250 BCE) states:

The upthrust on a body immersed in a fluid equals the weight of the fluid it displaces.

Mathematically:

$\boxed{U = \rho_{\text{fluid}} \times V_{\text{displaced}} \times g}$

The critical word is displaced: $V_{\text{displaced}}$ is the volume of fluid that the body excludes, which equals the body's submerged volume — not its total volume, if it is only partially submerged.

Derivation Sketch

Consider a cylinder of height $h$ and cross-section $A$ submerged vertically in a fluid of density $\rho_{\text{fluid}}$ . The pressure at the top of the cylinder is $p_1 = \rho_{\text{fluid}} g h_1$ (where $h_1$ is the depth of the top below the fluid surface); the pressure at the bottom is $p_2 = \rho_{\text{fluid}} g h_2$ (with $h_2 = h_1 + h$ ). The pressure on the sides cancels by symmetry. The net upward force is the area-times-pressure difference:

$U = (p_2 - p_1) A = \rho_{\text{fluid}} g (h_2 - h_1) A = \rho_{\text{fluid}} g h A = \rho_{\text{fluid}} g V$

This equals the weight of the fluid that would have occupied the volume now taken by the cylinder. The argument generalises to any submerged shape by decomposing the body into thin vertical columns.

Worked Example 7 — Submerged Stone

A stone of volume $V = 2.0 \times 10^{-4}$ m $^3$ is fully submerged in water. Find the upthrust.

$U = \rho_{\text{water}} \times V \times g = 1000 \times 2.0 \times 10^{-4} \times 9.81 = 1.962 \text{ N}$

If the stone has mass $0.60$ kg (true weight $W = 5.886$ N), its apparent weight when fully submerged is:

$W_{\text{apparent}} = W - U = 5.886 - 1.962 = 3.924 \text{ N}$

This "loss of weight" of about $33\%$ is why lifting a rock from a pool floor is easier than from dry ground.

Worked Example 8 — Floating Ship

A ship has total mass $5.0 \times 10^7$ kg. What volume of seawater ( $\rho = 1025$ kg m $^{-3}$ ) does it displace when floating freely?

For floating equilibrium, upthrust equals weight:

$\rho_{\text{sea}} V_{\text{displaced}} g = mg \Rightarrow V_{\text{displaced}} = \frac{m}{\rho_{\text{sea}}} = \frac{5.0 \times 10^7}{1025} = 4.88 \times 10^4 \text{ m}^3$

The ship displaces a volume of seawater whose weight equals the ship's weight. This is why hollow steel ships float despite steel being 7-8 times denser than water — their average density (including the enclosed air) is less than seawater's.

Floating and Sinking — The Condition

A body fully immersed in a fluid will:

Sink if $\rho_{\text{body}} > \rho_{\text{fluid}}$ (weight exceeds maximum upthrust).
Float at the surface if $\rho_{\text{body}} < \rho_{\text{fluid}}$ , partially submerged.
Be in neutral buoyancy (suspended at depth) if $\rho_{\text{body}} = \rho_{\text{fluid}}$ .

For a floating body in equilibrium, the fraction submerged equals the ratio of densities:

$\frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{body}}}{\rho_{\text{fluid}}}$

Proof: at floating equilibrium, $\rho_{\text{fluid}} V_{\text{submerged}} g = \rho_{\text{body}} V_{\text{total}} g$ , rearranging to give the above.

An iceberg ( $\rho_{\text{ice}} = 917$ ) in seawater ( $\rho_{\text{sea}} = 1025$ ) has $917 / 1025 \approx 0.89$ , so about 89% submerged — hence the proverbial "tip of the iceberg" (only 11% visible).

The human body has average density $\approx 1010$ kg m $^{-3}$ — just denser than fresh water (1000) but just less dense than seawater (1025). This is why floating in seawater is much easier than in a freshwater pool. With lungs full of air ( $\rho_{\text{body}} \approx 990$ ), most people can float in either; after exhaling fully, sinking begins.

flowchart TD
  A[Body of density rho_body in fluid of density rho_fluid] --> B{Compare densities}
  B -->|rho_body greater than rho_fluid| C[Body sinks fully submerged]
  B -->|rho_body equals rho_fluid| D[Neutral buoyancy at any depth]
  B -->|rho_body less than rho_fluid| E[Body floats partially submerged]
  C --> F[Apparent weight = W - U where U is upthrust on whole body]
  E --> G[Submerged fraction = rho_body over rho_fluid]
  E --> H[Apparent weight at surface = 0 net force]

Worked Example 9 — Archimedes' Crown Problem

Per the apocryphal Syracusan-bath story, King Hiero II commissioned a gold crown but suspected the smith had alloyed it with cheaper silver. Archimedes' challenge was to determine the crown's purity without melting it down.

Suppose a crown of mass $1.00$ kg has apparent weight in water of $8.80$ N. Is it pure gold (density $19320$ kg m $^{-3}$ )?

True weight in air: $W = mg = 1.00 \times 9.81 = 9.81$ N.
Upthrust in water: $U = W - W_{\text{apparent}} = 9.81 - 8.80 = 1.01$ N.
Volume of crown (from Archimedes' principle): $V = U / (\rho_{\text{water}} g) = 1.01 / (1000 \times 9.81) = 1.029 \times 10^{-4}$ m $^3$ .
Density of crown: $\rho = m / V = 1.00 / (1.029 \times 10^{-4}) = 9714$ kg m $^{-3}$ .

This is only half the density of pure gold — the crown is clearly heavily alloyed with a lighter metal (perhaps silver, $\rho_{\text{Ag}} = 10500$ kg m $^{-3}$ , or copper). Had it been pure gold, the upthrust would have been $U = \rho_{\text{water}} V g = \rho_{\text{water}} (m/\rho_{\text{Au}}) g = 1000 \times (1.00 / 19320) \times 9.81 = 0.508$ N, giving an apparent weight of $9.81 - 0.508 = 9.30$ N.

Archimedes is said — apocryphally — to have leapt from his bath shouting "Eureka!" ("I have found it") on realising that the volume of an irregularly shaped object could be measured by the volume of water it displaces, and hence its density determined without melting. The story is almost certainly legendary, but the principle is rigorous and has been the foundation of buoyancy physics for 23 centuries.

Density, Pressure and Archimedes' Principle

Density, Pressure and Archimedes' Principle

Density

Worked Example 1 — Copper Block

Worked Example 2 — Density of an Alloy

Worked Example 3 — Average Density of a Hollow Object

Pressure

Pressure is Inversely Related to Area

Pressure in a Fluid at Depth

Worked Example 4 — Pressure at the Bottom of a Swimming Pool

Worked Example 5 — Pressure on Submarine Hull

Worked Example 6 — The Hydraulic Press

Upthrust and Archimedes' Principle

Derivation Sketch

Worked Example 7 — Submerged Stone

Worked Example 8 — Floating Ship

Floating and Sinking — The Condition

Worked Example 9 — Archimedes' Crown Problem

Pressure-Depth Graph

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