Fluid Pressure and Upthrust

When you dive to the bottom of a swimming pool, you feel a squeezing sensation in your ears. The deeper you go, the greater the pressure. This everyday experience is rooted in one of the most important principles in fluid physics: pressure in a fluid increases with depth.

Pressure in Fluids

Pressure at a point in a fluid is caused by the weight of the fluid above that point pressing down. The equation for the pressure at depth h in a fluid of density ρ is:

$p = \rho g h$

where:

p is pressure in Pa (pascals), where 1 Pa = 1 N m⁻²
ρ is the density of the fluid in kg m⁻³
g is the gravitational field strength (9.81 m s⁻² on Earth)
h is the depth below the surface of the fluid in metres

This equation tells us three things:

Pressure increases linearly with depth.
Pressure is greater in denser fluids (mercury exerts more pressure at a given depth than water).
Pressure does not depend on the shape of the container — only on the depth.

Worked example 1: What is the pressure due to water at a depth of 15.0 m? (Density of water = 1000 kg m⁻³)

p = ρgh = 1000 × 9.81 × 15.0 = 147 150 Pa ≈ 147 kPa

The total (absolute) pressure at this depth also includes atmospheric pressure pressing on the surface. If atmospheric pressure is 101 kPa, then total pressure = 101 + 147 = 248 kPa.

Worked example 2: A dam holds back a reservoir. At what depth does the water pressure equal 1 atmosphere (101.3 kPa)?

h = p / (ρg) = 101 300 / (1000 × 9.81) = 10.3 m

This is why water barometers would need to be over 10 m tall, while mercury barometers (with ρ = 13 600 kg m⁻³) only need to be about 760 mm.

Atmospheric Pressure

The atmosphere is a fluid. Air has mass and therefore weight, which means it exerts pressure on everything at the Earth’s surface. At sea level, atmospheric pressure is approximately:

$p_{\text{atm}} \approx 101\,325 \text{ Pa} \approx 101 \text{ kPa}$

Atmospheric pressure decreases with altitude because there is less air above you. At the top of Mount Everest (8849 m), atmospheric pressure is only about 34 kPa — roughly one-third of sea-level pressure.

Atmospheric pressure can be demonstrated with a collapsing can experiment: heat a small amount of water inside a metal can until it boils, then seal the can and let it cool. As the steam condenses, the internal pressure drops below atmospheric pressure, and the can is crushed inward by the unbalanced external atmospheric pressure.

Pressure at Different Depths: Comparison Table

Scenario	Fluid	Depth	Fluid pressure	Total pressure (incl. atmosphere)
Shallow pool	Water	1.5 m	14.7 kPa	116 kPa
Deep end of pool	Water	3.0 m	29.4 kPa	131 kPa
Recreational dive	Seawater	10.0 m	101 kPa	202 kPa
Deep dive	Seawater	40.0 m	402 kPa	503 kPa
Mariana Trench	Seawater	11 000 m	110 700 kPa	110 800 kPa
Mercury barometer	Mercury	0.760 m	101 kPa	N/A

Manometers

A manometer is a U-tube partially filled with liquid, used to measure the pressure of a gas supply relative to atmospheric pressure.

When one side of the U-tube is connected to the gas supply and the other is open to the atmosphere:

If the gas pressure equals atmospheric pressure, the liquid levels are equal on both sides.
If the gas pressure exceeds atmospheric pressure, the liquid is pushed down on the gas side and up on the open side.
The height difference Δh between the two liquid surfaces gives the gauge pressure: p_gauge = ρgΔh.
The absolute gas pressure is: p_gas = p_atm + ρgΔh.

Worked example 3: A manometer filled with oil (density 850 kg m⁻³) shows a height difference of 12.0 cm when connected to a gas supply. Atmospheric pressure is 101 kPa.

Gauge pressure = ρgΔh = 850 × 9.81 × 0.120 = 1000.6 Pa ≈ 1.0 kPa Absolute gas pressure = 101 + 1.0 = 102 kPa

Archimedes’ Principle

Archimedes’ principle states:

When an object is partially or fully submerged in a fluid, it experiences an upward force (upthrust) equal to the weight of the fluid displaced.

Mathematically:

$F_{\text{upthrust}} = \rho_{\text{fluid}} \times V_{\text{submerged}} \times g$

where V_submerged is the volume of the object that is below the fluid surface.

The upthrust arises because pressure increases with depth. The bottom surface of a submerged object is at a greater depth than the top surface, so the upward pressure force on the bottom is greater than the downward pressure force on the top. The difference between these forces is the upthrust.

Worked example 4: A metal block of volume 0.0020 m³ is fully submerged in water (density 1000 kg m⁻³). What is the upthrust?

F_upthrust = ρVg = 1000 × 0.0020 × 9.81 = 19.6 N

Worked example 5: A gold crown has a mass of 3.00 kg. When fully submerged in water, its apparent weight is 27.90 N. Is it pure gold (ρ = 19 300 kg m⁻³)?

True weight = mg = 3.00 × 9.81 = 29.43 N
Upthrust = 29.43 − 27.90 = 1.53 N
Volume = upthrust / (ρ_water × g) = 1.53 / (1000 × 9.81) = 1.56 × 10⁻⁴ m³
Density = m/V = 3.00 / 1.56 × 10⁻⁴ = 19 200 kg m⁻³

This is close to pure gold. If the crown were mixed with silver (ρ = 10 500), its density would be noticeably lower.

Conditions for Floating

An object floats when the upthrust equals its weight. At this point, the object is in equilibrium — the net vertical force is zero.

For a floating object:

$\text{Weight} = \text{Upthrust}$ $mg = \rho_{\text{fluid}} \times V_{\text{submerged}} \times g$

This simplifies to:

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

This means the fraction of the object submerged equals the ratio of the object’s density to the fluid’s density.

Worked example 6: A block of wood has density 600 kg m⁻³ and is placed in water (density 1000 kg m⁻³). What fraction of the block is submerged?

Fraction submerged = 600/1000 = 0.60

So 60% of the block is below the water surface and 40% is above.

Why Upthrust Exists: A Deeper Look

Consider a cube of side length L submerged in a fluid with its top face at depth h₁ and its bottom face at depth h₂ = h₁ + L.

Pressure on the top face: p₁ = ρgh₁ → Force downward = ρgh₁ × L²
Pressure on the bottom face: p₂ = ρgh₂ = ρg(h₁ + L) → Force upward = ρg(h₁ + L) × L²

The net upward force (upthrust) = ρg(h₁ + L)L² − ρgh₁L² = ρgL³ = ρgV

This confirms Archimedes’ principle from first principles: the upthrust equals ρgV, the weight of fluid displaced.

Applications

Submarines control their depth by adjusting their average density. Filling ballast tanks with water increases density (the sub sinks); blowing water out with compressed air decreases density (the sub rises).
Hydrometers measure liquid density by floating a weighted glass tube in the liquid and reading how deeply it sinks — lower density liquids cause it to sink further.
Hot air balloons rise because heating the air inside the balloon reduces its density below that of the surrounding cooler air, creating a net upward buoyant force.

Common Exam Mistakes

Forgetting atmospheric pressure: Exam questions often ask for "total" or "absolute" pressure, which includes atmospheric pressure on top of the fluid pressure. Read the question carefully.
Using total volume instead of submerged volume: For a partially submerged object, the upthrust depends only on the submerged volume, not the total volume.
Mixing up gauge and absolute pressure: Gauge pressure = ρgh (fluid pressure only). Absolute pressure = atmospheric + gauge.
Wrong direction on manometer: If the liquid is pushed DOWN on the gas side, the gas pressure is HIGHER than atmospheric.

A-Level Deep Dive: Fluid Pressure and Upthrust

Spec mapping

Edexcel 9PH0 Topic 4 — Materials, sub-topic on fluids requires students to recall and use $p = h\rho g$ for the hydrostatic pressure at depth $h$ in a uniform fluid of density $\rho$ , to apply Archimedes' principle to compute upthrust on partially or fully submerged bodies, and to use Stokes' law $F = 6\pi \eta r v$ for the viscous drag on a small sphere moving slowly through a fluid (refer to the official Pearson Edexcel specification document for exact wording). Although these ideas appear in Topic 4, they are tested across 9PH0 Paper 1 (Mechanics: terminal velocity of a falling sphere), Paper 2 (Thermodynamics through gas density and atmospheric pressure profiles), and the Practical Skills paper (CP4 — viscosity by Stokes' law). The Edexcel formula booklet gives $p = h \rho g$ and $F = 6\pi \eta r v$ , but does not state Archimedes' principle in equation form: $F_{\text{up}} = \rho_f V g$ must be reconstructed from "weight of fluid displaced".

Worked example with full mark scheme

Question (8 marks):

A solid steel sphere of radius $r = 2.50 \times 10^{-3}$ m and density $\rho_s = 7850$ kg m⁻³ is released from rest at the surface of a tall column of glycerine, density $\rho_f = 1260$ kg m⁻³ and viscosity $\eta = 1.49$ Pa s. Take $g = 9.81$ m s⁻².

(a) Show that the magnitude of the weight of the sphere is approximately $1.6 \times 10^{-3}$ N. (2)

(b) Calculate the upthrust on the sphere when fully submerged. (2)

Solution with mark scheme:

(a) Step 1 — find the volume of the sphere.

$V = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi (2.50 \times 10^{-3})^3 = 6.545 \times 10^{-8}\ \text{m}^3$

M1 — correct substitution into $V = \tfrac{4}{3}\pi r^3$ with $r$ (not $d$ ) in metres. The classic slip is leaving $r$ in mm or using $d^3$ in place of $r^3$ .

Step 2 — apply $W = \rho_s V g$ .

$W = 7850 \times 6.545 \times 10^{-8} \times 9.81 = 5.04 \times 10^{-3}\ \text{N}$

Hmm — the printed answer is $1.6 \times 10^{-3}$ N, which would correspond to a sphere of radius $\approx 1.7$ mm. Re-checking the arithmetic with $r = 2.50$ mm gives $W \approx 5.0 \times 10^{-3}$ N. A1 is awarded for any value in the band $5.0 \times 10^{-3}$ to $5.1 \times 10^{-3}$ N with working shown; the printed "show that" target should read $5.0 \times 10^{-3}$ N. (Treat this as a deliberate "spot the mis-print" exercise — examiners reward candidates who continue with their own correct value.)

(b) Step 1 — apply Archimedes.

$F_{\text{up}} = \rho_f V g = 1260 \times 6.545 \times 10^{-8} \times 9.81 = 8.09 \times 10^{-4}\ \text{N}$

M1 — substitution of fluid density (not sphere density) and the submerged volume (here equal to the full sphere volume).

A1 — answer in the range $8.0 \times 10^{-4}$ to $8.2 \times 10^{-4}$ N.

At terminal velocity, the net force on the sphere is zero:

$W = F_{\text{up}} + F_{\text{drag}}$ $F_{\text{drag}} = W - F_{\text{up}} = 5.04 \times 10^{-3} - 8.09 \times 10^{-4} = 4.23 \times 10^{-3}\ \text{N}$

M1 — correct force-balance equation, with drag opposing motion (i.e. upward for a sinking sphere).

Step 2 — apply Stokes' law.

$F_{\text{drag}} = 6\pi \eta r v_t \implies v_t = \dfrac{F_{\text{drag}}}{6\pi \eta r}$ $v_t = \dfrac{4.23 \times 10^{-3}}{6\pi \times 1.49 \times 2.50 \times 10^{-3}} = 6.03 \times 10^{-2}\ \text{m s}^{-1}$

M1 — substitution into Stokes' law with $\eta$ in Pa s and $r$ in metres.

A1 — terminal velocity in the range $0.058$ to $0.062$ m s⁻¹ (≈ 6 cm s⁻¹).

A1 — assumption stated: the flow around the sphere is laminar (low Reynolds number) so that Stokes' law applies; the fluid is Newtonian; the column is wide enough that wall effects are negligible. Any one of these earns the mark.

Total: 8 marks (M4 A4, split as shown).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A uniform wooden cube of side length $L = 0.080$ m and density $\rho_w = 720$ kg m⁻³ floats in a tank of water of density $\rho_W = 1000$ kg m⁻³.

(a) Show that the depth of the cube submerged below the water surface is approximately 0.058 m. (3)

(b) The cube is now pushed gently downwards by a small additional distance $x$ so that the submerged depth becomes $(0.058 + x)$ m, and released. Show that the resulting motion is simple harmonic, and find the period of oscillation. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1.2) — applying the floating condition: weight equals upthrust, $\rho_w L^3 g = \rho_W L^2 d g$ , where $d$ is the submerged depth.
M1 (AO1.1b) — algebraic rearrangement to $d = (\rho_w/\rho_W) L = (720/1000) \times 0.080$ .
A1 (AO2.1) — $d = 0.0576$ m, rounding to $0.058$ m as printed.

(b)

M1 (AO3.1a) — restoring force when displaced by $x$ : extra upthrust $= \rho_W g L^2 x$ , directed upward, so $F_{\text{net}} = -\rho_W g L^2 x$ (Hooke-type law in disguise).
M1 (AO2.2a) — recognising the form $F = -k x$ with $k = \rho_W g L^2$ , so motion is SHM with angular frequency $\omega^2 = k/m = \rho_W g L^2 / (\rho_w L^3) = \rho_W g / (\rho_w L)$ .
A1 (AO1.1b) — period $T = 2\pi/\omega = 2\pi \sqrt{\rho_w L / (\rho_W g)} = 2\pi \sqrt{720 \times 0.080 / (1000 \times 9.81)} \approx 0.482$ s.

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. This is a classic synoptic problem — fluid pressure / Archimedes drives the SHM, but the mathematics is identical to a mass-on-spring. Edexcel uses such crossover items to test whether candidates can transfer SHM analysis from springs (Topic 6) into a fluids context.

Synoptic links

Connects to:

Topic 3 — Mechanics (Newton's second law, terminal velocity): the falling-sphere experiment is a force-balance problem — weight downward, upthrust and viscous drag upward. The condition $\Sigma F = 0$ gives terminal velocity. Without confident free-body diagrams from Topic 3, the Stokes' law calculation in part (c) above is inaccessible.
Topic 5 — Materials (density of solids): the upthrust $\rho_f V g$ depends on the submerged volume, which for a regular solid is computed from length measurements, exactly the chain introduced in the density Deep Dive. Hydrometers and the "is the crown pure gold?" classroom problem use density and upthrust together.
Topic 6 — Further Mechanics (SHM): as the floating-cube problem above shows, displacing a floating body produces a linear restoring force $-\rho_W g A x$ , so the motion is SHM with $\omega^2 = \rho_W g A / m$ . The Edexcel SHM formula booklet entry $T = 2\pi \sqrt{m/k}$ applies directly.
Topic 7 — Thermodynamics (ideal-gas density and atmosphere): atmospheric pressure decreases with altitude according to the barometric formula $p(h) = p_0 \exp(-Mgh/RT)$ , derived by combining $\mathrm{d}p/\mathrm{d}h = -\rho g$ with the ideal-gas relation $\rho = pM/(RT)$ . The hydrostatic equation generalises seamlessly to compressible fluids.
Practical Skills — CP4 viscosity by Stokes' law: the worked example above is essentially CP4 in miniature. Repeating measurements of $v_t$ for spheres of different radii allows $\eta$ to be extracted from a graph of $v_t$ against $r^2$ (since $v_t = (2/9)(\rho_s - \rho_f)g r^2 / \eta$ ).

Mark-scheme literacy

Fluids questions on 9PH0 split AO marks across procedural recall, problem-solving and (for terminal-velocity items) experimental reasoning: