Hydrostatic Pressure: Full Derivation

1. Force balance on an element

Take a small box of fluid, area A and thickness dz, with z measured upward. Vertical equilibrium demands the pressure force on the bottom face balance the pressure force on the top face plus the element's weight:

Divide by A dz and let dz → 0:

Pressure falls with height, at a rate set by the fluid's weight per unit volume. Nothing more is assumed than equilibrium — this is Newton's second law with zero acceleration.

2. Constant density: P = ρgh

For a liquid, ρ is constant and the equation integrates immediately. Measuring depth h downward from a free surface at atmospheric pressure p₀:

Two consequences follow. Pascal's law: pressure at a point acts equally in all directions (a tetrahedral element balance shows the same p on every face). Shape independence: pressure depends only on depth, never on the container's shape — the hydrostatic paradox.

3. Worked example — a mercury manometer

A U-tube manometer with mercury (ρ = 13 600 kg/m³) shows a height difference of 24 cm between its arms. The gauge pressure being measured is

Try variations with the hydrostatic pressure calculator.

Key takeaway

dp/dz = −ρg is the z-momentum Navier–Stokes equation with all motion removed. Statics is dynamics with u = 0.

4. Extended worked examples

Example 1 — pressure on a dam. Water depth 30 m: gauge pressure at the base p = ρgh = 998 × 9.81 × 30 = 2.937×10⁵ Pa ≈ 293.7 kPa — about 2.9 atmospheres. The pressure grows linearly with depth, so the average over the wall is half the base value, and the total thrust per metre width is F = ½ρgh² = 0.5 × 998 × 9.81 × 900 = 4.41×10⁶ N — 4.4 meganewtons pushing on every metre of dam. The linear profile also puts the resultant at h/3 from the bottom, which is why dams are thick at the base and why the overturning analysis uses the h/3 lever arm.

Example 2 — the barometer. Torricelli's mercury column: atmospheric pressure supports height h = p₀/ρg = 101 325/(13 546 × 9.81) = 0.762 m ≈ 762 mm. Do the same with water: h = 101 325/(998 × 9.81) = 10.35 m — the reason suction pumps cannot lift water from wells deeper than ~10 m, a fact that stumped Renaissance engineers until Torricelli explained it in 1643.

Example 3 — a submarine hull. At 200 m depth in seawater (ρ = 1 025): p = 1 025 × 9.81 × 200 = 2.01×10⁶ Pa ≈ 20 atm gauge. On a hatch of area 1 m² that is a 2-meganewton load — the mass of about 200 cars. Hull design, viewport thickness and rescue-depth limits all fall straight out of ρgh.

Example 4 — the hydraulic press (Pascal's principle at work). A force of 50 N on a 2 cm² master piston creates Δp = 50/2×10⁻⁴ = 2.5×10⁵ Pa, transmitted undiminished throughout the fluid. On a 200 cm² slave piston this delivers F = 2.5×10⁵ × 0.02 = 5 000 N — a 100× force multiplication, paid for by the piston moving 100× less distance. Car lifts, brakes and excavators are this example with plumbing.

5. Common misconceptions

"A wider container means more pressure at the bottom." Pressure depends on depth alone: a lake bed and a narrow standpipe filled to the same level share the same p. The force on the base differs (F = pA), and the difference is carried by the container walls — the resolution of the hydrostatic paradox.

"Pressure pushes downward." Pressure at a point is isotropic — it pushes on any surface, in whatever direction that surface faces. The upward pressure on a diver's chin equals the downward pressure on their scalp at the same depth. Buoyancy exists precisely because pressure pushes up on the bottom of a body harder than down on its top.

"Gauge and absolute pressure are interchangeable." p_abs = p_gauge + p_atm. Using gauge pressure in the ideal-gas law, or absolute pressure where a formula expects gauge (cavitation checks!), is among the most frequent unit-style errors in the subject. When a result matters, write which one you mean.

"ρgh applies to the atmosphere." Only over small height changes. Air's density falls with altitude, so the constant-ρ integration fails; the correct treatment is Example P4 below.

6. Where this shows up

Manometry remains the calibration backbone of pressure measurement — a U-tube converts pressure to a length, and length is the easiest quantity on Earth to measure well. Medicine is full of hydrostatics: blood pressure is quoted in mmHg because early sphygmomanometers were mercury manometers; an IV bag must hang ~1 m above the arm to overcome venous pressure (ρgh ≈ 10 kPa ≈ 75 mmHg); intracranial and intraocular pressures are managed in the same units. Civil engineering sizes dams, lock gates, retaining walls against ½ρgh² thrusts. And geophysics runs on the same equation: the 3.6 GPa at the bottom of the Mariana Trench and the lithostatic pressure gradient in boreholes (~22–25 kPa per metre of rock) are both dp/dz = −ρg with the appropriate ρ.

7. Practice problems

P1. Gauge pressure at the deep end of a 4 m pool. Solution: 998 × 9.81 × 4 = 39.2 kPa ≈ 0.39 atm — your ears notice.

P2. A U-tube contains water topped by 12 cm of oil (ρ = 850) in one arm. Find the water-level difference between arms. Solution: pressure balance at the oil–water interface level: ρ_oil g(0.12) = ρ_w g·Δh → Δh = 850×0.12/998 = 0.102 m = 10.2 cm.

P3. What absolute pressure does a diver feel at 25 m in seawater? Solution: p = 101.3 + 1 025×9.81×25/1000 = 101.3 + 251.4 = 352.7 kPa ≈ 3.5 atm — each 10 m adds about one atmosphere.

P4. Treat the atmosphere as isothermal ideal gas: ρ = pM/RT, so dp/dz = −pMg/RT. Solve, and find p at 5 000 m. Solution: p = p₀e^{−z/H} with scale height H = RT/Mg ≈ 8 000 m; p(5 000) = 101.3 × e^{−0.625} = 54.2 kPa — matching the thin air mountaineers breathe at that altitude.

8. Going deeper

The full static balance ∇p = ρg says pressure gradients align with gravity — so constant-pressure surfaces are horizontal. In a rotating frame, add centrifugal force and those surfaces become paraboloids: the mirror of a spinning telescope's mercury pool, and the reason a stirred teacup's surface dips in the middle. Replace gravity with any conservative body force and the same machinery yields the shape of self-gravitating planets (hydrostatic equilibrium is the definition of "planet" the IAU uses) and the pressure structure of stars, where dp/dr = −ρ(r)g(r) balanced against radiation is the first equation of stellar astrophysics. The humble manometer equation, scaled up, is holding the Sun together.

9. Historical context

Hydrostatics is antiquity's one complete branch of physics: Archimedes' On Floating Bodies (~250 BCE) states the buoyancy principle with full mathematical rigour, two millennia before Newton. Simon Stevin (1586) resolved the hydrostatic paradox — demonstrating that a thin tube of water can burst a barrel, since pressure depends on height alone — with public experiments that scandalised common sense. Torricelli (1643) invented the barometer and correctly read it as the weight of the atmosphere; Pascal had a barometer carried up the Puy de Dôme (1648) to watch the mercury fall with altitude, killing the "nature abhors a vacuum" doctrine with 800 metres of mountain, and articulated the transmission principle that bears his name. The SI unit of pressure honours him; the derivation in these notes is the algebra those experiments were begging for.

10. Another way to see it: energy, and buoyancy for free

Define the potential Φ = p + ρgz for an incompressible static fluid; the balance dp/dz = −ρg says exactly that Φ is constant everywhere. Pressure is thus a stored energy density: lifting fluid trades ρgz for p. This energy reading explains the hydraulic press honestly — pressure transmits undiminished because a quasi-static, incompressible connected fluid has one Φ, and forces scale with the areas that sample it.

Buoyancy now costs one integral. Submerge a body of volume V; the surrounding fluid presses on its surface with the same pressure field it would exert on fluid occupying that space. The net force is ∮(−p)n dA = −∫∇p dV = ∫ρg dV upward — magnitude ρ_fluid gV: Archimedes' principle, derived rather than decreed. The buoyant force is applied at the centroid of the displaced volume (the centre of buoyancy), which is why ship stability is a geometry problem about where that centroid moves as the hull tilts.

11. Frequently asked questions

Why doesn't the atmosphere fall down or fly away? It is falling — permanently, in balance. Gravity pulls air down; the pressure gradient (high below, low above) pushes up; dp/dz = −ρg is the truce. The exponential profile of P4 is what that truce looks like for a compressible gas.

Why can't I drink through a 12-metre straw? Sucking only removes pressure above the liquid; the atmosphere below does the lifting, and it can support at most p₀/ρg ≈ 10.3 m of water. Torricelli's limit is also why surface pumps on deep wells must push from below rather than suck from above.

Does blood pressure depend on posture? Strongly — pure ρgh. Standing, the ankle arteries sit ~1.3 m below the heart: +ρgh ≈ 13 kPa ≈ 100 mmHg on top of cardiac pressure, which is why varicose veins are a lower-leg disease and why fighter pilots wear G-suits that squeeze the legs.

Is pressure a vector pointing down? No — a scalar. Force needs a surface: dF = −p n dA, along the surface normal, whatever its direction. The gradient of pressure is the vector, and in statics it points straight down at magnitude ρg.

Gauge or absolute — which do instruments show? Most industrial sensors and tyre gauges read gauge (zero at atmosphere); barometers and vacuum gauges read absolute. The 100 kPa difference is a classic source of silent factor-of-two-ish errors near atmospheric pressure.

12. Further practice

P5. A rectangular lock gate 6 m wide holds 8 m of fresh water on one side, 3 m on the other. Net force? Solution: F = ½ρgw(h₁² − h₂²) = 0.5×998×9.81×6×(64−9) = 1.62×10⁶ N ≈ 1.6 MN toward the shallow side.

P6. An iceberg (ρ = 917) floats in seawater (ρ = 1 025). What fraction is submerged? Solution: equilibrium ρ_ice V = ρ_sw V_sub → V_sub/V = 917/1025 = 89.5% — the familiar nine-tenths, straight from §10's derivation.

P7. A mercury manometer connected across an air duct shows Δh = 24 mm. Find Δp. Solution: Δp = (ρ_Hg − ρ_air)gΔh ≈ 13 546×9.81×0.024 = 3 189 Pa ≈ 3.19 kPa (the air-column correction is 0.01%).

P8. How deep in seawater does absolute pressure reach 10 atm? Solution: 9 atm gauge = 912 kPa = 1 025×9.81×h → h ≈ 90.7 m — the working ceiling of surface-supplied commercial diving, set by this arithmetic.

13. Worked exam problem

Problem. A vented cylindrical storage tank holds 12 m of oil (ρ = 850 kg/m³). (a) Gauge pressure at the base? (b) A circular inspection hatch of diameter 0.5 m is centred 2 m above the base; find the force on it. (c) Absolute pressure at the base?

Solution. (a) p = ρgh = 850×9.81×12 = 100.1 kPa gauge — coincidentally almost exactly one atmosphere of oil. (b) Hatch centre depth = 10 m: p_c = 850×9.81×10 = 83.4 kPa; for a small hatch the centre pressure is the average, so F = p_cA = 83 400×π×0.25²... A = π(0.25)² = 0.1963 m²; F = 16.4 kN — the weight of a large car on a dinner-plate-sized cover, and the reason hatch bolts are specified by this exact calculation. (c) p_abs = 100.1 + 101.3 = 201.4 kPa. One linear law, three engineering answers.

14. Key takeaways

Static pressure obeys dp/dz = −ρg: it grows linearly with depth for liquids, depends on depth alone (not container shape), acts equally in all directions, and transmits undiminished through connected fluid (Pascal). Keep gauge and absolute pressures labelled; integrate with variable ρ for atmospheres; and remember that buoyancy is this equation applied around a closed surface — Archimedes derived, not decreed.

15. Where to go next

Set the fluid moving and the same pressure joins velocity and height in the Bernoulli equation; the general force balance it came from is the momentum equation behind Navier–Stokes. The hydrostatic pressure calculator runs depth problems with live steps, and the buoyant force calculator applies §10's derivation directly.