Pressure depends only on depth.

Vessels of different shapes joined at the base fill to one level; a draggable probe reads P = Pa + ρgh as absolute and gauge.

Find the depth in water at which absolute pressure is 2 atm (≈10 m) and show it is the same under every vessel.

A wider container gives more pressure at the bottom; mixing gauge and absolute pressure.

• Pascal's observation: in a fluid at rest, any two points at one common height share one common pressure.
• Proof with a tiny right-angled prism of fluid: the normal forces balance (F_b sin θ = F_c, F_b cos θ = F_a) and the face areas obey the same relations (A_b sin θ = A_c, A_b cos θ = A_a), so P_a = P_b = P_c.
• So pressure at a point acts equally in every direction; like other stresses it has no direction. The force on any surface inside a resting fluid is normal to that surface, whatever its orientation.
• A horizontal bar of fluid in equilibrium must have equal pressures at its two ends. Unequal pressures in a horizontal plane would drive a flow, so without flow the pressure is uniform across any horizontal plane.
• A vertical cylinder of fluid, base area A and height h: (P₂ − P₁)A = mg and m = ρhA, so P₂ − P₁ = ρgh.
• With point 1 at an open surface, P₁ = P_a (atmospheric) and P = P_a + ρgh. The excess P − P_a = ρgh at depth h is the gauge pressure.
• Neither the area nor the shape of the container appears in P = P_a + ρgh: only the height of the fluid column matters.
• Hydrostatic paradox: vessels A, B and C of different shapes, joined at the bottom by a horizontal pipe, fill to the same level although they hold different amounts of water, because the pressure at the bottom is the same under each.
• Swimmer 10 m below a lake surface (ρ = 1000 kg m⁻³, g = 10 m s⁻²): P = 1.01 × 10⁵ + 1000 × 10 × 10 = 2.01 × 10⁵ Pa ≈ 2 atm, double the surface value. At 1 km depth the rise is about 100 atm, which submarines must be built to withstand.

Teaching this idea? Put the simulation on your own page or LMS with this code.