Hydrostatic pressure and communicating vessels

Pressure near the water surface differs from pressure at the bottom of a pool. How can familiar ideas of mass, gravity and area explain the difference? Start by imagining a vertical column within the water.

A deeper layer must support more liquid above it against gravity, so pressure in a resting liquid increases downwards. Depth h is the vertical distance from the free surface to the point considered, measured in metres. The length of a sloping route to that point is irrelevant. Moving sideways at the same depth within one connected, uniform liquid does not change pressure; otherwise liquid would have a reason to flow.

An imaginary vertical column of height h and area S has volume Sh, mass ρSh and gravitational force ρShg. The lower layer must support this force in addition to the action from above. Divide by area: pж = ρShg/S = ρgh. Area cancels because widening the column increases both mass and supporting area together. Thus pж = ρgh gives the pressure increase due to the liquid column, not the total pressure at the point.

Double depth at unchanged density: the column above the point is twice as tall, and the added pressure doubles. At the same depth, a denser liquid gives a larger increase; at h = 0 the increase vanishes. Use ρ in kg/m³, g in N/kg and h in m: the product has units Н/м² = Па. We assume a uniform, resting, nearly incompressible liquid and practically constant g over the height considered.

At h = 0, ρgh is zero, but air still presses on an open surface. Total pressure is therefore p = p₀ + ρgh, where p₀ is the pressure above the surface. Doubling depth doubles only ρgh, not necessarily the whole sum. For the same p₀, ρ and depth, pressure does not depend on container shape. A narrow neck does not remove pressure below; sloping walls also transmit forces.

Connect two containers below the surfaces of the same liquid. If the levels differ, the taller column produces greater pressure at the connection and liquid flows until that difference vanishes. With equal pressure above both surfaces, equilibrium requires equal levels, even if the containers have different widths. Width changes how much water is needed to raise a level, not the relation ρgh. We neglect capillary effects in very narrow tubes. Different pressures above the surfaces can sustain different levels.

Apply greater pressure to one side of a U-tube than the other. The liquid level falls on that side and rises on the other. Flow stops when the vertical level difference Δh provides the required pressure difference: Δp = ρgΔh, using the manometer liquid's density. Equal pressures give equal levels. A liquid manometer compares two pressures. If one side is open to air, its reading gives pressure above atmospheric pressure, not total pressure.

Find the pressure due to a water column at a depth of 3 m: ρ = 1000 кг/м³, g = 10 Н/кг. Air pressure above the water is 100 kPa.

Pressure due to the water alone: pж = 1000·10·3 = 30 000 Па = 30 кПа.

Total pressure includes the air: 100 + 30 = 130 кПа.

The unknown width of the pool is unnecessary: under these conditions, pressure depends on depth, not shape.

Water-column pressure: 30 кПа; total pressure: 130 кПа.

The increase ρgh arises because lower layers support the weight of upper layers. The column area cancels in the calculation.

Total pressure is p₀ + ρgh. Zero depth does not mean zero pressure.

Levels in connected containers and manometers settle so that pressures at the connection balance.

Air also produces pressure. Next we will measure it and use pressure transmission to operate a hydraulic machine.