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Fluid Mechanics · Fluid Statics

Hydrostatic pressure and force on surfaces

Calculate pressure at depth in a liquid and the resultant force and its line of action on a vertical wall or gate.

45 minYear 1-2Prerequisites: Equilibrium and support reactions

By the end of this topic you will be able to

  • Calculate gauge and absolute pressure at a given depth.
  • Find the resultant force of a liquid on a vertical rectangular surface.
  • Locate the centre of pressure and calculate the overturning moment.

Pressure in a still liquid

In a fluid at rest, pressure acts equally in all directions and increases with depth because of the weight of the fluid above:

p=ρghp = \rho g h

where ρ\rho is the density (1000 kg/m³ for fresh water), g=9.81g = 9.81 m/s² and hh the depth. This is gauge pressure: pressure above atmospheric. Absolute pressure adds atmospheric pressure (about 101 kPa).

The pressure does not depend on the shape of the container, only on the vertical depth.

Force on a vertical wall

A dam, tank wall or sluice gate has pressure that grows linearly from zero at the free surface to ρgh\rho g h at the base, so the load is a triangle. For a rectangular wall of height hh (fully submerged to the surface) and width bb, the resultant is the area of the triangle multiplied by the width:

F=12(ρgh)hb=12ρgh2bF = \tfrac{1}{2}(\rho g h)\,h\,b = \tfrac{1}{2}\rho g h^2 b

The force acts at the centroid of the triangle, the centre of pressure, at depth 2h/32h/3 below the surface, or h/3h/3 above the base:

yb=h3y_b = \frac{h}{3}

It acts below the centroid of the wall (at h/2h/2) because pressure is greater at depth.

Interactive modelChange the depth, width or liquid. The pressure triangle and resultant update.

This interactive model needs JavaScript. The theory, worked example and problems on this page work without it.

Full screen

Practice problems

  1. Find the gauge pressure at 30 m depth in the sea (1025 kg/m³).
    Show answer
    p=1025×9.81×30=301.7p = 1025\times9.81\times30 = 301.7 kPa. Absolute pressure is about 403 kPa.
  2. A rectangular tank wall is 3 m high, 5 m wide, full of water. Find the force on it.
    Show answer
    F=12×9.81×32×5=220.7F = \tfrac12\times9.81\times3^2\times5 = 220.7 kN, acting 1.0 m above the base.
  3. A reservoir level rises from 4 m to 6 m behind a dam. By what factor does the force per metre length increase?
    Show answer
    (6/4)2=2.25(6/4)^2 = 2.25 times.
  4. Why does the dam wall thickness increase toward the base?
    Show answer
    Pressure, and hence the load, increases with depth, so the wall must resist more at the base.

Key formulae

Pressure at depth h
p=ρghp = \rho g h
Resultant force on a vertical rectangle
F=12ρgh2bF = \frac{1}{2}\rho g h^2 b
Centre of pressure (above the base)
yb=h3y_b = \frac{h}{3}
Overturning moment about the base
M=F·h3M = F\cdot\frac{h}{3}
Lecturer notes: suggested plan, activity and assessment

Suggested plan (45 min). 10 min: demonstration with a bottle drilled at three heights (water jets reach different distances); 10 min: derivation of p=ρghp = \rho g h; 15 min: triangular load, force and centre of pressure; 10 min: model.

Activity. Ask which fluid exerts more force at the same depth: mercury or water (and by what factor).

Discussion. Why is the centre of pressure below the centroid?

Assessment idea. A gate hinged at the top, find the force needed at the base to hold it shut.