Hydrostatic pressure and buoyancy

Engineering · Second-year university engineering · about 10 min

A liquid at rest pushes on surfaces because of its pressure, and that pressure increases with depth. The same pressure variation produces an upward force on submerged objects. Understanding both effects helps you analyse tanks, gates, floats and equipment placed in water.

The key idea

For a stationary liquid of uniform density, the gauge pressure at vertical depth hh below a free surface is

pg=ρghp_{\mathrm{g}}=\rho gh

Here pgp_{\mathrm{g}} is pressure above surface pressure, ρ\rho is fluid density, gg is gravitational acceleration and hh is measured vertically downwards.

Draw a tall transparent tank with a horizontal free surface at the top of a liquid and a vertical left tank wall. Show four small horizontal pressure arrows pointing from the water into the left wall at successively deeper positions, with lengths strictly increasing from top to bottom. Draw a downward vertical depth dimension from the liquid surface to the lowest sampled point and label it h. Beside the tank place a small graph of gauge pressure p_g on the horizontal axis against vertical depth h on the downward vertical axis, with a straight line beginning at zero surface gauge pressure and moving to the right as depth increases. Indicate density rho in the fluid and gravitational acceleration g as a downward arrow. No numeric values.
Gauge pressure grows linearly with vertical depth below the free surface.

Pressure at a specified depth

A static fluid applies pressure perpendicular to any small surface element. In a liquid of constant density, pressure increases linearly with vertical depth: p=p0+ρghp=p_0+\rho gh, where p0p_0 is the pressure at the free surface. If the surface is open to the atmosphere, pg=ρghp_{\mathrm{g}}=\rho gh measures gauge pressure and pabs=patm+pgp_{\mathrm{abs}}=p_{\mathrm{atm}}+p_{\mathrm{g}} gives absolute pressure. At the same horizontal level within one connected stationary liquid, pressures are equal. Do not use the sloping distance along an inclined wall in place of vertical depth.

Why a body experiences buoyancy

Pressure on a submerged object's lower surface is generally greater than on its upper surface, producing a net upward pressure force. Archimedes' principle gives FB=ρfgVdispF_{\mathrm{B}}=\rho_{\mathrm{f}}gV_{\mathrm{disp}}, where VdispV_{\mathrm{disp}} is the volume of fluid displaced and ρf\rho_{\mathrm{f}} is the fluid density. For a fully submerged solid, displaced volume equals its outside volume. For a floating body, only the submerged portion displaces liquid. Buoyancy depends on displaced fluid weight rather than directly on the object's own weight.

Floating, sinking and force balance

Take upwards as positive for a vertical force balance: Fnet=FB−WF_{\mathrm{net}}=F_{\mathrm{B}}-W if no other vertical forces act. An unrestrained object in equilibrium floats when buoyancy equals its weight; a denser-than-water object may sink if its weight exceeds the maximum buoyancy available from its full volume. For a flat vertical gate, remember that fluid pressure changes across its height, so the pressure force must be integrated or found using the appropriate area and centroid depth. A single pressure value at a point is not automatically the total pressure force on a large surface.

Worked example: A fully submerged rectangular block

A solid rectangular block measuring 0.30 m×0.20 m×0.10 m0.30\,\mathrm{m}\times0.20\,\mathrm{m}\times0.10\,\mathrm{m} is fully submerged in fresh water. Its centre is 2.50 m2.50\,\mathrm{m} vertically below the open free surface. The block weighs 80.0 N80.0\,\mathrm{N}. Take water density as 1000 kg/m31000\,\mathrm{kg/m^3} and g=9.81 m/s2g=9.81\,\mathrm{m/s^2}. Find the gauge pressure at its centre, the buoyancy force, and the net vertical force if no other forces act.

A water tank with a flat horizontal free surface and a small rectangular solid block entirely below it. Draw the block with its 0.30 m side horizontally across the page, 0.10 m side vertically, and 0.20 m side receding into the page in perspective. Draw a dashed horizontal line through the block centre and a vertical depth dimension from the water surface to this centre, labelled h = 2.50 m. At the block centre draw an upward arrow labelled F_B and a downward arrow labelled W = 80.0 N. Both arrows originate from the centre, and the downward arrow is modestly longer. Label water density and gravitational acceleration beside the tank. Do not write numerical buoyancy force, net force or gauge pressure.
The block experiences upward buoyancy and a downward weight.
Show the solution

1. Calculate gauge pressure at the centre.

pg=ρgh=(1000)(9.81)(2.50)=24525 Pa=24.5 kPap_{\mathrm{g}}=\rho gh=(1000)(9.81)(2.50)=24525\,\mathrm{Pa}=24.5\,\mathrm{kPa}

2. Find the displaced water volume. The block is fully submerged.

Vdisp=(0.30)(0.20)(0.10)=0.00600 m3V_{\mathrm{disp}}=(0.30)(0.20)(0.10)=0.00600\,\mathrm{m^3}

3. Calculate the buoyancy force.

FB=ρgVdisp=(1000)(9.81)(0.00600)=58.86 NF_{\mathrm{B}}=\rho gV_{\mathrm{disp}}=(1000)(9.81)(0.00600)=58.86\,\mathrm{N}

4. Apply a vertical force balance, taking upwards as positive.

Fnet=FB−W=58.86−80.0=−21.14 NF_{\mathrm{net}}=F_{\mathrm{B}}-W=58.86-80.0=-21.14\,\mathrm{N}

Answer: pg=24.5 kPa\boxed{p_{\mathrm{g}}=24.5\,\mathrm{kPa}} at the centre, FB=58.9 N\boxed{F_{\mathrm{B}}=58.9\,\mathrm{N}} upwards and Fnet=21.1 N\boxed{F_{\mathrm{net}}=21.1\,\mathrm{N}} downwards.

Common mistakes

Measuring depth along a sloping surface.

Use vertical distance below the reference free surface.

Adding atmospheric pressure when a gauge pressure is requested.

Use pg=ρghp_{\mathrm{g}}=\rho gh for gauge pressure and add surface pressure only for absolute pressure.

Using the solid density in the buoyancy equation.

Buoyancy uses displaced fluid density ρf\rho_{\mathrm{f}}, not object density.

Using the whole object volume when it is only partially submerged.

Use the immersed volume to determine displaced fluid weight.

Check your understanding

1. What is the gauge pressure at 3.00 m3.00\,\mathrm{m} depth in water with ρ=1000 kg/m3\rho=1000\,\mathrm{kg/m^3} and g=9.81 m/s2g=9.81\,\mathrm{m/s^2}?

2. A submerged object displaces 0.00200 m30.00200\,\mathrm{m^3} of fresh water. What buoyancy force acts if ρ=1000 kg/m3\rho=1000\,\mathrm{kg/m^3} and g=9.81 m/s2g=9.81\,\mathrm{m/s^2}?

3. An object floats freely in equilibrium and weighs 50 N50\,\mathrm{N}. What is the buoyancy force?

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