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Buoyancy calculations

Calculate upthrust with Archimedes' principle, find how much of a floating body is submerged, and find the reading on a support holding a body under water.

  • A-Level H2 Physics topic extensions

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Upthrust comes from the extra pressure of a fluid on the bottom of a body compared with its top. Working out that pressure difference gives a simple rule for the size of the upthrust, Archimedes’ principle, which explains why ships float and why things feel lighter under water.

Pressure increases with depth

Picture a vertical column of still fluid of density ρf, height h and cross-sectional area A. The fluid below must hold up the column’s weight, ρfAhg, as well as the force from the pressure on its top. So

p_bottomA = pₜₒₚA + ρfAhg ⇒ p_bottom-pₜₒₚ = ρfgh

Pressure increases with depth by ρfgh.

Archimedes’ principle

A body in the fluid has more pressure on its lower surface than on its upper surface, so the fluid pushes up on it harder than it pushes down. Adding up all these pressure forces gives the upthrust. The result is Archimedes’ principle: the upthrust equals the weight of the fluid displaced.

U = ρfgVd

ρf is the density of the fluid, not of the body, and Vd is the volume of fluid displaced: the whole volume of a fully submerged body, or only the submerged part of a floating one. Convert volumes to m³: 1 cm³ = 10⁻⁶ m³.

Floating

A body floating at rest has its upthrust equal to its weight. For a body of average density ρₒ and volume Vₒ,

ρfgVd = ρₒgVₒ ⇒ Vd/Vₒ = ρₒ/ρf

The fraction of the body below the surface equals the ratio of its density to the fluid’s. A body denser than the fluid would need a fraction greater than 1, so it sinks.

Guided practice 1

A floating ice cube

About 5 min

Problem

An ice cube of side 2.00 cm floats upright in tea, with 0.200 cm above the surface. The density of ice is 9.20 × 10² kg m⁻³. Find the density of the tea.

Find the density

Unit: kg m⁻³

Hints

Hint 1: submerged fraction

The submerged height is 2.00-0.200 = 1.80 cm, so the submerged fraction of the volume is 1.80/2.00.

Show solution step by step

  1. Submerged fraction

    Reason

    For a cube floating upright, the volume fraction equals the height fraction.

    Working

    Vd/Vₒ = 1.80/2.00 = 0.900

  2. Density of the tea

    Working

    ρf = ρₒ/0.900 = 920/0.900 = 1.02 × 10³ kg m⁻³

Check your understanding 1

A submarine stays still, fully submerged, with its engines off. What must be true about its average density?

Show answer

Its upthrust balances its weight, and it displaces its whole volume V: ρfgV = ρₒgV, so its average density equals the density of the seawater.

Holding a body under water

A dense body held still under water by a string from above has three forces on it: weight down, and upthrust and tension up. So T = W-U. The string’s tension, which a spring balance would read, is often called the apparent weight. The body’s mass and true weight have not changed; the water is simply supporting part of the weight.

A supporting string can pull from above or belowA dense body submerged in a fluid is held by a vertical string from above: weight points down, while tension and upthrust point up. A low-density submerged body is held by a string anchored below: upthrust points up, while tension and weight point down. In both cases all forces on the body balance.String from aboveString anchored belowDense body at restBuoyant body at restTUWUWTT = W − UT = U − W
Scroll across the figure to read all labels.
Draw the string's actual direction. For support from above, T + U = W. For an anchor below, U = W + T. Arrow lengths show illustrative force balances for each isolated body; no numerical force values are specified.

Worked example 2

Upthrust and tension on a submerged object

Extension

Problem

A 0.180 kg metal object of density 8.00 × 10³ kg m⁻³ hangs from a string, fully submerged in a liquid of density 8.00 × 10² kg m⁻³. Taking g = 9.81 m s⁻², find the upthrust and the tension in the string.

Worked solution

  1. Volume of the object

    Method

    Use the object’s mass and density.

    Reason

    A fully submerged object displaces its own volume of liquid.

    Working

    V = 0.180/(8.00 × 10³) = 2.25 × 10⁻⁵ m³

  2. Upthrust

    Reason

    Archimedes’ principle uses the density of the liquid.

    Working

    U = (8.00 × 10²)(9.81)(2.25 × 10⁻⁵) = 0.177 N

  3. Tension

    Working

    T = mg-U = (0.180)(9.81)-0.177 = 1.59 N

Try it yourself 3

Apparent weight in water

Minimal support

Problem

A 2.0 kg object of volume 1.5 × 10⁻³ m³ hangs from a spring balance, fully under water of density 1000 kg m⁻³. Taking g = 9.81 m s⁻², what does the balance read?

Find the reading

Unit: N

Hints

Hint 1: three forces

Weight acts down; the upthrust and the balance’s pull act up.

Show solution step by step

  1. Weight and upthrust

    Reason

    The reading is what remains of the weight after the upthrust.

    Working

    W = 2.0(9.81) = 19.6 N and U = (1000)(9.81)(1.5 × 10⁻³) = 14.7 N

  2. Reading

    Working

    T = 19.62-14.72 = 4.9 N

Common mistakes

  • Using the density of the body instead of the density of the fluid in U = ρfgVd.
  • Using the whole volume of a floating body instead of the submerged volume.
  • Forgetting to convert cm³ to m³.
  • Thinking the apparent weight is a change in the body’s mass or true weight.
Syllabus and review details

No official syllabus alignment is listed for this lesson.