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Buoyancy

Intuition

A submerged object is squeezed by pressure on all sides — but pressure grows with depth, so the push on its bottom beats the push on its top. The imbalance is a net upward force. Archimedes' genius was to notice its exact size: replace the object by fluid and that fluid would float in equilibrium — so the upward force on anything occupying that space equals the weight of the displaced fluid.

Formal statement (Archimedes' principle)

\[\boxed{\,F_\text{buoy} = \rho_\text{fluid}\, V_\text{displaced}\, g\,}\]

directed opposite to gravity. Consequently:

  • \(\rho_\text{object} < \rho_\text{fluid}\) → floats (rises until the submerged volume displaces exactly its weight)
  • \(\rho_\text{object} > \rho_\text{fluid}\) → sinks
  • \(\rho_\text{object} = \rho_\text{fluid}\) → neutrally buoyant

Worked example: the iceberg

Ice (\(\rho_\text{ice} = 917\ \text{kg m}^{-3}\)) floats in seawater (\(\rho_\text{water} = 1025\ \text{kg m}^{-3}\)). What fraction is submerged?

Float condition — buoyancy carries the full weight:

\[\rho_\text{water} V_\text{sub}\, g = \rho_\text{ice} V_\text{ice}\, g \implies \frac{V_\text{sub}}{V_\text{ice}} = \frac{\rho_\text{ice}}{\rho_\text{water}} = \frac{917}{1025} \approx 89\%\]

Only about a ninth of an iceberg shows. ("Tip of the iceberg" is quantitatively fair.)

Worked example: the melting ice cube

An ice cube floats in a glass of water. As it melts, does the water level rise, fall, or stay put?

While floating, the cube displaces a water volume weighing exactly its own weight. When it melts, it becomes that same weight of water — which occupies exactly the volume it was displacing. The level does not change. (Sea-level rise comes from land ice and thermal expansion, not from melting sea ice.)

Physical interpretation

Buoyancy is not a new force — it is the resultant of hydrostatic pressure integrated over the body's surface:

\[\mathbf{F}_\text{buoy} = -\oint_S P\, d\mathbf{A} = -\int_V \nabla P\, dV = \rho_\text{fluid}\, g\, V\, \hat{k}\]

using \(\nabla P = \rho_\text{fluid}\,\mathbf{g}\) inside the fluid. The divergence-theorem step is why the result depends only on displaced volume, never on shape.

Common mistakes

  • Using the object's density in the buoyancy formula. \(F_\text{buoy}\) depends on the fluid's density and the displaced volume only.
  • Using total volume for a floating body. Only the submerged part displaces fluid.
  • Thinking buoyancy disappears for sunk objects. A rock on the seabed still feels \(\rho_\text{fluid} V g\) upward; it's just smaller than the rock's weight.

Knowledge graph position

Prerequisites: Pressure, Hydrostatic equilibrium. Leads to: Rayleigh–Taylor instability.

Quiz

Q1 (computational). A 60 kg person floats in the Dead Sea (\(\rho \approx 1240\ \text{kg m}^{-3}\)). What volume of water do they displace?

Answer

\(V = m/\rho_\text{fluid} = 60/1240 \approx 0.048\ \text{m}^3\) (48 L) — noticeably less than in fresh water (60 L), which is why floating there feels so easy.

Q2 (conceptual). A boat carrying a boulder floats in a small pond. The boulder is thrown overboard and sinks. Does the pond level rise or fall?

Answer

Falls. In the boat, the boulder displaced its weight in water (\(V = m/\rho_\text{water}\), large). On the bottom it displaces only its volume (\(V = m/\rho_\text{rock}\), smaller since \(\rho_\text{rock} > \rho_\text{water}\)).

Q3 (multiple choice). Two solid cubes of identical volume, one steel and one wood, are fully submerged and held in place. The buoyant force is:

  • (a) larger on steel (b) larger on wood (c) equal (d) zero on the wood
Answer

(c). Same fluid, same displaced volume ⇒ same buoyant force. What differs is the net force once weight is included.