Archimedes Principle Calculator

Archimedes' principle says an immersed body is pushed up by a force equal to the weight of the fluid it displaces:

$F_b = \rho_{\text{fluid}}\, g\, V_{\text{disp}}$

Symbols and SI units:

• $F_b$ — buoyant force, newtons (N), directed vertically upward
• $\rho_{\text{fluid}}$ — density of the fluid, kilograms per cubic metre (kg/m³); fresh water is $1000$ kg/m³, seawater about $1025$ kg/m³
• $g$ — $9.81$ m/s²
• $V_{\text{disp}}$ — volume of fluid displaced, cubic metres (m³)

The buoyant force depends on the fluid's density, never the object's. A lead block and a wooden block of the same volume, both fully under water, feel exactly the same upthrust — they behave differently only because their weights differ.

When it applies: a body in a fluid at rest, in a uniform gravitational field.

The assumption people forget: $V_{\text{disp}}$ is the submerged volume, which equals the whole object's volume only when it is fully under.

A submerged object's apparent weight is what a scale reads once buoyancy is subtracted:

$W_{\text{apparent}} = mg - F_b$

Comparing densities settles what happens:

• $\rho_{\text{object}} > \rho_{\text{fluid}}$ — it sinks
• $\rho_{\text{object}} < \rho_{\text{fluid}}$ — it floats, partly out of the fluid
• $\rho_{\text{object}} = \rho_{\text{fluid}}$ — it hovers, neutrally buoyant

For a floating body the buoyant force exactly balances the weight, $F_b = mg$, and the submerged fraction follows:

$\frac{V_{\text{disp}}}{V_{\text{object}}} = \frac{\rho_{\text{object}}}{\rho_{\text{fluid}}}$

Ice at $917$ kg/m³ in seawater at $1025$ kg/m³ therefore floats with about $89\%$ of its volume under — the origin of the iceberg cliché.

The assumption people forget: the floating formula uses average density, hull and trapped air included, which is why a steel ship floats.

• Using the object's density in $F_b = \rho gV$ — the density in the formula is the fluid's.
• Using the full volume for a partly submerged object — only the submerged part displaces fluid.
• Mixing litres and cubic metres — $1\ \text{L} = 10^{-3}$ m³, so $2.0$ L is $2.0 \times 10^{-3}$ m³.
• Reporting buoyant force in kilograms — it is a force, so newtons.
• Thinking depth matters — for an incompressible fluid the upthrust is the same at $1$ m and $100$ m down, because the displaced volume is unchanged.
• Forgetting $g$ when converting a displaced mass to a force — $2$ kg of displaced water gives $19.6$ N, not $2$ N.
• Using water's density in another fluid — oil, alcohol and mercury all give very different results.