Archimedes' principle ($F_b={\rho }_{f}g{V}_{\mathrm{displaced}}$)

Archimedes' principle states that buoyancy equals the downward gravitational weight, or weight, of the displaced fluid. In a resting fluid of constant mass density ${\rho }_f$ under uniform gravitational acceleration $g$, its magnitude isHere $V_{\mathrm{displaced}}$ is the volume from which the body excludes the fluid: the whole external volume for a fully immersed solid body, and just the immersed volume for a partly immersed one. The body's own mass does not enter this buoyancy formula. ^[ref]

A short proof outline for a fully immersed body is to imagine replacing it by exactly the fluid it displaces. The surrounding hydrostatic pressure is unchanged. That replacement fluid is at rest, so its upward pressure force must balance its downward force from gravity. The original body receives the same inward pressure forces on the same closed surface, hence the same upward buoyancy.

Equivalently, let $\Omega$ be the displaced region, $\partial \Omega$ its closed surface and ${\mathbf{e}}_z$ the upward unit direction. The divergence theorem, applied component by component to the surface integral, and hydrostatic pressure giveThis also explains why horizontal forces cancel even for an irregular shape. A spatially constant surface pressure contributes zero to the closed-surface integral. If the mass density varies with depth, replace ${\rho }_f{V}_{\mathrm{displaced}}$ by the mass ${\int }_{\Omega }{\rho }_{f}dV$ of the displaced fluid.

The closed-surface argument assumes that the fluid applies pressure over the entire boundary of a fully immersed body. A body sealing against a tank bottom may have no fluid beneath it; then the lower-face pressure force used in the proof is absent, and a supporting contact force must be treated separately. For partial immersion, the usual buoyancy formula assumes the exposed surface receives the common atmospheric pressure, neglects the much smaller buoyancy from the air, and neglects surface tension.

For the $0.10\mathrm{m}$ cube in the stipulated water, full immersion displaces $0.001{\mathrm{m}}^3$, so $F_b=1000\times 9.81\times 0.001=9.81\mathrm{N}$. If its uniform mass density is $600\mathrm{kg}{\mathrm{m}}^{-3}$, its mass is $m=600\times 0.001=0.600\mathrm{kg}$ and its downward force from gravity is $mg=5.886\mathrm{N}$. To hold it fully immersed at rest requires an additional downward forceOn release from rest it initially moves upward; its eventual floating equilibrium has a smaller displaced volume. A same-size cube of uniform mass density $1200\mathrm{kg}{\mathrm{m}}^{-3}$ has mass $1.200\mathrm{kg}$ and downward force from gravity $11.772\mathrm{N}$. Full immersion still gives $9.81\mathrm{N}$ of buoyancy, leaving an initial net downward force of $11.772-9.81=1.962\mathrm{N}$. Subsequent motion also involves drag.