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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.

Buoyancy is the resultant upward force exerted by a resting fluid on an immersed body in a downward gravitational field. The fluid pushes inward, perpendicular to each part of the body's surface. Because hydrostatic pressure increases with depth, the upward push from below exceeds the downward push from above. Opposite horizontal forces cancel; buoyancy is the remaining vertical force. A sinking body still receives this upward support. ^[ref]

More precisely, pressure $p$ on a small surface of area $dA$ with outward unit normal vector $\mathbf{n}$ exertsThe minus sign says the force points into the body. On a cube, the top force points down, the bottom force points up, and the side forces cancel in pairs because opposite sides have the same pressure at each depth. This is a surface traction; gravity on the body itself is a separate body force. ^[ref]

For a computed example, stipulate resting water with constant mass density ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$ and uniform gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. A fully immersed cube has side $L=0.10\mathrm{m}$, top depth $h_t=0.20\mathrm{m}$ and bottom depth $h_b=0.30\mathrm{m}$. Each horizontal face has area $A=L^2=0.010{\mathrm{m}}^2$, and its volume is $V=L^3=0.001{\mathrm{m}}^3$. Its top and bottom gauge pressures areThe corresponding forces relative to the common surface pressure are $p_{g,t}A=19.62\mathrm{N}$ downward and $p_{g,b}A=29.43\mathrm{N}$ upward. Therefore the net buoyancy isUsing absolute pressures adds the same $p_{\mathrm{surface}}A$ to both face forces, leaving their difference unchanged. Zero gauge pressure at the surface does not mean zero absolute pressure. Increasing both depths by the same amount also leaves buoyancy unchanged in this constant-mass density model.

Archimedes' principle turns this calculation into a rule for bodies of any shape. It explains floating equilibrium and lets us infer an immersed body's volume from the difference between its force from gravity and the force needed to support it. In particular, buoyancy depends on the displaced volume and the surrounding mass density, whereas the body's own mass density decides how its force from gravity compares with that support.

A freely floating body at rest has no supporting contact or applied holding force. Neglecting surface tension and the buoyancy of air, its upward buoyancy balances its downward force from gravity:For a body of uniform mass density ${\rho }_{\mathrm{body}}$ and total volume $V_{\mathrm{body}}$, its mass is $m={\rho }_{\mathrm{body}}{V}_{\mathrm{body}}$. Substitution into Archimedes' principle givesPartial immersion is possible when ${\rho }_{\mathrm{body}}<{\rho }_f$. A nonuniform body uses its mean mass density $m/V_{\mathrm{body}}$ instead; for a hollow body this volume must count the enclosed space that excludes water. This explains how the amount of displaced water can support a vessel whose walls alone are denser than water. ^[ref]

For the $600\mathrm{kg}{\mathrm{m}}^{-3}$ cube in water of mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$, the displaced fraction is $600/1000=0.60$: sixty percent of its volume lies below the surface. With a horizontal top and bottom, the immersed depth $d$ satisfies $V_{\mathrm{displaced}}=Ad$, soIts remaining height is $0.040\mathrm{m}$, and the buoyancy is $1000\times 9.81\times 0.000600=5.886\mathrm{N}$, exactly balancing its downward force from gravity.

A floating density instrument. At equilibrium, Archimedes' principle gives $\rho V_{\mathrm{displaced}}=m$ for fixed instrument mass $m$, so it displaces more volume in a less dense liquid. A graduated narrow stem converts this volume change into a visible immersion reading. Calibration specifies a temperature and a reading convention for the meniscus; a sample's other dissolved substances can affect a sugar-calibrated scale.

An iceberg is a detached mass of freshwater ice floating in the ocean.

For a vertical-sided floating iceberg of height $H$, Archimedes' principle equates displaced water mass to ice mass. Its draft is ${\rho }_w^{-1}{\int }_0^H{\rho }_i(z)dz$ and its freeboard is the remaining height above water. A depth-dependent mass density must be integrated; using the densest layer throughout overestimates the draft.

Floating ice with submerged volume fraction $f$ has density $f{\rho }_l$ by Archimedes' principle. Its volume change on melting is therefore $v_l-v_s=(1-1/f)/{\rho }_l<0$. An enthalpy cycle gives $L_f=L_s-L_v$ for quantities consistently expressed per mass at the same temperature. Treating these latent heats and specific volumes as constant and integrating the Clausius-Clapeyron relation gives the displayed pressure estimate. Using a vaporization latent heat extrapolated from a different temperature carries that approximation into the inferred fusion heat.

Take $z$ positive downwards, and let the surface pressure be $p_0$. Static force balance gives the hydrostatic pressure equation $dp/dz=\rho g$. HenceOn a submerged body's surface, pressure traction is $-p\mathbf{n}$, where $\mathbf{n}$ points out of the body. The divergence theorem gives the resultantThus the buoyancy is upward with magnitude $\rho gV$, proving Archimedes' principle. The constant surface pressure makes no contribution because ${\int }_{\partial V}\mathbf{n}dS=0$.

For a floating iceberg of total volume $V$, the displaced seawater volume is $V-V_I$. Neglecting air buoyancy, equality of weight and buoyancy gives ${\rho }_{I}gV={\rho }_{w}g(V-V_I)$. Consequentlywith $0<{\rho }_I<{\rho }_w$, as required for partial flotation.

Measure $z$ downward from the iceberg surface. Its mass density is ${\rho }_i(z)=300+12z$ for $0\le z\le 50$ m and $900\mathrm{kg}{\mathrm{m}}^{-3}$ below that. The mass per unit horizontal area isBy Archimedes' principle, the draft is $d=(m/A)/{\rho }_w\simeq 248.78$ m. The iceberg freeboard is therefore $300-d\simeq 51.22$ m. The specified snow-to-ice density is reached atThe more porous snow and firn are consequently above water: atmospheric warming, meltwater infiltration, compaction and refreezing can modify them. At the waterline and below, dense ice is exposed to oceanic heat transfer and wave erosion, which can undercut the berg and cause breakage. If the porous transition were submerged, water infiltration and ocean heat could act directly on that material, changing the erosion and collapse mechanisms.

Take a body floating at rest in a homogeneous fluid, with uniform gravity, no tether or other support, and negligible air buoyancy. Imagine replacing the submerged part by the same volume of the surrounding fluid. That imaginary fluid parcel would be at rest: the surrounding hydrostatic pressure would balance its gravitational weight. The actual body experiences the same pressure traction over the submerged boundary, so the resulting buoyant force is upward with magnitudeThe body's vertical force balance therefore gives Archimedes' principle in the requested floating case:The argument determines displaced submerged volume, not necessarily the total volume of a partly emerged body.

The pressure resultant acts through the centre of buoyancy, the centre of mass of the imaginary displaced fluid. Let $G$ be the body's centre of gravity and $B_c$ the centre of buoyancy. The two remaining external forces are equal and opposite vertical forces. If their lines of action were distinct, they would form a nonzero couple and the body would turn. Rotational equilibrium consequently requiresThis is a condition for equilibrium, not sufficient evidence of its stability. Tilting changes the displaced shape and hence the position of the centre of buoyancy; the resulting torque determines whether the body returns towards its original orientation. A centre of gravity above the centre of buoyancy does not by itself settle that floating-body stability question.

A hydrometer is a sealed floating instrument with a weighted bulb and a narrow graduated stem. The ballast keeps the stem upright. By Archimedes' principle, its gravitational weight equals the buoyancy:It therefore sinks farther into a less dense liquid, displacing more volume, and floats higher in a denser one. The stem makes a small volume change readily visible as a change in immersion. If its cross-sectional area is $A$, then $V_{\mathrm{displaced}}=V_0+Az$ over a uniform part of the stem. A density scale is consequently not generally uniform in $z$.

Float the clean instrument freely in a sufficiently deep sample, without its touching the wall or bottom. Remove adhering bubbles, allow the temperature and motion to settle, and read the liquid surface at eye level using the meniscus convention specified for that instrument. A calibration for a particular temperature or reference density cannot be transferred without correction: both the sample and the instrument expand. An ordinary hydrometer measures density or specific gravity; a special scale for a particular chemical solution does not make it chemically selective. In particular, dissolved substances other than glucose can affect the density of urine.

At coexistence, the two phases have equal chemical potential. For a single component, $d\mu =-sdT+vdp$, with entropy and volume measured on the same molar or mass basis. Equating the phase derivatives along coexistence gives $\Delta vdp=\Delta sdT$. Reversible latent heat satisfies $L=T\Delta s$, henceHere $p$ and $T$ are coexistence pressure and absolute temperature, $L$ is latent heat for transforming phase 1 to phase 2, and $\Delta v=v_2-v_1$ uses the same amount of material as $L$.

For vaporization, neglect liquid molar volume compared with ideal-vapour volume $RT/p$. The ideal-vapour Clausius-Clapeyron approximation becomesUsing the normal boiling reference $T_b=373.15\mathrm{K}$, $p_b=1.01325\times 10^5\mathrm{Pa}$ and the supplied constant $L_e$ givesThis deliberately constant-latent-heat extrapolation does not reproduce the supplied triple-point pressure exactly. It predicts the liquid-vapour branch, metastable against ice below the triple-point temperature; it is not a precision sublimation-pressure prediction.

The phase diagram has solid, liquid and vapour regions meeting at the triple point. The sublimation and vaporization boundaries have positive slopes, whereas the ordinary ice–water boundary slopes negatively because ice has larger specific volume. The liquid–vapour line terminates at a critical point; the original schematic below emphasizes topology and marks the supplied triple point.

For the melting-pressure estimate, Archimedes' principle gives ${\rho }_i=0.90{\rho }_w=900\mathrm{kg}{\mathrm{m}}^{-3}$, soConvert the molar evaporation heat using water molar mass $M\simeq 0.0180\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, rounded from the tabulated value $18.0153\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$.^[ref] An enthalpy cycle through vapour then givesThis uses the question's temperature-independent evaporation heat all the way to the triple point; no replacement experimental fusion heat is inserted. The melting pressure inferred from flotation and latent heats, on holding $L_f$ and $\Delta v$ constant, isusing $T_t=273.16\mathrm{K}$ and $T=270.15\mathrm{K}$. Linearizing the logarithm about the triple point gives $53.4\mathrm{MPa}$ instead of $53.7\mathrm{MPa}$. Both are estimates within the stated constant-property approximation; the negative melting slope makes a positive pressure necessary to lower the melting temperature.

Archimedes' principle states that the resultant hydrostatic pressure force on a body equals the weight of the fluid displaced, directed opposite to gravity. Let $V_d$ be the displaced fluid region, with outward normal $\hat{\mathbf{n}}$ from the body into the fluid. Since the fluid pressure force is inward,Hydrostatic balance is $\nabla p={\rho }_f\mathbf{g}$, with downward gravitational acceleration $\mathbf{g}$. Thereforefor uniform fluid density and gravity. Uniform atmospheric pressure gives no net force on a closed surface. For a floating body, the same displaced-volume construction includes the free surface; any significant air buoyancy can be included as a second displaced fluid.

A freely floating body of mass $m$ satisfies $mg={\rho }_{f}g{V}_d$, so its displaced volume is $m/{\rho }_f$. A homogeneous body of density ${\rho }_b<{\rho }_f$ floats with submerged fraction ${\rho }_b/{\rho }_f$; if ${\rho }_b>{\rho }_f$, hydrostatic buoyancy alone cannot support it. The buoyant force acts through the centre of buoyancy, the centroid of the displaced volume for constant density, while weight acts through the centre of mass. Their relative positions determine rotational equilibrium and stability. For a fully submerged body, a centre of mass below the centre of buoyancy gives a restoring torque; a floating body's changing displaced shape additionally requires the metacentric height criterion. Surface tension, externally driven flow and contact forces can alter the simple force balance.

A large gas volume can reduce the effective mass density of the water surrounding the hull. In the simple quasihydrostatic model for the buoyancy of an aerated liquid, gas void fraction $\alpha$ givesBy Archimedes' principle, a boat displacing volume $V_{\mathrm{disp}}$ then has buoyant force ${\rho }_{\mathrm{mix}}g{V}_{\mathrm{disp}}$. To support the same boat mass $m$, it needs a larger displaced volume $m/{\rho }_{\mathrm{mix}}$, so its draft increases. If that exceeds the available watertight displacement, it can lose flotation or become swamped:Uneven aeration and strong plume motion can also disturb stability. This explains a possible hazard, not a claim that any bubble plume must sink a boat: upward water flow, turbulence and the distribution of bubbles can exert substantial additional forces.

Let the cylinder length be $\ell =0.020\mathrm{m}$, cross-sectional area $A$, and the fluid mass density ${\rho }_f$. At floating equilibrium, Archimedes' principle givesFor a small downward displacement $z$ from this equilibrium, the displaced fluid volume increases by $Az$. The additional upward buoyant force is ${\rho }_{f}gAz$, and Newton's second law becomesThis is simple harmonic motion, with angular frequency $\omega =\sqrt{2g/\ell }$ and oscillation periodThe result for vertical oscillations of a floating cylinder is independent of its cross-sectional area and fluid mass density because those quantities cancel using the known immersion fraction. It neglects drag, surface effects and the extra fluid inertia from the motion; the cylinder must remain partly immersed throughout the small motion.

A vertical cylinder of constant cross-sectional area $A$, partly immersed to depth $h_0$ in a fluid of mass density ${\rho }_f$, satisfies $mg={\rho }_{f}gA{h}_0$ by Archimedes' principle. A small extra downward displacement $z$ increases the buoyant force by ${\rho }_{f}gAz$, so $m\ddot{z}=-{\rho }_{f}gAz$. This is simple harmonic motion with the displayed angular frequency and oscillation period $2\pi \sqrt{h_0/g}$. The ideal model neglects fluid added inertia, drag and surface effects; it requires partial immersion throughout the motion.