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Buoyancy Force (Archimedes' Principle) Calculator

Archimedes' principle: the upward buoyant force on a body immersed in a fluid equals the weight of the fluid it displaces — F_b = ρ · V · g. Pick a fluid preset (fresh water, seawater, mercury, air…) or enter a custom density, set the submerged volume and gravity, and instantly see the buoyant force in N, kgf and lbf. Add the object's mass and the tool also compares weight versus buoyancy, classifies the result as floats / sinks / neutrally buoyant, and reports the fraction submerged for a free-floating body.

Buoyant force F_b

Archimedes' principle: the upward force on a body immersed in a fluid equals the weight of the displaced fluid.

Verdict

Add the object's mass to see whether it floats, sinks or stays neutrally buoyant.

Object weight W
Net force (upward is positive)
Effective object density
Fraction submerged when free-floating

Equation

F_b = ρ_fluid · V_submerged · g (N)

This is the hydrostatic buoyant force for a fluid at rest. Waves, turbulence, surface tension and gas compressibility are not modelled.

Formula

F_b = ρ_fluid · V_submerged · g // buoyant force in newtons (N) Displaced mass = ρ_fluid · V_submerged // equals F_b / g (kg) Net force = F_b − m_object · g // > 0 rises; = 0 neutral; < 0 sinks Fraction submerged when floating = ρ_object / ρ_fluid // valid when ρ_object < ρ_fluid

Frequently asked

Why does a steel ship float when a solid steel nail sinks?

It depends on the body's average density (mass ÷ total volume including internal voids), not on what it's made of. A solid steel nail has the density of steel (≈ 7850 kg/m³), well above water's 1000 — so it sinks. A steel ship is mostly enclosed air. Divide the ship's total mass by the volume enclosed up to the waterline (steel hull plus all internal air spaces) and the average density falls to roughly 200–400 kg/m³, far below water. Archimedes' principle only cares about displaced fluid volume; by shaping the steel into a hull, the ship displaces a far greater volume of water than the same mass of steel would as a solid block, generating enough buoyant force to support its weight.

Why can a non-swimmer float effortlessly in the Dead Sea?

The Dead Sea is roughly 33–34 % salt by mass (compared with ~3.5 % for normal seawater), giving it a density of about 1240 kg/m³ — much higher than fresh water (1000) or ocean water (1025). A human body averages roughly 985–1010 kg/m³ depending on body composition, well below the Dead Sea's density. The equilibrium fraction submerged is ρ_body / ρ_fluid ≈ 985/1240 ≈ 79 %, so you sit much higher in the water than in a pool — about 21 % of the body emerges, enough to read a book on your back. The same logic explains why divers wear extra weight in seawater versus fresh water, and why swimmers sit visibly higher in ocean swells than in a freshwater pool.

Does the buoyant force keep growing once an object is fully submerged?

For a rigid object in a (practically) incompressible liquid: once it is fully submerged, the buoyant force stays essentially constant with depth — F_b = ρ · V · g, and neither V nor ρ changes. The absolute pressure increases with depth, but buoyancy is the *difference* between pressure on the bottom and top of the body; that pressure difference times area is fixed and equals the weight of displaced fluid, independent of depth. Two exceptions matter in practice: (1) compressible objects (a balloon, a diver's wetsuit air pockets, a submarine ballast tank) shrink at depth — V decreases, so buoyancy decreases; (2) the fluid itself is weakly compressible at extreme depths (deep ocean) — ρ rises slightly, very slightly increasing buoyancy. For typical engineering and textbook problems these effects are negligible.

Do hot-air balloons and helium balloons also rely on buoyancy?

Yes — the formula is identical, just with air as the fluid. At sea level, air density is ρ ≈ 1.225 kg/m³, so each cubic metre of displaced air generates F_b = 1.225 · 1 · 9.81 ≈ 12 N of lift — enough to support about 1.225 kg. A hot-air balloon heats its internal air to ~100 °C, dropping its density to ~0.95 kg/m³; the ~0.28 kg/m³ difference yields about 0.28 kg of lift per cubic metre. A typical 2200 m³ balloon thus lifts roughly 600 kg — enough for a basket, fuel, and 2–4 passengers. Helium is even lighter at 0.179 kg/m³, giving about 1.05 kg of lift per cubic metre — that's why small helium balloons can carry tags or cameras, and why weather-balloon launches use just 1–2 m³ of helium or hydrogen to lift instrumentation to 30 km altitude.

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