How Buoyancy Works: Archimedes' Principle, and Why Ice Floats With 10.5 % Above Water

The force comes from pressure increasing with depth

Buoyancy is not a separate force of nature; it is what is left over when you add up the pressure on every face of a submerged object. Pressure in a fluid grows with depth as ρgh, so the bottom of a block is pushed up harder than the top is pushed down. The sideways pushes cancel by symmetry, and the vertical imbalance is exactly ρ_fluid × V × g — the weight of the fluid that used to occupy that space. That derivation is why the shape of the object is irrelevant and why only the displaced volume appears in the formula.

The practical version is apparent weight. A one-kilogram aluminium block has a volume of 1/2700 = 370.37 cm³ and weighs 9.807 N in air. Submerge it in fresh water and it displaces 0.37037 L, receiving 1000 × 0.00037037 × 9.80665 = 3.632 N of lift, so a scale hanging beneath the surface reads 6.175 N — as if the block had a mass of 0.630 kg. Nothing has been removed from the block; the water is simply carrying 37 percent of its weight. That fraction, 1000/2700, is again just the ratio of densities.

Why the submerged fraction is just a density ratio

A floating object is in equilibrium, so the buoyant force equals its weight: ρ_fluid × V_sub × g = ρ_object × V_total × g. Cancel g and rearrange, and the submerged fraction V_sub/V_total is ρ_object/ρ_fluid — no other quantity survives. Ice at 917 kg/m³ in seawater at 1,025 kg/m³ gives 0.8946, so 89.5 percent is under water and 10.5 percent stands proud. Do the same ice in fresh water at 1,000 kg/m³ and 91.7 percent is submerged, which is why a lake iceberg shows noticeably less of itself than a sea one.

The word doing the work is average. A steel hull floats not because steel is light but because the hull encloses air, and the average density of steel plus enclosed air is far below that of water. A 10,000 kg vessel floats by displacing 10,000/1,025 = 9.76 m³ of seawater, and it will sit lower in a river than in the sea because fresh water is less dense and more volume must be pushed aside to find the same weight. Load it and it settles further; that is the entire physics behind a Plimsoll line on a ship's side.

What the simple picture leaves out

The density ratio rule assumes the fluid is uniform and the object rigid and non-absorbent. Seawater is not uniform: 1,025 kg/m³ is a typical open-ocean surface figure, and real values run from roughly 1,020 in a brackish estuary to over 1,028 in cold, salty polar water. Sea ice is not the same as pure ice either — it traps brine and air, so its density spans a wide band and a real floe's freeboard is not simply the 10.5 percent that pure ice gives. Quote a density with the conditions attached, or the third significant figure is fiction.

Two more caveats are worth stating because competing explanations skip them. Floating is a question of average density, not of material, so the rule applies only if the object is fully wetted and does not soak up the fluid — a porous stone that gradually fills with water changes its own average density and can float then sink. And surface tension is a separate effect entirely: a steel needle or a water strider can rest on water whose density is far below theirs, held by the surface rather than by displacement. Whenever the object is small enough that surface tension matters, Archimedes alone will give the wrong answer.

Frequently asked questions

Why is an iceberg said to be nine-tenths under water?

Because the ratio of densities is close to nine-tenths, not because of a rule of thumb. Pure ice at 917 kg/m³ in seawater at 1,025 kg/m³ gives 917/1,025 = 89.5 percent submerged, so 10.5 percent shows. In fresh water the same ice is 91.7 percent under. The nine-tenths figure is a rounding of the seawater case, and it drifts as soon as the water is warmer, fresher or saltier than assumed, or the ice contains trapped brine and air rather than being pure.

Does the buoyant force depend on how deep the object is?

No, as long as the object is fully submerged and the fluid's density does not change with depth. Pressure grows with depth on the top face and the bottom face alike, and only the difference between them matters — a difference set by the object's own height, not by how far down it sits. A sealed can gets the same 9.81 N per litre one metre down as fifty metres down. The exceptions are real fluids that are compressible, and objects that are themselves compressible: a diver's wetsuit and lungs shrink with depth, reducing the volume displaced and therefore the lift, which is precisely why buoyancy control gets harder the deeper you go.

Is there buoyancy in air as well as in water?

Yes — air is a fluid and the same formula applies, just with a density around 1.2 kg/m³ instead of 1,000. That is why a helium balloon rises and why a precise mass measurement in air needs a buoyancy correction. The effect is small but not always negligible: a litre of air weighs about 0.012 N, so an object of one litre volume reads roughly 1.2 g light on a laboratory balance compared with its true mass in vacuum. Metrology labs correct for this routinely; for a kitchen scale it disappears far below the display's resolution.