Applying Archimedes’ principle: calculations and practical examples

Two possible scenarios

When an object is immersed in a fluid, two scenarios arise:

1. The object floats: it sinks until the buoyant force exactly balances its weight. Only part of its volume is submerged.
2. The object sinks: even when fully submerged, the buoyant force remains less than its weight. The object continues to sink.

At equilibrium (when the object is floating), we have: Pi = P, that is, rho_fluid × V_submerged × g = m_object × g

Practical example: the iceberg

An iceberg (ρ approximately 920 kg/m³) floats on seawater (ρ approximately 1025 kg/m³). Only a small part is above the surface: approximately 10 per cent of the total volume is visible, whilst the rest (90 per cent) is submerged. This is the origin of the expression ‘the tip of the iceberg’.

Practical example: the submarine

A submarine can float, sink or remain stable at depth thanks to ballast tanks which it fills with water or air. By altering its submerged volume and mass, it changes the balance between buoyancy and weight.

And in the air?

Air is a fluid too! A hot-air balloon floats in the air because the hot air inside the balloon is less dense than the cold air outside: ρ_hot_air < ρ_cold_air.

A common pitfall to avoid

Never say that “Archimedes’ buoyancy depends on the weight of the object”. It depends on the fluid (rho) and the submerged volume (V), full stop. The weight of the object only comes into play when determining whether this buoyancy is sufficient to make it float.