Buoyant force: Archimedes’ principle

Immersion exposes all wetted exterior faces to water pressure. For a rectangular body, lateral forces cancel in pairs, while greater depth gives higher pressure below. The remaining force points upward. A string therefore supports less of the object’s weight. Whether the object floats freely follows only after comparing buoyancy with weight and any additional forces.[1]

$F_A=\rho_{\mathrm{fluid}}V_{\mathrm{displaced}}g$ applies to approximately uniform fluid density and constant g in a hydrostatic model. Displaced volume is space the surrounding fluid cannot occupy because of the body. For a partly submerged sealed body, use only its submerged part. For full immersion, use its entire displacing external volume.[1]

Body mass does not appear directly in the buoyancy-magnitude formula. Together with weight and other forces, it nevertheless determines net force and therefore acceleration. Two fully immersed sealed bodies of equal external volume experience the same hydrostatic buoyancy in the same uniform fluid even if their masses differ. Whether they are currently moving upward or downward additionally depends on their previous motion. Only release from rest allows initial acceleration to determine initial motion direction.

Consider a fully immersed cuboid with horizontal area A and height H. Its bottom is H deeper than its top, so bottom pressure exceeds top pressure by ρgH. Pressure forces on symmetric opposite sides cancel. Top pressure acts downward and bottom pressure upward. This hydrostatic pressure difference arises in a stationary fluid from the weight of the fluid above.[1][2]

For arbitrary fully surrounded shapes, integrate pressure forces over the surface. An imaginary fluid filling the displaced volume would have its weight balanced by those pressure forces in hydrostatic equilibrium. The same pressure distribution on the body therefore yields buoyancy equal to the weight of the replaced fluid volume.[1]

This reasoning needs fluid pressure on the relevant surfaces. If a body seals against the bottom and liquid cannot reach underneath, the usual full buoyancy balance may fail. Examine contact forces and the actual pressure distribution explicitly.

One litre of water modelled at 1000 kg/m³ has mass 1 kg and weight 9.81 N for $g=9.81\,\mathrm{m\,s^{-2}}$. Kilograms and newtons are not interchangeable. The same displaced mass has a different weight where g differs.

Newton’s second law requires the sum of all forces on the selected body. For a stationary supported body underwater, tension and buoyancy together balance weight. The force meter measures tension, not a changed body mass. The subsequent floating example instead uses free equilibrium.[1][3]

Without other forces, a fully immersed object accelerates upward when $F_A>F_G$, and downward when $F_A<F_G$. Its motion also depends on initial velocity. An already sinking object can continue sinking while upward acceleration slows it down. Equality gives zero acceleration in this model but does not automatically imply stable equilibrium. With drag, a sinking object may later reach constant velocity although buoyancy alone remains smaller than weight.

Compare fluid density with average density of the entire displacing body, including sealed cavities. Steel-wall density alone does not determine a sealed hull’s behaviour. Buoyancy balance also does not guarantee stability against capsizing; centre-of-mass and centre-of-buoyancy positions during small rotations matter.

The principle applies to gases too. Original model check: a sealed envelope of total mass 0.80 kg displacing 1 m³ of air at assumed ambient density 1.20 kg/m³ has buoyancy 11.772 N. Weight is 7.848 N, leaving 3.924 N upward without other forces. Total mass includes envelope, contents and payload.

A body at an interface may displace two liquids or liquid and air. Counting only water omits the other buoyancy contribution. Neglecting air is often a good approximation but must match required accuracy. Continuous stratification replaces the sum with an integral.

$F_A=\rho Vg$ describes the hydrostatic pressure contribution. Moving fluids can generate additional drag and lift; aerodynamic wing lift is not merely displaced air weight. Vessel acceleration, fluid compression and surface tension may require further model terms.

A body with open cavities does not automatically displace its entire geometric envelope volume. If fluid enters, identify the regions genuinely excluded. Likewise, apparent weight reduction gives $F_A=F_G-S$ only when motion, other support forces and instrument conditions match the assumed balance.

Buoyancy depends on the density and volume of displaced fluid. Partial immersion uses only the immersed volume. Whether an object accelerates, is held or floats freely follows from the full force balance with weight and any additional forces.