Buoyancy & Archimedes' Principle

Take an upright block with horizontal faces of area A, its top at depth h₁ and its bottom at h₂ = h₁ + H. The fluid pushes down on the top with (P₀ + ρ_f g h₁)A and up on the bottom with (P₀ + ρ_f g h₂)A. Pushes on opposite side walls are equal at every depth and cancel in pairs. The net force is ρ_f g (h₂ − h₁)A = ρ_f g H A, upward. Put in numbers: a block with 0.20 m × 0.20 m faces and H = 0.10 m, top face 0.50 m under fresh water. Gauge pressures are 4905 Pa on top and 5886 Pa underneath, a difference of 981 Pa, and 981 × 0.040 = 39.2 N. Lower it until the top is 5.50 m down: now 53 955 Pa and 54 936 Pa, the same 981 Pa apart, the same 39.2 N. Each face there carries about 2.2 kN of gauge force, yet the body only feels the difference. Both P₀ and the depth have cancelled, and what is left, HA, is the volume of the block.

For a propeller or a potato the pressure sum is an ugly surface integral, and it is never needed. Imagine the body gone and its place filled with more of the same fluid. That parcel would stay at rest, so the fluid around it must push up on it with exactly its weight, ρ_f g V_sub. Pressure at each point of the boundary depends only on depth, not on what lies inside, so the real body receives the identical force. The parcel does not rotate either, so the force acts along the vertical through the parcel's centre of mass: the centroid of the displaced volume, called the centre of buoyancy B. Two consequences. The force belongs to the fluid and the geometry, so one litre of lead, of aluminium or of cork held under water each receive 1000 × 9.81 × 1.0 × 10⁻³ = 9.81 N. And only the immersed part counts. In air, at 1.20 kg m⁻³, the same litre receives 0.012 N, which is why air buoyancy is dropped for a brick and is the whole story for a balloon.

Hang a body from a string and lower it into a liquid. At rest, T + F_B = mg, so the string reads the apparent weight Wₐₚₚ = mg − F_B = (ρ_b − ρ_f) g V. An aluminium casting of mass 0.540 kg and density 2700 kg m⁻³ has V = 2.00 × 10⁻⁴ m³ and weighs 5.30 N; under water F_B = 1.96 N and the string reads 3.34 N. Run it backwards and two weighings become a density measurement: W − Wₐₚₚ is F_B, F_B/(ρ_f g) is the volume, and ρ_b = ρ_f W/(W − Wₐₚₚ) needs no ruler at all. The third law finishes the accounting. If the water pushes the casting up with 1.96 N, the casting pushes the water down with 1.96 N, and a balance under the beaker reads 1.96 N more the moment the casting goes under. The string has shed exactly what the balance has gained; the total support for beaker, water and casting has not changed.

If ρ_b is below ρ_f, full immersion gives F_B greater than mg and the body rises. As it breaks the surface V_sub shrinks, and it settles where ρ_f g V_sub = mg, which is V_sub/V = ρ_b/ρ_f. Pine at 520 kg m⁻³ floats 52.0% submerged in fresh water and 50.7% in seawater at 1025 kg m⁻³. Here ρ_b means the average density, hollow space included, which is how steel at 7850 kg m⁻³ makes a ship. A 2.0 × 10⁷ kg vessel displaces 2.0 × 10⁷ kg of water wherever it goes: 1.95 × 10⁴ m³ at sea, 2.00 × 10⁴ m³ in a river, so it must displace 2.5% more volume, and ride deeper, once it leaves the sea. Loads work the same way. For a hull with vertical sides and waterplane area A, an extra mass Δm needs an extra Δm/ρ_f of displaced water, so the draft grows by Δm/(ρ_f A): 45 000 kg of cars on a sea ferry with A = 600 m² sinks it 45 000/(1025 × 600) = 0.073 m. Push such a hull down a further distance x and buoyancy exceeds weight by ρ_f g A x, a restoring force, so floating equilibrium is stable against bobbing.

Weight acts at the centre of gravity G, buoyancy at B. In equilibrium they are equal and share one vertical line; tilt the body and they may form a couple. For a fully submerged body B is fixed in the body, and the couple rights it only if G lies below B, which is why a submarine carries its ballast low. A floating body is subtler. Heel it and the displaced volume changes shape, so B slides towards the side that dipped. The vertical through the new B crosses the body's centreline at the metacentre M, and the couple is a righting one if M is above G. For small tilts BM = I/V_sub, where I is the second moment of the waterplane area about the hull's long axis, an area property in m⁴ and not a mass moment of inertia. A box hull of length L, beam w and draft d has I = Lw³/12 and V_sub = Lwd, so BM = w²/(12d). A barge with w = 4.0 m and d = 0.80 m has B 0.40 m above the keel and BM = 1.67 m, so M sits 2.07 m up and G may ride well above B in safety. A uniform cube of side s and half the water's density has B at s/4 and BM = s/6, putting M at 0.417s, below G at 0.500s: it cannot float flat-face-up and rolls over until it finds a tilted position that is stable.

The swap argument assumed fluid pressing on every part of the wetted surface. Seat a flat drain plug of area 0.010 m² in the floor of a tank under 2.0 m of water, with air rather than water beneath it, and there is no upward push at all: the water holds it down with 1000 × 9.81 × 2.0 × 0.010 = 196 N. It assumed hydrostatic pressure, so in a lift accelerating upward with acceleration a the pressure gradient becomes ρ_f (g + a); buoyancy grows by the factor (g + a)/g, exactly as the force needed to support the body does, and the floating fraction stays put, while in free fall the pressure gradient vanishes and bubbles stop rising. It assumed one fluid density over the height of the body: across an oil-on-water interface each layer contributes the weight of what it displaces, and a balloon climbing through thinning air stops where its average density matches the local air. Small bodies can be carried by surface tension instead, as a steel needle laid on water is. And once the body moves, drag and the fluid that must be accelerated around it join the force balance alongside F_B.