Buoyancy and Floating Equilibrium

A floating object is in equilibrium when the buoyant force equals its weight, which means that it displaces a mass of fluid equal to its own mass, and the fraction of its volume submerged is ρ_object ÷ ρ_fluid; the floating is stable if the center of buoyancy and center of gravity are arranged so that a tilt produces a restoring torque. In brief: the buoyant force acts at the center of buoyancy, the centroid of the displaced fluid. For small tilts, a floating body is stable if the metacenter (the point where the line of buoyant force meets the axis after tilting) is above the center of gravity. Ships, icebergs and hot-air balloons illustrate equilibrium in different fluids.

A ship floats upright and returns to the vertical after rolling, while a badly loaded boat may capsize. This guide explains the conditions for floating equilibrium and stability, treats ships, icebergs and balloons, compares floating objects in a table, works through numerical examples and corrects common mistakes. The basic principle of Archimedes was introduced in an earlier article.

What is the condition for floating?

The buoyant force on a floating body, F_b = ρ_fV_subg, balances its weight mg. Hence ρ_fV_sub = m: the mass of the displaced fluid equals the mass of the body, and the fraction submerged is V_sub ÷ V = ρ_obj ÷ ρ_f. If the average density of the object is greater than that of the fluid, it sinks; if the densities are equal, it has neutral buoyancy and stays at any depth (submarines, fish with swim bladders). For a ship, the “average” density includes the air in the hull. The weight of fluid displaced by a floating ship equals its total weight (displacement tonnage), including cargo.

When is the floating stable?

The weight acts at the center of gravity G of the body, and the buoyant force at the center of buoyancy B, the centroid of the displaced volume. In equilibrium, both are on the same vertical line. If the body tilts, B moves toward the lower side, because the shape of the submerged part changes. For a small tilt, the vertical line through the new B intersects the axis of the body at the metacenter M. If M is above G, the weight and the buoyant force form a couple that returns the body to the upright (stable); if M is below G, the couple increases the tilt (unstable) and the body capsizes. The distance GM is the metacentric height. Ships are designed with ballast low in the hull, to lower G, and a wide hull, to raise M. Loading cargo high on a ship raises G and reduces the stability. A submerged body (a submarine) is stable only if G is below B. Another danger is the free surface effect: liquid sloshing in partly filled tanks shifts the center of gravity and can capsize a ship.

What are some applications?

• Ships and the Plimsoll line: lines on the hull show the safe draft in different waters (fresh, salt, tropical, winter).
• Icebergs: ice (917 kg/m³) in sea water (1,025 kg/m³) floats with about 89 percent of its volume below the surface.
• Hydrometers float lower in less dense liquids, and give the density from the depth reached.
• Balloons and airships float in the air when the weight of the displaced air equals the total weight; hot air at 100 °C has a density of 0.946 kg/m³, compared with 1.204 kg/m³ for air at 20 °C.
• Submarines change their buoyancy by pumping water in and out of ballast tanks.

Floating objects compared

Worked example: block, iceberg, ship and balloon

1. Wooden cube. A cube of side 0.10 m and density 600 kg/m³ floats in fresh water. Mass = 600 × 0.0010 = 0.60 kg. The submerged volume = m ÷ ρ_f = 0.60 ÷ 1,000 = 6.0 × 10⁻⁴ m³, so the cube sinks to a depth of 6.0 cm (fraction 0.60).

2. Iceberg. The fraction below water = 917 ÷ 1,025 = 0.895. For an iceberg of total volume 1.0 × 10⁶ m³, the volume above the surface is 0.105 × 10⁶ = 1.05 × 10⁵ m³.
3. Ship loading. A ship has a waterplane area of 1,000 m² (assumed constant near the waterline). Loading 500 tonnes (5.0 × 10⁵ kg) in sea water: the extra displaced volume = 5.0 × 10⁵ ÷ 1,025 = 488 m³. The ship sinks by Δh = 488 ÷ 1,000 = 0.49 m. Moving from sea water into fresh water, the draft increases by about 2.5 percent, because the density is lower (1,000 versus 1,025).
4. Hot-air balloon. A balloon with an envelope volume of 2,800 m³ filled with air at 100 °C, in air at 20 °C. Lift = (ρ_out − ρ_in) V = (1.204 − 0.946) × 2,800 = 723 kg. This supports the envelope, basket, burner, fuel and passengers.
5. Stability. A rectangular barge 20 m long, 8.0 m wide and 2.0 m deep floats with a draft of 1.0 m. The center of buoyancy is at 0.50 m above the bottom. The metacentric radius for rolling is BM = I ÷ V_sub = (20 × 8.0³ ÷ 12) ÷ (20 × 8.0 × 1.0) = 853 ÷ 160 = 5.33 m, so the metacenter is 0.50 + 5.33 = 5.83 m above the bottom. If the center of gravity of the loaded barge is at 1.0 m above the bottom (G below M), the barge is stable, with GM = 4.8 m.

How was buoyancy applied?

Archimedes’ treatise On Floating Bodies (about 250 BCE) laid the foundation. Pierre Bouguer introduced the metacenter in 1746, which allowed the stability of ships to be calculated. Samuel Plimsoll campaigned in the 1870s for the load line on ships to prevent overloading, which became law in Britain in 1876. The Montgolfier brothers flew the first hot-air balloon in 1783.

Which mistakes are common?

Mistake 1: Saying that a floating object displaces its volume of fluid

It displaces its weight of fluid, which corresponds to the submerged volume only.

Mistake 2: Thinking that floating requires a hollow hull

Any object with an average density lower than the fluid floats. Hollow hulls reduce the average density of steel ships.

Mistake 3: Ignoring stability

An object can float but be unstable and roll over, if the center of gravity is too high.

Mistake 4: Using the same draft for all waters

A ship floats lower in fresh water than in sea water.