Buoyancy and stability of floating bodies

Why it matters

Buoyancy decides how deep a hull, pontoon or float sits, and the float in a carburettor bowl or fuel-tank sender works on exactly this principle. Stability analysis tells you whether a floating body returns upright after a disturbance or capsizes, which is why loading cargo high on a ship or barge is dangerous. GATE problems on metacentric height are short, formula-driven and easy marks once the geometry is clear.

Key ideas

• Archimedes' principle. A body wholly or partly immersed in a fluid experiences an upward force equal to the weight of fluid it displaces. The force acts through the centre of buoyancy B, the centroid of the displaced volume, not the centroid of the body.
• Floating condition. In equilibrium the buoyant force equals the body's weight, so the displaced volume is V_d = m/ρ_fluid. A body whose average density is less than the fluid floats with a fraction ρ_body/ρ_fluid of its volume submerged (for a uniform body).
• Submerged bodies (submarine, balloon): B is fixed relative to the body because the displaced shape does not change when tilted. They are stable only if B lies above the centre of gravity G, neutral if B and G coincide, unstable if B is below G.
• Floating bodies. When a floating body heels through a small angle, the shape of the displaced volume changes and B moves sideways towards the deeper side. The vertical through the new B cuts the body's original axis of symmetry at the metacentre M.
  • Stable: M above G (GM > 0) — the buoyancy and weight form a restoring couple.
  • Neutral: M coincides with G (GM = 0).
  • Unstable: M below G (GM < 0) — the couple overturns the body.
  • A floating body can be stable even with G above B, as long as M is above G. This is why ships with a high G can still be stable.
• Metacentric radius BM = I/V_d, where I is the second moment of area of the waterline plane about the tilt axis. A wide waterline raises M strongly (BM grows with width cubed), so beamy hulls resist rolling.
• Choice of axis. For a rectangular barge, rolling about its long axis uses I = L·B³/12 (B is the beam). This gives the smaller BM and governs stability.
• Large GM vs small GM. Large GM makes a stiff vessel that rolls quickly and uncomfortably; small GM gives a slow, gentle roll but less reserve. Passenger ships aim for a moderate GM.
• Experimental GM. Shift a known weight w across the deck by x and measure the heel angle θ.
• The analysis holds for small angles (roughly up to 10°); large-angle stability needs the righting-arm (GZ) curve.

Formulas

F_B = ρ_f·g·V_d F_B buoyant force (N), ρ_f fluid density (kg/m³), V_d displaced volume (m³), g = 9.81 m/s².

BM = I / V_d BM metacentric radius (m), I second moment of the waterline area about the tilt axis (m⁴). Rectangle of length L along the axis and breadth b: I = L·b³/12. Circle of diameter D: I = π·D⁴/64.

GM = BM − BG = KB + BM − KG GM metacentric height (m); K keel or base; KB, KG heights of B and G above the base (m). If G lies below B, BG is negative and GM = BM + |BG|. Small-angle stability: GM > 0 stable.

GM = w·x / (W·tan θ) Experimental: w moved weight (N), x distance moved (m), W total weight including w (N), θ heel angle.

T = 2π·k / √(g·GM) Period of free roll (s); k radius of gyration about the roll axis (m). Small oscillations.

Worked examples

Example 1 (standard) — floating timber block. Given: a block 2 m long, 1 m wide and 0.8 m high, relative density 0.6, floats in water with its 0.8 m side vertical. Check its stability for rolling about the long axis.

1. Depth of immersion: weight = buoyancy → 0.6 × 0.8 = 0.48 m.
2. Displaced volume: V_d = 2 × 1 × 0.48 = 0.96 m³.
3. Heights above base: KB = 0.48/2 = 0.24 m; KG = 0.8/2 = 0.40 m.
4. Waterline I about the long axis: I = L·b³/12 = 2 × 1³/12 = 0.1667 m⁴.
5. BM = I/V_d = 0.1667/0.96 = 0.1736 m.
6. GM = KB + BM − KG = 0.24 + 0.1736 − 0.40 = 0.0136 m. Answer: GM ≈ 0.014 m > 0, so the block is stable, but only just.

Example 2 (GATE level) — limiting height of a floating cylinder. Given: a solid cylinder of diameter D and height H, relative density S = 0.8, floats in water with its axis vertical. Find the largest H/D for stable equilibrium.

1. Immersion depth = S·H = 0.8H; KB = 0.4H; KG = 0.5H; BG = 0.1H.
2. Waterline is a circle: I = π·D⁴/64. V_d = (π·D²/4) × 0.8H.
3. BM = I/V_d = D²/(16 × 0.8H) = D²/(12.8H).
4. For stability, GM = BM − BG ≥ 0 → D²/(12.8H) ≥ 0.1H → H²/D² ≤ 1/1.28.
5. H/D ≤ 1/√1.28 = 0.884. Answer: H/D must not exceed about 0.884. In general H/D ≤ 1/√(8·S·(1 − S)).

Example 3 (inclining test and roll period). A ship weighing 20 MN heels 1.5° when a 50 kN load is moved 6 m across the deck. GM = 50 000 × 6/(20 × 10⁶ × tan 1.5°) = 300 000/523 690 = 0.573 m. A vessel with k = 4 m and GM = 0.8 m rolls with T = 2π × 4/√(9.81 × 0.8) = 8.97 s.

Common mistakes

• Taking B as the centroid of the whole body instead of the centroid of the displaced volume.
• Using I about the wrong axis — for a long barge use the smaller I (rolling about the long axis), I = L·b³/12.
• Using the body volume instead of the displaced volume in BM = I/V_d.
• Assuming G must be below B for stability; that is the condition for fully submerged bodies only.
• Forgetting that a lower centre of gravity (ballast) raises GM, and loads high on deck lower it.
• Using a fluid density of 1000 kg/m³ when the problem says sea water (about 1025 kg/m³).

For GATE ME

Expect: depth of immersion of a floating block or cylinder, metacentric height of a block or barge, limiting dimensions for stability, the stability of a submerged body, and inclining-experiment calculations. Conceptual MCQs test the conditions for stable, neutral and unstable equilibrium for floating versus submerged bodies. Practise drawing K, B, G and M on a sketch before writing any equation.

Quick check

1. A cube of side 1 m floats half-submerged in fresh water. What is the buoyant force?
2. What is the stability condition for a fully submerged body?
3. Which second moment of area is used in BM = I/V_d?
4. Does a larger GM give a shorter or longer rolling period?
5. A barge has KB = 1 m, BM = 1.04 m and KG = 1.5 m. Find GM. Answers: 1. 1000 × 9.81 × 0.5 = 4905 N. 2. Centre of buoyancy above the centre of gravity. 3. That of the waterline plane about the tilt axis. 4. Shorter. 5. 1 + 1.04 − 1.5 = 0.54 m.