Buoyancy and stability of floating bodies

Why it matters

Buoyancy decides whether a float valve, hydrometer, pontoon, buoy or ship floats and at what draft; stability decides whether it stays upright when a wave, wind gust or shifting load tilts it. The same metacentric-height check is used for barges carrying machinery, floating platforms and liquid-level sensors that use floats.

Key ideas

Buoyant force. Pressure on the bottom of a submerged body is larger than on its top, so the net pressure force points upward. Integrating gives Archimedes' principle: the buoyant force equals the weight of the fluid displaced, and acts vertically upward through the centre of buoyancy B, the centroid of the displaced volume.

• A fully submerged body displaces its whole volume; a floating body displaces just enough fluid to equal its own weight.
• A body floats if its average density is less than the fluid's; for a floating body, the fraction submerged equals ρ_body/ρ_fluid.
• The apparent weight of a submerged body is its true weight minus the buoyant force.

Equilibrium types. A body is in equilibrium when its weight W (acting at the centre of gravity G) and buoyancy F_B (acting at B) are equal, opposite and on the same vertical line. After a small tilt:

• Stable: the couple formed by W and F_B rotates the body back.
• Unstable: the couple tilts it further.
• Neutral: no couple; it stays in the new position.

Fully submerged bodies (submarines, balloons). B is fixed relative to the body because the displaced shape does not change with tilt. Such a body is stable only if B lies above G.

Floating bodies and the metacentre. When a floating body heels through a small angle, the immersed shape changes: a wedge emerges on one side and an equal wedge is immersed on the other, so B moves sideways to a new position B′. The vertical through B′ meets the original centre line at the metacentre M. For small angles M is fixed, and:

• Metacentric height GM = distance from G to M.
• GM > 0 (M above G): stable. GM < 0: unstable. GM = 0: neutral.
• A floating body can be stable even when G is above B, because B shifts toward the immersed side. This is why ships can carry heavy machinery above the waterline.

What sets BM. BM = I/V, where I is the second moment of the waterplane area about the axis of tilt and V the displaced volume. A wide waterplane (beam) raises M strongly, since I varies with the cube of the width. Always check tilting about the axis with the smallest I (for a long body, rolling about its length), because that gives the smallest GM.

Practical trade-off. A large GM gives a stiff body that rights quickly but rolls with a short, jerky period, uncomfortable for passengers and hard on cargo; a small GM gives a gentle roll but little reserve against capsizing. Ships are designed for moderate GM. Loading high on deck, free liquid surfaces in partly filled tanks and ice on superstructure all reduce GM.

Measuring GM — inclining experiment. Move a known weight w across the deck by a distance x and measure the steady heel angle θ. The overturning moment w·x is balanced by the righting moment W·GM·tanθ.

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.

V_d / V = ρ_body / ρ_f (floating body)

• Fraction of volume submerged (–).

BM = I / V_d

• BM = metacentric radius (m), I = second moment of the waterplane area about the tilt axis (m⁴), V_d = displaced volume (m³). Small angles of heel. Rectangle of length L and width b tilting about the length: I = L·b³/12. Circle of diameter D: I = π·D⁴/64.

GM = BM − BG = KB + BM − KG

• K = keel (base) point; KB, KG = heights of B and G above the base (m). If G is below B, BG is negative and GM = BM + |BG|.

Righting moment = W · GM · sinθ

• W = weight of the body (N), θ = angle of heel. Small angles.

GM = w · x / (W · tanθ) (inclining experiment)

• w = shifted weight (N), x = distance moved (m), W = total weight including w (N).

T = 2π · k / √(g · GM)

• T = period of free rolling (s), k = radius of gyration about the roll axis (m).

Worked examples

Example 1 — floating block (standard). A wooden block 2 m long, 1 m wide and 0.5 m deep (relative density 0.6, uniform) floats in fresh water with its 0.5 m side vertical. Find the draft and the metacentric height for rolling about the long axis.

1. Draft: d = (ρ_body/ρ_f)·H = 0.6 × 0.5 = 0.30 m.
2. Displaced volume: V_d = 2 × 1 × 0.3 = 0.6 m³.
3. KB = d/2 = 0.15 m; KG = H/2 = 0.25 m (uniform block).
4. I about the long axis: I = L·b³/12 = 2 × 1³/12 = 0.1667 m⁴.
5. BM = I/V_d = 0.1667/0.6 = 0.2778 m.
6. GM = KB + BM − KG = 0.15 + 0.2778 − 0.25 = 0.1778 m. Answer: draft 0.30 m; GM ≈ 0.178 m, so it is stable. (About the short axis I = 1 × 2³/12 = 0.667 m⁴ and GM ≈ 1.01 m, which is far more stable.)

Example 2 — upright cylinder (GATE level). A solid uniform cylinder of height H = 1 m and relative density s = 0.8 floats in water with its axis vertical. Find the minimum diameter for it to float stably in that position.

1. Draft = sH, so KB = sH/2 and KG = H/2.
2. BM = I/V_d = (π·D⁴/64)/(π·D²·sH/4) = D²/(16·s·H).
3. Stability needs GM > 0: D²/(16sH) + sH/2 − H/2 > 0, which gives D² > 8·s·(1 − s)·H².
4. D_min = H·√(8 × 0.8 × 0.2) = 1 × √1.28 = 1.131 m. Answer: D must exceed about 1.13 m. A tall thin cylinder of the same material would roll over and float on its side.

Example 3 — inclining experiment. A ship of total weight 50 MN heels by 2° when a 300 kN load is moved 8 m across the deck. Find GM, and the roll period if the radius of gyration is 6 m.

1. GM = w·x/(W·tanθ) = 300 × 10³ × 8 / (50 × 10⁶ × tan 2°) = 2.4 × 10⁶/(1.746 × 10⁶) = 1.375 m.
2. T = 2π·k/√(g·GM) = 2π × 6/√(9.81 × 1.375) = 37.70/3.672 = 10.3 s. Answer: GM ≈ 1.37 m; roll period ≈ 10.3 s.

Common mistakes

• Using I about the wrong axis. Stability is governed by the smaller second moment of the waterplane, usually about the longitudinal axis.
• Using the total volume of the body instead of the displaced volume in BM = I/V_d.
• Applying the floating-body rule (M above G) to a fully submerged body; there B must lie above G.
• Putting G at mid-height for a body that is not uniform, or ignoring added loads when locating G.
• Taking I of the whole body cross-section instead of the waterplane area (the cut at the free surface).
• Forgetting that BG is negative when G lies below B, which changes the sign in GM = BM − BG.
• Using the formulas for large heel angles; BM = I/V_d is valid only for small angles.

For GATE ME

Expect numericals on the draft of a floating block or cylinder, the fraction submerged, tension in a cable anchoring a submerged buoy, the apparent weight of a body in water, and the metacentric height of rectangular pontoons or cylinders. Conceptual MCQs test the stability conditions for floating versus submerged bodies, the meaning of M, and the effect of beam on stability. Practise deriving the D/H condition for a floating cylinder and setting up KB + BM − KG cleanly.

Quick check

1. What fraction of an ice block (relative density 0.92) floats below the water surface?
2. Where does the buoyant force act?
3. A submarine is fully submerged. What is the condition for stable equilibrium?
4. Why does widening a pontoon increase its stability?
5. A floating body has KB = 0.4 m, BM = 0.5 m and KG = 1.0 m. Is it stable?

Answers: 1. 0.92 (92 %); 2. at the centroid of the displaced volume (centre of buoyancy), vertically upward; 3. the centre of buoyancy must lie above the centre of gravity; 4. BM = I/V and I grows with the cube of the width, so M rises; 5. no — GM = 0.4 + 0.5 − 1.0 = −0.1 m, so it is unstable.