Buoyancy and Floating Stability
Whether it floats is a density question. Whether it stays upright is a geometry question, and a body can pass one and fail the other.
Skip to the animationBuoyancy is not a separate force but the unbalanced result of pressure growing with depth, and it equals the weight of displaced fluid — but whether a floating body *stays upright* is a second and entirely different question, settled by the metacentre rather than by density.
Where the upward force comes from
There is no law of buoyancy separate from hydrostatics. The bottom face of a submerged block is deeper than its top, so it is pushed harder — upward. The side faces cancel in pairs. What remains is a net upward force.
- 1The pressure difference between top and bottom is Δp = ρg × (block height).
- 2Multiply by the face area: F = Δp·A = ρg × height × A.
- 3Height × A is the block's volume, so F = ρgV.
- 4ρgV is exactly the weight of the fluid that volume displaced.
Nothing about the body's material appears anywhere in that derivation. Only the volume it occupies. Any shape works, because any shape is a stack of thin blocks and the argument survives the sum.
Whether it floats
A floating body displaces exactly its own weight of fluid, settling at whatever draught makes those equal. Whether that is achievable at all depends on one comparison.
| Condition | Result | Example |
|---|---|---|
| ρ_body < ρ_fluid | Floats, partly out of the water | Timber, a ship's hull with its air |
| ρ_body = ρ_fluid | Neutrally buoyant, hovers anywhere | A trimmed submarine, a fish with its bladder set |
| ρ_body > ρ_fluid | Sinks | A solid steel bar |
"Steel is denser than water" is true and irrelevant. A ship's average density — hull plus enclosed air — is what counts. Hole the hull, the air is replaced by water, the average crosses the threshold, and the outcome reverses.
Whether it stays upright
Floating and being stable are separate questions, and a body can pass the first while failing the second. Two points decide it.
- Centre of gravity G
- Where the body's own weight acts. Fixed within the body, and it moves only if the loading changes.
- Centre of buoyancy B
- The centroid of the *submerged volume*. It moves when the body heels, because the underwater shape changes.
- Metacentre M
- Where the vertical through the shifted B crosses the body's original centreline.
- Metacentric height GM
- The distance from G to M. Positive means stable; it is the number naval architects quote.
- 1Upright, G and B lie on the same vertical line and there is no moment.
- 2Heel the body: the submerged shape changes, so B migrates toward the immersed side.
- 3G stays where it was, so weight and buoyancy now form a couple.
- 4If M is above G, the couple is restoring and the body rights itself.
- 5If M is below G, the same couple is overturning and the heel accelerates.
Why more stability is not simply better
| Large GM (stiff) | Small positive GM (tender) | |
|---|---|---|
| Righting action | Fast and forceful | Slow and gentle |
| Roll period | Short — snaps back and rolls again | Long and comfortable |
| Risk | Cargo shifts, crew injuries, structural fatigue | Little reserve if it heels far |
| Typical of | A wide, shallow barge | A tall, narrow container ship high on fuel |
GM also changes during a voyage. Burning fuel from low tanks raises G and reduces stability, which is why ballast is actively pumped rather than set once at departure.
The large-angle limit, and free surface
GM is only the initial slope of the righting-lever curve and describes the first few degrees. Further over, the lever GZ rises, peaks, and falls back to zero at the *angle of vanishing stability* — beyond which the couple capsizes rather than rights.
Loose water in a part-full tank runs to the low side as the vessel heels, effectively raising G and shrinking the whole curve. This free-surface effect is why tanks are subdivided by longitudinal bulkheads, and why a part-full tank is more dangerous than a full one.
The numbers you will be asked for
- Archimedes' principle
F_B = ρ_fluid · g · V_displaced
- Flotation condition
ρ_fluid · V_displaced = m_body
- Metacentric height
GM = I / V_displaced − BG
I is the waterplane's second moment of area
- Righting moment
M = W · GM · sinθ
small angles only
- Roll period
T ≈ 2π·k / √(g·GM)
large GM gives a short, violent period
Watch it work
Check yourself
question 1 / 4
One question at a time. Pick an answer to see why it is right or wrong, then move on — there is no score to keep and nothing is saved.