Topic 2 of 10 — free theory
Buoyancy & Flotation
Archimedes' principle, floating bodies, and stability of submerged and floating objects.
Take the free 5-question mini-quiz ↓
Archimedes' principle
A body immersed in a fluid feels an upward buoyant force equal to the weight of the fluid it displaces. That is the whole principle — everything else is bookkeeping about which volume counts:
Fb = γfluid · Vdisplaced
| Fb | buoyant force, acting vertically upward through the centroid of the displaced volume |
| Vdisplaced | volume of fluid displaced = submerged volume of the body only |
Floating equilibrium: W = Fb
| W | total weight of the floating body |
For a floating body, the displaced volume is whatever sits below the waterline — not the whole body. For a fully submerged body, it is the entire volume.
Stability — will it stay upright?
Floating is not enough; the body must float stably. A small tilt shifts the centre of buoyancy; the body is stable if the resulting moment rights it. The test is the metacentric height:
GM = MB − GB stable if GM > 0
| G | centre of gravity of the body |
| B | centre of buoyancy (centroid of submerged volume) |
| M | metacentre — where the tilted buoyant-force line crosses the body's centreline |
| MB = I0/Vsub | distance from B to M; I0 is the second moment of the waterplane area about its centroidal axis |
In words: the metacentre M must sit above the centre of gravity G. Broad, shallow hulls are stable; tall, narrow ones need checking.
Advanced extension: initial stability
For a small heel angle θ, write the metacentric height from a common vertical datum:
GM = KB + BM − KG, BM = Iwaterplane/Vdisplaced
The small-angle righting moment is approximately W·GM·sinθ. Positive GM produces a restoring moment; zero is neutral; negative is unstable.
Geometry trap: Iwaterplane is the second moment of the area cut by the water surface, about the heel axis. It is not the solid body's mass moment of inertia.
Worked example Draft of a loaded barge
Given:
- Rectangular barge: 20 ft long × 10 ft wide × 6 ft deep.
- Total weight (barge + cargo): 60 kips.
- Floating in fresh water, γ = 62.4 lb/ft³.
Solution:
- Equilibrium: W = γ · Vsub. The submerged volume is L × B × d, where d is the draft.
- 60,000 = 62.4 × 20 × 10 × d → d = 60,000 / 12,480 = 4.81 ft.
- Freeboard (dry depth above water) = 6.00 − 4.81 = 1.19 ft.
Answer: Draft ≈ 4.81 ft; freeboard ≈ 1.19 ft.