---
title: "Buoyancy and stability of floating bodies"
subject: "Fluid Mechanics"
chapter: 2
key: "semester_4/fluid_mechanics_mechanical:buoyancy-metacentric-height-stability"
source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/buoyancy-metacentric-height-stability
site: BCT Engineering
---

# Buoyancy and stability of floating bodies

Fluid Mechanics · Chapter 2: Fluid Statics

- Asked in: year not given, 8 marks
- Video: [Fluid Mechanics Lesson 03D: Buoyancy and Stability](https://www.youtube.com/watch?v=AG4Au3ldb54) — John Cimbala
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/buoyancy-metacentric-height-stability
- Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-2

## Question

State Archimedes' principle. Define centre of buoyancy, metacentre and metacentric height. Derive an expression for the metacentric height of a floating body, and state the conditions of stable, neutral and unstable equilibrium for floating and for submerged bodies.

## Short answer

Archimedes’ principle: buoyant force = weight of displaced fluid. Centre of buoyancy B is centroid of displaced volume; metacentre M is intersection of buoyancy line with original vertical for small heel; GM = I/V – BG. Floating stability: stable if GM>0 (M above G); submerged stability: stable if B above G.

## Answer

#### Archimedes' principle and definitions
**Archimedes' principle:** a body immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced. $F_B = \rho g V_{disp}$.

- **Centre of buoyancy (B):** centroid of the displaced volume of fluid; the line of action of $F_B$ passes through it.
- **Metacentre (M):** the point where the line of action of buoyancy, for a small angle of heel, cuts the original vertical centre line of the body.
- **Metacentric height (GM):** distance between the centre of gravity G and the metacentre M.

#### Expression for metacentric height
When a floating body heels through a small angle $\theta$, the shape of the displaced volume changes: a wedge emerges on one side and an equal wedge is immersed on the other. B shifts to a new position $B'$.

```text
        M
        |\
        | \
   G ---|--\---
        | B  B'
   ~~~~~~~~~~~~~~~~ water line
```

The horizontal shift of the centre of buoyancy is found from the moment of the wedge buoyancy about the centre line. For small $\theta$, the moment of the wedge volumes equals $I\theta$, where $I$ is the second moment of area of the waterline plane about the longitudinal axis of tilt. Then

$$
V \cdot BB' = I\theta \quad \text{and} \quad BB' = BM \cdot \theta
$$

so

$$
BM = \frac{I}{V}
$$

where $V$ is the volume of fluid displaced. Hence

$$
GM = BM - BG = \frac{I}{V} - BG
$$

$BG$ is positive when G is above B, and negative when G is below B. The metacentric height is taken as positive when M is above G.

#### Conditions of equilibrium
| Condition | Floating body | Submerged body |
|---|---|---|
| Stable | M above G ($GM > 0$) | B above G |
| Neutral | M coincides with G ($GM = 0$) | B coincides with G |
| Unstable | M below G ($GM < 0$) | B below G |

For a stable floating body, the weight and buoyancy form a **righting couple** $W \cdot GM \sin\theta$ that restores the body to its original position. For an unstable body the couple is an overturning couple.

For a submerged body, the shape of the displaced volume does not change when the body tilts, so B stays fixed relative to the body and there is no metacentre in the same sense; stability needs B above G.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The centre of buoyancy is:
   A. the centre of gravity of the body
   B. the centroid of the displaced volume of fluid
   C. the point where the weight acts
   D. the metacentre of the body
   Answer: B. The buoyant force acts through the centroid of the displaced fluid volume. (Key point: Archimedes' principle and centre of buoyancy)

2. The derivation gives BM = I/V. What do I and V represent?
   A. I = second moment of area of the waterline plane about the tilt axis; V = volume of fluid displaced
   B. I = moment of inertia of the whole body; V = total volume of body
   C. I = area of the waterline; V = weight
   D. I = draught; V = velocity
   Answer: A. The wedge moment equals I theta, so V x BB' = I theta, giving BM = I/V with GM = BM - BG. (Key point: Metacentric height expression)

3. A floating body is in stable equilibrium when:
   A. B is above G
   B. M is below G
   C. M is above G (GM > 0)
   D. G coincides with M
   Answer: C. With GM > 0, the weight and buoyancy form a righting couple W GM sin(theta) that restores the body. (Key point: Stability of floating body)

4. A fully submerged body is stable when:
   A. B is above G
   B. B is below G
   C. M is above G
   D. B coincides with G
   Answer: A. For a submerged body the displaced shape does not change on tilting, so stability needs B above G; B coinciding with G is neutral. (Key point: Stability of submerged body)

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