---
title: "Metacentre and laboratory determination of height"
subject: "Fluid Mechanics"
chapter: 3
key: "semester_3/fluid_mechanics:metacentre-importance-lab-determination"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/metacentre-importance-lab-determination
site: BCT Engineering
---

# Metacentre and laboratory determination of height

Fluid Mechanics · Chapter 3: Hydrostatics

- Asked in: 2073 Shrawan, 7 marks; 2072 Chaitra, 6 marks (asked 2 times)
- Video: [Fluid Mechanics Lab - Metacentric Height](https://www.youtube.com/watch?v=PXaEm5EyNhw) — KLS GIT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/metacentre-importance-lab-determination
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-3

## Question

Explain the metacentre with appropriate diagram. How do you determine the metacentric height of a rectangular vessel in laboratory (steps)?

## Short answer

The metacentre M is where the buoyancy line after a small heel meets the original vertical axis; metacentric height GM determines stability (positive GM = stable). In the lab, float a pontoon, shift a weight w by distance x, measure heel angle θ, and compute GM = w x / (W tan θ), repeating for several x.

## Answer

#### Metacentre
When a floating body is tilted by a small angle, the centre of buoyancy $B$ shifts to a new position $B'$ because the shape of the displaced volume changes. The vertical line through $B'$ meets the original vertical axis of the body at a point $M$, the **metacentre**. The distance $GM$ is the **metacentric height**.

```text
        upright               heeled
         |  M                  M
         |                    /|
         G                   / G
         B                  B'  W
   weight W down, buoyancy W up through B'
```

- If $M$ is above $G$ ($GM>0$): a righting couple $W\cdot GM\sin\theta$ acts, so the body is **stable**.
- If $M$ is below $G$: an overturning couple acts, so the body is **unstable**.
- If $M$ coincides with $G$: **neutral** equilibrium.

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

where $I$ is the second moment of the waterline area about the axis of tilt and $V$ is the displaced volume. A larger $GM$ gives greater stability but a shorter rolling period.

#### Laboratory determination of $GM$ of a rectangular vessel (pontoon)
**Apparatus:** a rectangular pontoon, a water tank, a movable weight $w$ on a cross-bar, a plumb line and a graduated scale to read the tilt angle.

**Steps:**
1. Weigh the pontoon with its fittings; its total weight is $W$.
2. Float it in the water in the tank and check that it floats level, with the plumb line at zero on the scale.
3. Place the movable weight $w$ at the centre of the cross-bar, then shift it a measured distance $x$ to one side.
4. The pontoon heels; read the angle $\theta$ from the plumb line on the scale (or find $\tan\theta$ from the deflection).
5. Repeat for several positions $x$ on both sides, and for different values of $w$.
6. Compute $GM$ for each reading, then take the average:

$$
w\,x=W\,GM\tan\theta\ \Rightarrow\ GM=\frac{w\,x}{W\tan\theta}
$$

7. As a check, calculate the theoretical value $GM=\dfrac{I}{V}-BG$ with $I=\dfrac{LB^{3}}{12}$ for the rectangular waterline, and compare it with the experimental value.

## Check yourself

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

1. What is the metacentre M?
   A. The point where the vertical through the new centre of buoyancy cuts the original vertical axis
   B. The centre of gravity of the body
   C. The new centre of buoyancy after tilting
   D. The centre of the waterline area
   Answer: A. B shifts to B' on tilting; the vertical through B' meets the original axis at M. GM is the metacentric height. (Key point: Definition of metacentre)

2. What happens if M is above G?
   A. An overturning couple acts
   B. A righting couple W GM sinθ acts and the body is stable
   C. Neutral equilibrium
   D. The body capsizes
   Answer: B. M above G (GM > 0) gives a restoring couple; M below G is unstable; M at G is neutral. (Key point: Stability condition)

3. In the pontoon experiment, a weight w is moved a distance x and the heel angle θ is read. How is GM found?
   A. GM = W x / (w tanθ)
   B. GM = w tanθ / (W x)
   C. GM = w x / (W tanθ)
   D. GM = W / (w x tanθ)
   Answer: C. Heeling moment w x equals the righting moment W GM tanθ, where W is the total weight. (Key point: Lab formula)

4. Why are readings repeated for several x on both sides and different w?
   A. To compute GM for each reading and average, then compare with the theoretical I/V - BG
   B. To increase the pontoon weight
   C. To change the position of the metacentre deliberately
   D. To avoid using the plumb line
   Answer: A. Averaging reduces error; the theoretical value uses I = L B^3 / 12 for the rectangular waterline. (Key point: Experimental procedure)

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