## What is Metacentre & Metacentric height?

*Metacentre is the point at which the line of action of the upthrust will meet the normal axis of a floating body, when the body is tilted by a small angle from its equilibrium position. The distance between the metacentre point ( M ) and centre of gravity ( G ) , is called metacentric height.*

Knowing of the metacentric height of any floating body is very important, because –

- Metacentric height ascertain the stability of a floating body like ships, buoys etc.
- It is essential during designing of floating vessels.

### Metacentre

Consider about a body floating in a liquid as shown in figure.

Let, ( G ) is the *centre of gravity* of the body and ( B ) is the *centre of buoyancy*.

For *equilibrium* of the body, ( B ) and ( G ) must lie on same vertical line called normal axis.

Suppose, the body is given a tilt by a small angle as shown in figure (B). This tilting will produce following effects –

- Because, the shape of displaced fluid changes, the
*centre of buoyancy*will also shift to a new position at ( B_1 ) - The line of action of upthrust i.e.
*buoyant force*is now vertical through the point ( B_1 ) - The vertical through the point ( B_1 ) will intersect the normal axis of the body at a point ( M ) .

This point ( M ) is called the ** metacentre**. It is the point which decides about the stability of any floating body. The height between the metacentre point ( M ) and centre of gravity ( G ) of the body is called the

**metacentric height.**## Equilibrium of floating bodies

There are two conditions for equilibrium of a floating body –

- Weight of the liquid displaced must be equal to the weight of the body.
- The centre of gravity of the body and the centre of buoyancy must lie on the same vertical line.

### Stability of floating bodies

Stability of floating body depends upon –

- Weight of the body.
- Weight of the displaced liquid by the body.
- Position of metacentre and metacentric height.

Consider about a body floating in a liquid as shown in figure.

Let, the body is slightly tilted from its *equilibrium* position. Then, centre of body and the centre of buoyancy do not lie on the same vertical line and the weight ( W ) of the body and up thrust ( U ) will form a *couple* having rotating tendency. Three conditions are possible as follows –

#### Stable equilibrium

Condition for stable equilibrium is shown in figure ( B ). Here, Metacentre point ( M ) lies above the point of *centre of gravity* ( G ) .

- On giving a small tilt to the floating body, the
*weight*of the body ( W ) and the up thrust ( U ) will form a*Righting couple.* - This righting
*couple*or restoring couple will bring the body back to its equilibrium position. - So, the body is said to be in
*stable equilibrium.*

#### Unstable equilibrium

Condition for unstable equilibrium is shown in figure ( C ). Here, Metacentre point ( M ) lies below the point of *centre of gravity* ( G ) .

- On giving a small tilt to the floating body, the weight of the body ( W ) and the up thrust ( U ) will make the body tilt further and the body will not be restored to its earlier position.
- So, the body is said to be in
*Unstable equilibrium.*

#### Neutral equilibrium

In this case, Metacentre point ( M ) lies just at the point of *centre of gravity* ( G ) . When the metacentre and the centre of gravity coincides, the body will remain in equilibrium in any position in which it is floating. This happens with spherical shaped bodies.

- If the body is tilted, it will remain in equilibrium in the new position.
- In the new position also the weight of the body ( W ) and the up thrust ( U ) will remain in the same position as before.
- So, a body floating in this way is said to be in
*Neutral equilibrium.*

## Methods to find Metacentric height

Metacentric height of a floating body can be determined by two ways –

- By analytical method.
- By experimental method.

### To find Metacentric height by analytics

The position of metacentre, ( M ) with respect to the position of the *centre of buoyancy* ( B ) can be determined by analytical method as follows.

Consider about a cubical body floating as shown in figure ( A ).

Let, the body is tilted by a small angle ( \theta ) in a clockwise direction as shown in figure (B).

The immersed section has now changed from ( pqrs ) \ \text {to} \ ( p_1q_1rs ) . Consequently, the centre of buoyancy ( B ) has shifted to ( B_1 ) .

The effect of tilting is to move an immersed wedge of water ( pop_1 ) to the position ( qoq_1 ) . This apparent movement of the water wedge has resulted in shifting the centre of buoyancy from ( B ) \ to \ ( B_1 ) .

Since, the volume of water displaced by the floating body has not changed, we have –

\text {Area} \ ( pop_1 ) = \text {Area} \ ( qoq_1 )

Now let, the length of the body is ( l ) . Consider a thin transverse section of length ( dl ) as shown in figure ( C ).

Let –

- ( b ) is breadth of the floating body.
- ( V ) is volume of water displaced by the immersed part of body also called
*wetted volume***.** - ( dV ) is volume of water displaced by the slice, i.e., volume of water displaced by dl length of the body.
- ( I ) is
*moment of inertia*of a horizontal section of the body at water surface about the longitudinal axis. - ( dI ) is moment of inertia of the slice section considered about the longitudinal axis.
- ( g ) is
*centre of gravity*of the triangular prism ( pop_1 ) . - ( g_1 ) is centre of gravity of the triangular prism ( qoq_1 ) .
- ( w ) is
*specific weight*of water.

Then, by geometry of figure –

( g g_1 ) = \left ( \frac {2}{3} \right ) b

Also, by *Archemede’s principle* –

\text {Weight of body} = \text {Weight of water displaced by the immersed volume of body} = w V

Again –

- Weight of the slice of length \quad ( dl = w dV )
- Volume of the triangular wedge is \quad \left ( \frac {1}{2} \right ) \times \frac {b}{2} \times \theta dl \left ( \frac {b}{2} \right ) = \frac {b^2}{8} \theta dl
- Therefore, weight of triangular wedge is \quad \left ( \frac {w b^2 \theta dl}{8} \right )

Let, ( M ) is the metacentre. Taking moments about point ( M ) , we get –

\text {Moment due to moving the centre of gravity of triangular prism of water from} \ g \ \text {to} \ g_1 = \text {Moment due to moving the centre of up thrust of water from} \ B \ \text {to} \ \ B_1 .

- Therefore, \quad \left ( \frac {w b^2 \theta dl}{8} \right ) \left ( \frac {2 b}{3} \right ) = ( w dV ) ( BB_1 )

Or, \quad w \left ( \frac {dl b^3}{12} \right ) \theta = w dV . BB_1

Since, ( \theta ) is small so, \quad BB_1 = BM \theta

- Therefore, \quad w \left ( \frac {dl b^3}{12} \right ) \theta = w dV . BM . \theta

But, \quad \left ( \frac {dl b^3}{12} \right ) = dI

- Therefore, \quad w \left ( dI \right ) \theta = w dV . BM . \theta

Or, \quad dI = BM . dV

Above relation holds good for each slice of the floating body. Hence, integrating for the whole length of the body, we will get –

I = BM V

- Therefore, \quad BM = \left ( \frac {I}{V} \right )

Thus, metacentric height \quad MG = ( BM - BG )

- In this way metacentric height of a floating body can be determined analytically.

### To Find Metacentric height by experiment

The position of metacentre, ( M ) with respect to the position of the *centre of buoyancy* ( B ) can be determined by experiment as follows.

Consider about the figure of a floating ship.

- A small known weight ( w ) is placed on one side of the ship.

A

*simple pendulum*is made consisting of a heavy weight ( W ) suspended by a long cord. Pendulum is placed in the ship at a suitable location.- The position of the bob of the pendulum is marked and noted.

Let, ( l ) is the length of the simple pendulum. Now, the weight ( w ) is made to move from the position A to the position B as shown in figure.

Let, this *displacement* is ( x ) . Since the string of the pendulum will remain plumb, the angle of heel ( \theta ) can be measured by the apparent deflection of the pendulum.

Let, the apparent horizontal displacement of the bob of the pendulum is ( GG' = y ) .

Then, \quad \tan \theta = \left ( \frac {y}{l} \right )

The *moment* caused by the weight of the bob equals to the moment caused by moving ( w ) from A to B. Now, taking moments about the point ( M ) –

- Moment caused by weight of bob is \quad ( W \times GG' ) = ( W y ) = ( W GM \tan \theta )
- Moment caused due to shifting of weight from A to B is \quad ( w x )

Therefore, for equilibrium we have –

W GM \tan \theta = w x

Therefore, Metacentric height will be –

GM = \left ( \frac {w x}{W \tan \theta} \right )

By knowing the values of ( w ), \ ( W ), \ ( x ) \ \& \ ( \theta ) the value of metacentric height is calculated from above relation.

See numerical problems based on this article.