## Friday, 16 February 2018

We were discussing the basic definition and derivation of total pressurecentre of pressurebuoyancy or buoyancy force and centre of buoyancy in our previous posts.

Today we will see here the analytical method to determine the meta-centric height with the help of this post.

### Analytical method to determine the meta-centric height

As we know that meta-centre is basically defined as the point about which a body in stable equilibrium will start to oscillate when body will be displaced by an angular displacement.

Let us consider a body which is floating in the liquid. Let us assume that body is in equilibrium condition. Let us think that G is the centre of gravity of the body and B is the centre of buoyancy of the body when body is in equilibrium condition.

In equilibrium situation, centre of gravity G and centre of buoyancy B will lie on same axis which is displayed here in following figure with a vertical line.

Let us assume that we have given an angular displacement to the body in clockwise direction as displayed here in above figure.

Centre of buoyancy will be shifted now towards right side from neutral axis and let us assume that it is now B1.

Line of action of buoyancy force passing through this new position will intersect the normal axis passing through the centre of gravity and centre of buoyancy in original position of the body at a point M as displayed here in above figure. Where, M is the meta-centre.

As we have already discussed the term Meta-centric height i.e. the distance between the meta-centre of the floating body and the centre of gravity of the body. Therefore, MG in above figure will be the meta-centric height.

Angular displacement of the body in clockwise direction will cause a wedge-shaped prism BOB’ on the right side of the axis to go inside the water as displayed in above figure. This wedge will indicate the gain in buoyant force to the right of the axis. This gain in buoyant force will be indicated by dFB acting vertically upward through the centre of gravity of the prism BOB’.

Similarly, there will be one identical wedge-shaped prism AOA’ on the left side of the axis to go outside the water and we have also displayed it in above figure. This wedge will indicate the respective loss in buoyant force to the left of axis. This loss in buoyant force will be indicated by equal and opposite force dFB acting vertically downward through the centre of gravity of the prism AOA’.

There will be two equal and opposite couple acting on the body. Let us find first these two equal and opposite couple.

Let us analyze these two forces dFB. These two forces will form one couple which will tend to rotate the body in counter-clockwise direction.

Another equal and opposite couple will be developed due to displacement of centre of buoyancy from B to B1.

Let us consider one small strip of thickness dx at a distance x from the centre O, at right side of the axis, as displayed in above figure.

Area of the strip = x θ dx
Volume of the strip = x θ dx L
Where L is the length of the floating body
Weight of the strip = ρ g x θ L dx

Similarly, we will consider one small strip of thickness dx at a distance x from the centre O, at left side of the axis and we will have the weight of the strip ρ g x θ L dx.

Above two forces i.e. weights are acting in the opposite direction and therefore there will be developed one couple.

We will discuss another term i.e. Conditions of equilibrium of a floating and sub-merged bodies in our next post.
Do you have any suggestions? Please write in comment box.

### Reference:

Fluid mechanics, By R. K. Bansal