We were discussing the basic
definition and derivation of total pressure; centre of pressure, buoyancy 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 B

_{1}.
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 dF

_{B}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 dF

_{B}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 dF

_{B}. 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 B

_{1}.
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

Image
Courtesy: Google

In the equation of BM, first rho is of body and second is of liquid. Then how is it cancle out?

ReplyDeletePositive site, where did u come up with the information on this posting?I have read a few of the articles on your website now, and I really like your style. Thanks a million and please keep up the effective work. https://brandsocial.me/celebs/leila-rahimi/

ReplyDelete