We continue with our discussion on force on
a plain surface submerged in a fluid. And
we take up the same example that we do in
the last class and we will try to find out
the force on these surfaces by direct integration
without going into the standard expression
for force. So, here if we recall what is the
objective of this problem is to solve that
what is the force that is required to keep
this gate stationary.
And to do that what we require is to find
out the moment of the distributed forces acting
on this semicircular plate with respect to
the heat point flow that was one of the objectives
for writing the equation of equilibrium.
So, to do that what we can take? We can take
small element so when you take a small element
here you have a, consider a radial and circumferential
elements simultaneously. So, we can consider
a small element shaded like this what is the
specification of the small element? It is
located or centered around r, theta that means
we consider that this is located at a radial
location of r and angular location of theta.
The angle subtended by the element is d theta.
The radial location is small r which is the
radial location of the element. The radial
width of the element is dr. So, what is the
differential area that is being represented
by 
the element, no question I just shaded element
the shaded area. So, it is roughly like a
rectangle one of the sides is rd theta another
side is dr. So, rd theta dr. What is the force
that acts on this element due to pressure?
What is the pressure acting on this element?
It is located to the local height of the element.
What is this local height? So, H is the total
height, -r sin theta. So the force acting
on this df =H-r sin theta* rho* g* r d theta
dr. What is the movement of this force with
respect to o? So this into r sin theta so
if we can break it up into two terms multiply
these by r sin theta. So, it will be H r square
sin theta –r cube sin square theta * rho
g d theta dr.
Now, the total moment of this distributed
force with respect to the .o is integral of
this dM. Now, when you consider the integral,
the integral is with respect to both theta
as well r. It is a double integral. So, you
may evaluate the integral with respect to
r first or theta first. It is irrelevant.
Let us say that you want to evaluate the integral
with respect to r first. So, you have first
the integral with respect to r. So, when you
have integral with respect to r small r varies
from 0 to R.
So, the first term becomes this is r square
dr that means r cube/3. This we are integrating
with respect to R sin theta –this is r cube
dr so R to the power 4/4 sin square theta
d theta. Now, integral with respect to theta.
What is the limit of theta? 0 to pi. The remaining
work is very simple I need not complete this
one. There are very integrals and you can
complete yourself. So, once you complete these
integral.
You will get an expression for resultant movement
of the distributed force with respect to 0
in terms of HR and of course rho and g. And
this resultant movement =F*R so that will
tell you what is the value of A. We have seen
that why it is so because all other forces
acting through it will pass through o so they
will have no movement with respect to o. So,
this is just like these are the two counter
acting.
So, this example shows that whenever you have
a force on a surface it is not necessary that
you have to go by the formula that we have
derived you may as well take a small element
consider pressure distribution over the small
element and find out what is the resultant
force due to that pressure distribution resultant
movement and so on by fundamentally integrating
into over the entire area without going into
the formula.
Next what we will consider is force on a curve
surface. Now can you tell me what is the fundamental
difference or what do you expect the fundament
difference to be as compared to the force
on a plain surface.
How you expect force on a curve surface to
be different conceptually? Force on a curve
surface, again what is this surface? This
is immersed in a fluid at rest that we are
not repeating. Let us say the surface is something
like this. Again if you see this surface is
actually something like this and there is
some fluid on one side and that fluid is exerting
some force on the surface. So, this is again
edge view of the surface.
Earlier we were concentrating on force on
a surface like this which is a plain surface
now it is a curve surface. So, you can understand
the geometrical difference that is why fundamentally
in terms of basic mechanics how do you expect
these to be different as compare to that for
force on a plain surface. Force direction
changes as you move from one point to the
other point that means it is not a system
of parallel forces.
So, the plain surface has to deal with a system
of parallel forces so you can treat it like
a scalar addition problem or a scalar summation
problem. Where as in this case it is having
the pressure always acting normal to the boundary.
Normal to the boundary is a direction that
is changing from one point to the other and
therefor the resultant force is being dictated
by a varying direction of the surface.
So you no more have a system of parallel elements
which is giving rise to the resultant. So,
whenever you are adding here to make the resultant
force it has to be a true vector addition.
It is not just by adding it up in a scalar
way. Integration is nothing but summing up
the individual components and you could see
that a very simple integration give the same
result for force on a plain surface.
For a curve surface therefor fundamentally
the principle the same that is you take element
of the curve surface to find out what is the
force acting on it. The force acting on it
will be normal to it. You may break it up
into components horizontal and vertical component
in this way for each of the elements you can
find out the horizontal and vertical components
algebraically sum them up and make the vector
addition.
So, this is something which is very trivial.
Now, it is possible sometimes to reduce the
calculation a little bit by taking some help
from the concept of force on a plain surface.
Let us see how? Just consider that there is
a fluid column like this. This is a volume
of fluid which we are considering to be located
within the projected part of this curved surface.
So, in this end of the curve surfaces are
projected to the free surface.
Whatever volume is contained within that contained
volume is enclosed by this dotted line. Now,
let us say that we are interested on the forces
which are acting on these volume of fluid.
So, these volumes are now having two plain
surfaces. Three plain surfaces but forget
about the top surface. Just for the time being
consider the side surfaces. So, if you consider
the volume say A, B, C, D the surfaces AB,
and CD are plain surfaces.
And let us see that what are the forces which
are acting on this plain surfaces? So, we
are essentially trying to draw a free body
diagram on the volume element which is enclosed
by this dotted line. What are the forces which
are acting on this? So, when you have the
left phase there is some fluid towards the
left of it that exerts a force due to pressure.
What will be the direction of that force.
Let us say we set up coordinates like this
x and y.
So, what will be the direction of that force?
x, so we call it a horizontal force assuming
that x is a horizontal direction. Let us say
that if H or FH1 is the horizontal force acting
from the left towards this element. Similarly,
if you have some fluid again on this side
there will be a horizontal force FH2 acting
from the other side towards this remember
this force is due to pressure. It is compressive
in nature so whatever is the fluid element
located on the other side.
It is having a tendency to compress it. And
that is how the sense of these vectors are
there. There are the horizontal components
of the forces on these sides. What are the
additional forces on this element? It should
have its own width. So, whatever water or
fluid is contained here that will have its
weight say W. Any other force? There is a
reaction between the surface and the fluid
and that reaction again is likely to have
two components.
One is the horizontal component and other
is the vertical component. So, let us say
that it has horizontal component FH and a
vertical component FV. What are these? These
are the components of the reaction forces
exerted by the curve surface on the fluid.
Now, the fluid is in equilibrium. So, when
the fluid is in equilibrium you must have
resultant force along x =0 so you have FH+FH1-FH2=0.
That means FH = - FH1-FH2 then resultant force
along y =0. What it means? Fv –W=0 that
means Fv=W. From this apparently very simple
calculation we come up with a very interesting
result that is if you have a curve surface
still whatever forces which are acting on
it you may resolve it into two components.
For the horizontal component it is basically
the resultant force on the horizontal projections
or on the projection of these curves.
Ends of the curves on a vertical plan or so
called horizontal components of the forces
on vertical projections of the ends of the
curve. That means if you have a curve like
this when you consider the end this end when
you consider its projection the projection
is just like this. So, we are basically having
to consider a horizontal component of force
for the left. A horizontal component of force
for the right and the difference of these
two are actually giving the horizontal component
of net force.
For the vertical component it is just the
weight of the fluid that is being contained
within these extended volume. So, if you somehow
can calculate the weight of the fluid and
that is as good as calculating the volume
of the fluid. Because then you can use the
density to calculate the weight of the fluid.
So, whatever fluid is contained here within
these doted lines, weight of that fluid is
the vertical component of the force.
And whatever the horizontal component of force
or whatever is the force because of pressure
on the projections taken from the sides of
these curves that contributes to the resultant
horizontal force. And remember that these
are the forces exerted by the curve surface
on the fluid. We are interested in the other
thing the opposite thing that is what is the
force on the curve surface. So, by Newtons
third law those are just negative of this
one.
So –FH and –FV are the forces which are
exerted by the fluid on the curve surface.
Let us take an example to see that how we
calculate it.
Let us say that we have a long circular cylinder
and you have the free surface of the fluid
in this way. So, on the left side this is
the free surface on the right this is the
free surface and the solid material is a circular
cylinder long its length is perpendicular
to the plain of the . We are interested to
find out what is the resultant horizontal
and vertical component of force exerted by
the fluid on this . So, what we will do is
we will keep some names of or we will keep
some markers to important of the surface.
Say the four important points A, B, C, and
D and we will consider the forces acting on
this curve surface one-by-one. So, if let
us consider the force acting on AB, first
let us consider the vertical components of
forces then we will do the horizontal component.
So, for vertical component what we want? We
should raise projections from the end of the
surface till reaches the free surface. Whatever
is the volume of the fluid contained within
that it is the weight of that.
So, for the part AB it is like this. Next,
let us consider the part DC so when you consider
the part AB and you know the vertical component
of force is it downward or upwards? It is
downwards from common sense it is clear that
the pressure is being exerted in such a way
that it’s vertical component is downwards.
Now, consider BC. So, for BC what we do again
we raise the projection from one end it is
like this.
From C we raise the projection up to the free
surface so whatever is the volume that is
contained within this. So, what volume is
contained within this? This is the volume
that is contained between BC and its projected
parts up to the free fluid surface. So, this
is an imaginary volume of fluid and what will
be the direction of the vertical component
of the force acting on it? Upwards or downwards?
How do you make it out just see that if you
have BC like this you will have the pressure
acting on it this way its vertical component
will be upward. So, it is such a distributed
pressure over BC. So, look into the fundamental
origin that will give you the guideline whether
the resultant force is upward or downward
on that part of the surface. Now, when you
consider this two together you can see the
common part.
Which is shaded once it has come downwards
another it has come upwards. So, they have
canceled out so what remains is the fluid
equivalent to the volume of half of the cylinder
for this half part. So, what is the vertical
component of force acting on say A, B, C.
It is nothing but the equivalent to the weight
of the volume of half of this cylinder. Weight
of what? Weight of the fluid equivalent to
the volume of the half of the cylinder.
So, that means as if it has displaced up to
it equivalent to its volume and that is exerting
it an upthrust. This is nothing but the Archimedes'
principle that you have learnt in high school
physics. So, in effect what is happening when
the solid is being immersed in a fluid it
tends to displace a volume of fluid which
is equivalent to the volume which is immersed
and that tends to exert an up thrust, net
up thrust and that up thrust is nothing.
But same as the weight of the displaced volume
of fluid by that particular volume of solid.
Now, if you come CD so for CD how you calculate
the force vertical component of the force?
So, you extend it up to the free surface so
for this part the free surface is up to the
level shown. So, it is nothing but the volume
of this much. So is it upwards or downwards?
It is upwards. So, if you have the pressure
acting in this way its vertical component
will be upwards.
So, the resultant is upwards with what magnitude
if r is the radius of this cylinder? So, what
is the volume corresponding to the left part
that is pi R square/ 2*L where L is the length
of the cylinder. Then this part is pi r square/
4 * L. This is the volume that multiplied
by the density of the fluid. Not density of
cylinder but density of the fluid * g and
acting upwards. So, this is the resultant
vertical component upwards.
How can you calculate that what is the location
through which the resultant of these force
passes? So, it will definitely be passing
through the centroid of the displaced volume
and then it boils down to the calculation
of the centroid of the displaced volume. We
are not going into that you can do it by simple
statistics. How do you calculate the horizontal
component of force? So, again you have something
in the left and something in the right.
So, when you have this A, B, C on A, B, C
what is the horizontal component of force?
Let us say that is FH1 just following this
symbol and on CD there is a horizontal component
of force say that is FH2. So, FH1 is what?
What is its projected area on which you are
considering that because this is now equivalent
to force on a plain surface. So, 2R is the
height and L is the length.
So, its projection on the side view is 2R
is its height because 2R is the height of
the cylinder and L length perpendicular to
the plain of the figure. So, if you recall
what is the resultant force flow of the fluid
g into A, what is A? 2R*L that is the equivalent
projected area*AC. What is AC? AC is just
R so when you say HC, HC is the location of
the centroid of the projected area not the
curved area.
So, it is the location of the centroid of
the projected area from the free surface.
So, that is FH1. Is it towards right or left?
It is towards right because again you can
see that its horizontal component is towards
right. So, this is towards right and then
on CD what is FH? First of all it is acting
towards left, its horizontal component. So,
we put a minus sign again rho g. What is the
projected area?
Into R*L* R/2 which is the location of the
centroid of the projected area vertical location
from the free surface. So, this is FH1 –FH2
that is FH hence you can calculate what is
the resultant. So, these are two individual
components you can find out the resultant
by vector addition so that is trivial. Let
us consider a second example.
Let us say that you have a curved gate instead
of a plain gate you have a curve gate and
there is fluid on one side of it. Say you
have fluid on this side there is a free surface
of the fluid given by this and this fluid
tries to exert some force on this curve surface
and there is some balancer so there is some
external force which is applied here to keep
it in equilibrium and the location of this
external force is given by this H and the
depth to which this fluid is filled up is
D.
So, why such problems are practically important?
Let us look into maybe 1 or 2 practical cases
to see that why such effects are important?
So, we will look into some example applications.
If you clearly see this is like a dam. So
there are forces which are exerted by water
and these force components are like these
are quite heavy or large forces huge forces
which are acting and the structure must be
strong enough to sustain it.
And it is very common and to save us from
floods or other calamities there are many
occasions where there are reservoirs in which
these water supply the store. So, until or
unless the rainfall is very, very sevier it
retains its height. But once the rainfall
is so strong it cannot retain its height so
some of the water has to be released. And
when the water is not released it is under
static condition and it has to be calculated
on the bases of fluid statics condition.
The resultant forces and the situation of
equilibrium of course when the water is very,
very dynamic then we do not consider the fluid
statistic. But you consider even the dynamic
effect of water as it on the surface. So,
you can see the shape of the so called dam
it is not really a plain surface but it is
a curve surface. It is like an arch type.
There are fluids at different ends and it
is important to calculate that what is the
resultant force that is there.
So, that is the motivation behind solving
such a problem that it gives you a clue of
how to design may be dams or sluice gates
where across which you have different fluid
elements which are exerting forces. So, here
on one side you have some fluid element as
say water on another side say atmosphere.
So, on the right side there is water in this
example and we want to calculate what are
the resultant force so what we will do we
may solve this in again two ways.
One is by looking into horizontal and vertical
components of forces according to the principle
that you have just seen. Or maybe just by
direct integration of the forces on elements.
So, if we do that the second one is little
bit may be easy to begin with. So, let us
say that we have considered a small element
here at a location x, y. For idealization
let us say that this curve surface where its
projection is considered.
It had the equation x=ay which is the equation
of this curve. So, at the x, y if you consider
a small element say of size dl then what is
the resultant force due to pressure on dl.
First of all how to specify this dl? Say at
x, y we can consider a dl which is comprising
of the resultant some displacement along x
to dx and some displacement along y say dy.
So dl may be written as good as square root
of dx square + dy square.
So, if you draw a magnified figure if this
is dl it is like sum of dx along x dy along
y. If this is a small part of the curve this
is approximately tangent to the curve at this
point. At that x, y and so you can consider
that this angle theta is in net the local
slope of the curve at that point where it
is aligned with the tangent to the curve at
x, y. So, we know what is dl if you know what
is the width of the fluid or what is the width
of the gate say w is the width of the gate.
Then w into dl is the area on which the fluid
pressure is acting. So, what will be the direction
of the action of the fluid pressure? It is
acting in this way normal to the surface.
So, that will have two components one is the
horizontal component and another is the vertical
component. So, how do you calculate the horizontal
component and the vertical component? First
of all what is the force because of this.
It is P *dl into W, where p is the pressure
at x, y so how it is related to the depth
capital D? So, how can you write express p
in terms of the depth D? rho g* so the height
of this is D-y from the pre-surface. This
is the resultant force but if we break it
up into component that is what is our objective
and we should break it up into components.
Because, we cannot just integrate it like
this.
Because the direction of such force on each
element is changing so we cannot just algebraically
or scalar add. We should take out extract
individual components and then add up the
components. So, when we take the horizontal
component, what is the horizontal component
because of this? So, this 
is the resultant force 
so the dl makes theta with the horizontal.
So, normal to the dl should make an angle
theta with the vertical.
So, the horizontal force is p dl w* sin theta.
Now, if you look into this figure dl sin theta
is nothing but dy. So, p*w*dy so let us call
this as dFH because this is just on a differentially
small area. So, rho g w D-y dy. How can you
calculate the vertical component of the force?
So, calculating horizontal component straight
forward you just integrate it. So, when you
integrate it let us try to write the complete
expression.
I will leave the integral on you but at least
let us write the complete expression. So,
FH is integral of dFH. What should be the
limit over which you integrate? Y=0 to y=D.
Something very interesting it shows as if
this is independent of the function of the
graph x=ay cube. It does not matter what is
x. And intuitively it is supposed to be that
way because when you take the projection of
the surface in the side plain.
Does not matter how it is curved because the
projection will always be straight. So you
just required to know the extent of the depth
and you can verify that this will be nothing
but the projection on the surface which is
basically surface projected on a vertical
plain. And the same horizontal component of
force will come out that exercise. What about
the 
vertical component of the force? So, dFv so
it is pdl w cos theta.
So, rho g D-y *W, what is dl cos theta? dx.
So, now you may express either x in terms
of y or y in terms of x. The function is given
and if we integrate you will be getting the
total vertical force. So, Fv is 
so tx you can write as say 3 ay square dy
and you can now take the limit from y=0 to
y=D. We will not go into the integration in
details because it is a very simple integration
and it is not worth to waste time just on
that.
But we will focus on something which is bit
more important. That is let us say that we
want to find out the same vertical component
of force but using the method of the weight
of the fluid that is contained within that
projected field. One way to see that whether
it is correct or not is that if we find out
that weight and we come up with an expression
it should be same as this one. So, that you
can of course find out the way.
And check that the expression at the end is
the same as what you get out of this integral.
But even before that how you qualitatively
assess it. So, what was our method? You project
from the corners vertical lines which meat
the free surface or its extended form. So,
one line projected here and another it becomes
a point and it should be the shaded volume
of . If I were in your place the first question
that would have come to my mind is there is
no fluid here.
So, how can you claim that weight of this
shaded volume of fluid will give you the vertical
component. So, just think in this way. Let
us say that this is a curve surface and let
us consider that there is volume of fluid
one on this side another on this side. If
that was the case then you can see that in
a static condition it is naturally in equilibrium
because what is the force due to pressure
from one side? The same is the force due to
the pressure on the other side.
So, the two sides are keeping it in equilibrium.
Now, when you do not have a fluid here that
means that this part is minus it is subtracted.
So, it is a condition equivalent to a deviation
from equilibrium because of a lack of presence
of this rather than the real presence of this.
So, the deviation from equilibrium is because
of this equivalent volume of element which
is in the upper part.
So, if you calculate that volume of fluid
and find the corresponding weight of that
you will see that you will get it exactly
the same as this one. And you can clearly
tell that what should be the direction of
vertical component of force acting on this.
Upwards or downwards? Upwards because the
vertical component of this one is like upwards
and directed in this way. So, it is not apparently
the vertical component of this shaded volume
of fluid.
Because it will appear that if there is a
volume of fluid here its weight should act
downwards. You have to remember it is an imaginary
volume of fluid and it is just giving you
the equivalent volume that needs to be employed
for calculation of vertical component of force.
the exact sense of the force has to be determined
from the physical meaning of these type. So,
if you subtract a kind of volume like this
it is as good as extra upward force.
Because something is missing some weight is
missing from the top. Otherwise also from
the direction of pressure itself it will follow.
So, either way whenever you are calculating
either by the fundamental method of finding
force components or individual elements summing
the vector, summing them up that means summing
up scalar forms in terms of the x and the
y or the horizontal or the vertical components
or finding out the vertical and the horizontal
components.
By the alternative method that we have seen.
Whatever is it you must asset it in the correct
sense from the physical condition and that
will not always be dictated by the rule based.
It will come from the consideration of where
is the volume of fluid that is present is
exerting the force and what is the sense of
that force when it is exerting a pressure
on the . So, that consideration should give
you the proper sense of the vertical component
of the force.
Now, we have considered force components on
plain and curve surfaces. We have seen that
there are some simple ways by which we can
evaluate these force components. Now what
we have assumed is that when the surface is
put in the fluid the surface is in equilibrium
and that equilibrium is not disturbed. But
if there is a slight tilt because of whatever
reason then that equilibrium may be disturbed.
And if that equilibrium is disturbed what
will happen we will have to understand. So,
we have to now go through the concept of stability
of floating and submerged bodies.
Let us say that first we consider submerged
bodies. So, submerged body is something which
is completely immersed in the fluid and floating
means a part is at the top I means above the
free surface and the below the free surface.
So, let us take an example. Let us say that
we have this kind of a body likes like a parachute
type. The top is light and bottom is heavy
because of the added mass.
Initially it is completely submerged and the
resultant forces whatever are acting on this
you may write in terms of the buoyancy force
and the weight. So, the buoyancy force will
be acting through some point and the weight
will be acting through some point. So, the
bouncy force is based on the volume which
is submerged not the mass. So, when you consider
the volume that is submerged the greater portion
of the volume is at the top.
So, may be the resultant buoyancy force act
through this. But the mass is more concentrated
towards the bottom so the weight may be is
more concentrated of the resultant force due
to the weight distribution is passing through
the point which is g, or the center of gravity
that is located somewhat below. And the resultant
buoyancy force is now acting through some
point B. So, what is the point B?
So, B is the so called center of buoyancy
that means whatever is the location of the
centroid of the displaced body. So, it is
fully a geometrical concept whereas when you
have Ge this depends on the distribution of
mass over the body. So, this is something
where for equilibrium you have FB=W it is
in equilibrium. Now, let us say that you have
slightly tilted it. So, when you have slightly
tilted it, it has a deformed not deformed
but deflected configuration like this.
So, its axis has got tilted from the original
vertical one may be because of some disturbance.
Now, the entire body is within the fluid therefor
the location of the center of buoyancy force
and center of gravity relative to the body
does not change because the entire body is
within the fluid itself. So, if you have say
this as G, this still remains as g if you
have this as B, it still remains as B.
Because it is already totally within the fluid
so its volume distribution within the fluid,
its mass distribution everything it does not
change. So, you have FB acting like this you
have W acting like this. Only thing what has
now changed is that no more FB and W are collinear.
So when they are not collinear they will still
be equal and opposite forces but not passing
along the same line so it will creat a couple
moment.
So, what will that couple moment try to do?
So, if you see the sense of this couple moment
what it will try to do? So it will try to
create a rotation like this which is shown
in the figure. If you look into the senses
of the forces and these rotation what they
will try to do? It will try to bring it back
to its original position or configuration.
So, we call it a restoring movement. On the
other hand if g was above B then this would
have been downward.
And this should have upwards and that would
have tried to increase the angular displacement
even further. So, what is the whole mark of
a stability, stable equilibrium that is if
you have a slight displacement it will try
to come back or be restored to its original
configuration. So, this type of situation
ensures that it takes to come back to its
original configuration. However if g was above
me it would have tried to increase the angular
displacement even further not restoring.
But helping the disturbance. So, in that way
it will be unstable equilibrium. What would
be the situation if B and G are coincident
somehow so wherever it is there still it will
be a collinear say you have somehow an arrangement
where you have this as B, the same point is
G. So, this is B this is as good as G. So,
whatever is the weight and the buoyancy they
are always acting along the same line no matter
whether it is tilted or not.
So, wherever it is tilted it will locally
attain equilibrium and that equilibrium is
known as neutral equilibrium. So, the stability
of submerged bodies the equilibrium condition
depends on the relative location of B with
respect of G. So, what we can summarize from
this if B is above G what it will imply. It
will imply stable equilibrium. If B is below
G it is unstable equilibrium and if B coincides
with G that is neutral equilibrium.
So, far so good but we have to remember that
submerged bodies are not the only types of
bodies that we need to consider many practical
examples are cases of floating bodies like
ships. So, a part is within the fluid and
the part is outside the fluid. So, what will
be the situation for that let us take an example.
Let us consider a floating body. So, when
you consider a floating body 
let us see that how is it different first
from a submerged body.
So, let us say that this is a free surface
of a fluid. This is the body which is floating
something like a boat shape or similar to
that. If you consider its intersection with
the free surface of the fluid that is the
sectional view of the intersection with the
fluid. Let us say that it is something like
this. So, this part which we have drawn is
like this one whatever we have drawn that
plain is like this. Now, this we are assuming
that this is having an axis of symmetry.
Say this is the axis of symmetry now let us
say that this is tilted. Let us see what happens
this is tilted? We will assume that this is
tilted very slightly because when you test
the stability we just give a small displacement
and see how it responds to the small disturbance.
So, we just tilt it like this 
and it comes to this configuration. So, whatever
the line of interface before now say that
line of interface has with relative to the
body whatever was the line say AB.
Now say it becomes A prime, B prime. If it
is symmetrical with respect to the axis over
which it tilts you will see that one interesting
thing has happened. What is that interesting
thing? Some new part has gone down into the
fluid. Some new part has come up and if it
is very symmetric these two parts are of same
volume. So, the volume that was earlier immersed
is the still the same. But the distribution
of the volume has changed.
So, once the distribution of volume has changed
what has happen? The center of buoyancy has
changed. So, there is no sanctity with respect
to the location of the center of buoyancy
that is very, very important. So, for the
submersed body when you have the location
of the center of buoyancy relative to the
body it does not change whereas when you have
a floating body depending on its tilted configuration.
There will be an extent of dominance of one
side of the body relative to its immersed
condition with respect to the other and accordingly
there will be a bias. There will be a preferred
side across which the center of buoyancy will
be moving. So, the center of buoyancy cannot
be one of the fixed parameters with respect
to which we may decide whether it will be
stable or not that is the first thing. So,
the stability criteria for submerged bodies
will not work.
So, whenever we are trying to learn something
new we have to understand that why are we
trying to learn it afresh. I mean if the same
criteria for submerged bodies would have worked
we would not have not have gone into this
exercise. So, first we are getting that motivation
that how or where is the difference? The difference
again I sum up is like this the center of
buoyancy location is now not fixed with respect
to the body but it goes on evolving as the
body is tilting.
So, let us say that the center of buoyancy
now comes to this possible say CB. We expect
that it would be coming towards this direction
because it now tilted towards the right. So,
more part of body is now into the fluid towards
the right. So, you have the center of buoyancy
in this way. So, you have the resultant buoyancy
force like this. The center of gravity is
something which is fixed with reference to
the body that does not change.
So, if the center of gravity earlier was say
relative to the body here. Let us say that
the center of gravity is still here. So, the
weight of the body is like this. So, now again
you can see that there is a couple moment
and whether it is restoring or helping it
depends on that if it is extended where it
will meet the axis. If it meets down of phase
1 then it’s one way. If it meets above g
it is the other way. So, where it meets the
axis that point is known as metacentre.
So, in our next class we will see that what
is the consequence of this meta center and
how the location of the meta center will dictate
the stability under this condition. So, it
is not the center of buoyancy that is important
here but the location of the metacentre relative
to the body is what is going to decide whether
the body should be stable or not for a floating
body. That we will take up in the next class.
Thank you.
