Good afternoon, I welcome you all to this
session of Fluid Mechanics. Last class, we
were discussing the force Hydrostatic force
exerted on a plane surface submerged in a
static expansion of fluid. And what we have
finally concluded that if there is a free
surface of fluid, and a plane surface; plane
inclined surface is submerged in this expansion
of fluid. Then, the pressure force is exerted
perpendicular to the plane surface at each
and every point. And the pressure intensity
at any point as we know that it is equal to
rho times, g times the height of that point
or the depression of that point as the vertical
depth of that point from the free surface.
So, the pressure intensity varies at different
points depending upon their depression from
the free surface.
And finally, we recognized that the total
hydrostatic force due to this hydrostatic
pressures on any one side of the plane surface
is equal to the pressure intensity at this
center of area of the surface times the total
area. That means, it is equivalently at the
same area if it is placed horizontally at
a depth equal to the depth of the center of
area of the inclined surface. Then the pressure
which could have been exerted that is the
uniform pressure equal to rho times g times
the height from the free surface that is the
height of the centroid for the inclined surface.
So, therefore, we are concluded that for an
any inclined surface. The total pressure force
is equal to the area of the surface times
the pressure intensity at the centroid of
the area or center of area.
And we also recognize the another fact that
the pressure center. That is the point on
action of these results and pressure forces
passes through a point which is below the
center of area. That means, it is more at
a more depth than that of the center of area
which is obvious because pressures are increasing
with the depth. And the center of pressure
can be found out by a simple expression that
the movement of area about an axis which we
take as the intersection of the plane surface
with the free surface, we take an axis perpendicular
to this. So, movement of area of this surface
about this axes divided by the area, and the
coordinates of the center of pressure. So,
this we recognize in the last class.
Now, today we will be discussing the hydrostatic
pressure forces or hydrostatic forces on submerged
curve surface. If the surface is curve then
how you will find out the Hydrostatic pressure
force?
Now, let us see in general here. Let us first
write hydrostatic 
force on a curved surface. Now, let us see
here in general a curved surface. So, this
is a curved surface you can see which in general
has got curvature in both the directions,
and which is emerged in a fluid whose free
surface is this one in an expansion of fluid.
Now, in this case, we consider a three dimensional
system, because of these curvature in two
directions. We consider the axes o x, o y
at the free surface. That means, this x y
coordinate plane is at the free surface, and
oz axis we consider vertically downward. Now
it is true that at each and every point on
the surface the pressure force intensity.
Or the magnitude is equal to rho times the
density of the fluid which is constant. In
case of liquid times the g times its vertical
depth from the free surface which we denote
by the z coordinate. That means, rho g z,
but the difference between this with that
of a plane surface is that each and every
point the force is normal to the surface which
are not parallel.
In a case of a plane surface what happened?
That force for example, there could have been
a plane surface the force on each elemental
area is normal to this surface and they are
parallel to each other. So, therefore, a scalar
summation of the elemental force was possible,
which we did by simple integration of an elemental
force throughout the area. But in case of
a curved surface since, the normal at each
end point of the surface which is the direction
of the pressure forces changes. Because of
the curvature is simple scalar summation.
That means, a simple integration will not
be possible.
So, therefore, the general procedure of simple
basic mechanics is that. In that case, we
will have to take an elemental force. And
we will have to fix certain reference coordinate
directions in which we will take the components
and some of the components in a particular
direction. And we will find out the components
of the resultant forces in those directions.
And then again vectorially some of the components
to find out the resultant force, it is as
simple as like this from elementary mechanics.
So, now we concentrate here, we will see to
do that we recognize an elemental area d A
which is at a vertical depth z. We just project
this area here, and this is so near that is
the z in the y z plane. So, z represents the
vertical height or depression from the free
surface of this area d A. Now, you see the
hydrostatic force due to pressure is acting
on this surface. If we consider the force
acting by the liquid on this side I have told
on one side we will consider. Other side we
will consider as an open to atmosphere, then
this pressure force is normal to this area
d F. Now, this force can be resolved into
3 directions with respect to this frame of
reference o x, o y and o z accordingly d F
x is the x direction x component, d F y is
the y component and d F z is the z component.
Now, if we assume that the normal to this
surface makes an angle alpha with the x coordinates
makes an angle beta. Let this is beta; this
big angle normal to this with y axis is beta
and normal to this with z axis. That means,
this angle as gamma; that means, alpha beta
and gamma are the angles made by the normal
to this area with x axis, with y axis; that
means; this angle and with z axis. Then we
can tell simply that d F x is equal to what?
d F x is d F. Now, d F is d A into rho g z.
If we consider that this area is at a vertical
that z from the free surface. So, simply rho
g z times, what we can write? d A cos alpha.
What is d F Y? rho g z d A cos beta. And what
is d F z? That is rho g z d A cos gamma.
Now, you see this d A cos alpha physically
represents the projected area of d A on y
z plane; that we denote as d A x. That means
it is the projected area of elemental area
d A on a plane whose perpendicular is x; that
means, on y z plane. So, therefore, we can
replace d A cos alpha by d A x. And we can
write d F x is equal to rho g z d A x. Similarly,
we can write d F y as rho g z. Similarly,
A cos beta will be the projected area of d
A on a plane, whose perpendicular is the y
direction; that means, it is on the plane
x z.
So, that we represents as d A y. And similarly,
d F z will represent rho g z times d A cos
gamma which is the projected area of d A on
a xy plane. That means, the plane perpendicular
to z direction which is so near to this area
is d A z, d A y area is not shown here it
is difficult to show in this two dimensional
paper. Now, we will see that if we now integrate.
Now, the different direction component forces
of these elementary forces, if it is made.
Now, we can make a scalar sum. That means,
now, we can tell F x as integration of d F
x double integration, because this is equal
to rho g z d A x.
Therefore, you see rho g z d A x. Now, you
can recall this can be simply written considering
a force on a plane surface that this is rho
g z c into A x, rho g is common. If we take
rho g out, then the double integral z d A
x is made over the entire area Ax. Where A
x represents the projectional area of this
entire curved surface area A. That means,
it is the projectional area on y z plane that
is the plane perpendicular to x axis of the
entire curved surface area A x. And z c represents
the center of area of that projectional area.
Or you can tell the projection of the center
of area on the curve surface on y z plane,
both are same.
Therefore, we see simply F x is equal to again
I am writing it is very interesting result
rho g z c A x. In a similar fashion I can
write F y by integrating this d F y as rho
g z d A y double integration. Which becomes
is equal to rho g z c in to A y. Where z c
is again the center of area the vertical distance
or the coordinate of the center of area projected
on x z plane. And A y is the projected area
of the entire area or entire curved surface
on the x z plane that is the plane perpendicular
to y. And similarly, we can write d F z is
equal to integration of rho g z d A z. Now,
here we stop, we cannot write anything more
fore here. Now, two very interesting result
comes out from this expression for F x and
this expression for F y. Now, we see, the
x component of the hydrostatic force on the
entire area is equivalent to the hydrostatic
force on a vertical plane surface which is
parallel to y z plane.
And area of the surface equals to the projected
area of these curve surface on the y z plane.
Similarly, the y component of force on this
curved surface is equal to the hydrostatic
force of a plane surface area which is parallel
to x z plane. And whose area is equal to the
projection area of this surface on x z plane.
Therefore, we can conclude from here that
the hydrostatic force in any horizontal direction
of a curve surface equals to the hydrostatic
force exerted on a plane surface which is
perpendicular to that direction. And whose
area is equal to the projected area of the
plane on that particular plane perpendicular
to this direction.
So, again I am telling the conclusion is like
that, if there is a curved surface, sometimes
it helps us in solving problems. That if there
is a curved surface submerged in a liquid,
then the hydrostatic pressure force in any
horizontal direction, it may be x, it may
be y any way you can orient in any horizontal
direction is equal to the force on a plane
surface. Whose area is equal to the projected
area of that surface in a plane perpendicular
to the direction? That means, if we consider
a plane perpendicular to the direction, horizontal
direction in which we are finding out the
net component of the total hydrostatic pressure
force on the curved surface.
Then if we imagine a plane perpendicular to
that direction then this is equal to the hydrostatic
pressure force of a plane surface, which is
the projection of this curved surface on that
plane, which is perpendicular to that horizontal
direction. So, this is a very important conclusion
we get. Now; obviously, if we now define,
want to define the center of pressure for
these x component of force separately. We
can find out p which is given by z p. If we
consider this as the centroids so, this will
be line on the same line vertical line. Because
I have told that it may not be displayed.
That means, y c and y p will be equal, because
for a plane surface the centroid A l axis.
Or the axis to the center of area is such
that the areas are the movements of the areas
on both sides are zero about this.
And, since the hydrostatic pressure force
does not vary with hydrostatic pressure forces
vary only with the area. So, that this is
not displaced in along this line from the
center of area. But this is displaced here,
because the hydrostatic pressure force varies
with the d A that we have already recognized
earlier. So, if we that way find that p is
the center of pressure with a vertical depression
or coordinate z p. Then we know from our earlier
discussion that z p in this case is given
by this integral. We can write I about this;
this is the y and z plane. So, this will be
about the y axis, y y divided by a x into
z c. That means, the movement of area of the
projected area on y z plane about the o y
axis, divided by this total area a x that
is the projected area of the curved surface
in the y z plane divided by the area projected
area a x times the z c.
That of course, we can change that again by
parallel axis theorem by transforming this
second movement of area about o y axis to
another axis parallel to o y through c that
you can do as well in case similar as we did
in case of plane surface. So, this is the
similar way we can find out the pressure center
for that is the point of application of F
y by applying the same equation. That is the
second movement of area of the projected area
on x z plane. That means, that will be A y,
Ay about x axis divided by this. That means
I can write this well I can write this similarly,
that z p point of application of F y as integration
of I y y that is not the integration; this
is I y y that takes the integration.
So, I can write this way that I; this is I
x, because this is x, I x x of the area A
y times the z c. That means, where A y is
the area projected area of the surface on
x z plane. And its second movement of area
about this x axis divided by the area itself
is z c. That means, we will treat as simple,
simply as pressure forces on plane surface
which are the projected surfaces on y z and
x z plane. Now, what about the vertical force?
Now, you see the vertical force component
is this. This is not d F z; this is F z now
F z. Now, let us write this F z is integration
of d F z; that means, rho g z d z.
So, now, what is z d z you see that if this
is z d A z. Now, what is z d a z? d a z is
the projectional area on x y plane which means
that if we consider a prismatic fluid element
over the d A up to the free surface d a z
is the cross sectional area. That means the
projected area that means the area perpendicular
to the z directions projected area on x y
plane. Therefore, this quantity simply represents
rho g, if we take out; this z d a z simply
represents volume. That means, d v first I
write this is the d v cut I will use always
v cut as volume, because v will be using velocity
afterwards. So, this is the elemental volume
of this prismatic fluid element.
So, this is simply rho g v. That means, we
can tell that the z component of the hydrostatic
pressure force over an elemental area is equal
to the weight of the fluid above that elemental
area up to the free surface. And for the entire
surface it is the weight of the fluid that
is content in the region vertically above
the surface up to the free surface. That means,
if we make the vertical projections from the
surfaces up to the free surface then the bulk
of the liquid which is content within this
vertical projection from the curved surface
to the free surface, the weight of that liquid
is equal to the hydrostatic z component of
the hydrostatic force.
So therefore, next is that this force the
point of action of this force is through the
center of volume of this bulk of fluid which
is content vertically above the surface up
to the free surface. And it is little complicated
to find out the pointer application. So, resultant
force is very simply we just vectorially add
F x square plus F y square plus F z square.
But the pointer application of this force
is difficult in a sense that we can find out
the force F x direction. And we can find out
the application of F y direction. And the
meeting point cut with point we know, and
if we draw a vertical line through it. Then
we will see this is the actual force will
pass through this point which will also go
through the center of volume.
So, this I will explain while solving a problem.
So; however, we can find out that this is
the resultant force whose direction can be
found out if I know the magnitudes of F x
F y F z. And whose value the magnitude is
found out in this way by vector summation
of F x F y and F z. And whose pointer application
is found out by finding out the pointer application
of all the three forces. So next, I summarize
the thing again by telling this. For a curved
surface, the net hydrostatic pressure forces
are found first by taking its components in
refer directions in any horizontal direction.
The hydrostatic pressure force is equal to
the force exerted on a plane surface which
is the projected surface on a plane which
is perpendicular to that direction. That means,
if you want to find out the hydrostatic pressure
force in x direction. We will take the projection
of this curved surface on y z plane that is
the plane perpendicular to x direction. And
treat the plane surface that is the projected
surface in a plane surface in y z plane as
the vertical surface. And then we find out
the hydrostatic pressure force in that plane.
And that will represent the hydrostatic pressure
force in the horizontal direction x direction
or y direction for the curved surface.
And the vertical component if we take one
of the coordinate axis in the vertical direction;
obviously, we will have to take the vertical
component of the pressure force is equal to
the total weight of the fluid bulk of the
fluid which is content within a region created
by vertically projecting from the surface
up to the free surface. That means, that is
the weight of the liquid vertically above
the surface up to the free surface. The bulk
of the liquid that is content the weight of
that is equal to the hydrostatic z component
of the hydrostatic pressure forces. So, this
way we can find out the pressure force in
a curved surface.
Now, few things I must tell you. This is all
about the pressure force and a curved surface.
And there may be some situations where the
free surface may not be defined properly.
For example, this is a case that in an expense
of water there is a card gate or surface like
that. Now, this is the free surface which
may not be extended in this direction. Now,
what happens? This is not a submerged surface,
but see that each and every point this pressure
is atmospheric pressure p g. But each and
every point there is a pressure force due
to the height of this.
So, there is a net pressure force in this
direction that if no. But to apply this formula
we can do it analytically. But we want to
find out the z component of the hydrostatic
pressure force. And if we try to apply this
concept that it is the weight of the liquid
above this surface. If we are told to find
out only this part, that what is the magnitude
vertical component at the hydrostatic force?
Then what we do is commonsense that we take
an imaginary free surface. That means you
extend the free surface, as if it is being
submerged with this as free surface. Same
thing, we can find out the submerged free
surface. Hydrostatic force means if this is
the surface I just take a projection. So,
this is the weight of this liquid.
So, from geometry if I can find out the weight
of this liquid easily I can tell this is the
vertical component of the hydrostatic forces.
Sometimes the situation is like that there
is a pressurized container there is no free
surface a pressurized container. This is the
pressurized container where the fluid is at
some pressure. So, the pressure intensity
there is with its depth. And here at the top
surface the pressure is p 1. So, pressure
is here p 1 but at this point the pressure
is p 1 plus the hydrostatic pressure due to
height. This is a container.
What is the force acting on this? So, you
can go with the analytical expression by integration.
But if we have to use this concept if we have
told to find out the vertical component of
the hydrostatic forces. Then what we do? This
we cannot treat as an submerged surface below
the free surface of the water. But we can
transfer that by imagining a free surface
whose here we will go up to a vertical height
is very simple p 1 h is equal to p 1 by rho
g. That means, we create some height which
is equal h p 1 by rho g and make it that is
as if this is the free surface.
And we consider that this is emerged in the
free surface. Then what is the hydrostatic
z component of the hydrostatic pressure forces?
This volume of the liquid, weight of this
volume of the liquid; that means, we can create
a free surface and consider it as if it is
submerged on this free surface. So, only thing
is that the direction will be like this if
this is a free surface. And it is submerged
then if we consider the force on this surface
this side. It will be in this direction here,
because of this pressurize chamber it will
be in this direction.
So, this will be the free surface, because
pressure here will be this pressure plus this
height, which will be same as this vertical
line. That means to extrapolate from the pressure
a free surface this is known as imaginary
free surface concept. Sometimes we have to
make to find out the vertical component of
forces easily. If we feel that geometrically
it is easier to find out the weight of the
liquid vertically above the curved surface
up to the free surface. Then an imaginary
free surface depending upon the situation
we can think off.
Now, we come to the next part Buoyancy, what
is Buoyancy? Now I come to buoyancy. Probably,
this you have heard at your school level,
there will be a little recapitulation of your
things which you have done at your school
level buoyancy. Now, you know that if a body
is totally or partially submerged. If totally
submerged or partially emerged, then the body
experience will be in a liquid or even in
a fluid body experience is upward trust, where
from this upward trust comes.
Now, this phenomenon is known as buoyancy.
And this upward force that a body is a experienced
we are also experienced, because we are totally
submerged that emerged in fluid air. So, we
are always experiencing a upward trust. So,
it is a fact that a body if it is emerged
in a fluid then it experiences totally emerged
or partially emerged it experiences an upward
force. This phenomena is buoyancy probably
at a very little let; that means, in the class
7 or 8 level we have read. And this phenomenon
is known as buoyancy and the force is known
as Buoyancy force.
So, where from this upward force comes? Please
tell; where from this net upward force comes?
Where from it comes? Where from this force
come? Tell me the force come from a force
concept, not because on another physics, then
I will as write that that physics creates
a force. Force comes from some force concept.
Why these force come? This is nothing, but
the distribution net resultant of the hydrostatic
pressure forces. Nothing else the body is
submerged or partially emerged. So, there
is hydrostatic pressure forces distribution
around the body. And then net defect of the
hydrostatic pressure forces is a vertical
upward force which is the buoyant force it
comes from the distribution of hydrostatic
pressure force.
Now, let us see what happened? That let us
consider a free surface of water free surface.
And let us consider a body like this, let
us consider a body totally emerged or submerged
body. So, what will happen? The pressure force
is will act on the body like this p. Though
I am writing it as p, but do not misunderstand
that this p is same everywhere, because this
p depends upon it height from the free surface.
So, on the surface of the body pressure forces
and experience because of the fluid that is
the surface force we have discussed.
Now, you tell me the common sense tells that
the resultant force due to this surface force
pressure forces in any horizontal direction
becomes 0. This is because; this is; obviously,
found out this balances each other; that means,
horizontal force is 0. Now, if we apply our
earlier knowledge that the resultant force
component in any horizontal direction on a
curved surface is equal to the projected area
on a plane perpendicular to that direction.
If we apply that concept now this curved surface
is a close curve surface for a body and whose
projected area in any plane is 0.
So, therefore, the concept tells that the
net hydrostatic force in any horizontal direction
is 0, but it is not so, in the vertical direction.
Why you see; obviously, vertical direction
the pressure force at these point and pressure
force at these points can never balance because
they differ in the height. Because this is
z pressure increases with the increasing z.
Now, if we take a prismatic liquid, a prismatic
liquid whose cross sectional area is d A z.
And in such away that these point the upper
surface is at a height from the or a depth
from the free surface z 1.
While this lower surface, this lower surface
of the prismatic fluid is at a depth z 2.
Then we can write that the pressure vertical
component of the pressure forces on this prismatic
element of the surface of the body is on this
surface is rho g z 1 in to d a z. Similarly,
this is acting, let this is the d F 1. So,
I can write d F 1. Similarly the upward force
at the bottom base of this prism in the vertical
direction is d F 2 which is equal to d F 2
rho g is very simple we have done earlier
a d a j. So, therefore, this is more than
this. So, therefore, a net force acting d
F 2 minus d F 1, let d F vertically upward
in the prismatic sys fluid is rho g z 2 minus
z 1 into d a z. Now, this becomes is equal
to rho g z d a z where z is the height that
means the height of the prismatic. That means,
this is simply rho g z d v.
So, therefore, if we integrate then we get
F z is equal to triple integration. That means,
over the volume rho g d v; that means, we
get rho g v. So, therefore, we get the idea
that the net hydrostatic force pressure force
in any horizontal direction is 0. But in the
vertical direction there is a force which
x upward in the body. Whose magnitude equal
to rho g times the volume of the solid body
the entire volume of the solid body? Now,
you see that when solid body exists here under
submerged condition; that means, it has displaced
the liquid in this portion.
So, therefore, we can tell that the same volume
of the liquid has to be displaced by the solid.
So, that solid makes its space. So, that is
why this v is sometimes refers to as displaced
volume displaced; that means, the volume displace
and displace volume always will be volume
of the body immerged in the fluid. In case
of a submerged body which is totally immerged
in the fluid. This displace volume of is its
total volume because it has to occupy by its
total volume. But when the fluid is partially
submerged that partially immerged not fully
immerged. Then this displace volume is the
volume of the solid that is within the fluid;
that means, that part of the volume. So, this
is referred technically as displaced volume.
So, therefore, we come to the conclusion that
the buoyant force that is the buoyancy force
which is the net upward force. And it is as
a consequence of the hydrostatic pressure
force surrounding at on the surfaces of the
body is equal to equal in magnitude to the
weight of the displaced volume of the fluid,
weight of the displaced volume of the fluid
by the solid. Or you can tell it is equal
to the weight of the fluid whose volume corresponds
to the volume immerged in the fluid volume
of the solid immerged in the fluid.
That means, in case of a submerged volume
total volume of the solid. In case of a partial
immerge solid it is the volume that is variant
different. That means, it is equal to the
volume displaced weight of the fluid whose
volume equal to the displaced volume of the
fluid by the solid. And this phenomenon was
discovered by Archimedes which is the starting
point of the fluid mechanics. And that is
why this phenomenon is known as Archimedes
principal. This phenomena is known as let
us write with pride the name of this great
man Archimedes principal. So, this is the
buoyancy.
Now, we come to the equilibrium. Now, we will
discuss the fourteen equilibrium of submerged
bodies. First, we consider equilibrium of
submerged bodies. Now, if a body is submerged
in a liquid. So, you see a vertical upward
force that is the buoyancy force. Whose magnitude
is the weight of the total weight of the fluid
whose volume is equal to the volume of the
solid, there is acting upward. What is another
force acting? That is the weight of the body.
So, one thing you will have to write equilibrium
of submerged unconstraint bodies. That means,
the body should not be constant otherwise,
there is no other force acting the body is
kept free. Another vertical another force
which is another force acting is the weight
vertically down words. There is no other force
because in a static free there is no friction
force. So, only the friction forces which
gives rise to buoyancy forces another is the
weight. So, therefore, if the weight is more
than the buoyancy force the body will go down
sink and sink. So, there is no equilibrium
it will only rest if there is a surface solid
surface.
Where the body can be kept at this, because
of the reaction force exerted by the solid
surface we will balance the net value of the
weight minus the buoyancy. But if it so happen?
The density of the fluid; the density of the
solid, and the volume of the body is such
that buoyant force equals to the weight of
the body.
That means weight of the body equals to the
buoyancy. That means, the density is same
can only happen if the densities are same.
In that case, what happens? The weight and
the buoyancy are equal in magnitude. And if
they become collinear and if they become also
collinear; that means, vertical with the same
pointer application on the same vertical line.
Then the body will be in perfect equilibrium
condition under submerged state. So, therefore,
the condition equilibrium of submerged bodies
here we can write unconstrained. That means
free that means, only weight and buoyancy
is acting.
So, now you see what happens? Let us consider
a body which is like this, let us consider
a body like this. Now, the condition of equilibrium
is that the weight acting from its C G, let
this is a C G. And buoyancy force acting through
the center of buoyancy B, let us call B or
let us call as G they have to be equal W is
equal to B. And they should be collinear.
So, they should be in the same vertical line.
But there may be a displacement between G
and B along the vertical line G may be upward
than the B or may be down to b that depends
upon the distribution of mass, relative distribution
of mass and weight. If weight is uniformly
distributed, mass is uniformly distributed
throughout the volume. Then of course, G and
B coincides, but not necessarily in general
they have to be coincide, but even if they
are not coincident.
But they should the collinear. So, this is
the condition of equilibrium of a submerged
body. Now, as you know in mechanics there
are 3 types of equilibrium; one is stable
equilibrium, another is neutral equilibrium,
another is unstable equilibrium. So, we will
discuss this stable, unstable and neutral
equilibrium in the next class. So, this is
the condition that the weight and the buoyancy
force have to be collinear for equilibrium.
But whether it will be stable equilibrium
or unstable and neutral equilibrium will discuss
in the next class. That there are 3 types
of equilibrium will define these 3 classes
of equilibrium as you already know in your
mechanics. Then we will find conditions under
which the body will be in stable equilibrium
under submerged condition or neutral equilibrium
or in unstable equilibrium. The next class
well
Thank you.
