Good morning, I welcome you all to this session
of Fluid Mechanics. And last class, we were
discussing about the stability of floating
bodies. And we ultimately came to the conclusions
that if the meta centre is above the centre
of gravity, then a floating body is instable
equilibrium. If the meta centre coincides
with the centre of gravity, it is in neutral
equilibrium. And if the meta centre is below
the centre of gravity, the floating body is
on stable equilibrium. This, we discussed.
Now we see that also we recognized or we derived
an equation representing the meta central
hide in terms of the dimension of the body;
that is the geometrical shape of the body
and its dimensions. And do if we recall the
equation is like that the distance between
the centre of buoyancy to the meta centre,
along the old vertical line containing the
centre of buoyancy, centre of gravity and
the meta centre is equals to the moment of
area of the plane of rotation, above an axis
perpendicular to the plane of rotation divided
by the immersed volume.
Now, we just see one interesting thing that
when let us consider a floating body, let
us consider a ship; let us consider a ship
like this a floating. Now, we see that if
this is the meta centre a stable condition,
if this is the centre of gravity, and this
is the centre of buoyancy, this is under stable
condition. That means, under a tilted condition,
under a tilted displaced condition, the ship
looks like this; ship looks like this. And
we have recognized one thing, that this is
M, this is the mu B centre of buoyancy B dash
and this is the G, through which the vertical
force W acting through which M. Now, we see
in this case, the restoring moment which acts
in this direction opposite to the direction
of tilt is equal to this distance into W or
H B. Let us call W and this distance can be
written for a small angle theta G M sin theta,
this is the restoring moment.
Now, if I equate this with the moment of inertia
and the angular acceleration from the conservation
of angular momentum, the theorem of conservation
of angular momentum. We can write this is
equal to the momentum inertia of these body
mass momentum inertia of these body about
the plain of or the axis of rotation axis
of rotation times into d square theta d t
square. So, this is the angular acceleration;
that means, theta is the angle of wheel at
any time instantaneous angularly at any time
with a negative sign. Because this is a retarding
one because this is a couple is a restoring
couple. So, therefore, the role is reducing;
that means, if we consider the rolling of
the ship we can write the equation for the
angular position with respective to time by
this considering the theorem of conservation
of angular momentum.
For small angle theta, sin theta can be replaced
as theta and if, if I write this these equation
is ultimately transferred to d theta d square
theta d d square plus W into G M G M is the
meta centric height divided by I into theta
is 0. So, therefore, we see that we have simple
harmonic motion for the angular displacement.
That means for the rolling of the ship, a
simple harmonic motion whose time period t
can be written has twice pi times under root
of 1 by this; that means, under root of I
divided by W into GM, this you know from a
simple mathematic. So, this is the time period
of root. So, one very interesting things comes
out from this that time period of role is
inversely proportional to the meta centric
height; that means, if we meet the meta centric
height B; that means, they give a better chance
of stability then the time period will be
very small. And time period very small means
rolling will be fast which can cause discomfort
to the passengers.
So, in ships carrying passengers sometimes
the meta centric height is may deliberately
low at little sacrifice of the stability to
increase the period of as to increase the
period time period; that means, for better
comfort the time period is increased. But
in one ships it is other way the meta centric
height is made as large as possible to have
a much better chance of stability. Whereas,
the time period of role becomes very if this
is very high so this is very small. So, time
period of role becomes very small. So, comfort
is sacrificed at the cost of stability, but
for ships carrying passengers sometimes the
comfort is not sacrifice that much with the
respect of stability. Because stability is
also first primary thing is the stability.
But what elements we should give for GM, if
you make GM very high with a better factor
of the safety sometimes it effects the, causing
the uncomfort to the passengers.
So, is this the key equation for the role?
That is all regarding this meta centric height
problem, but few things I must tell you at
this moment. Before concluding these buoyancy
thing, one thing you must know that for a
floating body even if G is above the centre
of buoyancy it is stable. Why? Centre of buoyancy
gets a chance to sit in the direction of the
tilt. So, when centre of buoyancy shift direction
of tilt because of the changing in the inners
volume, not total volume is distribution that
stability is improved. Similarly, if the gravity
also tries to change in the direction of this
tilt it will give an adverse effect of stability
that you can very well understand. So, therefore,
the gravity the centre of gravity is not allowed
to change in the direction of the tilt.
So, this may happen in a case when some cargo
may move in the ship or sometimes it happens
if the floating body contains inside a liquid.
So, when it is giving a tilt the free surface
of the liquid will be horizontal. So, some
liquid volume will be moving towards the tilt.
So, therefore, the centre of mass will be
shifted towards the tilt. So, that will give
an adverse effect to the stability. So, therefore,
the, if the floating body is contain liquid,
these are separated into a number of large
compartments large number of compartments
large sorry large number of compartment. So
that the free surface almost remains the same
there is a liquid change in the distribution
of mass and if there are some solid particles
which wants to move for example, small cargos
in a big ship that has been prevented.
So that gravity centre of gravity it is not
allowed to move in the direction of the tilt
well. Now, after this I will just solve few
examples. Let us see, which is very important,
first is example one and an inclined tube
manometer. Let us see the example one, an
inclined tube manometer as shown in figure.
I will show the figure afterwards reads 0
an inclined tube manometer as shown in figure
reads 0. When A and B are at the same pressure
this A and B that requires the figure. So
that is why figure some A and B, you just
recall it when at the same pressure an inclined
tube manometer shown as shown in figure reads
the, the reservoir diameter is 50 millimeter
as you know an inflective manometer is a big
reservoir and a small tube and that of inclined
tube which is inclined 5 millimeter which
is very small compared to the d z. For theta
is equal to 30 degree and gage fluid with
S is equal to 0.332, that is the specific
gravity S is specific gravity of the fluid.
Find P A minus P B, that is the pressure at
A and pressure at B and the difference of
this as a function of reading R. R is the
reading, that I will show in the figure how
much will the error be if the deflection of
the liquid level in the reservoir were neglected.
So, this is all right, again I am reading
an inclined tube manometer as shown in figure
reads 0, when A and B are at the same pressure.
The reservoir diameter is 50 millimeter and
that of inclined tube 5 millimeter that is
the inclined tube. For theta is equal to 30
degree and gage fluid with S 0.332; that means,
the gage fluid is specific gravities 0.332.
Find the difference of pressure between A
and B, this is P A minus P B as a function
of reading R that is the manometer reading.
How much will the error be, if the deflection
of the liquid level in the reservoir were
neglected? So, the advantage of liquid level
reservoir to be neglected is that we can make
the reservoir as I have told earlier not of
transfer in material it may be costly. So,
only we can make the inclined tube transparent.
So, this problem is well understood. Now,
with this problem.
Let us see, what exactly the problem tells.
Let us draw this is the sorry, this is the
inclined tube and this is the system A. Let
this is the point A, where the pressure is
exerted be and let this is the point B, where
the pressure is exacted P B; that means, A
and B this 2 open ends of this 2 sides, one
is the reservoir and other is the short link
are connected to 2 systems is pressure P A
and P B. Well let us consider, the initial
level of the liquid be here, let us consider
the initial level of the liquid when there
were no pressure A and B, P A and P B equal
that is initial; that means, initial level
means, if the manomate.
This is the manometric, full of manometric
liquid, then this will be at the same level;
that means, the reserve it is written in the
problem you see that when A and B are at the
same pressure; that means, the system pressures
are same they are not connected to A and B
which are varying impression either way you
can understand. The reading figure reads 0;
that means, inclined tube manometer reads
0; that means, here the reading is 0 this,
we measure like this. Now, what happens, when
this pressure is applied when A and B, it
is applied to that when P A is greater than
P B, then what happens this limb will be decreased.
And here this will be increased; that means
the manometric fluid.
That means this is the manometric fluid. So,
this reading we tell it as R, which is graduated
along, this tilt this reading we tell it as
R, this is the reading. Now, we have to find
out P A minus P B as a function of R, which
is very simple. Now, what we have to do? We
have to find out, this is the level, now this
is the. So, we are finding out the pressure
at this point. So, let us consider a hydrostatic
equation, we are writing at this level from
both the sides. Let us consider this deflection
as Y and this is definitely. So, this one
is definitely, if this be with theta, this
is R sin theta. So, actually truly speaking,
we will have to use hydrostatic equation at
this level of the liquid.
So, original level of the liquid is not very
important. Now, we have to find out the difference
in level both the limbs like a simple, you
tube manometer from this side, what is this
P A is equal to now here the height, we are
not considering, consider a system comprisable
fluid gas which gives the pressure state to
this P A. Because you know the height of the
fluid in case of a gas used density is low
is neglected. So, it is not a liquid extractive
system pressure which maybe here is just impressed
on this surface. So, it is simply P, which
is equated from this side P B plus this height;
plus this height of the manometric fluid;
that means, plus R sign theta plus Y. Well
R sign theta plus Y into rho into G where
rho is the density of the manometric fluid
here. Now, what is Y? Y is from the continuity.
That means the amount of liquid which is been
depressed in the reservoir has gone here.
So, therefore, one can write that this Y into
A is equal to where A is the area of this.
Well is equal to this R, which has moved along
this pi R into A; that means, this equals
the same volume of the liquid which has come
down here has gone up there. So, with this
we can write P A minus P B is equal to what
R sin theta, if we take common, 1 plus, what
is this Y is equal to R A by A; that means,
A by A sin theta. Any question? A simple mathematics
at school level, but the concept is that at
this point. We are equating the pressure from
both the limbs as we do in case of a manometer
U tube manometer it is a modified version
of an U tube manometer.
Now, therefore, now we see that without doing
that. If we have a not a transparent material
with the reservoir only this part is transferring,
we do not know this level we only read 0 at
its initial level what is the problem statement.
Then we only see the deflection; that means,
the movement of this liquid in the inclined
tube and only read R and try to find out P
A minus P B as only R sin theta into row G,
then this is the actual write A and this is
the measured.
So, if we measure only the movements from
0 to this that initially, it was 0 and we
moved measured the only R reading that is
the what is that reading movement of the liquid
in that tube. Then this will be the reading
and it is common sense just you see from this
2 that the difference, this difference between
this 2 will be very less provided A is very
less than A. If A by A is very small than
1 sin theta is not very small in that case,
because it usually varies between 10 20 30.
So, if A by A is very small. So, this part
is negligible this 2 will be equal. Now, the
problem is to find out the error. So, if you
define error as let this is delta P actual
and this is delta P measured then.
If we define the error e, well then e is equal
to in terms of percentage delta P actual minus
delta P measured divided by delta P actual
this is the percentage error 100. If you make
that; that means, this becomes this can you
see the delta P actual minus delta P measure.
So, if you make that this will be 1 by 1 plus
A by a, a simple algebra if you make 1 by
A plus; that means, delta P makes actual minus
delta P measure . So, therefore, remunerated,
it will be R sin theta A by a sin theta. And
denominator, it will be delta P actual R sin
theta 1 plus A by a sign theta R sin theta
will cancel. And the, we again make the numerator
1 by dividing A by A sin theta.
So that in denominator we will get that is
the simple algebra. Now, if you make the substitution
here, A is this 50 millimeter and this diameter
is 5 millimeters small. So, this whole square
is at the diameter into sin theta half. So,
this comes out to be into 100 well and this
will be 1.96 percent, any difficulty? So,
this is the value of delta P A, this is the
value of delta P M. This is the freed mechanic
after that it is a class 8 level algebra.
So, you can find out the delta P A minus delta
P M by delta P A into 100 becomes equal to
this. So, if you put the values, you will
get this 1.96 percent think that is nothing
great only the concept is here. Now, come
to this the second problem which is very interesting.
Second problem; well second problem as you
see it is the problem related to the pressure
on a carte surface. Pressure on a simple surface
is very easy, if you have any problem you
just ask me. Do not make any noise in the
class. You just ask me, is there any problem,
difficulty? Please. Then now you see this
problem tells like this example 2; calculate
the force if required to hold the gate in
a closed position. There is a gate which is
a sector of a circular R; that means, it is
a, this is 0.6 meter; this is 0.6; this is
0.6. So, this has to be understood from the
figure. So, this sector of a circular R, this
gate is 1.2 meter wide, which I had forgotten
to write 1.2 meter wide; that means, the gate
is like this, this is plain in the perpendicular
direction. This is the plain of its, plain
of the oil plain of the gate.
So, in this direction perpendicular direction
the gate is 1.2 meter wide; that means, it
is a sector of a cylindrical surface, 1 sector
quadrant, 1 quadrant of a cylindrical surface
not sector. I will tell 1 quadrant of a cylindrical
surface used is that is 0.6. So, this is connected
to a closed box with oil and water at pressure
how do you know. Because if you attach a manometer,
this manometric fluid is mercury, this manometric
fluid is mercury Hg. So, with the manometric
fluid, we get a deflection; that means, this
is open to atmosphere this part is open to
atmosphere so; that means, there is a pressure
in these oil and water. So, what is due to
this pressure in the oil throughout? So, a
pressure force is exerted on the carte gate
which is a quadrant of a cylindrical surface,
what is the force of this.
So, what should be our first duty to do please
tell me, what should be our first duty to
do in this problem? To find out, please tell
pressure at this point, top point. Let this
pressure at the top point is P, we should
find out the pressure at this top point. So,
what to do, you write the manometric equation
to find out the pressure at the top point;
that means this part is the water which you.
So, if we write the manometric equation at
this plain, what we get from this side P plus.
What we will write the pressure due to this
height this column of oil; that means, rho
G H, what is rho density? Oil S means specific
gravity; that means, 0.8 into 10 to the power
3 is the rho into G 9.81 rho G into H. What
is h? 0.6 rho GH, rho G H plus the height
of the water column equivalent to 0.6 here
0.2 here and also 0.6 here; that means, 0.6
all water column plus 0.2 plus.
Again 0.6, you come to this point from this
straight plus again 0.6 as it is given in
the problem this deflection is 0.6 times the
density of the water is 10 to the power 3
H rho G rho H g. Whatever you write H rho
into G 9.8 here, H is 0.6 rho is this and
g is this H rho G. So, starting from this
point P that is pressure P we come here. Equate
from side is equal to it is atmospheric pressure
0. If we want to find out the force due to
the pressure above the atmosphere, because
this side atmospheric pressure force is acting.
So, therefore, we take atmospheric pressure
as 0. So, 0 plus only this column of liquid
0.6 and manometric fluid is mercury, whose
density is rho is 13.6 to the power 3 H rho
into G 9.8 imply. So, if you equate this equation,
if you solve this equation P will come equal
to some 60, well 61.61. You check that 61.61
into 10 to the power 3 Newton per meter square
well; that means, 61.61 kilo Newton per meter
square.
So, this is the P, now next part is to find
out. So, this we know the P, now we should
find out the pressure on this carte surface.
Let us apply our idea that what is the horizontal
force on this carte surface. Please tell?
What is the horizontal component of force
in the carte surface? Yes, if we make a projection
of this area like this the horizontal force
which is acting on this projected surface,
projected plain surface is the horizontal
component of the force acting on the carte
surface. So, what is the horizontal? What
is the force acting on the horizontal plain
surface, please tell?
What is the horizontal force acting on the
plain surface? That is the pressure at the
centroid. So, centroid is at the middle; that
means, 0.3 meter from the surface. So, what
is the pressure force at the 0.3 meters? What
is the pressure intensity; that means, this
P plus rho G H rho is 0.8 10 to the power
3 rho G into 0.3. That is the pressure at
its centroid; that means, centroid of the
centre of area of the plain surface, which
is a plain surfaces uniform plain surface,
whose depth is or height is 0.6 meter. So,
that is 0.3, so at the middle the centre of
the area this is the pressure intensity.
So, then the force is acting not at the middle,
this is the centre of the area. Into the area,
what is the area now? Area is 0.6 into 1.2
very good. So, this will come out to the horizontal
component 46.05 into kilo Newton, 46.05 kilo
Newton. Now, the next part is the vertical
force. How you find out the vertical force,
please tell me? The vertical force will be
equivalent to the weight of the liquid above
this carte surface. Please do not talk, please
tell me whether it is all right or no. If
you do not understand you please ask me, Yes
please. Density of finding F is at an average
force that is active on the gate, not average
force. So, you have not attended the class,
the thing is that the horizontal component
of the force is equal to force acting on a
plain area which is oppositional area in to
that direction.
So, the force acting on this plain area consider
a vertical area is equal to the pressure at
the centroid, no question of average times
its area. If we integrate the force component,
you will get like that which has been already
told in the earlier classes. So, do not ask
the silly question, that it is an average
that it is the total force acting which is
the integrated force of all elemental force
components on the surface which finds out,
which is found out to be the pressure intensity
at the centroid times the area. The simple
formula was derived in the class earlier.
So, therefore, in any plain surface in submerge
surface the total force acting on the plain
area is equal to the pressure intensity at
it centroid times, its area this has been
derived so, no question of average and all
those things.
Now, what is the vertical component of force?
The vertical component of forces act on this
surface, if we find out the weight of the
liquid above the carte surface that also,
we derived in the earlier classes, those who
have attended the earlier classes they knows
them. If you have not attended you please
see earlier lecture. So, this was derived
vertical component is equal to the weight
by magnitude, weight of the surface. This
is the easier way to do weight of the liquid
above the surface up to the free side; that
means, we will have to consider for example,
a free surface. Let us make a gap, because
space is not there as, if there is free surface
whose height is h. This H; this H you can
see this H.
If this is the height of the free surface
which is extrapolated imaginary free surface
that also was discussed in the earlier class.
This becomes equal to this simply pressure
P here by rho G; that means, this simply equal
to pressure. What is this pressure? 61.61
into 10 to the power 3 divided by the rho;
that means, this we are having, we are finding
out that imaginary free surface. That means,
if the oil is allowed to go; that means, without
this chamber there is a pressure; that means,
oil free surface could have been extended
to a height h from here, which is equal to
P by rho g. That means, 0.8 10 to the power
3, it is very simple 9.81. And that becomes
equal to that exactly becomes equal to 7.85
meter; that means, we get an h is equal to
7.85 meter. Now, it is very simple, that we
can tell now that it is the weight of this
liquid vertically above the surface as; that
means, weight of this part of the liquid up
to the free surface.
So, we will have to find out the weight of
this liquid, the weight of this liquid. So,
how to find out the weight of this liquid,
please tell me, how to find out the weight
of this liquid? That is yes, you can find
out the volume weight of this liquid means
the volume of this liquid, volume of this
parallelepiped, whose area this area, this
into this and multiplied by this minus the
volume of this cylindrical part. So, therefore,
we can write a F v is equal to rho, rho is
0.8 10 to the power 3 better we write rho
and into the volume. What is the volume, volume?
You can write that this is equal to 7.85 plus
0.8; that means, volume is 7.85 plus 0.6 very
good into minus this side, 1.2 is same minus
Pi into 0.6 whole square by 4. Because this
is the quadrant of a circular surface quadrant
of a cylindrical surface, this side 1.2; that
means, this into 1.2.
So, this is the volume of this liquid, 7.85
plus 7.85 plus 0.6 into 0.6; that means, this
side is 0.6 into 0.6, oh 7.5 very good; very
good; very good. I am very happy; very happy,
7.85 into 0.6, then it comes meter square
minus 5.6 whole square by 4 very good, into
1.2, very simple thing. So, we get the value
of F d. So, what is F d? F d is 45.08 kilo
Newton, this divided by 10 to the power 3.
If you make, this will be kilo Newton. Now,
the question is that there is a hinge here,
let the point is o. Now, we have to find out,
to find out F, the question was, what is the
force F required to hold the gate? So, therefore,
to find out a, we will have to take the movements
of the forces above well, movement of all
the forces. So, therefore, we must know the
point of application of the forces, now one
catches that do not try to find out in this
type of problem.
The point of application of this horizontal
force and the vertical force rather the catches
like that as because this is the cylindrical
path. All the forces will be perpendicular
to the element and this perpendicular always
passes through the radius; that means, these
are the radial linewhich are always perpendicular
to. So, by which, we can tell that the resultant
force is a line, which is perpendicular to
this and passes through the separate origin
here this origin. So therefore, this is another
radial line, because all the force components
act along the radial line they may not be
parallel, but the radial line.
So, therefore, the intelligent part of this
work is to assume that one radial line is
the line of action for the resultant force
F. F R for example, resultant pressure forces
F R, let the angle makes this resultant force
with the vertical with theta with theta, then
you take simply movement, then you take simply
movement. But before that you have to find
out what is the value of this theta? What
is the value of this theta, this theta value?
You can find out tan theta is equal to F H
by F V.
Simple with that F H that is the h component
divided by F H theta and F H and F B are known.
And if you write, if you use this value of
F H and F B, we will see this is 46.05 divided
by 45.08 and this becomes equal to 1.02 and
theta is 45 point, this you check 5 7. Now,
when this theta is known the resultant force
is specified. Now, I can take movement about
the hinge o to find it. So, F into points,
what is this 0.6 is equal to the resultant
force F R. They are in the opposite direction
into the perpendicular distance from the hinge
point to the line of action of the resultant
force, which will be equal to 0.6 sin theta;
that means 0.6 sin theta well, 0.6; 0.6 sin
theta; 0.6 sin theta is this. So, this gives
you F equal to 46.01 kilo Newton. All of you
have understood. Now, we come to the third
problem, well third problem is a very simple
problem a uniform wooden cylinder has a specific
gravity of 0.6.
A uniform wooden cylinder has a specific gravity
of 0.6. Find the ratio of diameter quick?
Find the ratio of diameter to length of the
cylinder? Find the ratio of diameter to the
length of the cylinder? So, that it will just
float upright in a state of neutral equilibrium
in water probability. One of the simplest
problem in meta centric height, a uniform
wooden cylinder has a specific gravity of
0.6. Find the ratio of diameter to length
of the cylinder. So, that it will just float
upright in a state of neutral equilibrium
in water well. So, this is a very simple problem,
oh God again, a uniform wooden cylinder consider
a wooden cylinder specific gravity of 0.6.
Find the ratio of diameter to length of the
cylinder? So, that it will just float upright
in a state of neutral equilibrium, these are
the pertaining points.
Now, what is this? You consider a wooden cylinder
and it is floating in the equilibrium, S is
equal to specific gravities 0.6. Let us consider
the height of the wooden cylinder as h; that
means, so if this be Y, what will be the value
of Y. The height of the floating part uniform
wooden cylinder, Y will be how much 8 9 level
thing 0.68. How, because weight is balanced
by the weight, where is the weight X? Weight
X by h by 2; that means this is the centre
of gravity through which the W acts. And this
led this point is o base point o G is equal
to 0.5 h. Now, centre of buoyancy acts at
Y by 2, this is the centre of buoyancy, this
is A A B.
So, from the equilibrium consideration, W
is equal to A B, what is W pi d squared. d
is the diameter of the cylinder times h into
density of the cylinder rho’s is equal to
the buoyant force. Buoyant force is the weight
of the h rho into G of course, G is there
G is equal to the weight of the displaced
volume. What is the displaced volume pi d
square by 4 into Y into rho of sorry, into
rho of water into G. Now, rho C is equal to
0.6 rho of water. So, from which we get Y
is equal to 0.6 h, it is very simple plus
line 9 level, it is now. So, when 0.6 is the
specific gravity Y is 0.6 h.
So, therefore, B is the centre of buoyancy
for the uniform cylinder, it will be at the
middle of this Y. That means, 0 B is equal
to 0.3 h; that means, this distance is 0.3
h and 0 to G, the distance is 0.5 h. Meta
centre will be above the centre of gravity
for stable equilibrium and for neutral equilibrium,
it is coinciding. Let us discuss in general
M is above this. So, what is the value of
B M, we know B M is I by Y. What is the concept
of I? I is the movement of area of the plain
of rotation. If I take a sectional view, I
will see a circle that is the circular cross
section. And we have to take the movement
of area of this circle about this axis that
is the axis perpendicular to the plain of
rotation. So, this axis we will take plain
of rotation. So, about this axis, what is
the movement of area I, I is equal to pi d
4 by 6 d 4.
So, this is the movement of area of this circle,
circular area of diameter D. So, this is an
another diameter; that means, the movement
of area of a circle about its diameter times
the volume, what is the beamers volume, please
tell me? pi d square by 4 into this one; that
means, 0.6 h. So, this is equal to B. So,
what is M G or G M, B M minus B G? Well B
M minus B G so, B M minus B G understand,
this B M minus B G. Now, B G is what is B
G? Let us write here B G is; B G is o G minus
o B. Yes, 0.6 h into pi d square by 4 y is
0.6 h. So, Y is 0.6 h, it is immerged. When
you are correct, it is immerged volume that
is why pi d square by 4 h, not h 0.6 h. So,
B G is 0.2 h and B M is this one.
So, therefore, next part is very simple, that
is g m is equal to pi. So, now, you cancel
it; that means, d square by, then it is 16
into 0.6 that is 9.6 into h minus 0.2 h. This
is, this has to be greater than 0 for a stable
equilibrium for neutral equilibrium. According
to the problem, this is 0. So, this gives
d by h equals to 1.386, from this you get
a value d by h is equal to 1.386. So, this
is a very simple problem of meta centric height.
Please tell me, whether there is any problem,
today I think we have completed this second
chapter. Of course, the time is short; otherwise
I could have given a closure lecture today.
So, next class, I will be giving a short closure
lecture of what we have covered in this section
and we will pass on or we will start the next
section kinematics of fluids.
Well, thank you.
