Welcome back to the video course on fluid
mechanics. In the last lecture in fluid statics
we were discussing about the buoyant forces
and then Archimedes principles and later issues.
So in the Archimedes principle you have seen
that, when a body is immersed in a fluid.
Either wholly or partially, it is buoyant
or lifted up by a force which is equal to
the weight of the fluid displaced by the body.
So this we have seen then we have seen that
we can equate the buoyant force to the weight
of liquid displaced and then we can apply
this theory to many engineering problems.
The point through which the buoyant force
acts as going through the center of buoyancy
and the buoyant force passes through the centroid
of the displaced volume. So that also we have
seen in the last lecture and then we can get
to the buoyant force to the weight of the
liquid displayed. So that means the fb is
equal to gamma into the volume so this volume
displaced to be liquid. So this theory we
can apply to many engineering problems and
also say equipments like hydrometer which
is used for the specific determination is
based up on this principle.
Now based up on the buoyant force and the
Archimedes principle we will discuss the stability
and equilibrium in fluid statics. So as we
have discussed earlier the position floating
body or a submerge body or a body can be either
in stable equilibrium or unstable equilibrium
or say neutral equilibrium.
So a stability of a submerged or a floating
body meant the tendency for the body to return
to the original upright position after it
has been displaced slightly. So here in this
slide you can see that here in container there
is water and then say here you can see there
is a body and then we are just putting a small
force on this. Then say we are observing what
will happen whether it will restart back or
it will overtone, so that is what will discuss
in the stability of a submerged or a floating
body. So as I mentioned so here if you have
water in container so we can put a body here.
It is floating you can see that when we give
an overtone what is whether a moment or force
is applied. So what happens to the body whether
it is coming back or it is returning? So that
is what we are discussing here. So as far
as the stability and equilibrium is concerned
conditions of equilibrium can be stable or
unstable or neutral so a body is said to be
in a state of stable equilibrium.
If a small angular displacement of the body
sets up a couple that tends to oppose the
angular displacement of the body thereby tending
to bring the body back to its original position.
So here in this slide you can see here there
is a fluid and then you can see there is a
body and then a force is applied and then
a moment is given. So it is the couple tends
to oppose the angular displacement of the
body and tending to bring in back.
So here in this small experiment here you
can see there is water in a container and
then body is floating. So here I am just giving
a small returning movement like this but you
can see that still the body is returning to
its original positions by just making it stale.
So this kind of equilibrium when a body is
if the small angular displacement of the body
sets of a couple that tends to oppose the
angular displacement so this kind of equilibrium
is called stable equilibrium.
So we have already seen in the center of gravity
and center of buoyancy so generally a body
is said to be in stable equilibrium if center
of gravity is below the center of buoyancy.
So here in this slide you can see here the
center of gravity see and then center of buoyancy
just above that so if cg center of gravity
below the center of buoyancy then the body
is said to be equilibrium and then there will
be a restoring couple so that it will be stable.
Then, unstable equilibrium a body said to
be in a state of unstable equilibrium if a
small angular displacement of the body set
a couple that tends to further increase in
the angular displacement of the body thereby
not allowing the body to restore its original
position. So this is so called the unstable
equilibrium. So here we just see a small experiment
here. So here there is a bodies floating I
am just putting a small weight up on this
floating body.
And then what happens is if I am giving a
returning moment to this body here so if say
as we have seen it is a stable equilibrium
means if it is turning back to its original
position but here when if it is returning
the returning couple is formed and then it
is going down. So like this you can see now
it is the body is said to be unstable equilibrium.
Since it has gone from the previous position
to a new position and then it is say in returning
couple took place and body is not allowed
to distort its original positions.
So this is so called the unstable equilibrium.
So if we observe in experiments the center
of gravity and center of buoyancy. So when
a body is floating or submerging a quantity
of liquid will be equivalent quantity of liquid
will be displaced so that center of buoyancy
through where that how much liquid is displaced
so that if that center of buoyancy is below
the center of gravity then you can see that
it will be generally unstable equilibrium.
So a stable equilibrium is when a center of
buoyancy above the center of gravity and a
body is said to be unstable equilibrium if
the center of buoyancy is below the center
of gravity as shown in this figure do that
returning couple is formed and then the floating
body becomes unstable.
A third category of stability and equilibrium
is neutral equilibrium say a body is said
to be in a state of neutral equilibrium if
a small angular displacement of the body does
not set a couple of any kind and therefore
the body adopts the new position given to
it by the angular displacement without either
returning to its original positions or increasing
the angular displacement.
So in this case a body is there so what happens
is it is when a small displacement is given
if a body does not set a couple of any kind
and therefore the body adopts the new position
as given like this. A body whether a floating
body or submerged body in fluid static and
say that if the body is stable equilibrium
or the body is unstable equilibrium or the
body is in neutral equilibrium.
Especially, when we are using in the vessels
and bought learning the stability of the body
is very important. So we will be using some
of this principle as far as to determine whether
the body is stable or unstable in many of
the practical engineering field. Now related
to this say stability and also the buoyant
force and then the center gravity and center
of buoyancy the one important term is called
the metacenter for floating bodies so what
is metacenter.
Here in this slide you can see that there
is a small body or a small vessel is say floating
in a floating condition and then this is the
center of gravity or the body is at g here
and then since it is partially submerged the
center of buoyancy is at point b here in this
figure and the body weight is indicated here
also w and then the buoyant force is there
so that is fb acting in the opposite direction
so since it is in stable equilibrium it is
in a condition fb is equal to the weight of
the body. So this is the situation.
And now we are just giving a small displacement
like this in this second figure we are giving
a small displacement to the floating body
partially floating body like this a small
displacement is given. So then you can see
that this line is partially shifted in this
direction so that the new position of the
body is inclined position like this and then
we can see that say due to this effect say
the weight of the liquid this based this slide
changes and then it is say the position changes
and then the center of buoyancy shifting from
b to b one.
So this new position is b one and then you
can see that when if you put a vertical from
b one and then if you draw a line from say
the center of buoyancy to the old position
center of gravity then it those two lines
will coincide at position m. So this is called
the metacenter and this you can see that here
small angle is formed due to which will be
equal to this angle here.
Metacenter we can define as metacenter is
the point of intersection between action of
floating body passing through the point b
the center of buoyancy and g the center of
gravity and the vertical line passing through
the new center of buoyancy. So for an equilibrium
a body in equilibrium in floating or submerge
condition we are giving a small team or small
displacement and then you can see that the
center of buoyancy shifting so the metacenter
is the point of intersection between the action
of floating body passing through the points
b or g the old positions and then vertical
liner passing through the new center of buoyancy.
So this is generally the metacenter and the
metacenteric height is generally used to see
whether a body is in stable equilibrium and
also how much vessels like ship or above how
much weight it can take without any unstable
problem. This is important of this metacenter
so the metacenter height is the distance between
the center of gravity g and metacenter m.
So in the previous slide we have seen so here
this distance between this line and this line
is intersection is called metacenter and the
metacenter height is the distance between
g to m so here this distance gm is called
the metacenter height. So metacenter height
is the distance between the center of gravity
and the metacenter m.
So the metacenter floating bodies it is very
important we have to determine the metacenter
and we have to see what is the position of
the metacenter that we can determine the weather
bodies in stable conditions or unstable conditions
Here if m the metacenter lies above g so that
righting moment may be w into w the weight
of the floating body gm the metacenter height
into sin theta.
So this to make the body is said to be equilibrium
is stable the body is stable equilibrium if
this gm is positive and if m lies below g
that means say in the previous slide. So here
if this m lies below g when we are giving
a large displacement and due to other problems
it can lie below g then there will be a overturning
moment and then you can see that again this
will be this overturning moment will be w
the weight of the body multiplied by the gm
into sin theta and then the equilibrium is
unstable and here gm is negative.
So with respect to the metacenter also we
can say that whether there is stable equilibrium
or whether there is unstable equilibrium.
A stable equilibrium is when gm is positive
or when m lies above g it is the body said
to be stable equilibrium and if m lies below
g then the body said to be in unstable equilibrium
and in the case if m coincides with g the
body is said to be neutral equilibrium so
that metacenter and the center of gravity
of body coincides then we say that the body
is said to be in neutral equilibrium.
So if we determine the metacenter position
and the metacenter height we can determine
whether the body is stable equilibrium or
unstable equilibrium or in a neutral equilibrium.
So in the next slides let us see how we can
determine this metacenteric height
So here there is a floating body like this
floating vessel.
You can see there is a weight put on one side
of the vessel and then we are moving this
vessel from one position say this is old position
of the vessel we are moving this vessel this
body on the vessel the small body on the vessel
from one position to another so that there
will be a small displacement will be taking
place and then we what will determine the
metacentric height.
So metacentric height as we have discussed
the height of we can determine if angle of
tilt a theta that means with respect to the
position of this body, there is a small tilt
theta caused by moving load then we can determine
it is a distance non x across the deck is
measured so that is the metacenteric height.
So here in this case the overturning moment
of load this load p is moving from one side
to another and if x is the distance from the
old position to the new position then overturning
moment which the moment of the load is equal
to p into x.
Then the next figure shows more clearly about
this. So to determination to determine the
metacenteric height so this is the floating
body and small weight is placed. so when it
is in this position it is in stable equilibrium
and there is not tilt and when we are shifting
this small lot to one side you can see that
there will be a tilt for the displacement
and there will be therefore the floating body
like this and then you can see that a small
angle tilt you can see this is the new position
of this for horizontal line.
So there will be a small angle of tilt and
then small w is the weight of this body here
and w is the weight of the floating body which
initial pass through the center of gravity.
So g and then if b is the buoyancy center
of buoyancy initially then if there is w is
placed like in this figure then you can see
that buoyant force is equal to weight of the
body, that is a stable equilibrium. Now we
are moving this slightly to one side so that
we want to determine the metacentric type.
If gm is the metacentric height and the w
is the total weight is equal to mg including
p that means the p is the load which is moving
and this mg is the weight of the floating
body. So the writing moment we can write w
into gm into theta is very small and theta
is the angle of tilt with the horizontal so
in equilibrium in tilted position we can write
the writing moment is equal to overturning
moment that means we are assign is the body
is equilibrium we can write writing moment
is equal to the overturning moment.
So we can write w into gm the metacentric
height into theta is equal to p into x so
where p is the load placed over the floating
body and x is the distance between old position
and new position. So that metacentric height
we will get gm is equal to p into x by w into
theta so where this w is the total weight
of the body including the small weight p.
So the true metacenteric height is value of
gm as theta tends to zero. So here this theta
tends to zero here this w is the p which is
integrate earlier and capital w is the weight
of the floating body and this small w is the
or is that is equal to p. So this will be
the metacentric true metacentric height is
the value of gm as theta tends to zero. So
here you can see theta here in this slide
so when that tends to zero to metacentric
height. So for this small theta this derivation
is valid so that we can write gm is equal
to either p in the previous figure or in this
figure w is px by w tan theta or when theta
is more we can write w theta.
So as we have seen here in this slide. So
this derivation is with respect to the previous
figure here and then this same thing we can
write gm is equal to w small w x where this
small w is the weight of the body small body
on the vessel and then w is the total weight
so gm is equal to wx by w theta or when it
is theta is more we can write w theta as indicated
in this slide.
Now like this we can determine the metacentric
height so now just we will discuss a small
example.
To show how we can determine - we will discuss
the small metrical example to determine the
metacentric height. So here you can see here
n this paper here. There is a floating body
like this and there is a small weight is placed
like here. So we want to determine the metacenteric
height. So the problem is the weight of the
ship is 150 Newton the ship boats or one side
are filled with water and weight 0.85 atoms
Newton. The mean distance from the center
of ship to the boat is 8.5 meter that means
this distance from the center of the ship
to this boat which is placed here is 8.5 meter.
If the angle of displacement of the plb line
theta is three degree fifteen minutes calculate
the metacentric height. So this boat here
is placed here is big ship here and a small
boat is placed that 8.5 meter find in center
line of the ship. So since this ship is placed
at a distance 8.5 meter small tilt therefore
the ship so that tilt is given as 3 degree
15 minutes.
So we want to determine the metacentric height.
So this problem here the ship is given and
here the small boat is placed 8.5 meter and
then initially you can see if say the central
line it is in stable equilibrium the boat
is not considered then the center of gravity
is here and then the weight of the ship is
w which is given as 150. So here this with
respect to this position of the boat is g
is since the small weight is w is added so
g is shifting slightly to g dash. So the metacentric
height as per the previous derivation you
have seen earlier we can determine metacentric
height is gm is equal to w the small w that
means the weight of the boat w into this distance
x divided by w tan theta where here this small
w is weight of the boat.
So that is equal to 0.85 and so here as we
have seen that w is capital w is total weight
of the ship plus the boat so the capital w
is equal to 150 the weight of the ship 0.85
is the weight of the boat. So total capital
w is equal to one 50.85 and then x is equal
to 8.5 meter and theta is told given as 3
degree 15 minutes. So metacenteric height
is gm is equal to wx divided by capital w
tan theta S here the gm the metacentric height
is equal to w is 0.85 multiplied by 8.5 divided
by 150 .85 that is the w is the total weight
the 150.85 into say tan theta tan 3 degree
15 minutes 0.05. So, the 150.85 multiplied
by 0.568 from which we will get the metacentric
height as pint eight four three meter.
So like this simple example we can determine
the metacentric height by placing a small
weight and then we are trying to determine
the metacentric height. Similarly, as you
have seen in the previous slides so in a very
similar way we can determine the metacentric
height for the floating bodies. The next with
respect to this metacentric height we want
to determine the position of metacentric relating
to the center of buoyancy. So here now this
paper here we want to determine the position
of metacenter.
Relating to the center of buoyancy, so that
is the problem. So here you can see there
is a big vessel and then the old position
the position is horizontal and when there
is a small tilt you can see that the new position
here is same. This a aa indicate this horizontal
line indicate the old position and d to the
small tilt theta the new position is say here
the inclined line. So ac is the original water
line ac is the original water line and a dash
c dash is the new water line and then b is
the center of buoyancy in equilibrium condition
so here you can see this b this is the center
of buoyancy equilibrium condition ang g is
the center of gravity.
So a small tilt theta is given and that is
the angle of tilt and then with respect to
this the vessel is slightly tilted towards
the right and then a dash c dash is the displaced
position for this vessel. Now we want to determine
the position of metacenter relating to the
center of buoyancy so that is the problem.
For small angle theta as shown in this figure,
we can write dm which is the distance b can
be written as this b dash that b dash divided
by theta when the angle is very tilt angle
of tilt is very small; we can write bm is
equal to b dash divided by theta. So total
weight of the fluid displaced unchanged. Since,
we are giving a small tilt only with respect
to this vessel here.
So the tilt is very small so due to that we
can see that the total weight of the fluid
displaced is unchanged and then the weight
of the wedge we can with respect to this the
weight of the wedge with respect to the fluid
displace this aoa dash is equal to weight
of the wedge c dash oc dash coc dash. So aoa
dash is equal to coc dash and now to determine
the position of the metacenter relate to the
center of buoyancy.
We will consider a small area a like in this
figure place a small area is considered in
this figure here a small area is considered
and it is at distance x from the center line
o.
So this is the center line and we are considering
a small area at a distance x from oo. Now
with respect to this when the vessel is tilted
you can see that the volume set by a that
means the small area which we have seen here
the small area when that vessel is tilted
the volume set out by a is equal to d dash.
So here this is the position of the small
area which we are considering and this d dash
distances this between the old position new
positions. d to d dash so d dash multiplied
by a so that can be written as that is equal
to the area small area into a distance x into
the angle theta so the volume set out by is
equal to ax into theta.
So we are considering a small area now you
sum up all such volumes and multiplying specific
weight og if the liquid you can see that here
in the figure the weight of the aoa dash is
equal to say aoa dash is equal to sigma x
equal to 0 to ao rho g ax theta.
Similarly the weight of the wedge coc dash
we can write as sigma c is equal to 0, x is
equal to 0 rho g ax theta. Now since as we
have seen here the weight of the wedge a so
a dash is equal to cos dash so that we can
equate this weight of the wedge aoa dash is
equal to weight of the wedge coc dash. So
we can equate so that you will get here rho
g theta into sigma x equal to 0 ax equal to
ax is equal to rho g theta sigma x equal to
0 to co ax.
So that we can find you will write sigma ax
is equal to 0 where the small area which h
we have consider and x is the distance from
the center line of the vessel. Now this we
can see that these sigmas ax it is the first
moment of area of water line plane about the
center line o. So this sigma ax is the from
the center line o it is the first moment of
area so that we can say that axes must pass
through the centroid of the waterline so that
sigma ax is the first moment of area of water
line plane about oo.
Now, couple produced by moment of wedge aoa
dash to coc dash. Since we are providing a
small tilt here and the vessel is tilted so
when it is the wedge couple produced by moment
of wedge aoa dash to coc dash equal to the
couple due to moment of this the resultant
force r the reaction force r from b to b dash
so the couple produced by moment couple produced
by moment of wedge aoa dash to coc is equal
to the couple due to the moment of this reaction
r where r is equal to w.
So due to the moment of r from b to b dash
the new position of the center of buoyancy
so hence the b dash we can calculate. Now
we will take the moment of the weight of fluid
said out of bay area at about oo that here
the moment of weight of fluid said out by
area at about oo is equal to rho g ax theta
into x since you have take in a moment rho
g ax into theta into x. Now we can write the
total moment due to the altered displacement
as rho g theta sigma ax square total moment
due to the altered displacement.
So here you can see that there is a displacement
here. So with respect to this we can write
total moment due to the altered displacement
is equal to rho g theta sigma ax square. So
here you can see the sigma ax square is I
which is the second moment of area waterline
plane about the center line.
So this is second moment of area of water
line is I so that sigma ax square is equal
to I. Finally, we can write the total moment
due to the altered displacement is equal to
rho the density multiplied by g acceleration
into gravity into theta into i. So the total
moment of due to the altered displacement
is equal to rho g theta into i. So moment
due to the moment of r finally we can write
r into bb dash is equal to rho g into v into
bb dash, where v is the displaced volume so
rho g into v into bb dash where v is the volume
of the liquid displaced. Now with respect
to this finally, we can find out the bm that
means the metacenter relating to the center
of buoyancy that is what we are trying to
find out.
Finally you can equate rho g into bb dash
is equal to rho g theta into i . Finally we
get bb dash is equal to this rho g is canceled
from this side and then bb dash is equal to
theta the angle of displacement into i the
second moment of inertia divided by the volume
of the liquid displacement.
So bb dash is here you can is the new the
distance from the old position of center of
buoyancy due to the new position bb dash is
equal to theta the angle of displacement and
i is the second moment of the area and v is
the volume of liquid displacement the volume
the liquid displaced so bb dash is equal to
theta I by v and then finally we get bm that
is what we are looking for bm is the distance
from the center of buoyancy to the metacenter
bm is equal to bb dash divided by theta or
this is equal to I by v where I is the second
moment of area of waterline plane about oo.
Finally we get bm is equal to bb dash divided
by theta that is equal to I by v so this distance
bm is called as the metacentric radius.
So like this we can determine the position
of metacenter relating to the center of buoyancy.
So what we are doing here is say we are considering
in a small tilt and with respect to how much
liquid is displaced and then the new position
and then taking the moment and finally we
get this bm that means the distance from the
center of buoyancy to the metacentric distance
as the second moment of area of waterline
about oo divided by the vole of the liquid
displaced.
So that is what we are getting as called the
metacentric radius so like this we can determine
the position of metacenter relating to the
center of buoyancy. Now with respect to the
center of buoyancy and then with respect to
the metacenter which we are discussed one
more metrical example. So here is the problem.
Here you can see there is a floating vessel
and then a small load is placed over that
so we want to determine the maxim weight of
the center of gravity of this load that means
the placed load above the vessel. So that
the vessel remaining stable equilibrium. So
the problem statement is as shown in figure
a cylindrical buoy two meter in diameter and
one point five meter height and weighing fifteen
kilo in Newton flow in salt water of density
one thousand kilo gram per meter cube.
Its center of gravity is point five meter
from the bottom if a load 3 Kilo Newton is
placed on the top. Calculate the maxim height
of the center of gravity of this load above
the bottom if buoy remain in stable equilibrium?
So here a vessel or a buoy is there as shown
in this figure. So it is a cylindrical in
shape and its diameter is 2 meter and then
its height is 1.5 meter height and then is
15 Kilo Newton and it is floating on a salt
water of density 1025 kilogram per meter cube
and initially it is stable and then its center
of gravity is 0.5 meter its center of gravity
is which is the center of gravity g so it
is 0.5 meter from the bottom.
Now we have placing a load three Kilo Newton
just on the center to the buoy likes this.
A new load is placed three Kilo Newton on
the top. We want to calculate the maxim height
of the center of gravity that means the position
of when we place a load new load the center
of gravity is shifting from g to g dash. So
you want to determine we want to calculate
the maxim height of the center of gravity
of this load when from the bottom in buoy
remains in stable equilibrium. So this is
the problem you want to determine the new
position of the center of gravity.
So here all the problem dimensions are given
and the various loads here the buoy weight
is 15kilonewton and a new placed load is 3kilonewton.
So here let g be the center of gravity of
the buoy as shown here. This g is the center
of gravity of the buoy and then g one is the
center of gravity of the load. So that load
is placed just on the line with symmetric
to the buoy. So that g and g one is the same
straight line and g one is the center of gravity
of the load and then g dash is the combined
here. Now load is placed so the center of
gravity is shifting so g dash is the combined
center of gravity with respect to the load
and then buoy that means a 15kilonewton is
the weight of the weight of the buoy and 3
kilo Newton is the load. So with respect to
the new position of the combine center of
gravity let it be g dash and the z is the
depth of immersion of buoy that means say
this is the position of the sea water.
So the buoy is immerse that to a depth of
z as shown in this figure z it is not given
we have to find this z so z is the depth of
immersion of the buoy. So with respect to
this the problem statement now the solution
let v is the volume of the salt water displaced
since we are placing buoy and then extra large
there will be a liquid will be displaced So
let v be the volume of the salt water displaced.
Finally, we can find out the buoyancy force
is equal to weight of the salt water displaced.
We have seen earlier the buoyancy is the weight
of the liquid displaced. So here in this case
the buoyancy force is equal to the weight
of the salt water displaced. Since the buoy
is placed on sea so this buoyancy force is
equal to the weight of the salt water displaced
so that is equal to rho the density of the
salt water and multiplied g the acceleration
due to gravity into the volume of the salt
water displaced.
So that is equal to rho into g into v. Finally
you can write what we said is in this figure
z is the depth of the immersion of the buoy.
So we can write the buoyancy rho into g into
pie d square d is the diameter of the buoy
so pie d square z is the reposition of the
buoy multiplied by the depth of immersion
of the buoy. Finally buoyancy force is rho
g pie square by four into z now for equilibrium.
Here buoyancy force is equal to the combine
weight, Now we have the buoy weight of the
15 kilo Newton and then we have putting an
extra load of 3 kilo Newton. So the combined
weight is equal to fifteen plus three so that
18 kilo Newton that is equal to the buoyancy
force.
So you can equate w plus w one is equal to
w one is the extra added weight w is the weight
of buoy. So that is equal to rho into g into
pie by four d square into z. So we want to
determine this z that means the depth of immersion
of buoy.
So z is equal to four times w plus w one divided
by rho into g into pie d square. So once we
substitute all the values capital w is the
weight of buoy is equal to fifteen and small
w one is the added load is three. So 4 into
15 plus 3 then we say it is Kilo Newton we
want to determine in terms of Newton. So we
have multiplied by 1000 divided by rho g is
the salt water with respect to the density
1025 into g 9.81 and d square is say pied
square to meter is the diameter of the buoy
so pie in to two square so that will give
z as 0.57 meter so we get the depth of immersion
of the buoy as point five seven meter.
Now we want to determine the center of buoyancy.
So in this figure now we can see that this
z is older determine as this 0.57 meters so
z is 0.57 meters. So ob is equal to half of
this z so that we can write this ob is equal
to half of z so that is equal to point two
eight five meter.
So the buoy and load is stable equilibrium.
So as far as the problem statement the buoy
and load is in stable equilibrium and the
metacenter m coincide with the combine center
of gravity g dash. So the metacenteric height
in the case g dash is equal to zero.
Since it is stable equilibrium say this metacenter
m coincide with the combined cg g dash so
m and g dash coincide, so that we can write
mg dash g dash m is equal to 0 and now bg
dash this the new position of this center
of gravity g dash from to that position from
g dash g equal to bm that means since the
g dash is equal to the met center m.
So bg dash is equal to bm or bg dash is equal
to bm is equal to as per the formula which
we have derived is equal to second moment
of the area I divided by v so the moment of
immersion I divided by v the volume of that
displacement f the liquid.
So here I is equal to pie d to the power four
d to the power four divided by sixty four
pie d to the power four divided by 64 and
the volume of displaced liquid is pie d square
z by 4. Finally, bg dash is equal to bm is
equal to this d square divided by 16 into
z. So d is power two so 2 square divided by
16 into 0.57, which is the z. We have determined
we have finally we get bg dash is equal to
bm is equal to 0.438 meter.
Now we want to determine the position g dash
the new due to the added weight the new position
g dash we want to determine. So the position
of g dash is say from the bottom of the buoy
that is equal to z dash. This is the new position
of z dash is equal to half z that means up
to b this ob plus BG dash. So ob is equal
to half z.
So z dash is the position of g dash the new
center of gravity position z dash is equal
to half z plus bg dash so half z is already
determine a point two eight five meter plus
bg dash. We have already determined here as
0.438 meter we have already determine here
this is bg dash 0.438 meter. Finally position
of g dash z dash is equal to 0.285 plus 0.438
is equal to 0.723 meter.
Now this value of this mater one say here
from this figure, you can see that say to
the center of gravity of the extra load g
one that z1 from the bottom of the buoy this
z1. We want to determine so value of z1 is
corresponds to the z dash that means with
respect to the new center of gravity we can
take moment about the point o.
So here you can take moment about this particular
point o so that we can determine this z1 so
w1 into z1 where w1 is the weight of the buoy.
So, w1 into z1 plus 0.5 w is equal to w plus
w1 z dash so z1. We finally get as w plus
w1 into z dash minus 0.5 w divided by w1.
So here in this gates the total weight is
fifteen of the buoy and three Kilo Newton
of the extra load placed so it is eighteen
ten to the power in terms of Newton multiply
z dash. We have already determined the z dash
as point seven two three meter into point
seven two three meter into 0.723 minus 0.5
w is the weight of the buoy.
So 0.5 into 15 into 10 to the power three
divided by w1 is the extra load 3 into 10
to the power three. Finally, we get z1 that
means from the bottom of the buoy to the position
of the center gravity of the extra load placed
as 1.838 meter.
So here finally we have determine in this
figure we have to determine z as 1.838 meter
and we have to determine the position of g
dash this z dash. We have to determine as
0.723 meters. So this gives the solution of
the problem. Now before closing this chapter,
we will just have an overview of what we have
studied the fluid statics, a brief summary.
What we have studied in fluid static?
As I mentioned at the beginning of this lecture
the fluid statics is what we are studying
what is happening in the fluid is at rest
say with respect to the forces at in the pressure
and then buoyancy then all this things what
we have studied. We have studied in fluid
statics principles the fundamentals and theories
have been introduced as far as this fluid
statics is concerned and then, we have studied
about the pressure measurement.
So we have seen that the various say metrologies
like given by using manometer then, we have
seen the automatic gauge equipments like Borden
gauges. We have seen the pressure measurement
using the various equipments and then we have
also discussed the forces on submerged surfaces
like inclined surfaces horizontal surfaces
vertical surfaces then curved surfaces.
We have seen how we can determine the forces
on submerged bodies and then we have seen
how we can determine the resultant forces
say when the body is either in submerged floating
body in horizontal position or vertical position
or inclined position. How we can determine
the resultant force and then correspondingly
how we can determine the center of pressure
and then we have discussed about the buoyant
forces and we have also seen the Archimedes
principle and then how we can determine the
center of buoyancy and then with respect to
its various applications we have seen in this
fluid statics.
Finally at the end of the chapter we have
seen with respect to buoyant force say when
a body is submerged or when a body is a floating
it can be stable equilibrium unstable equilibrium
neutral equilibrium.
So this also we have seen and then we have
seen with respect to the position of the center
of gravity position of the center of buoyancy
how we can define whether the body is stable
or the whether the body is unstable or whether
the body is in neutral equilibrium and then
finally, we have defined term metacenter.
There is small tilt is given an then whether
it is in stable equilibrium which is time
to come back so with respect to the small
tilt given we have also seen what is metacenter
and also we have seen how we can determine
the metacentric height.
So this are some of the important topics we
have discussed in fluid statics and as we
have seen in this lecture there are a large
number of applications as far as the fluid
is static is concerned starting from with
respect to the pressure measurement say we
have seen, how we can with respect static
fluid how we can determine the pressure measurement
and including from the how the doctor is determine
the human pressure the body pressure? How
the application is there and then also we
have seen with respect to the Pascal law how
the application side like a hydraulic lift
how it is lifting it is put a small force
it is making it’s a trying to say rise larger
bodies.
So that also we have seen as applications
as in some of the previous slides we have
seen. So like this a large number of applications
are there as far as the fluid statics is concerned.
So fluid statics is one of the important branch
of fluid mechanics the advantage here, we
can say, since the fluid is say not in moving
since it is just static conditions. So the
analysis is much easier.
So the application but large number of obligation
here we have a man can car or hydrolic leaves
so like that various applications are there
as far as, The fluid statics is concerned
and then also the ships and the movement of
ships and boats with respect to this stability
very importance. So the stability the stable
the equilibrium and then also the metacenter
height which is we have discussed it has also
got large number of applications as far as
say this fluid statics concerned.
Fluid statics is one of the important branches
of fluid mechanics and here over analysis
is much easier compared to fluid dynamics.
In this case in the fluid is moving the fluid
is just static so that, we can easily determine
the various forces by just taking and just
using the equations of equilibrium like sigma
fx is equal to 0 sigma fy is equal to 0 or
sigma fz is equal to 0 that means the force
in xyz direction. Since it is static and we
can determine sigma fx equal to 0 fy is equal
to 0 fz is equal to 0 and also, we can take
a moment as we have doing in mechanics most
of this principles used in mechanics directly
applicable as for as the fluid statics concerned.
Various applications as we have seen here
in various slides where large number of applications
and fluid statics one of the most important
branch of fluid mechanics and advantage here
is the analysis much easier compared to the
dynamic fluid dynamics which will be discussing
the coming lectures. So with this, we are
finishing the topic on fluid statics and next
we will be discussing kinematics of fluid
and then fluid dynamics before going to the
turburant fluent laminate fluent and other
theories.
