with this background now we will move into
more general considerations for equilibrium
of fluid elements here whenever we were discussing
the surface tension force we were assuming
that the pressure is being distributed in
a particular way and we were intuitively using
some concept of high school physics that ah
if you have a depth of h then what should
be the variation in pressure because of that
depth of h now we will look into it more formally
so we will go into the understanding of fluid
statics 
we start with an example or a fluid element
which is in static equilibrium and we take
an example of a two dimensional fluid element
just for simplicity we have taken such examples
earlier what these example signify ah that
you have a uniform width in the other direction
perpendicular to the plane of the board
let us say that delta x and delta y are the
dimensions of this fluid element remember
that this fluid element is at rest when this
fluid element is at rest that means we are
sure that it is non deforming because ah deforming
fluid element is definitely not a fluid element
at rest and when it is non deforming we are
clear that there is no shear which is acting
on it that means there is only normal force
which is acting on all the phases of the fluid
element so we can designate the state of stress
on each phase of the fluid element by pressure
so let us try to to do that let us say that
we are only writing forces along the x direction
just for simplicity similar equations will
be valid for the y direction when you have
ah say the left phase under consideration
just like this let us say that p is the pressure
on the left phase and the force corresponding
to that is p into delta y into say one which
is the width of the fluid element when you
come to the opposite phase we are bothered
about these phases right now because we are
only identifying the forces acting along x
because we will write equation of equilibrium
along x ah not that forces are not there on
the other phases so this is not the complete
free body diagram it only just shows the forces
along x direction
so if the pressure here is p what should be
the pressure here that is under question will
it be p will it be something different from
p in general we are not really committed to
what are the other forces which are acting
on it there may be any other body force which
is acting on it along x and y so if the pressure
here is p the question is will the pressure
here be p or something else in general special
case of course it may also be p here but we
are talking about a more general consideration
mathematically speaking what question we are
trying to answer we have a function here say
p we want to find out the value of the function
at a different location say this location
is x we are interested to now find out the
value of the function at x plus delta x in
terms of what is the value of function at
x the function here is p that means we want
to see that hat is p at x plus delta x in
terms of what is p at x and that we can easily
do by using a taylor series expansion so that
we will do we will write this as this as p
at x plus the first order partial derivative
of p with respect to x into delta x plus and
so on
there are infinite number of terms but as
you take delta x very small may be you may
neglect the higher order terms in comparison
to the dominating term and the gradient keeping
that in mind that we are treating with cases
where delta x delta y are very small so delta
x delta y all tending to zero so this will
become from the expression that we have here
what we can write this will be p that will
be the pressure here we will keep this in
mind so later whenever we encounter any function
we will use the taylor series expansion to
identify what is the change that is taking
place across different phases of fluid elements
becausethat we will have to do very commonly
many of our analysis so this multiplied by
the area on which it is acting is the force
due to pressure on this phase
let us say that there is a body force which
is also acting on the fluid element so the
body force let us say that b x is the body
force per unit mass acting along xso b x is
body force per unit mass along x so what will
be the total body force which is acting on
this along x fist you have to find out what
is the mass of the fluid element what is that
it is ah the density times the volume of the
element that is delta x into delta y so this
is the mass of the fluid element that times
the body force per unit mass gives the total
body force along x so these are the forces
which are acting on the fluid element
now let us try to answer another question
are this still the force only forces which
are acting if the fluid element is under rigid
body motion that is the fluid element is moving
like a rigid body there is no internal deformation
but as a whole it is just like a solid that
is getting displaced that may be displaced
linearly or angularly but it it is having
a motion but the motion is a rigid body motion
if that is the case then are these the only
forces see what forces we have identified
surface forces we have identified body forces
we have identified so the question was down
to that are these the only surface forces
even if the fluid is under rigid body motion
the answer is what see what is the difference
between a fluid element at rest and fluid
element are under rigid body motion the only
difference that when it is under rigid body
motion it might be having like a velocity
acceleration and so on but in terms of the
surface forces which are acting if the fluid
element is non deforming then there is no
shear component of force so for a non deforming
fluid element there is no difference between
the surface forces which are presented in
this diagram and the surface forces which
are there when it is say moving with a acceleration
so these type of forcing description is equally
valid if the fluid is under rigid body motion
so ah we ah the identify this situation not
just at fluid under rest but also rigid body
motion we will see such examples were ah the
rigid body motion of the fluid will ah be
very interesting like you may have a rotation
of a fluid element like a rigid body and we
will see that what kind of ah situation it
creates so broadly this is also studied ah
under the category of fluid statics not because
it is a static condition but in terms of the
characteristic of the fluid the deformable
nature is not highlighted here and that is
why we may use broadly similar concepts and
we will learn these concepts together under
the same umbrella because they are very very
related in one case it will it may have an
acceleration in other case it may not be otherwise
it is very very similar
so let us say that it is under rigid body
motion and therefore let us say it has some
acceleration along x say a x is the acceleration
which is there along x so we can write the
newtons second law of motion for the fluid
element and when you do that what do we get
the resultant force which is acting along
x is equal to the mass of the fluid element
time x times acceleration along x so it is
p into delta y minus the other term which
is there on the opposite phase plus rho into
delta x into delta y into b x is equal to
rho into delta x into delta y that is a mass
times the acceleration along x delta x into
delta y will get cancelled from both sides
these are small but not equal to zero these
are tending to zero so you can cancel from
both sides
at the end what final expression you will
get so this will be the expression which relates
the pressure radiant with the body force that
is acting and if there is any acceleration
that acts on the i mean that the fluid element
is having similar expressions are valid for
the motion along y so we are not repeating
it again with this kind of a general idea
so this is a very general expression this
general expression just considers that there
is a body force and the fluid is having some
acceleration in a particular directions subjected
to the body force but its a non deformable
fluid element
with that understanding we will try to identify
that what is the variation of pressure just
due to the effect of gravity as a body force
in a fluid element at rest we consider that
there is a free surface of a fluid this is
a symbol in fluid mechanics that will be using
to designate a free surface these are triangle
with two horizontal lines very short horizontal
dashes or lines at the bottom this is a kind
of a technical representation of a free surface
we consider that ah we are interested about
some depth usually the direction in which
depth varies ah is typically taken as z direction
this is just a common notation in most of
the books that ah the vertical direction across
which the gravity ah is acting ah of course
the opposite to the action of gravity because
gravity u will be vertically downwards and
opposite to that is considered as a z axis
just as a common notation
so let us try to write this kind of equation
for this fluid which is at rest ah it is it
is of a substantial depth so we are interested
to find out what is the pressure at this point
which is at a depth h from the free surface
this there is no acceleration of this fluid
it is under absolute rest so the a x or here
a z term will be zero so you will have minus
this one when you have the z direction you
also have a horizontal direction like x and
for x you can write again similar expression
so if you write it for x what is the body
force which is acting along x there is no
body force which is acting along x because
only body force to which this is subjected
is the gravity so there is no body force along
x and there is no acceleration that is it
is having along x
so the second expression is even more simple
but it gives as a very important insight what
is that that if you are not having a body
force along a particular direction and the
fluid is under rest then pressure doesnt vary
with in the same fluid along that direction
that means for a horizontal ah along a horizontal
line you are not having any pressure variation
in a continuous fluid system and this is one
of the basic principles that we use for measurement
of pressure differentials as you have seen
in examples of manometers earlier so this
is something which is of great consequence
but it is a obvious conclusion what we get
from the first equation so we let us try to
replace the b z what is what is b z
g
see this is the z direction and this is acting
in the opposite direction so this
minus g
minus g so you have from this and since pressure
is not varying with x so you can write this
as d p d z in place of partial derivative
because it is now just a function of a single
variable so you can write this as minus d
p d z is equal to row g so d p is minus rho
z d z so if you want to find out what is the
pressure difference between say two points
a and b so you have to integrate it with respect
to the z variation from say a to b when it
is a say we take our reference such that the
origin is located here that means at a z is
equal to zero at b it is minus h so you can
write p b minus p a is equal to minus integral
of now it is important to see that what is
the length scale that we are considering over
which this variation is taking place if this
h is quite large there may be a significant
variation in density over it just like consider
the atmosphere which is above the surface
of the earth as you go more and more above
the surface of the earth you expect the density
to change because the temperature changes
and so on and ah therefore the density in
many cases may not be [treatez/treated] treated
as a constant so if it is treated as a constant
then it it it can only come out of the integral
similarly you also are probably working on
ah length scale over which g is not changing
if if you are taking a large height like if
for people who are dealing with atmospheric
sciences for them the length scales are large
lengths scale over which you may have even
a change in acceleration due to gravity but
if you consider that such a situation is not
there just for simplicity so if rho into g
is a constant that is the [bet/best] best
way to say because a very ah tough mathematician
will say that i dont care whether rho is varying
whether g is varying i am happy in bringing
this term out of the integrals so long as
rho into g is a constant so may be ah mathematicians
way of looking into it is that rho varies
in a particular way g varies in a particular
way but those variation effects gets cancelled
out somehow so that rho into g is a constant
may be very hypothetical but for our case
to bring it out of the integral the product
being a constant that will serve our purpose
so then what you have then you have p b minus
p a is equal to rho g h that means p b is
equal to p a plus rho g h which is your very
well known expression
now important thing is that see we are not
expressing the pressure at b just in an absolute
sense we are expressing it relative to the
pressure at a many times this pressure at
a say say this is atmosphere so that may be
taken as a reference so if this is taken as
a reference as p equal to zero as an example
so whatever is the atmospheric pressure say
we call it zero that means any other pressure
we are expressing relating to the atmospheric
pressure so then in that case p b is the pressure
relative to p atmosphere if p equal to zero
is the atmosphere it is not definitely equal
to zero but if you if you have p equal to
zero we i mean that is that is just the choice
of your reference so that you expressed any
other pressure in terms of that as a reference
so any pressure which is expressed in term
of that atmospheric pressure that is a relative
way of expressing the pressure it is not that
always you have to express relative to atmospheric
pressure but atmospheric pressure being a
well known standard under a given temperature
so reference with respect to atmospheric pressure
is something which is a very standard reference
that we many times use so reference pressure
relative to atmosphere so we are talking about
a reference where the reference is the atmosphere
then what ever pressure is there at any other
point we call that as a gauge pressure so
this is just a terminology so gauge pressure
means that any pressure relative to atmospheric
pressure so that means it is nothing but p
absolute minus p atmosphere just the difference
between the absolute pressure and atmospheric
pressure that is as good as taking the atmospheric
pressure as zero reference and mentioning
the pressure relative to that
so this is a very simple exercise but from
this we learn something what we learn something
so whenever we have an expression we should
keep in mind what are the assumptions under
which it is valid so ah we will develop the
habit of not using any formula like a magic
formula this is very very important formula
based education is very bad so whenever you
have a formula and you want to us e it try
to be assured that it is valid for the condition
in which it you are applying if not exactly
but at least approximately so when you are
using pressure equal to h rho g what are the
what are the assumptions under which it is
valid
so obviously rho into g is a constant and
there is no other body force which is acting
on it and fluid is at rest that is these are
the assumptions that ah are there with such
a simple expression now with these kind of
a concept one may utilize device one may utilize
this type of concept in making devices for
measuring atmospheric pressures just like
you have barometers whenever we will be learning
ah a concept we will try to give examples
of measurement devices which try to utilize
those concepts as all of you know a barometer
may be utilized to measure the atmospheric
pressure
so how it is there you have say inverted tube
and this inverted tube is say put in a bath
of some fluid say mercury and ah let us say
that it is they are up to this much height
now this much of height is there there are
various which are acting on these so one is
you have we are of course neglecting the local
surface tension effect and the capillary formation
you must keep in mind that ah as this radius
becomes smaller and smaller the effect of
the curvature may be more and more important
because surface tension effect will be more
and more important and there may be significant
errors in reading because of that
now if we just neglect that effect for the
time being then you have atmospheric pressure
acting from this side if you assume that there
is a vacuum here there is a big question mark
whether there will be vacuum or not but let
us for the sake of simplicity assume that
there is a vacuum then what ever pressure
is there ah which is acting from this side
that is balanced by the height of the liquid
column which is there on the top so ah from
that you can get an estimation for what is
the pressure here let us say that p is the
atmospheric pressure so p into the surface
area on which it is acting is the force that
is being sustained by the weight of the liquid
column so that is nothing but what
h rho g into area
so it is like h rho g that into the area and
area gets cancelled from both sides so you
get this p if it is vacuum as the h rho g
but if it is not a vacuum let us say that
there is some pressure here which is the vapour
pressure of the fluid which is occupying this
and it is common that such vapour pressure
will be there why because if it is a saturated
liquid it is likely to have its own vapour
on the top of that and that that will always
exerts some pressure so it is never a vacuum
in in an ideal sense so we can say that p
minus p vapour is actually what is being balanced
by this weight so that is the h rho g
so if there is the vapour pressure you cannot
just use h rho g for the estimation of the
atmospheric pressure but you have to make
a correction because of the presence of the
vapour and that is a function of the temperature
because vapour pressure varies with temperature
very commonly the ah mercury is one of the
fluids that is used for this purpose and why
mercury is being used obviously because it
is quiet dense it will not occupy a very large
height for representing the atmospheric pressure
if you use any other fluid it may occupy a
great height so it it may be an unmanageable
device unmanageably long device also the vapour
pressure of mercury is quite small in in most
of the temperature ranges and therefore this
correction is not that severe these two are
the important reasons there are many other
reasons which are ah always into the picture
when you ah when you select a fluid for measurement
of a pressure like like in a barometer
a barometer is a very interesting device ah
we have discussed something quite seriously
but ah i would just like to ah i mean share
a kind of a story associated with barometer
a very well known story and ah i i am sure
that many of you have heard about it long
time back ah in in a high school examination
there was a question it was a physics examination
and the question was that ah how can you measure
the height of a building using a barometer
now although all of you or most of you have
heard about this story we will try to get
a moral out of that story and we will try
to keep that in mind whenever we are going
to learn something
so what the student replied ah in in the answer
the answer was that ah you just have a thread
you connect the barometer with the thread
go to the roof of the house just drop that
barometer with the thread and then ah like
ah the total height that the total height
of the thread that is required to bring the
barometer to the ground
[laughter]
maybe plus whatever is the portion of the
barometer will give the height of the building
[laughter]
now as in most of the examination systems
this student was given a big zero
[laughter]
and ah the expectation was it was justified
that why he was given a big zero because it
was expected that the answer should reveal
some basic concept in physics it was a physics
examination but these doesnt reveal any basic
concept in physics so he was given a zero
but then ah the student went for an arbitration
it was a democratic system even in that time
[laughter]
so the student said that no ah ah like i mean
let my answer be reviewed so there was a panel
the panel said that ok may be you were not
aware that what kind of expectation that we
are having from your answer so we give you
another chance so you think about a solution
for this question ah which ah reflects your
understanding in physics and we will evaluate
you from that so student said that let let
me be given some time so ah he was given five
minutes to think about that so he was thinking
for five minutes and when he was thinking
for five minutes ah and he was still not coming
up with a answer then ah like the evaluators
were very happy that he he might be failed
again so they said that no you you could not
come up with an answer so we are sorry then
he said that no actually there are lots of
answers have come to my mind so i am not sure
that what should i say and thats why i was
not giving a response and then he was asking
permission that ah i mean am i going to be
allowed to speak of that remaining answers
ah then they said that fine i mean whatever
you have thought you just tell then he said
that ah i will what i will do is i will drop
the barometer from the top
[laughter]
of the building and i will measure the time
that it takes to reach the ground
[laughter]
and h equal to half g t square so from the
time ah ii can measure the height it reflects
some understanding of physics but it is a
bit distractive because the barometer
[laughter]
ah like a it it may be damaged and like and
so on then you said that no ah then if you
want a different answer may be what i will
do i will try to ah make a pendulum out of
the barometer
[laughter]
swing it one in the bottom ground level another
on the top level and we will measure the time
period
[laughter]
and these time period difference will give
the difference in g between the two heights
[laughter]
and since g is a function of height
[laughter]
it will tell us that what is the height difference
between the two and ah i mean still the examiners
were not happy and they but still they were
ready to pass him because these were like
a reflecting some of the basic concepts in
physics then he said that ah even if ah i
am given ah different opportunity what i will
do is i will climb ah across the stair case
and ah in the side of the stair case ah you
can just put the barometer one after the other
till
[laughter]
you traverse the entire height the number
of times you put it ah you multiply with the
length ah its a basic length measurement principle
so from that you get that and he said that
there are many ah answers which are coming
to my mind but given me a entire freedom what
i will do i will not really put this much
of effort i will go to the house master and
tell that see now i have a beautiful new barometer
for you i am giving it to you please tell
be what is the height of the house
[laughter]
and ah the house master will obviously tell
that because it is it is like a gift free
gift that the housemaster is having at the
end he said that perhaps you are not expecting
all these answers from me you are expecting
to me to give a very structured answer that
the barometer measures that atmospheric pressure
so from the difference in the level of atmospheric
pressure in the two heights we can easily
say that what is the difference in height
between the two and since this is the most
structured answer i hope that you will be
happy with that and then of course the evaluators
passed him and ah he was quite successful
and the name of the student is niels bohr
i mean who later on i mean ah like discovered
many many beautiful ah phenomena in physics
so the whole idea is that ah i would always
encourage you may be i mean none of our self
like niels bohr ah i mean we are not born
we those special abilities but at least whenever
you are having an opportunity to think of
solutions dont always go for a structured
solution try to think about different possibilities
whenever you are thinking about designing
a measurement principle whenever you are thinking
about a solving a particular problem just
try to think about various possibilities some
of the possibilities may not be ah very very
encouraging very very welcome but at least
these possibilities will give us some kind
of clue that what could be alternatives some
of the alternatives may be discarded they
may not be very smart but they will at least
give us a scope of thinking laterally and
thats how one may improve in science technology
and research and ah i mean such a simple example
like barometer ah i mean one is always reminded
of that kind of a story and i feel that it
it is it is not something too bad to share
with you
so what we will do is we will not ah go further
ahead today we will stop here in the next
class we will just make a plan of what we
will do in the next class now we have found
out a particular way in which you have a estimation
of variation in pressure because of a body
force which is acting and for fluid element
which may be at rest or subjected to acceleration
so we will utilize this principle to calculate
two important things one is if there is a
plane surface which is immersed in a fluid
what is the total force which is there on
the plane surface because of the pressure
distribution now you have realized that pressure
is like a distributed force because it varies
with the depth so at different depths you
have different pressures
therefore it will be like a simple statics
problem where you have a distributed force
on surface to find out what is the total force
which is acting if you have a curved surface
we will see that there are technique may be
a bit different but ah broadly we can utilize
some of the concepts of pressure distribution
on a plane surface even to calculate force
on curved surface so in the next class we
will ah we will see what are the forces on
plane and curved surfaces which are there
in a fluid at rest and ah then ah to see that
what are the consequences and will work out
some problems related to that so we stop here
today
thank you
