Lecture 8.2
Fluids
For a long time now we have been talking about classical mechanics, part of
physics which studies macroscopic motion of particle-like objects or rigid bodies. Using
different methods we have considered translational motion, rotation and equilibrium.
However, all this time we have been talking about solid substances only. You know, of
course, that solids are not the only form of matter existing in nature. Physics of fluids
studies liquids and gases. The area of physics, which studies motion of fluids, is known
as hydrodynamics. The area of physics, which studies fluids in equilibrium, is known as
hydrostatics. We shall discuss simple applications of the both.
Fluids include both liquids and gases. They both are of enormous significance for
our everyday life. They are used everywhere, starting from drinking water and air which
everybody needs for living, and ending up with different types of technical fluids
involved in almost any technological process. So, what are the main differences between
fluids and solids? In contrast with solids fluid can flow. It does not have fixed shape.
Usually fluid takes the shape of the boundary (container) which it has been placed in.
This happens because, as any other structure fluid consists of molecules. However, the
intermolecular bonds in fluids are much weaker than in solids. Solids have a crystalline
structure, where every molecule has its own place in the crystalline lattice. Intermolecular
bonds do not allow molecules to leave their places. They can only oscillate near fixed
positions. Fluids, on the other hand, do not have these strong intermolecular interactions.
Molecules can move relatively free under the influence of external forces. Liquid still has
some close-range order in its structure and it can maintain its continuity. Gases do not
even have that.
Since fluids are so much different from solids, we will need a different set of
variables to describe them. We already know that to find the correct coordinate system
sometimes means to solve the problem. It is also true about finding the correct set of
variables. In the case of a rigid body which has fixed shape and with the same properties
for each of its parts, we used mass and force as fundamental variables to solve
mechanical problems. However, when we are dealing with fluids, we have to remember
that they can change their properties throughout the volume. They do not even have a
fixed volume. Different parts of fluid can move with different speeds relative to each
other. This means, that instead of considering fluid as a whole, we have to study small
parts of this fluid. These parts can be treated as relatively uniform and moving with
certain speeds. This is why we have to introduce a different set of variables: instead of
mass, we shall use density; instead of force we shall use pressure.
We shall consider a small element of fluid which has volume V and mass m . In
this case the density of the element is going to be

m
.
V
(8.2.1)
At this point we have made an assumption of classical hydrodynamics that we can treat
fluid as if it has continuous density which changes smoothly in space rather than
considering fluid as complicated molecular structure. This means that volume V is
large enough compared to the size of one molecule, so it includes many molecules and
density  , introduced by means of the equation 8.2.1, is somewhat averaged density over
local volume V . Even though this picture is not perfect, especially when talking about
very small objects like small liquid droplets consisting of only a few molecules, but it
works fine for liquid in larger volumes. In fact, it is all right to use this approach even for
macroscopic droplets which we can be observed in the everyday life, such as rain
droplets. Since we are going to study only fundamentals of hydrodynamics, we shall go
even further and limit our attention by uniform fluids only. These fluids have the same
density everywhere throughout their volume, so

m
.
V
(8.2.2)
This is quite an accurate approximation, as long as we are only considering liquids and do
not study liquid-gas boundaries. Liquids are almost incompressible. However, it is not
accurate to treat gases this way. Equation 8.2.2 can only give us the average density of
gas or volume needs to be very small, so one can ignore effects of gravity.
SI unit for density is kilogram per meter cubed ( kg m3 ). Table 14-1 in the book
shows densities for some common substances.
If we consider a small element of fluid and insert a force sensor there, we can
measure the force, F , acting on the surface area, A , of the sensor. In this case we can
introduce pressure of fluid inside of this element as
p
F
.
A
(8.2.3)
Everything which we have said about local density is also true about pressure, which is
considered to be changing smoothly throughout the fluid. In ideal case, we can even treat
it as constant everywhere in fluid, if it has flat surface and uniform external force exerted
perpendicular to this surface, so that
p
F
.
A
(8.2.4)
This assumption is more accurate for gases in small volumes, where we can ignore the
influence of gravitational field (because of the small density of gas), while it is not very
accurate for liquids, where pressure changes with depth due to gravitational effects. It is
an experimental result that in uniform fluid pressure is a scalar, so equation 8.2.4 only
involves absolute value of force.
SI unit for pressure is Pascal, which is Newton per meter squared ( 1Pa  1N 1m2 ).
Other common units are related as
1atm  101
.  105 Pa  760torr  14.7 lb in2 .
1. Hydrostatics
As it was already mentioned, hydrostatics studies fluids at rest. Let us consider a
liquid which is placed into a tank and stays there at rest (see the picture)
Everybody knows that pressure in water gets larger with increase of depth. It may
become enormous at the on the bottom of the ocean. On the other hand pressure in air
gets smaller with increase of height. It is very low at the top of mountains. Let us see
what the reason for that is. We need to find pressure in fluid as function of the altitude.
We usually refer to this pressure as hydrostatic pressure. The y-axis in the picture is
directed upward, so the value of y-coordinate y1 at level 1 is larger than y2 at level 2. We
shall consider a liquid element in the form of the cylinder (it is shown by a different color
in the picture, even though it is still the same liquid). It has the same area A of the top
surface as of the bottom surface. This element is in static equilibrium, which means that
the vector sum of all the forces acting on it is zero. There are three forces acting on this


cylinder: the gravitational force mg in the downward direction, the force F1 in the

downward direction due to the liquid's pressure above the cylinder, and the force F2 in
the upward direction due to the liquid's pressure below the cylinder. The condition of
equilibrium gives us
F2  F1  mg ,
b
g
The mass of this cylinder can be calculated as m  V  Ah  A y1  y2 , where
h  y1  y2 is the cylinder's height and  is the density of liquid. With account of
equation 8.2.4 we have
b
g
p2 A  p1 A  A y1  y2 g ,
b
g
p2  p1  g y1  y2 .
(8.2.5)
The last equation relates pressure at two different levels. If we choose the ground level
with pressure p0 equal to the atmospheric pressure at the surface of this tank, then the
pressure p at any depth h is going to be
p  p0  gh .
(8.2.6)
We can also apply equation 8.2.5 to calculate the pressure of air at certain height, which
is going to be less than its value at the ground level. Since the gas density is much less
than the density of liquid, the pressure inside of gas can be considered as almost constant
in the case, when the size of the container h is small. So, the pressure at point in a fluid in
static equilibrium depends on the depth of that point but not on any horizontal dimension
of the fluid or its container.
Pressure, p, in the equation 8.2.6 is said to be absolute pressure. It consists of two
terms: the atmospheric pressure p0 and the gauge pressure gh due to the liquid itself.
Example 8.2.1 The U-tube contains two liquids: water of density  w  998 kg m3
is in the right arm and oil of unknown density  x is in the left arm. Measurement gives
l  135mm , d=12.3mm (see the picture). Find the density of this oil.
Since pressure at the same level is the same, we can calculate pressure at the level of
water-oil interface by two different methods. First using the oil arm which gives
b g
p  p0   x g l  d ,
Then, using the water arm which gives
p  p0   w gl .
After that we have
b g
 gbl  d g   gl ,
 bl  d g   l ,
kg
kg
   l bl  d g  998 135mm b135mm  12.3mmg  915 .
m
m
p0   x g l  d  p0   w gl ,
x
w
x
x
w
w
3
3
The above example gives us an idea of how one can build a device to measure
pressure, in particular a device which allows measuring the atmospheric pressure. Since
even small changes in this pressure can cause a lot of change in weather, this device is
very important. It is known as the mercury barometer, which was first introduced by
Italian scientist Evangelista Torricelli in the 17th century. This simplest barometer
consists of the long glass tube filled with mercury. It is closed at one end and open at the
other end, which is placed in the dish with mercury. The pressure at the dish level is
equal to the atmospheric pressure. On the other hand, this pressure is the same as the
gauge pressure of the column of mercury, so p0   m gh . Observing the height of the
column, we can find what the atmospheric pressure is. This is why the unit of pressure,
known as millimeter of mercury column, was introduced. Normal atmospheric pressure
gives the height of this column to be 760mm.
Exercise: Why do you think Torricelli used expensive and (as we now know)
dangerous substance of mercury instead of common water to build his barometer?
Another important principle, which is related to many applications, is the Pascal's
principle. This principle states that a change in the pressure applied to an enclosed
incompressible fluid is transmitted undiminished to every portion of that fluid and to the
walls of its container. The most useful application of the Pascal's principle is hydraulic
lever, shown on the picture below
If one applies an external force of magnitude F1 to the left smaller piston with the area
A1 it will produce change in the liquid’s pressure p 
F1
. This change will be the same
A1
everywhere inside of liquid. So, the pressure acting on the second larger piston from the
liquid will also change by the same amount. This means that it will be additional force
acting on the second piston from the liquid, which is
F2  pA2  F1
A2
.
A1
Since the area of the second piston is larger than the area of the first piston, the force
F2  F1 . So one can apply relatively small force at one side of the system and obtain a
very large force at the other side to lift some heavy object. You might think that this
would be impossible because of the law of conservation of energy. In fact, there is
nothing wrong with conservation of energy here. Even though the force gets larger, but
the second piston will move for a distance, d 2 , which is smaller than distance, d1 , for
which the first piston has moved. Indeed, this liquid is incompressible, so it can not
change its total volume, which means that changes of the volume are the same on both
sides d1 A1  d2 A2 ,
d 2  d1
A1
.
A2
In this case work performed by the second force is the same as the work performed by the
first force
W2  F2 d2  F1
A2
A
d1 1  F1d1  W1
A1
A2
and energy is conserved. So, a given force applied over a given distance can be
transformed to a grater force applied over smaller distance.
Example 8.2.2 In the hydraulic system shown on the picture before, the piston on
the left has a diameter of 4.5 cm and a mass of 1.7 kg. The piston on the right has a
diameter of 12 cm and a mass of 3.2 kg. If the density of the fluid is 750 kg/m3, what is
the height difference h between the two pistons?
Pressure at the same level in liquid is the same. This pressure can be found either as
the pressure of the piston on the left p1 or as the pressure of the piston on the right p2
plus the pressure of the liquid with height h. This means that p1  p2  gh . Taking into
account that pressure can be found as the force acting per unit of area, one has
m1 g m2 g

 gh ,
A1
A2
m1 m2

 h ,
A1 A2
h
FG
H
IJ
K
F
GH
I
JK
1 m1 m2
4 m1 m2
4


 2 
2
 A1 A2
 d1 d2
 750 kg m3
F 17. kg  3.2kg I  105
GH b0.045mg b012
J . m.
. mg K
2
2
Let consider another well-known effect which occurs in fluid, the buoyant force.
Everybody knows that if you put something heavy into water, you can lift it with much
less effort compared to that in the air. This is because water as well as any other fluid
exerts a buoyant force on any object placed in it. If we consider some body placed in
liquid, we can notice that pressures are different at the top and at the bottom of this body,
since we have already learned that pressure in liquid depends on depth only. This
pressure difference is the reason for the buoyant force. The force acting on the top surface
of the body downwards is F1  p1 A and on the bottom surface of the body upwards
F2  p2 A , so the net buoyant force acting on this body from the fluid is
b
g
Fb  F2  F1  A p2  p1  A f gh  m f g ,
(8.2.7)
where m f   f Ah is the mass of the fluid displaced by the body. So, we have arrived to
the Archimedes' principle: When a body fully or partially submerged in a fluid, a buoyant
force from the surrounding fluid acts on the body. This force is directed upward and has
a magnitude equal to weight of the fluid that has been displaced by the body.
The existence of the buoyant force is the reason why floating becomes possible.
Indeed, let us consider some object partly submerged into fluid and floating in it. If it
floats, it is in equilibrium, so the net force acting on this object in the vertical direction is
zero. On the other hand, the only two forces acting on it are gravitational force and
buoyant force, so
Fb  Fg  0,
m f g  mg  0.
This would only be possible if mass of the displaced fluid is equal to the mass of the
body. However, if the mass of the body is more than the mass of the displaced fluid then
the body will sink. If the body is completely submerged, then its volume equal V  V f
the volume of the displaced fluid and we have
m f  m,
 f V  V ,
 f  ,
so, if completely submerged body floats in fluid it has the same density as this fluid. If
the density of the body is less than the one of the fluid it will float but it will not be
completely submerged. If the density of the body is larger than the density of fluid, it
sinks.
Example 8.2.3. A geologist finds that a moon rock whose mass is 8.20 kg has an
apparent mass of 6.18 kg when it is submerged in water. What is the density of the rock?
First of all we have to understand what apparent mass is. This mass is measured by
some scale when the rock is submerged in water. The scale usually shows the tension
force not the real weight. When in the air this tension is equal to the weight of the object.
In the water this tension, T, will be equal to the apparent weight, T  ma g , where ma
stands for the apparent mass. Let us use Newton's second law for the rock submerged into
water. Since this rock is in balance, one has mg  T  FB  0 . According to Archimedes'
principle, the buoyant force FB   waterVg , so
mg  ma g   waterVg  0,
m  ma   waterV  0
Here V 
m  ma 
m
 rock
.
is the volume of the rock, so m  ma 
 water
m 0.
 rock
 water
m,
 rock
 rock
m

,
 water m  ma
 rock   water
m
kg
8.20kg
kg
 998 3
 4.05  103 3 .
m  ma
. kg
m 8.20kg  618
m
2. Hydrodynamics
So far we have been talking about stationary fluids. Now let us consider fluids in
motion. We shall only consider ideal fluids. There are several assumptions we will make
about these fluids. Those are the following
a) Fluid undergoes a steady (laminar) flow. The velocity of this fluid at any point
in space is not changing with time neither in magnitude nor in direction.
b) This fluid is incompressible (it is liquid rather than gas). It has the constant
value of density everywhere.
c) This flow is nonviscous. There is no friction between different layers of this
fluid.
If you ever used a garden hose, you know that speed of water emerging from this
hose depends on the area of the hose's cross section. The smaller the area (if you partially
close the hose) the higher the speed of the emerging water. Let us see how this speed is
related to the area of the cross section. Suppose, we have a tube of length L and area of its
cross section varies along the tube. It has area A1 at the beginning of the tube and liquid
flows with speed v1 through this cross section, and it has area A2 at the end of the tube
and liquid flows with speed v2 through that cross section. The liquid is incompressible
and the tube has a fixed volume. This means, that the amount (volume) of liquid entering
the tube at one end should be equal to the amount (volume) of liquid leaving the tube at
the other end. Let us find what volume of liquid is passing the tube's cross section for a
small time t . This volume can be calculated as V  Ax , where x  vt the distance
for which the fluid moves for that time. Since this volume is the same at both ends of the
tube, one has
V1  V2 ,
A1x1  A2 x2 ,
A1v1t  A2 v2 t ,
A1v1  A2 v2 .
So
Av  const .
(8.2.8)
This last relation is known as the equation of continuity, which states that the quantity
Rv  Av , the volume flow rate, remains constant for any tube of flow in ideal fluid.
Example 8.2.4 A room measures 3.0 m by 4.5 m by 6.0 m. If the heating and air
conditioning ducts to and from the room are circular and have a diameter of 30 cm, what
is the speed of the air flow in the ducts necessary for all of the air in the room to be
exchanged every 10 min? (Assume that the air’s density is constant.)
Since the density of the air is constant, one can write continuity equation in the
form v1 A1  v2 A2 , where the left side of the equation refers to the duct and v1 is the
unknown speed of air in the duct, while A1  d 2 4 is the area of the cross section for
the duct. The right side of the equation refers to the room, where v2 
a
is the speed of
t
the air in the room, which has distance a between walls and A2  bc is the area of the
cross section for this room. So
v1
d 2
4

4abc 4  3.0m  4.5m  6.0m
m
m
a
.

 115
 19
.
bc , v1 
2
2
t
min
s
d t
 0.30m 10 min
b
g
The content of the continuity equation is simple. It means conservation of matter in
the tube. Let us see what we can find from the other important conservation law, the law
of conservation of energy. We shall consider a tube, which is not necessarily horizontal,
so one end of the tube is elevated for height h1 and another end is elevated for height h2 .
Let us use the energy conservation in the form of work-kinetic energy theorem for the
fluid moving along the tube. In this case the work performed on this fluid is equal to the
change of the fluid’s kinetic energy, which is
W  K ,
W  K2  K1 ,
W
1
1
mv22  mv12 ,
2
2
where m  V is the mass of fluid, which enters the tube at one end and leaves it at the
other end for time interval t . The work done on this fluid element comes from the two
sources: from the gravitational force and from the external forces at both ends of the tube.
The work of gravitational force on the mass element m , when it moves from one end of
the tube to the other end, is taken with minus-sign change of its gravitational potential
energy, which is
b
g
Wg   mg h2  h1 .
Work is also done on the mass element by the force at the first end of the tube to push
liquid into the tube, this work is
W1  F1x  p1 A1x  p1V .
Another work is done by the liquid in the tube itself to push the fluid outside of the tube
at the other end, this work is
W2   F2 x   p2 A2 x   p2 V .
Combining all this together, we arrive to
1
1
mv22  mv12 ,
2
2
1
1
 mg h2  h1  p1V  p2 V  mv22  mv12 ,
2
2
1
1
p1V  mgh1  mv12  p2 V  mgh2  mv22 ,
2
2
m
1 m 2
m
1 m 2
p1 
gh1 
v1  p2 
gh2 
v2 ,
V
2 V
V
2 V
1
1
p1  gh1  v12  p2  gh2  v22 .
2
2
Wg  W1  W2 
b
g
This equation is known as Bernoulli's equation and it states, that quantity
(8.2.9)
1
p  gh  v 2  const
2
remains constant along the tube of flow.
Example 8.2.5 What is the lift (in Newtons) due to the Bernoulli's principle on a
wing of area 80m2 if the air passes over the top and bottom surfaces at speeds 340m/s and
290 m/s, respectively?
Let us apply Bernoulli's equation. One can ignore the thickness of the wing so the
top part of the wing and the bottom part are at the same height, which means
1
1
P1  v12  P2  v22 , where subscript 1 refers to the top and subscript 2 refers to the
2
2
bottom. So the pressure difference between the bottom part and the top part is
P  P2  P1 
1 2 1 2
v1  v2 . To find the lift force acting on the wing, one has to
2
2
d
i
1
multiply this pressure difference by the area of the wing, so F  AP  A  v12  v22 .
2
Taking into account that it is the air with density   129
.
we have F  80m2
129
. kg m3
2
kg
flowing around the wing
m3
F FG 340 mIJ  FG 290 mIJ I  163
GH H s K H s K JK .  10 N .
2
2
6