Fluids
I.
Pressure
Concept 1:
In a fluid, the forces are not located at a specific point like a rope
pulling on a block. Instead, the forces are distributed across an area of
the fluid. For this reason, it is easier to work with force per area
instead of force!! This is also true for solid objects like beams or
bridges when doing real world problems where forces are distributed
across the object.
F
F
Single Contact Force
A.
Distributed Force
Definition - P
The magnitude of pressure is defined as the force component
perpendicular to an infinitesimally small surface area divided by the
surface area.
P
B.
F
A
You should note that pressure has a direction (air pressure in a tire
pushes outward on the tire and not inward). However, we don’t
usually need the full power of vectors in the problems at this level
(much like doing 1-D problems) since the force on any object placed
in an ideal fluid is always perpendicular to the object’s surfaces. Thus,
textbooks often use pressure when they mean the “magnitude of the
pressure.”
Example: Bed of Nails Magic Trick
Most people think that a material tears/breaks when too great of a force is
applied to the object. Actually the material’s response is determined by both
the magnitude of the force and how the force is distributed. In other words, it
is the pressure that is important.
A layer of atoms in a material are held together by electrical forces between
the atoms. A simple model would be a group of kids in a line holding hands
as in “Rover-Red-Rover” shown as white circles below. The applied force
can be modeled as a group of kids (six kids – black dots shown below)
running into the line. Increasing the force would be like increasing the total
number of kids running into the line. If the force is spread out (each of the
kids runs into the line at a different place – first picture) the line may be able
to handle the force. However, if the kids all rune into the line at the same
place (second picture) then the same force may break through the line. A
common example of this phenomena is floating. When you lay out in the
water, you spread out your normal force, this means that each water
molecule has to apply less force to support you. However, if you try and
walk on water, your normal force is distributed upon a smaller area and the
water molecules are unable to apply enough force to offset your weight
before they are pushed aside (line breaks). This is the same principle behind
the bead of nails in which you distribute your force across many nails.
Low Pressure, Same Force
High Pressure, Same Force
II.
Density (actually mass density)
Concept 2: A fluid can change its shape depending on the vessel that
it is placed into. This makes it difficult to talk about the amount of
mass in a certain shape. Thus, we don’t talk about the mass of the free
body, but the mass per volume (density) of the free body.
A.
Definition - 
The density of a uniformly object is defined as the object’s mass
divided by the object’s volume.
ρ
M
V
Density is a useful concept even for solid objects!! The mass of a gold
ball depends on the size of the ball. However, the density of gold is
the same for any size ball. It is only a property of gold at a particular
temperature and pressure. Thus, the densities of materials are
tabulated in handbooks including the CRC and can be used to
distinguish between different materials.
III.
Increasing Pressure at A Depth
We know that the pressure applied by the water against the hull of a
submarine at great depths is very large and can even crush the
submarine if the submarine welds have the slightest flaw. We will use
Newton’s 2nd Law to derive the following equation for this important
example both as an example of how fluids are just an extension of the
past work this semester and because it leads to Pascal’s principle.
P  P0  ρ g h
A fluid of density  is shaded in the figure below. We now consider
the free body diagram for a cylindrical section of fluid that has a cross
sectional area A and thickness h.
y
A
h
x
IV.
Pascal’s Principle
A change in pressure applied to an enclosed fluid is transmitted
undiminished to every point of the fluid and to the walls of the
container.
A.
Submarine Example
The first term in our previous work demonstrates Pascal’s principle.
The pressure applied by air on the surface of the water, P0, is
transferred to the submarine at any depth. The second term is the
increase in pressure due to the weight of the water above the
submarine pushing down upon the submarine and the water
surrounding the submarine.
For the submarine, we see that the pressure is applied at all points
around the submarine in order to ensure that the net external force is
zero (i.e. it is not accelerating). We neglected this in our previous
FBD work in order to simplify the drawing. For the submarine, the
pressure would need to be greatest at the bottom since it must hold up
the weight of the submarine and the submarine has a finite size. Our
previous work suggests that this extra pressure must be the second
term in our equation which only depends on the fluid!! This will lead
us in the next section to Archimedes’s principle which was known
hundreds of years before Pascal’s principle.
B.
Pressure on a Vessel
If instead of a submarine, we consider an infinitesimally small unit of
volume then the pressure must be the same on all sides of the volume
element. This fact has some interesting consequences. Consider an
infinitesimal volume element of fluid next to the wall of a container as
shown below.
P
P
P
P
The pressure applied on the left side of the volume element by the
container must be the same as the pressure applied on the right side of
the element by the rest of the fluid since the volume element can not
accelerate. By Newton’s third law, this also means that the volume
element must apply the same pressure back up the container. Thus,
our formula in section III also gives us the pressure on the walls of a
vessel. Since the pressure on a dam increases at the bottom of the
dam, the thickness of the dam must be larger at the bottom.
C.
Example – Hydraulic Lift
A useful application of Pascal’s principle is the hydraulic lift.
Consider an enclosed fluid shown below. A downward pressure of P
is applied to a piston of area A1. This pressure is undiminished
through the fluid (Pascal’s principle) and pushes upward on a second
piston on area A2.
P
P
Same height
Using the definition of pressure, we have
P
F1 F2

A1 A 2
Thus, we see that the upward force on the second piston can be made
large than the downward force F1 by the ratio of the areas of the two
pistons.
 A2 
F2  
 F1
A
1


This is the technique in which an auto mechanic can push down on a
small piston and raise a car. It might seem sensible to think that the
forces on the pistons would be the same. However, the force is
increased because there are many more fluid atoms pushing on the
larger area piston than on the smaller piston!!
V.
Archimedes’s Principle
Any object completely or partially submerged in a fluid is buoyed up
by a force with magnitude equal to the weight of the fluid displaced
by the object.
Consider an object of mass mo that is partially or totally submerged
like the one shown below with its free body diagram.

B
y
x

W
F m
y
o
Ay
B  W  mo A y
The weight of the object can be related to the density of the object, o,
and the object’s total volume V by
W  mo g  ρo g V
Using Archimedes’s principle, the buoyancy force is related to density
of the fluid f, and the volume of the fluid displaced (this is the
portion of the volume of the object which is submerged) Vs.
B  m w g  ρ f g Vs
Substituting into our Newton II work we have
ρ f g Vs  ρ o g V  m o A y
Replacing mo in terms of the object’s density, we have
ρ f g Vs  ρ o g V  ρ o V A y
ρ f Vs  ρ o V g  ρ o V A y
Case I: Floating (Partially Submerged) Object
In this case the object is in equilibrium so Ay = 0. This gives us the
following condition:
ρ f Vs  ρ o V g  0
ρ f Vs  ρ o V   0
ρ f Vs  ρ o V
Vs  ρ o 
   1
V  ρf 
This says that the fraction of the volume of the object that is
submerged is equal to the ratio of the density of the object to the
density of the fluid. Since the largest fraction of the volume that can
be submerged is 100%, we also see that the density of the object must
be less than the density of the fluid if the object is to float..
Case II: Totally Submerged Object
In this case the volume of water displaced is equal to the volume of
the object.
Vs  V
Putting this condition in our Newton II results, we have
ρf g V  ρo g V  ρo V A y
ρ f
 ρ o g V  ρ o V A y
ρ f
 ρ o g  ρ o A y
If the object’s density if greater than the fluid then the object will
accelerate downward (sink) until it hits the bottom and an additional
force is added to our free body diagram. If the object’s density is less
than the density of the fluid, the object will accelerate upward to the
surface and will become only partially submerged as in case II. If the
density of the object and the fluid is the same then the object will be
in equilibrium and remain submerged at this depth.
VI. Continuity Equation
For an incompressible fluid the volume of fluid passing through any
size cross section of a pipe must be the same due to conservation of
mass. This condition can also be stated in a more useful form as
follows:
The product of the cross section and the component of the velocity of
the fluid perpendicular to the cross sectional area is a constant.
v  A  Flux  constant
This conserved quantity is called flux which is Latin for “water flow.”
Proof: Consider a fluid flowing through a pipe of varying cross sectional
area as shown below.
v1
v2
An amount of fluid of mass M enters the pipe on the left at speed v1.
Since the amount of fluid in the pipe can not increase, the same mass
of fluid must be pushed out on the right side of the pipe at speed v2.
We can write the mass of the volume of fluid entering the left hand
side over a time t as
M  ρ V  ρ A1 x 1
M  ρ A 1  v1 t 
We can write the mass of the volume of fluid exiting the right hand
side over a time t as
M ρ V ρ A2 x2
M  ρ A 2 v 2 t 
Equating our two results gives us the continuity equation as
ρ A 1  v1 t   ρ A 2  v 2 t 
A 1 v1  A 2 v 2
Q.E.D.
The continuity equation explains many interesting everyday events
including the fact that when water flows through a reduced cross
sectional area it will travel at a higher speed. This is why we reduce
the area of a garden hose to make a sprayer and why fast moving
rapids occur when a river becomes shallow!!! It is just a simple
consequence of conservation of atoms!!!
VII. Bernoulli’s Equation
Bernoulli’s equation is a consequence of applying conservation of
energy to a fluid.
Consider a fluid flowing through a pipe as shown below.
F2
x2
x1
y2
y1 F1
The work done on the fluid by an external force F1 in displacing the
fluid by x1 is
W1  F1Δx1  P1 A1 Δx1  P1 V
The work done on the fluid by an external force F2 as the fluid is
displaced by x2 is
W2  F2 Δx 2 Cos180    P2 A 2 Δx 2   P2 V
Assuming that there are no other un-conservative forces acting on the
fluid, we have from conservation of mechanical energy
Wnc  W1  W2  ΔK  U
P1 V  P2 V  K 2  K1  U 2  U1
The kinetic energy of the fluid at position 1 is given by
1
1
2
2
K 1  m v1  ρ V v 1 .
2
2
The kinetic energy of the fluid at position 2 is given by
1
1
2
2
K 2  m v2  ρ V v2 .
2
2
The gravitational potential energy at point 1 is given by
U1  m g y1  ρ V g y1 .
The gravitational potential energy at point 2 is given by
U2  m g y2 ρ V g y2 .
Substituting our results into our conservation of mechanical energy
equation and performing some algebra gives us Bernoulli’s equation.
1
1
2
2
P1 V  P2 V  ρ V v 2  ρ V v1  ρ V g y 2  ρ V g y1
2
2
1
1
2
2
P1  P2  ρ v 2  ρ v1  ρ g y 2  ρ g y1
2
2
1
1
2
2
P1  ρ v1  ρ g y1  P2  ρ v 2  ρ g y 2  constant
2
2
Bernoulli’s Equation
According to Bernoulli’s equation a faster moving fluid applies a
lower pressure upon an object than a slower moving fluid.
This and other interpretations of Bernoulli’s equation will allow us to
explain many interesting physical phenomena including curve balls,
lift on air plane wings, and why fluids flow down hill. We will discuss
some of these in class. What is important to remember is that no
matter what we call it, we are still simply analyzing systems using
energy analysis!!!