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AP Physics 2
UNIT 1 – FLUID MECHANICS
Fluid Mechanics - Hydrostatics
States of Matter
Before we begin to understand the nature of a Fluid we must
understand the nature of all the states of matter:
The 3 primary states of matter
- solid - Definite shape and volume.
- liquid -Takes the shape of its container, yet has a definite
volume.
- gas - Takes the shape and volume of its container.
Special "states
- Plasma, Bose-Einstein Condensate
Density
The 3 primary states have a distinct density, which is defined as mass per unit
of volume.
Density is represented by
the Greek letter, “RHO”, r
What is a Fluid?
By definition, a fluid is any material that is unable to withstand a
static shear stress. Unlike an elastic solid which responds to a
shear stress with a recoverable deformation, a fluid responds
with an irrecoverable flow.
Examples of fluids include gases and liquids.
Why fluids are useful in physics?
Typically, liquids are considered to be incompressible. That is once
you place a liquid in a sealed container you can DO WORK on the
FLUID as if it were an object. The PRESSURE you apply is
transmitted throughout the liquid and over the entire length of
the fluid itself.
Pressure
One of most important applications of a fluid is it's pressure- defined as a Force
per unit Area
Example
A water bed is 2.0 m on a side an 30.0 cm deep.
(a) Find its weight if the density of water is 1000 kg/m3.
(b) Find the pressure the that the water bed exerts on the floor. Assume that the entire
lower surface of the bed makes contact with the floor.
a) V  2  2  0.30 
m
m
r   1000  
V
V
W  mg  11760 N
1.2 m3
1200 kg
F mg 11760 N
b) P  


2
A
A
4m
2940 N/m2
Hydrostatic Pressure
Suppose a Fluid (such as a liquid) is at REST, we call this HYDROSTATIC PRESSURE
Two important points
• A fluid will exert a pressure in all directions
• A fluid will exert a pressure perpendicular to any surface it compacts
Notice that the arrows on TOP of the objects are smaller than at the BOTTOM.
This is because pressure is greatly affected by the DEPTH of the object. Since the
bottom of each object is deeper than the top the pressure is greater at the
bottom.
Pressure vs. Depth
Fatm
Fwater
mg
Fwater= Fatm + mg
Suppose we had an object submerged
in water with the top part
touching the atmosphere. If we
were to draw an FBD for this
object we would have three
forces
1. The weight of the
object
2. The force of the
atmosphere pressing
down
3. The force of the water
pressing up
Pressure vs. Depth
But recall, pressure is force per unit area. So if we solve for
force we can insert our new equation in.
F
P
Fwater  Fatm  mg
A
Note: The initial
PA  Po A  mg
m
r   m  rV
V
PA  Po A  rVg
V  Ah
PA  Po A  rAhg
P  Po  rgh
pressure in this case
is atmospheric
pressure, which is a
CONSTANT.
Po=1x105 N/m2
A closer look at Pressure vs. Depth
P  Po  rgh
Depth below surface
Initial Pressure – May or MAY NOT be atmospheric pressure
ABSOLUTE PRESSURE
P  rgh
Gauge Pressure = CHANGE in pressure or the DIFFERENCE in
the initial and absolute pressure
Example
a) Calculate the absolute pressure at an ocean depth of 1000
m. Assume that the density of water is 1000 kg/m3 and that
Po= 1.01 x 105 Pa (N/m2).
b) Calculate the total force exerted on the outside of a 30.0 cm
diameter circular submarine window at this depth.
P  Po  rgh
P  1x105  (1000)(9.8)(1000)
P  9.9x106 N/m2
F
F
F
P  2 

2
A r
 (0.30)
2.80 x 106 N
A closed system
If you take a liquid and place it in a system that
is CLOSED like plumbing for example or a car’s
brake line, the PRESSURE is the same
everywhere.
Since this is true, if you apply a force at one part
of the system the pressure is the same at the
other end of the system. The force, on the
other hand MAY or MAY NOT equal the initial
force applied. It depends on the AREA.
You can take advantage of the fact that the
pressure is the same in a closed system as it
has MANY applications.
The idea behind this is called PASCAL’S
PRINCIPLE
Pascal’s Principle
Another Example - Brakes
In the case of a car's brake pads, you have a
small initial force applied by you on the
brake pedal. This transfers via a brake line,
which had a small cylindrical area. The
brake fluid then enters a chamber with
more AREA allowing a LARGE FORCE to be
applied on the brake shoes, which in turn
slow the car down.
P1  P2
Fbrake pedal
Abrake pedal

Fbrake pad / shoe
Abrake pad / shoe
Buoyancy
When an object is immersed in a fluid, such as a liquid, it is buoyed UPWARD by a
force called the BUOYANT FORCE.
When the object is placed in fluid is
DISPLACES a certain amount of fluid. If
the object is completely submerged, the
VOLUME of the OBJECT is EQUAL to the
VOLUME of FLUID it displaces.
Archimedes' Principle
" An object is buoyed up by a force equal to the weight of the fluid displaced."
In the figure, we see that the
difference between the weight in
AIR and the weight in WATER is 3
lbs. This is the buoyant force that
acts upward to cancel out part of
the force. If you were to weight
the water displaced it also would
weigh 3 lbs.
Archimedes' Principle
FB  (mg ) FLUID m  rV
FB  ( rVg ) Fluid
Vobject  VFluid
Example
A bargain hunter purchases a "gold" crown at a flea market. After she gets home, she
hangs it from a scale and finds its weight in air to be 7.84 N. She then weighs the crown
while it is immersed in water (density of water is 1000 kg/m3) and now the scale reads
6.86 N. Is the crown made of pure gold if the density of gold is 19.3 x 103 kg/m3?
Fobject( air)  Fobject( water )  Fbuoyant
7.84  6.86  FB 
0.98 N
FB  (mg ) Fluid  r fluidV fluid g
V fluid 
0.0001 m3
Vobject 
0.0001 m3
massobject 
r object 
0.80 kg
mobject
Vobject

8000 kg/m3
NO! This is NOT gold as 8000<19300
Fluid Dynamics
Fluid Flow
Up till now, we have pretty much focused on fluids at rest. Now let's look at fluids in
motion
It is important that you understand that an IDEAL FLUID:
•Is non viscous (meaning there is NO internal friction)
•Is incompressible (meaning its Density is constant)
•Its motion is steady and NON – TURBULENT
A fluid's motion can be said to be STREAMLINE, or LAMINAR. The path itself is
called the streamline. By Laminar, we mean that every particle moves exactly
along the smooth path as every particle that follows it. If the fluid DOES NOT
have Laminar Flow it has TURBULENT FLOW in which the paths are irregular and
called EDDY CURRENTS.
Mass Flow Rate
Consider a pipe with a fluid moving within it.
The volume of the blue
region is the AREA times
the length.
A
L
v
Length is velocity times
time
Density is mass per
volume
A
L
v
Putting it all together
you have MASS FLOW
RATE.
What happens if the Area changes?
v2
L1=v1t
A2
The first thing you MUST
understand is that MASS is
NOT CREATED OR
DESTROYED!
IT IS CONSERVED.
L2=v2t
The MASS that flows into a region = The MASS that flows
out of a region.
v1
A1
Using the Mass Flow rate
equation and the idea that a
certain mass of water is
constant as it moves to a
new pipe section:
We have the Fluid Flow
Continuity equation
Example
The speed of blood in the aorta is 50 cm/s and this vessel has a radius of 1.0 cm. If the capillaries
have a total cross sectional area of 3000 cm2, what is the speed of the blood in them?
A1v1  A2v2
r v  A2v2
2
1 1
 (1) 2 (50)  (3000)v2
v2 
0.052 cm/s
Bernoulli's Principle
The Swiss Physicist Daniel Bernoulli, was interested in how the velocity
changes as the fluid moves through a pipe of different area. He especially
wanted to incorporate pressure into his idea as well. Conceptually, his
principle is stated as: " If the velocity of a fluid increases, the pressure
decreases and vice versa."
The velocity can be increased by pushing the
air over or through a CONSTRICTION
A change in pressure results in a NET
FORCE towards the low pressure
region.
Bernoulli's Principle
Funnel
Ping pong
Ball
Constriction
Bernoulli's Principle
The constriction in the Subclavian artery causes the
blood in the region to speed up and thus produces
low pressure. The blood moving UP the LVA is then
pushed DOWN instead of down causing a lack of
blood flow to the brain. This condition is called TIA
(transient ischemic attack) or “Subclavian Steal
Syndrome.
One end of a gopher hole is
higher than the other causing
a constriction and low
pressure region. Thus the air
is constantly sucked out of the
higher hole by the wind. The
air enters the lower hole
providing a sort of air recirculating system effect to
prevent suffocation.
Bernoulli's Equation
Let’s look at this principle
mathematically.
X=L
F1 on 2
-F2 on 1
Work is done by a section of water applying a force on a second
section in front of it over a displacement. According to Newton’s
3rd law, the second section of water applies an equal and
opposite force back on the first. Thus is does negative work as
the water still moves FORWARD. Pressure*Area is substituted
for Force.
Bernoulli's Equation
v2
A2
y2
L1=v1t
L2=v2t
v1
y1
A1
ground
Work is also done by GRAVITY as the water travels a vertical displacement
UPWARD. As the water moves UP the force due to gravity is DOWN. So the
work is NEGATIVE.
Bernoulli's Equation
Now let’s find the NET WORK done by gravity and the water acting on itself.
WHAT DOES THE NET WORK EQUAL TO? A CHANGE IN KINETIC
ENERGY!
Bernoulli's Equation
Consider that Density = Mass per unit Volume
AND that VOLUME is
equal to AREA time LENGTH
Bernoulli's Equation
We can now cancel out the AREA and LENGTH
Leaving:
Bernoulli's Equation
Moving everything related to one side results in:
What this basically shows is that Conservation of Energy holds true within a fluid and
that if you add the PRESSURE, the KINETIC ENERGY (in terms of density) and
POTENTIAL ENERGY (in terms of density) you get the SAME VALUE anywhere along a
streamline.
Example
1 atm = 1x105 Pa
Water circulates throughout the house in a hot-water heating system. If the water is
pumped at a speed of 0.50 m/s through a 4.0 cm diameter pipe in the basement
under a pressure of 3.0 atm, what will be the flow speed and pressure in a 2.6 cmdiameter pipe on the second floor 5.0 m above?
A1v1  A1v2
r12v1  r22v2
(0.04) 2 0.50  (0.026) 2 v2
v2 
1.183 m/s
1 2
1 2
Po  rvo  rgho  P  rv  rgh
2
2
1
1
3x105  (1000)(0.50) 2  (1000)(9.8)(0)  P  (1000)(1.183) 2  (1000)(9.8)(5)
2
2
P  2.5x105 Pa(N/m2) or 2.5 atm