AP Physics B
Fluid Dynamics
1
College Board Objectives
II. FLUID MECHANICS AND THERMAL PHYSICS
A. Fluid Mechanics
1. Hydrostatic pressure
Students should understand the concept of
pressure as it applies to fluids, so they can:
a) Apply the relationship between pressure, force,
and area.
b) Apply the principle that a fluid exerts pressure in
all directions.
c) Apply the principle that a fluid at rest exerts
pressure perpendicular to any surface that it
contacts.
2
Fluid mechanics, cont.
d) Determine locations of equal pressure in a
fluid.
e) Determine the values of absolute and gauge
pressure for a particular situation.
f) Apply the relationship between pressure and
depth in a liquid, DP = r g Dh
3
Fluid Mechanics, cont.
2. Buoyancy
Students should understand the concept of
buoyancy, so they can:
a) Determine the forces on an object
immersed partly or completely in a liquid.
b) Apply Archimedes’ principle to determine
buoyant forces and densities of solids and
liquids.
4
Fluid Mechanics, cont.
3. Fluid flow continuity
Students should understand the equation of
continuity so that they can apply it to
fluids in motion.
4. Bernoulli’s equation
Students should understand Bernoulli’s
equation so that they can apply it to fluids
in motion.
5
Unit Plan
• Read Chapter 10.1-10 re Fluid
Mechanics in Giancoli
• Assignment:
• Q/4,7,11
• P/5,6,11,12,14,16,25,33,35,
• and 44,46,48
6
10.1&2 Density &
Specific Gravity
• The mass density r of a substance is
the mass of the substance divided
by the volume it occupies:
m
r
V
unit: kg/m3
r for aluminum 2700 kg/m3 or 2.70 g/cm3
mass can be written as m = rV and
weight as mg = rVg
Specific Gravity: r substance / r water
7
Problem 10.5
5. (II) A bottle has a mass of 35.00 g when empty
and 98.44 g when filled with water. When filled
with another fluid, the mass is 88.78 g. What is
the specific gravity of this other fluid?
5. Take the ratio of the density of the fluid to that
of water, noting that the same volume is used
for both liquids.
rfluid  m V fluid mfluid 88.78 g  35.00 g
SGfluid 
SJ



 0.8477
r water  m V water mwater 98.44 g  35.00 g
fluid
8
• A fluid - a substance that flows and
conforms to the boundaries of its
container.
• A fluid could be a gas or a liquid;
however on the AP Physics B exam
fluids are typically liquids which are
constant in density.
9
An ideal fluid is assumed
• to be incompressible (so that its
density does not change),
• to flow at a steady rate,
• to be nonviscous (no friction between
the fluid and the container through
which it is flowing), and
• flows irrotationally (no swirls or
eddies).
10
10.3 Pressure
Any fluid can exert a force
perpendicular to its surface on the
walls of its container. The force is
described in terms of the pressure it
exerts, or force per unit area:
F
p
A
Units: N/m2 or Pa (1 Pascal*)
dynes/cm2 or PSI (lb/in2)
1 atm = 1.013 x 105 Pa or 15 lbs/in2
*One atmosphere is the pressure exerted on us
every day by the earth’s atmosphere.
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The pressure is the same in every
direction in a fluid at a given depth.
Pressure varies with depth. 
P = F = rAhg so P = rgh
A
A
12
A FLUID AT REST EXERTS
PRESSURE PERPENDICULAR TO
ANY SURFACE THAT IT
CONTACTS. THERE IS NO
PARALLEL COMPONENT THAT
WOULD CAUSE A FLUID AT REST
TO FLOW.
13
PROBLEM 10-9
9.
9.
(I) (a) Calculate the total force of the
atmosphere acting on the top of a table that
measures 1.6 m  2.9 m.
(b) What is the total force acting upward on the
underside of the table?
(a)
The total force of the atmosphere on
the table will be the air pressure times the area
of the table.

F  PA  1.013  10 N m
5
2
 1.6 m  2.9 m   4.7 10 N
5
(b) Since the atmospheric pressure is the same on the underside of the
table (the height difference is minimal), the upward force of air pressure is
the same as the downward force of air on the top of the table,
4.7 10 N
5
14
10.4 Atmospheric Pressure and Gauge
Pressure
p1
p1
p1
h
p2
h
p2
h
p2
• The pressure p1 on the surface of the water is 1
atm, or 1.013 x 105 Pa. If we go down to a depth
h below the surface, the pressure becomes
greater by the product of the density of the water
r, the acceleration due to gravity g, and the
depth h. Thus the pressure p2 at this depth is
p2  p1  rgh
15
In this case, p2 is called the absolute pressure -the total static pressure at a certain depth in a
fluid, including the pressure at the surface of the
fluid
The difference in pressure between the surface and
the depth h is gauge pressure
p2  p1  rgh
Note that the pressure at any depth does not
depend of the shape of the container, only the
pressure at some reference level (like the surface)
and the vertical distance below that level.
p1
p1
p1
h
p2
h
p2
h
p2
16
14.(II) (a) What are the total force and the
absolute pressure on the bottom of a swimming
pool 22.0 m by 8.5 m whose uniform depth is 2.0
m? (b) What will be the pressure against the side
of the pool near the bottom?
(a)The absolute pressure is given by Eq. 10-3c, and
the total force is the absolute pressure times the
area of the bottom of the pool.

P  P0  r gh  1.013  105 N m 2  1.00  103 kg m3

9.80 m s 2
  2.0 m 
 1.21 105 N m 2

F  PA  1.21 105 N m 2

7
22.0
m
8.5
m

2.3

10
N



17
(b) The pressure against the side of
the pool, near the bottom, will be the
same as the pressure at the
bottom,
P  1.2110 N m
5
2
18
10.5 Pascal’s Principle
• Pascal’s Principle -
if an external pressure
is applied to a confined fluid, the pressure at
every point within the fluid increases by that
amount. Applications: hydraulic lift and brakes
Pout = Pin
And since P = F/a
Fout = Fin
Aout
Ain
Mechanical Advantage:
Fout = Aout
Fin
Ain
19
Problem 10-10
10.(II) In a movie, Tarzan evades his captors by
hiding underwater for many minutes while
breathing through a long, thin reed. Assuming
the maximum pressure difference his lungs can
manage and still breathe is calculate the deepest
he could have been. (See page 261.)
10.The pressure difference on the lungs is the
pressure change from the depth of water
DP  r g Dh  Dh 
DP
rg

 133 N m 2 
 85 mm-Hg  

1
mm-Hg


1.00 10
3
kg m
3
 9.80 m s 
2
 1.154 m  1.2 m
20
10-7 Buoyancy and Archimedes’ Principle
This is an object submerged in a fluid. There is a
net force on the object because the pressures at
the top and bottom of it are different.
The buoyant force is
found to be the upward
force on the same volume
of water:
21
10-7 Buoyancy and Archimedes’ Principle
The net force on the object is then the difference
between the buoyant force and the gravitational
force.
22
10-7 Buoyancy and Archimedes’ Principle
If the object’s density is less than that of water,
there will be an upward net force on it, and it will
rise until it is partially out of the water.
23
10-7 Buoyancy and Archimedes’ Principle
For a floating object, the fraction that is
submerged is given by the ratio of the object’s
density to that of the fluid.
24
10-7 Buoyancy and Archimedes’ Principle
This principle also works in
the air; this is why hot-air and
helium balloons rise.
25
22. (I) A geologist finds that a Moon rock
whose mass is 9.28 kg has an apparent
mass of 6.18 kg when submerged in
water. What is the density of the rock?
22.
The difference in the actual mass and the apparent mass is the
mass of the water displaced by the rock. The mass of the water
displaced is the volume of the rock times the density of water, and
the volume of the rock is the mass of the rock divided by its density.
Combining these relationships yields an expression for the density
of the rock.
mactual  mapparent  Dm  r waterVrock  r water
r rock  r water
mrock
Dm

 1.00  10 kg m
3
3
mrock
r rock

 9.28 kg  6.18 kg
9.28 kg
 2.99 103 kg m 3
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24.(II) A crane lifts the 18,000-kg steel hull of a ship out of
the water. Determine (a) the tension in the crane’s cable
when the hull is submerged in the water, and (b) the
tension when the hull is completely out of the water.
24.(a) When the hull is submerged, both the buoyant force
and the tension force act upward on the hull, and so their
sum is equal to the weight of the hull. The buoyant force
is the weight of the water displaced.
T  Fbuoyant  mg 
T  mg  Fbuoyant  mhull g  r waterVsub g  mhull g  r water
mhull
r hull
 r water 
g  mhull g  1 

r

hull 
3
3


1.00

10
kg
m
5
5
 1.8  10 4 kg  9.80 m s 2   1 

1.538

10
N

1.5

10
N
3
3 
7.8  10 kg m 

27
24. (b)When the hull is completely out of the
water, the tension in the crane’s cable
must be equal to the weight of the hull.



T  mg  1.8 104 kg 9.80 m s 2  1.764 105 N  1.8 105 N
28
34.(III) A 5.25-kg piece of wood SG  0.50 
floats on water. What minimum mass of lead, hung from the
wood by a string, will cause it to sink?
34.For the combination to just barely sink, the total weight
of the wood and lead must be equal to the total buoyant
force on the wood and the lead.
Fweight  Fbuoyant  mwood g  mPb g  Vwood r water g  VPb r water g 
mwood  mPb 
mPb  mwood
mwood
r wood
r water 
mPb
r Pb
 r water 
 1
r
 wood   m
wood
 r water 
1  r 

Pb 
r water
 r water 
 r water 
 mPb  1 
 1 
  mwood 
r Pb 

 r wood 
 1

 1  1

1
 SG



0.50
 wood   5.25 kg 
  5.76 kg



1  1 
1 


 1  SG 
11.3



Pb 
29
10-8 Fluids in Motion; Flow Rate and the
Equation of Continuity
If the flow of a fluid is smooth, it is called streamline or
laminar flow (a).
Above a certain speed, the flow becomes turbulent (b).
Turbulent flow has eddies; the viscosity of the fluid is much
greater when eddies are present.
30
10-8 Fluids in Motion; Flow Rate and the
Equation of Continuity
We will deal with laminar flow.
The mass flow rate is the mass that passes a
given point per unit time. The flow rates at any
two points must be equal, as long as no fluid is
being added or taken away.
This gives us the equation of continuity:
(10-4a)
31
10-8 Fluids in Motion; Flow Rate and the
Equation of Continuity
If the density doesn’t change – typical for
liquids – this simplifies to
.
Where the pipe is wider, the flow is slower.
32
10-9 Bernoulli’s Equation
A fluid can also change its
height. By looking at the
work done as it moves, we
find:
This is Bernoulli’s
equation. One thing it
tells us is that as the
speed goes up, the
pressure goes down.
33
36.
(I) A 15-cm-radius air duct is used to replenish the
air of a room 9.2 m  5.0 m  4.5 m every 16 min. How
fast does air flow in the duct?
2.0 cm 2 .
36.
We apply the equation of continuity at constant
density, Eq. 10-4b. Flow rate out of duct = Flow rate into
room
Aduct vduct   r 2 vduct 
Vroom
t to fill
room
 vduct 
Vroom
 r t to fill
2
room

 9.2 m  5.0 m  4.5 m 
2
 60 s 
  0.15 m  16 min  

1
min


 3.1m s
34
• 39. (II) A 85 -inch (inside) diameter garden hose is used to
fill a round swimming pool 6.1 m in diameter. How long
will it take to fill the pool to a depth of 1.2 m if water
issues from the hose at a speed
0.40ofm s?
• 39. The volume flow rate of water from the hose,
multiplied times the time of filling, must equal the volume
of the pool.
 Av hose 
Vpool
t

 t
Vpool
Ahose vhose
  3.05 m  1.2 m 
2

2
"  1m

  0.40 m s
  12  85  


" 
 39.37  

 4.429  105 s

 5.1 days

 60  60  24 s 
4.429  105 s 
1day
35
40. (II) What gauge pressure in the water mains is
necessary if a firehose is to spray water to a height of
15 m?
40. Apply Bernoulli’s equation with point 1 being the water
main, and point 2 being the top of the spray. The
velocity of the water will be zero at both points. The
pressure at point 2 will be atmospheric pressure.
Measure heights from the level of point 1.
P1  12 r v12  r gy1  P2  12 r v22  r gy2 

P1  Patm  r gy2  1.00  103 kg m3


9.8 m s 2 15 m   1.5  105 N m2
36
Visit the follow website from
Boston University
• http://physics.bu.edu/~duffy/py105
.html
– For more information about (choose
from left panel)
•Pressure; Fluid Statics
•Fluid Dynamics
•Viscosity
37
At the website complete the
following:
1. Read and record important
equations and facts.
2. For each equation write the
quantity for each symbol
3. Write the unit for each
quantity (symbol ok)
38
Demonstrations to View
• http://www.csupomona.edu/~physic
s/oldsite/demo/fluidmech.html
39