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```Chapter 11
Fluids
Learning Objectives
FLUID MECHANICS AND THERMAL PHYSICS

Fluid Mechanics

Hydrostatic pressure
Students should understand the concept of pressure as it
applies to fluids, so they can:

Apply the relationship between pressure, force, and area.

Apply the principle that a fluid exerts pressure in all
directions.

Apply the principle that a fluid at rest exerts pressure
perpendicular to any surface that it contacts.

Determine locations of equal pressure in a fluid.

Determine the values of absolute and gauge pressure for
a particular situation.

Apply the relationship between pressure and depth in a
liquid, P = g h
Learning Objectives

Buoyancy
Students should understand the concept of buoyancy, so
they can:
 Determine the forces on an object immersed partly or
completely in a liquid.
 Apply Archimedes’ principle to determine buoyant
forces and densities of solids and liquids.
 Fluid flow continuity
 Students should understand the equation of continuity
so that they can apply it to fluids in motion.
 Bernoulli’s equation
 Students should understand Bernoulli’s equation so
that they can apply it to fluids in motion.
Mass Density
2. Pressure
3. Pressure & Depth in a Static Fluid
4. Pressure Gauges
5. Pascal’s Principle
6. Archimedes’ Principle
7. Fluids in Motion
8. The Equations of Continuity
9. Bernoulli’s Equation
10.Applications of Bernoulli’s Equation
11.Viscous Flow (AP?)
1.
Chapter 11:
Fluids
Section 1:
Mass Density
Fluids

Fluids are substances that can flow, such as liquids
and gases, and even some solids

In Physics B, we will limit our discussion of fluids to
substances that can easily flow, such as liquids and
gases
DEFINITION OF MASS DENSITY
The mass density of a substance is the mass of a
substance divided by its volume:
m

V
 : density ( greek letter rho)
m : mass
V : volume
SI Unit of Mass Density: kg/m3
Example 1 Blood as a Fraction of Body Weight
The body of a man whose weight is 690 N contains about
5.2x10-3 m3 of blood.
(a) Find the blood’s weight and (b) express it as a
percentage of the body weight.



m  V  5.2 103 m3 1060 kg m3  5.5 kg
(a)
(b)


W  mg  5.5 kg  9.80 m s 2  54 N
54 N
Percentage 
100%  7.8%
690 N
11.1.1. Which one of the following objects has the largest mass?
a) a gold solid cube with each side of length r
b) a brass solid sphere of radius r
c) a silver solid cylinder of height r and radius r
d) a lead solid cube with each side of length r
e) a concrete solid sphere of radius r
11.1.2. Liquid A has a mass density of 850.0 kg/m3 and Liquid B has a
mass density of 1060.0 kg/m3. Seventy-five grams of each liquid
is mixed uniformly. What is the specific gravity of the mixture?
a) 0.955
b) 0.943
c) 0.878
d) 0.651
e) 0.472
11.1.3. What volume of helium has the same mass as 5.0 m3 of
nitrogen?
a) 35 m3
b) 27 m3
c) 7.0 m3
d) 3.5 m3
e) 0.72 m3
Chapter 11:
Fluids
Section 2:
Pressure
Pressure
F
P
A
SI Unit of Pressure: 1 N/m2 = 1Pa
pascal
The force on a surface caused by pressure is always
normal (or perpendicular) to the surface. This means
that the pressure of a fluid is exerted in all directions, and
is perpendicular to the surface at every location
Example 2 The Force on a Swimmer
Suppose the pressure acting on the back
of a swimmer’s hand is 1.2x105 Pa. The
surface area of the back of the hand is
8.4x10-3m2.
(a) Determine the magnitude of the force
that acts on it.
(b) Discuss the direction of the force.
F
P
A

F  PA

3
F  1.2 10 N m 8.4 10 m
5
2
2

 1.0 103 N
Since the water pushes perpendicularly against the back of the
hand, the force is directed downward in the drawing.
Atmospheric Pressure

Atmospheric pressure is

Differences in atmospheric
pressure cause winds to blow
 Low
atmospheric pressure
inside a hurricane’s eye
contributes to the severe
winds and the development
of the storm surge
Atmospheric Pressure at Sea Level: 1.013x105 Pa = 1 atmosphere
Problem
Calculate the net force on an airplane window if cabin pressure
is 90% of the pressure at sea level, and the external pressure is
only 50% of that tat sea level. Assume the window is 0.43 m tall
and 0.30 m wide and atmospheric pressure at sea level is
100,000 Pa
11.2.1. A swimmer is swimming underwater in a large pool. The
force on the back of the swimmer’s hand is about one thousand
newtons. The swimmer doesn’t notice this force. Why not?
a) This force is actually smaller than the force exerted by the
atmosphere.
b) The force is large, but the pressure on the back of the hand is
small.
c) The force is exerted on all sides equally.
d) The swimmer is pushing on the water with the same force.
e) I do not know, but I’m sure I would feel that kind of force.
11.2.2. In a classroom demonstration, a physics professor lies on a “bed of nails.” The bed
consists of a large number of evenly spaced, relatively sharp nails mounted in a board so
that the points extend vertically outward from the board. While the professor is lying
down, nearly one thousand nails make contact with his body. Which one of the
following choices provides the best explanation as to why the professor is not harmed by
the bed of nails?
a) The nails are not as sharp as nails typically used in construction.
b) The professor is wearing special clothes that are not easily penetrated by nails.
c) The professor’s skin has been caliced after years of doing the demonstration, so nails no
longer penetrate the skin.
d) The professor’s weight is distributed over all of the nails in contact with the professor’s
body, so the pressure exerted a nail at any location is too small to penetrate the skin.
e) The force due to gravity on the professor is balanced by the upward force of the nails, as
explained by Newton’s third law of motion, so the professor doesn’t accelerate
downward.
11.2.3. Amanda fills the two tires of her bicycle to the pressure
specified on the side wall of the tires. She then gets onto her
bicycle and notices that the bottoms of the tires look flatter than
before she mounted the bicycle. What happens to the pressure in
the tires when she is on the bicycle compared to when she was off
the bicycle?
a) The pressure inside the tire increases.
b) The pressure inside the tire decreases.
c) The pressure inside the tire has the same value.
11.2.4. In snowy regions of the world, the local people may wear snow shoes
below their normal shoes or boots. These snow shoes have a much
larger area than a regular shoe or boot. How does a snow shoe improve
a hiker’s ability to walk across a snowy region?
a) The hiker’s weight is distributed over the area of the snow shoes, which
reduces the pressure on the snow below and minimizes sinking
into the snow.
b) The snow shoes increase the normal force of the snow on the hiker.
c) The snow shoes increase the upward pressure of the snow on the hiker.
d) The snow shoes compact the snow making it harder to sink into it.
e) The hiker’s weight is reduced by wearing large area snow shoes.
11.2.5. Helium gas is confined within a chamber that has a moveable
piston. The mass of the piston is 8.7 kg; and its radius is 0.013 m.
If the system is in equilibrium, what is the pressure exerted on the
piston by the gas?
a) 1.639 ×104 Pa
b) 8.491 × 104 Pa
c) 1.013 × 105 Pa
d) 1.606 × 105 Pa
e) 2.619 × 105 Pa
11.2.6. Two identical balloons, filled with unequal amounts of the same gas, are connected via
a pipe with a closed valve as shown. One balloon has five times the diameter of the
other. Which of the following will occur when the valve is opened?
a) Nothing will happen. Both balloons will remain the same size.
b) The small balloon will get larger and the large one will get smaller, but they will end up
with different sizes.
c) The small balloon will get larger and the large one will get smaller. They will end up with
equal sizes.
d) Most of the air in the smaller balloon will move into the larger balloon.
e) Most of the air in the larger balloon will move into the smaller balloon.
11.2.7. In which one of the following cases is the pressure exerted on
the ground by the man the largest?
a) A man stands with both feet flat on the ground.
b) A man stands with one foot flat on the ground.
c) A man lies with his back flat on the ground.
d) A man kneels with both knees on the ground.
e) A man stands with the toes of one foot on the ground.
11.2.8. Carol hangs a piece of stained glass artwork that she has just completed on
her window using a suction cup hanger. Which one of the following statements
best explains the force that holds the suction cup to the glass window?
a) There is a high amount of pressure between the glass window and the suction
cup.
b) There is a very low pressure between the glass window and the suction cup.
c) There is a smaller pressure on the suction cup due to the atmosphere than the
pressure between the suction cup and the glass window.
d) There is a greater pressure on the suction cup due to the atmosphere than the
pressure between the suction cup and the glass window.
e) Pushing the suction cup against the glass window causes a very strong chemical
bond to form between the glass window and the suction cup.
11.2.9. What is the force that causes liquid to move upward in a
drinking straw as a person takes a drink?
a) that due to a low pressure region caused by sucking
b) that due to the pressure within the liquid
c) that due to atmospheric pressure
d) that due to the person sucking on the straw
e) that due to friction forces within the straw
Chapter 11:
Fluids
Section 3:
Pressure & Depth in a Static Fluid
Pressure of a Fluid
F
y
 P2 A  P1 A  mg  0
P2 A  P1 A  mg
m  V
P2 A  P1 A   Vg
V  Ah
P2 A  P1 A   Ahg
P  Po  gh
Conceptual Example 3 The Hoover Dam
Lake Mead is the largest wholly artificial
reservoir in the United States. The water
in the reservoir backs up behind the dam
for a considerable distance (120 miles).
Suppose that all the water in Lake Mead
were removed except a relatively narrow
vertical column.
Would the Hoover Same still be needed
to contain the water, or could a much less
massive structure do the job?
Example 4 The Swimming Hole
Points A and B are located a distance of 5.50 m beneath the surface
of the water. Find the pressure at each of these two locations.
P  Po  gh
atmospheric pressure




P2  1.01105 Pa  1.00 103 kg m3 9.80 m s 2 5.50 m

 
P2  1.55 105 Pa


11.3.1. You are vacationing in the Rocky Mountains and decide to ride a ski lift to the
top of a mountain. As you go up, your feel your ears make a “pop” sound because
of changes in atmospheric pressure. Which way does your ear drum move, if at
all, during your ascent up the mountain?
a) inward, because the pressure outside the ear is larger than the pressure inside the ear
b) inward, because the pressure outside the ear is smaller than the pressure inside the
ear
c) No movement occurs. This is just the sound of air bubbles bursting inside the ear.
d) outward, because the pressure outside the ear is smaller than the pressure inside the
ear
e) outward, because the pressure outside the ear is larger than the pressure inside the
ear
11.3.2. Consider the mercury U-shaped tube manometer shown.
Which one of the following choices is equal to the gauge pressure
of the gas enclosed in the spherical container? The acceleration
due to gravity is g and the density of mercury is .
a) gc
b) gb
c) ga
d) Patm + gb
e) Patm  gc
11.3.3. An above ground water pump is used to extract water from a
well. A pipe extends from the pump to the bottom of the well.
What is the maximum depth from which water can be pumped?
a) 19.6 m
b) 39.2 m
c) 10.3 m
d) 101 m
e) With a big enough pump, you can extract it from any depth.
11.3.4. An aquarium has a length L, a width W, and a height H. Glass
of thickness d is used for the sides of the aquarium. If you were to
design a similar aquarium, except that it has a length 5L and a
width 5W, what thickness of glass would you need to use?
a) d
b) d 5
c) 5d
d) 10d
e) 25d
11.3.5. A water well is dug to a depth of 15 meters. A pump is used to
bring the water upward through a pipe, but the pump can only raise
the water about 7.5 meters. By which of the following methods
can the water be brought all the way to the surface?
a) Connect a second, identical pump in series with the first pump.
b) Connect a second, identical pump in parallel with the first pump.
c) Replace the pump with a more powerful pump.
d) Replace the pipe with a narrower pipe.
e) None of these methods will work.
Chapter 11:
Fluids
Section 4:
Pressure Gauges
Gauge Pressure



P = gh
 P: pressure (Pa)
 : density (kg/m3)
 g: acceleration due to gravity (m/s2)
 h: height of column (m)
This type of pressure is often called gauge pressure:
 Does not include the effect of atmospheric
pressure on top of the fluid
If the fluid is water, this is referred to as hydrostatic
pressure.
Barometers
P2  P1   gh
Patm   gh
Patm
h
g
1.0110
5

Pa
h
13.6 103 kg m 3 9.80 m s 2


h  0.760 m  760 mm

Manometers
P2  PB  PA
PA  P1   gh
absolute pressure
P2  Patm   gh



gauge pressure
Sphygmomanometer
11.4.1. The height of mercury in one barometer is 0.761 m above the
reservoir of mercury. If a second barometer is brought to the same
location that has a liquid with a density of 2650 kg/m3, what is the
height of the fluid in the second barometer?
a) 0.761 m
b) 2.02 m
c) 2.87 m
d) 5.03 m
e) 3.91 m
Chapter 11:
Fluids
Section 5:
Pascal’s Principle
PASCAL’S PRINCIPLE
Any change in the pressure applied
to a completely enclosed fluid is transmitted
undiminished to all parts of the fluid and
enclosing walls.
P2  P1   g 0 m 
F2 F1

A2 A1
 A2 
F2  F1  
 A1 
Example 7 A Car Lift
The input piston has a radius of 0.0120 m
and the output plunger has a radius of
0.150 m. The combined weight of the car and the
plunger is 20500 N. Suppose that the input
piston has a negligible weight and the bottom
surfaces of the piston and plunger are at
the same level. What is the required input
force?
 A2 
F2  F1  
 A1 
 0.0120 m 2
F2  20500 N 
 131 N
2
 0.150 m 
11.5.1. A fluid is completely enclosed in the system shown. As the piston is moved to the
right, which one of the following statements is true?
(a) Pushing the piston to the right causes the pressure
on the left side of the vertical cylinder to be larger than
that on the right side.
(b) Pushing the piston to the right causes the pressure
on the bottom of the vertical cylinder to be larger than
that on the top.
(c) Pushing the piston to the right causes the pressure throughout the vertical cylinder to
increase by the same amount.
(d) Pushing the piston to the right causes the pressure in one part of the vertical cylinder and a
corresponding decrease in another part of the cylinder.
(e) After the piston is pushed by a distance s and held in that position, the pressure will be the
same at all points within the fluid.
11.5.2. At an automotive repair shop, a hydraulic lift is used to raise
vehicles so that mechanics can work under them. In one lift,
compressed air with a maximum gauge pressure of 5.0 × 105 Pa is
used to raise a piston with a circular cross-section and radius of 0.15
m. What is the maximum vehicular weight that can be raised using
this lift?
a) 7.5 × 104 N
b) 6.0 × 104 N
c) 4.4 × 104 N
d) 3.5 × 104 N
e) 1.1 × 104 N
11.5.3. A woman stands on a platform with a diameter D1 that acts as a piston.
The combined mass of the woman and the piston is 75 kg. The piston
pushes downward on a reservoir of oil that supports a second platform with
a diameter D2 on which 12 women are standing. The combined mass of the
second platform and the 12 women is 780 kg. Both platforms are at the
same height during this demonstration; and they are at rest. What is the
ratio D2/D1?
a) 2.4
b) 3.2
c) 4.8
d) 10
e) 12
Chapter 11:
Fluids
Section 6:
Archimedes’ Principle
Archimedes’ Principle



A body immersed in a fluid is buoyed up by a force
that is equal to the weight of the fluid it displaces
When an object floats, the upward buoyant force
equals the downward pull of gravity
This buoyant force can float very heavy objects, and
acts upon object in the water whether that are
floating, submerged, or even sitting on the bottom of
the body of fluid.
FB
FW
Archimedes’ Principle
P2  P1   gh
FB  P2 A  P1 A  P2  P1 A
V  hA
FB   ghA
FB  
V g
displaced
fluid
Fbuoy  Vg
Floaters
If the object is floating then the
magnitude of the buoyant force
is equal to the magnitude of its
weight.
Example 9 A Swimming Raft
The raft is made of solid square pinewood.
Determine whether the raft floats in water and if
so, how much of the raft is beneath the surface.
FBmax   Vg   waterVwater g
Vraft  4.0 m 4.0 m 0.30 m   4.8 m



FBmax  1000 kg m3 4.8m3 9.80 m s 2

FBmax  47000 N
Wraft  mraft g   pineVraft g



Wraft  550 kg m3 4.8m3 9.80 m s 2

Wraft  26000 N  47000 N
It will float, but how much will be submerged?
Wraft  FB
W   water Ahg
h
W
 water Ag

26000 N
1000 kg m3 4.0 m4.0 m9.80 m s 2   0.17 m
Conceptual Example 10 How Much Water is Needed
to Float a Ship?
A ship floating in the ocean is a familiar sight. But is all
that water really necessary? Can an ocean vessel float
in the amount of water than a swimming pool contains?
11.6.1. A solid block of mass m is suspended in a liquid by a thread. The
density of the block is greater than that of the liquid. Initially, the
fluid level is such that the block is at a depth d and the tension in the
thread is T. Then, the fluid level is decreased such that the depth is
0.5d. What is the tension in the thread when the block is at the new
depth?
a) 0.25T
b) 0.50T
c) T
d) 2T
e) 4T
Heavy Objects can be moved with water

Check out these photo from the aftermath of
Hurricane Katrina!
11.6.2. A fisherman rows his boat out to his favorite spot on a pond,
where he decides to drop a heavy anchor from his boat. When he
drops the anchor into the pond, what happens to the level of the
pond?
a) The level of the pond decreases.
b) The level of the pond stays the same.
c) The level of the pond increases.
11.6.3. In the distant future, a base has been constructed on the moon.
Inside the base, a ball is dropped into a container of water. Given that
the acceleration due to gravity is about one sixth the value on the
Earth’s surface, which of the following statements correctly describes
the buoyant force on the ball?
a) The buoyant force is equal to 1/6 the weight of the displaced water.
b) The buoyant force is equal to 6 times the weight of the displaced
water.
c) The buoyant force is equal to the weight of the displaced water.
d) The buoyant force is equal to 1/6 the weight of the ball.
e) The buoyant force is equal to the weight of the ball.
11.6.4. A coin is dropped into a lake. As the coin sinks, how does the
buoyant force on the coin change?
a) The buoyant force decreases as the depth increases.
b) The buoyant force increases as the depth increases.
c) The buoyant force decreases as the speed increases.
d) The buoyant force decreases as the depth decreases.
e) The buoyant force has a constant value.
11.6.5. Three fourths the volume of a glass is filled with water. Ice is
then added until it reaches the top of the glass. The water level at
that point is h. Which one of the following statements concerning
the level of water after the ice melts is true?
a) The final level of water will be lower than h.
b) The final level of water will be at h.
c) The final level of water will be higher than h.
d) This cannot be answered without knowing the mass of ice added.
e) This cannot be answered without knowing the volume of ice added.
11.6.6. A boy is playing in the bathtub with a toy boat floating on
the surface of the water. His favorite rock is the cargo on the
boat. If the boy takes the rock from the boat and drops it into
the water, how will the water level of the bathtub change, if at
all?
a) The water level will rise.
b) The water level will fall.
c) The water level will remain the same.
Chapter 11:
Fluids
Section 7:
Fluids in Motion
Types of flowing fluids:
In steady flow the velocity of the fluid particles at any point is constant
as time passes.
Unsteady flow exists whenever the velocity of the fluid particles at a
point changes as time passes.
Turbulent flow is an extreme kind of unsteady flow in which the velocity
of the fluid particles at a point change erratically in both magnitude and
direction.
More types of fluid flow

Fluid flow can be compressible or incompressible.

Most liquids are nearly incompressible.

Fluid flow can be viscous or nonviscous.

An incompressible, nonviscous fluid is called an
ideal fluid.
When the flow is steady, streamlines are often used to represent
the trajectories of the fluid particles.
Making streamlines with dye
and smoke.
Chapter 11:
Fluids
Section 8:
The Equation of Continuity
When a fluid flows…

… mass is conserved
 Provided
there are no inlets or outlets in a stream
of flowing fluid
 The
same mass per unit time must flow
everywhere in the stream
The mass of fluid per second that flows through a tube is called
the mass flow rate.
The Equation of Continuity
m  V   A v
t
distance
m2
  2 A2 v2
t
m1
 1 A1v1
t
EQUATION OF CONTINUITY
The mass flow rate has the same value at every position along a
tube that has a single entry and a single exit for fluid flow.
1 A1v1  2 A2v2
SI Unit of Mass Flow Rate: kg/s
The Equation of Continuity
Incompressible fluid:
A1v1  A2v2
Example Problem
A Pipe of diameter 6.o cm has fluid lowing through is at 1.6 m/s. How
fast is the fluid flowing in an area of the pipe in which the diameter is 3.0
com? How much water flows through the pipe per second?
A1v1  A2v2
 r12v1   r22v1
2
v1r12
v2  2  1.6 m s  .03m   6.4 m s
r2
 .015m 
Problem
The water in a canal flows 0.10 m/s where the canal is 12 m deep
and 10 m across. If the depth of the canal is reduced to 6.5 m at an
area where the canal narrows to 5.0 m, how fast will the water be
moving through this narrower region?
A1v1  A2v2
11.8.1. At the site of a burning building, a firefighter is using a hose with a
radius of 0.032 m that has water flowing through it at a rate of 9.95 m/s.
At the end of the hose, there is a nozzle with a radius of 0.013 m. First,
calculate the speed of the water exiting the nozzle and use that to
determine the range, the maximum horizontal distance the water can
reach, if it is directed at an angle of 45 with respect to the horizon.
a) 6.2 m
b) 59 m
c) 120 m
d) 210 m
e) 370 m
11.8.2. A child has left an outdoor faucet open. An adult walks up and notices that
the diameter of the stream is 0.25d at the bottom and d at the top as it falls
vertically from the faucet. What is the physical explanation for this narrowing
of the stream?
a) The flow rate at the top of the stream is not sufficient to maintain a constant
cross-sectional area.
b) Atmospheric pressure is greater than the pressure within the stream of water.
c) The water has left the pipe that carried it and no longer maintains the shape of the
pipe.
d) The water accelerates as it falls and the cross-sectional area must decrease to
maintain a constant flow rate.
e) The stream of water is experiencing friction with the air as it falls and part of it
slows down while part of it falls at a constant speed.
11.8.3. The drawing shows a section of a pipe system in which an
incompressible fluid can either flow in or flow out various
channels. All of the pipes have the same diameter. The
direction and flow rates are indicated on the drawing. What is
the mass flow rate and direction of flow for the unknown pipe?
a) 0.2 m3/s
b) 0.4 m3/s
c) 0.6 m3/s
d) 0.8 m3/s
e) 1.0 m3/s
Chapter 11:
Fluids
Section 9:
Bernoulli’s Equation
Bernoulli's Theorem


The sum of the pressure, the potential energy per
unit volume, and the kinetic energy per unit volume
at any one location in the fluid is equal to the sum of
the pressure, the potential energy per unit volume,
and the kinetic energy per unit volume at any other
location in the fluid for a non-viscous,
incompressible fluid in streamline flow.
All other considerations being equal, when a fluid
moves faster, the pressure drops
The fluid accelerates toward the
lower pressure regions.
According to the pressure-depth
relationship, the pressure is lower
at higher levels, provided the area
of the pipe does not change.
W   F s  F s  PAs  P2  P1 V
P
F
A
F  PA
F  P A
Wnc  E  12 mv12  mgy1   12 mv22  mgy2 
P2  P1 V  12 mv12  mgy1   12 mv22  mgy2 
m

V
P2  P1   12 v12  gy1   12 v22  gy2 
P1  12 v12  gy1  P2  12 v22  gy2
BERNOULLI’S EQUATION
In steady flow of a nonviscous, incompressible fluid, the pressure, the
fluid speed, and the elevation at two points are related by:
P1  12 v12  gy1  constant
Problem
Knowing what you know about Bernoulli’s principle, design
an airplane wing that you think will keep an airplane aloft.
Draw a cross section of the wing
Problem

An above ground swimming pool has a hole of
radius 0.10 cm in the side1.0 m below the surface of
the water. How fast is the water flowing out of the
hole?
Problem

An above ground swimming pool has a hole of
radius 0.10 cm in the side1.0 m below the surface of
the water. How much water flows out each second?
Problem

An above ground swimming pool has a hole of
radius 0.10 cm in the side 1.0 m below the surface
of the water. How far does the water land from the
side of the pool if the hole is 1.0 m above the
ground?
Problem

Water travels through a 9.6 cm diameter fire hose
with a speed of 1.3 m/s. At the end of the hose, the
water flows out of a nozzle whose diameter is 2.5
cm. What is the speed of the water coming out of
the nozzle?
Problem

Water travels through a 9.6 cm diameter fire hose
with a speed of 1.3 m/s. At the end of the hose, the
water flows out of a nozzle whose diameter is 2.5
cm. If the pressure in the hose is 350 kPa, what is
the pressure in the nozzle in kPa?
11.9.1. Many tall buildings have reservoirs of water on the roofs. The surface of
these reservoirs are at atmospheric pressure. If all of the water in one tall
building were delivered from such a reservoir, would it be a good idea to
deliver water to all of the floors of the building from a single pipe connected
to the reservoir?
a) Yes, that would ensure a steady flow of water at constant pressure on all
floors.
b) Yes, that would ensure a steady flow of water at a slightly varying pressure
on all floors.
c) No, the pressure would be too large on the lower floors and exit faucets at
dangerous speeds.
d) No, the pressure would be too small on the lower floors and would not exit
faucets.
11.9.2. A curtain hangs straight down in front of an open window. A
sudden gust of wind blows past the window; and the curtain is
pulled out of the window. Which law, principle, or equation can
be used to explain this movement of the curtain?
a) the equation of continuity
b) Pascal's principle
c) Bernoulli's equation
d) Archimedes' principle
e) Poiseuille's law
11.9.3. Fluid is flowing from left to right through the pipe shown in the
drawing. Points A and B are at the same height, but the cross-sectional
areas of the pipe are different at the two locations. Points B and C are at
two different heights, but the cross-sectional areas of the pipe are the
same at these two locations. Rank the pressures at the three locations in
order from lowest to highest?
a) PA > PB > PC
b) PB > PA = PC
c) PC > PB > PA
d) PB > PA and PB > PC
e) PC > PA and PC > PB
11.9.4. An air blower is attached to a funnel that has a light-weight ball
inside as shown. How will the ball behave when the blower is turned
on?
a) The ball will roll around inside the funnel in
random directions.
b) The ball will bounce around inside the funnel
in random directions.
c) The ball will be drawn upward to the top of
the funnel and remain there.
d) The ball will remain stationary, unaffected by the flowing air.
e) The ball will spin rapidly as it moves slowly around inside the funnel.
11.9.5. A piece of paper is laying flat on a desk near a closed window. When the
window is opened, air rushes over the top of the desk and the paper is observed
to be lifted upward and then carried away by the wind. Which of the following
statements best describes why the paper was lifted from the desk?
a) The paper was attracted by the force of the wind.
b) The pressure of the moving air above the paper was greater than that of the air
between the paper and the desk top.
c) The pressure of the moving air above the paper was less than that of the air
between the paper and the desk top.
d) The weight of the paper was reduced by the wind blowing over it and the normal
force of the desk was then greater than the weight. The normal force pushes the
paper upward.
e) The wind pushed the side of the paper and lifted it upward.
Chapter 11:
Fluids
Section 10:
Applications of Bernoulli’s Equation
Conceptual Example 14 Tarpaulins and Bernoulli’s Equation
When the truck is stationary, the
tarpaulin lies flat, but it bulges outward
when the truck is speeding down
the highway.
Account for this behavior.
Example 16 Efflux Speed
The tank is open to the atmosphere at
the top. Find an expression for the speed
of the liquid leaving the pipe at
the bottom.
v2  0
P1  P2  Patm
P1  12 v12  gy1  P2  12 v22  gy2
y2  y1  h
1
2
v12  gh
v1  2 gh
Chapter 11:
Fluids
Section 11:
Viscous Flow
Flow of an ideal fluid.
Flow of a viscous fluid.
FORCE NEEDED TO MOVE A LAYER OF VISCOUS FLUID WITH
CONSTANT VELOCITY
The magnitude of the tangential force required to move a fluid
layer at a constant speed is given by:
F
Av
y
coefficient
of viscosity
SI Unit of Viscosity: Pa·s
Common Unit of Viscosity: poise (P)
1 poise (P) = 0.1 Pa·s
POISEUILLE’S LAW
The volume flow rate is given by:
R 4 P2  P1 
Q
8L
Example 17 Giving and Injection
A syringe is filled with a solution whose
viscosity is 1.5x10-3 Pa·s. The internal
radius of the needle is 4.0x10-4m.
The gauge pressure in the vein is 1900 Pa.
What force must be applied to the plunger,
so that 1.0x10-6m3 of fluid can be injected
in 3.0 s?
P2  P 

8LQ
R 4



8 1.5 10 3 Pa  s 0.025 m  1.0 10 6 m 3 3.0 s
 1200 Pa
 4.0 10 m 
-4
4

P1  1900 Pa
P2  P1  1200 Pa
P2  3100 Pa


F  P2 A  3100 Pa  8.0 10 m  0.25N
5
2
```