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AP PHYSICS B
Fluids
Teacher Packet
AP* is a trademark of the College Entrance Examination Board. The College Entrance Examination Board was not
involved in the production of this material.
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Fluids
Objective
To review the student on the concepts, processes and problem solving strategies necessary to
successfully answer questions on fluid mechanics.
Standards
Fluids is addressed in the topic outline of the College Board AP* Physics Course Description
Guide as described below.
II. Fluid Mechanics and Thermal Physics
A. Fluid Mechanics
1. Hydrostatic Pressure
2. Buoyancy
3. Fluid Flow Continuity
4. Bernoulli’s equation
AP Physics Exam Connections
Topics relating to fluids are tested every year on the multiple choice and in most years on the free
response portion of the exam. The list below identifies free response questions that have been
previously asked over fluids. These questions are available from the College Board and can be
downloaded free of charge from AP Central. http://apcentral.collegeboard.com.
2008
2007
2005
2004
2003
2002
Free Response Questions
Question 4
2008 Form B Question 4
Question 4
2007 Form B Question 4
Question 5
2005 Form B Question 5
Question 2
2004 Form B Question 2
Question 6
2003 Form B Question 6
Question 6
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Fluids
What I Absolutely Have to Know to Survive the AP* Exam
The study of fluids is separated into two distinct parts: hydrostatics (fluids at rest) and hydrodynamics
(fluids in motion). The static pressure due to an incompressible fluid depends primarily upon the depth of
the fluid. Pascal’s principle states that the pressure exerted at one location in a confined fluid is transmitted
undiminished throughout the fluid. Archimedes principle states that the buoyant force acting on an object in
a fluid is equal to the weight of the fluid displaced by the object. As a fluid flows through a pipe, the flow
rate through the cross sectional area is the same at any point in the pipe. Bernoulli’s equation relates static
pressure of a fluid to its dynamic (moving) pressure.
Key Formulas and Relationships
Volume: V = (length * width * height) = lwh
Unit: m3
Density: ρ =
m
V
Unit: kg/m3
Note: Density of water is 1000 kg/m3
Pressure: P =
F
A
Unit: N/m2 = 1 Pascal = 1 Pa
Note: 1 atmosphere of pressure is approximately equal to 1× 105 Pa
Gauge Pressure: Pgauge = ρ gh
Absolute Pressure: Pabsolute = P0 + Pgauge = P0 + ρ gh
Pascal’s Principal:
F1 F2
=
A1 A2
Buoyant Force: FBuoy = ρVg = Weight of Displaced Fluid
Unit: N
Volume Flow Rate: υ = Av
Unit: m3/s
Continuity Equation: v1 A1 = v2 A2
Bernoulli’s Equation: P + ρ gy +
1 2
ρ v = constant
2
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Fluids
Important Concepts
•
Pressure:
o Pressure is not a force.
o Pressures from fluids always generate forces that are perpendicular to the surface of
whatever the fluid is in contact with.
•
Static Pressure in a Fluid:
Remember from our study of forces and Newton’s Laws
that an object at rest must be in equilibrium. Consider the
dot pictured in the container of fluid to the right. For the dot
to be in equilibrium all the forces on it must cancel out. The
gravitational force pulls the dot down and must be canceled
by a force pushing upward on the dot.
Fupward on the dot = mg downward on the dot
mg
F
m
v
PA = ρ gV and V=Area × Height
PA=ρ gAh
P=ρ gh
F = PA and ρ =
The static pressure due to the fluid depends upon the depth of the fluid, the density of the
fluid, and the acceleration due to gravity. Note that the area does not make any difference!
h
The pressure is the same at the bottom of both of the containers pictured above because they
have the same depth of fluid. This means that at the same vertical location in a continuous
fluid, the pressure is always the same.
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Fluids
Consider the “U” shaped tube filled with a fluid and fitted with a piston pictured below.
Along the horizontal line the pressure in the fluid is the same: P = ρ gh1
piston
fluid
original height h1
horizontal line
Now the piston is pushed downward with a force toward the horizontal line thus moving the
fluid higher in the right hand side of the “U” shaped tube. This force increases the pressure
in the fluid at the horizontal line to a new value: P = ρ gh2
piston
Force
fluid
new height h2
horizontal line
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Fluids
o Absolute Pressure and Gauge Pressure
If there is anything on top of the fluid itself, the pressure in the fluid would need to be
greater to maintain equilibrium.
Mass and
atmosphere
atmosphere
Example 1) The atmosphere might be pushing down on the fluid from above. In this case the
pressure in the fluid would be:
P = Patmosphere + ρ gh
Example 2) The atmosphere pushes down from above and a mass is positioned on top of a
piston above the fluid. In this case the pressure in the fluid would be:
P = Patmosphere + Pdue to mass + ρ gh
P = Patmosphere +
mg
+ ρ gh
Apiston
To avoid as much confusion as possible, two terms have been developed to describe the
static pressure in a fluid: Absolute Pressure and Gauge Pressure.
Gauge Pressure is the pressure caused only by the fluid itself:
Pgauge = ρ gh
Absolute Pressure is the pressure caused by the fluid (ρgh) and anything on top of the fluid,
such as the atmosphere and any mass or piston (P0):
Pabsolute = P0 + Pgauge = P0 + ρ gh
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Fluids
•
In a confined fluid, a pressure exerted at one location is transmitted to all parts of the fluid. This is
called Pascal’s Principle.
P1 = P2
F1 F2
=
A1 A2
F2
F1
A2
A1
Pascal’s Principle is utilized in both hydraulic (liquid) and pneumatic (gas) devices where a small
force is used to push the fluid through a smaller cylinder into a larger area cylinder where the force
⎛A ⎞
is multiplied by a factor, the ratio of the cross sectional areas ⎜ 2 ⎟ .
⎝ A1 ⎠
Note that the force is amplified, but the distances through which the cylinders move are not the
same. The large cylinder moves only a short distance, whereas the small cylinder moves a greater
distance. The work output of a Pascal’s Principle device will equal the work input if there is no
friction.
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Fluids
•
Objects submerged in a fluid will receive a greater fluid pressure on the
bottom than on the top. This occurs because the bottom of the object is
deeper in the fluid.
Downward
Pressure
Pbottom > Ptop
ρ ghbottom > ρ ghtop
hbottom > htop
This difference in pressure produces a buoyant force that is directed
upward and is opposite the force of gravitational attraction.
Upward
Pressure
The equation for buoyant force is: FBuoy = ρ fluidVdisplaced fluid g
The density in the equation is for the fluid and the volume is for the amount of fluid displaced by the
object in the fluid. Note that ρ fluidVdisplaced fluid is the same as the mass ( m = ρV ) of the displaced
fluid. The buoyant force does not depend upon the weight or the shape of the submerged object, but
the weight of the displaced fluid. This was first discovered by Archimedes.
Archimedes Principle: The buoyant force will equal the weight of the fluid that is displaced by the
object.
FBuoy = ρ fluidVdisplaced fluid g = mdisplaced fluid g = Weight of Displaced Fluid
FBuoy
mg
When an object floats, the weight of the object is equal to the buoyant force (weight of
displaced fluid). The volume of the object is not necessarily equal to the volume of the
displaced fluid, however. See the figure above.
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Fluids
FBuoy
mg
When an object sinks, the weight of the object is greater than the buoyant force (weight of
displaced fluid). Hence, the volume of the object will equal the volume of the displaced
fluid as per the figure above.
•
The continuity equation states that the volume of an incompressible fluid entering one part
of a flow tube must be equal to the volume leaving downstream.
v1 A1 = v2 A2
(Note that the product of velocity and the cross sectional area of the flow tube is the volume
flow rate of the fluid in units of m3/s. The volume flow rate remains constant throughout all
parts of the flow tube.)
A1
A2
v2
v1
The important consequence of the continuity equation is that the velocity of the fluid is
slower in wider parts of the flow tube and faster in narrower parts of the flow tube.
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Fluids
•
Bernoulli’s principle applies the ideas of work and energy to dynamic fluid flow to derive
what is now called Bernoulli’s equation:
1 2
1
ρ v1 = P2 + ρ gy2 + ρ v22
2
2
y1 = y2 thus ρ gy1 = ρ gy2
P1 + ρ gy1 +
P1 +
1 2
1
ρ v1 = P2 + ρ v22
2
2
(Using Bernoulli’s equation is analogous to using the law of conservation of energy. Simply
identify two points on a streamline and equate them using Bernoulli’s equation.)
1
2
Example 1) Consider the flow tube above. Points 1 and 2 are both centered in the middle of
the flow tube. Bernoulli’s equation tells us:
P1 + ρ gy1 +
1 2
1
ρ v1 = P2 + ρ gy2 + ρ v22
2
2
The vertical location, densities of the fluid, and “g” at points 1 and 2 are the same. Thus,
they cancel each other out on both sides of the equation.
1 2
1
ρ v1 = P2 + ρ gy2 + ρ v22
2
2
y1 = y2 thus ρ gy1 = ρ gy2
P1 + ρ gy1 +
P1 +
1 2
1
ρ v1 = P2 + ρ v22
2
2
The Continuity equation tells us that the velocity of the fluid at point 2 must be greater than the velocity
at point 1. This implies that the pressure in the fluid decreases as the speed of the fluid increases.
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Fluids
Free Response
Question 1 (15 pts)
18 m
1m
2m
4m
1. A ship carrying cargo into a port is hit by a sudden storm that causes one of the cargo
containers to fall overboard and sink to the bottom of the ocean. The ocean depth where
the container sank is 18 m. The container itself has a mass of 2400 kg. 9000 kg of cargo
was loaded inside the container. The density of ocean water is 1025 kg/m3.
A. Calculate the gauge pressure on the top of the container.
1 point for the correct use of gauge
pressure.
(3 points max)
Gauge pressure does not include the 1 atm of pressure
above the ocean.
The top of the container is only 17 meters below the
surface of the ocean.
Δh = 18 m – 1 m = 17 m
Pgauge = ρgΔh = (1025 kg/m3)(9.8 m/s2)(18 m – 1 m)
Pgauge = 1.7 ×105 Pa
®
1 point for correct height of 17 m
1 point for the correct answer including
correct units and reasonable number of
significant digits
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Fluids
B. Calculate the total downward force exerted on the top of the container due to the
water.
(3 points max)
1 point for using absolute pressure
Total downward force implies that we need to use
the absolute pressure to determine the force.
1 point for the correct force equation
F = PabsoluteA
F = Pabsolute(length * width)
F = (1 atm + Pgauge)A
F = (100,000 Pa + 1.7 E5 Pa)(4 m)(2 m)
F = 2.2×106 N
1 point for the correct answer
including correct units and reasonable
number of significant digits
C. A rescue ship with a crane is sent to retrieve the cargo. Using a cable attached to the
top of the container, the ship hoists the container up at a slow constant rate. Assuming no
resistance due to the water, determine the tension in the cable.
(5 points max)
1 point for a statement that ΣF = 0
Since the velocity is constant, the container is in
dynamic equilibrium and ΣF = 0. There are three
forces acting on the container. Tension and Buoyant
force act upward and Weight acts downward.
ΣF = 0
FBuoy + FT = mg
FT = mg - FBuoy
FT = (mcontainer + mcargo)g - ρVg
FT = (mcontainer + mcargo)g - ρ(length * width * height)g
FT = (2400 kg + 9000kg) (9.8 m/s2) - 1025 kg/m3)(4
m)(2 m)(1 m)(9.8 m/s2) = 31,000 N
1 point for the correct expression
FBuoy + FT = mg
®
1 point for adding the masses
1 point for correct calculation of the
volume
1 point for the correct answer
including correct units and reasonable
number of significant digits
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Fluids
D. On the graph below sketch the tension in the cable as the container is lifted out of the
water at a steady rate.
Tension in
the cable
Container still
completely
submerged in
the ocean
Container
partially
submerged
Top of
container at
the surface
of the ocean
2 points max)
The tension is a constant 31,000 N while the
container is still in the water. While the container is
being lifted out of the water, the tension increases at
a constant rate as the buoyant force decreases. Out of
the water, tension equals the weight of the container
and cargo (110,000 N).
®
Bottom of
container at
the surface of
the ocean
Container lifted
completely out
of the ocean
1 point for correctly drawing the
constant tension regions while the
container is still submerged and after it
has been lifted from the ocean
1 point for correctly drawing a linear
transition between the two constant
regions
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Location
of the
container
Fluids
E. All of the cargo is removed and the container is placed back into the ocean where it
now floats. Calculate d the portion of the container below the waterline.
d
1m
2m
4m
1 point for the correct expressions
relating weight and buoyancy force
(2 points max)
For floating objects, buoyant force equals the weight.
Fg = FBuoy
1 point for the correct answer
including correct units and reasonable
number of significant digits
mg = ρ gV
m = ρ ( lwd )
m
2400kg
=
ρ lw 1025 kg 2m 4m
( )( )
m3
d = 0.29 m
d=
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Fluids
Question 2 (15 pts)
A large container of fluid sits on a table. It has an opening at the bottom that is sealed by
a plug. The opening has a cross sectional area of 1.8 × 10−3 m 2 . The density of the fluid is
900 kg/m3. The figure is not drawn to scale.
0.65 m
0.4 m
h
0.25 m
A. What is the net force on the plug?
(3 points max)
The net force on the plug will be the difference in
forces caused by the external air pressure and the
internal fluid pressure. Thus the net force is caused
by the gauge pressure.
Fnet = Ffluid – Fair
Fnet = PfluidA – PairA
Fnet = (Pfluid – Pair)A
Fnet = (Pgauge)A
Fnet = ρghA
Fnet = (900 kg/m3)(9.8 m/s2)(0.4 m)(1.8 ×10-3 m2)
Fnet = 6.35 N
®
1 point for finding the net force as the
difference between the force from the
fluid and the air
1 point for finding force from the
pressure: F = PA
1 point for the correct answer
including correct units and reasonable
number of significant digits
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Fluids
B. The plug is removed. Calculate the velocity of the fluid as it flows out of the opening.
(3 points max)
1 point for the correct application of
Bernoulli’s Equation
This is a dynamic fluids problem that can be solved
using Bernoulli’s Equation.
(P + ρgy + ½ ρv2)at the top of the fluid =
(P + ρgy + ½ ρv2)at the opening
Since the external pressure is the same at the top of
the fluid and at the opening (1 atm), the pressure
can be canceled out. The velocity of the fluid at the
top is very small and is assumed to be zero, Also,
y = 0 at the opening. This leaves us with this
equation:
(p + ρgy + ½ ρv2)at the top of the fluid =
(p + ρgy + ½ ρv2)at the opening
1 point for using Bernoulli’s Equation
to find: vopening = 2 gy at the top of
the fluid
1 point for the correct answer
including correct units and reasonable
number of significant digits
ρgyat the top of the fluid = ½ ρv2at the opening
gyat the top of the fluid = ½ v2at the opening
vopening = 2 gy at the top of
the fluid
m⎞
⎛
vopening = 2 ⎜ 9.8 2 ⎟ ( 0.4m )
s ⎠
⎝
m
v = 2.8
s
C. Calculate the volume flow rate of the fluid out of the opening.
2 points max)
1 point for determining volume flow
rate using υ = Av
Volume flow rate is equal to the velocity of the
fluid times the area of the opening:
υ = Av
⎛
⎝
υ = (1.8 × 10−3 m 2 ) ⎜ 2.8
υ = 0.005
m⎞
⎟
s⎠
1 point for the correct answer
including correct units and reasonable
number of significant digits
m3
s
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Fluids
D. The fluid lands on the floor 0.25 m away from the container. Determine how high h
the container is above the floor.
(4 points max)
1 point for the correct application of
kinematics in the x-direction: x = vxt
This is a 2-D trajectory problem from kinematics.
In the x-direction:
x = vx t
1 point for calculating the correct time
1 point for the correct application of
kinematics in the y-direction:
x 0.25m
t= =
= 0.09 s
vx 2.8 m
s
y = y0 + vy0t + ½ at2
In the y-direction:
y = y0 + vy0t + ½ at2
The initial y-position and initial y-velocity are both
zero.
1 point for the correct answer
including correct units and reasonable
number of significant digits
1 2
gt
2
1⎛
m⎞
2
y = ⎜ 9.8 2 ⎟ ( 0.09s )
2⎝
s ⎠
y = 0.04 m
y=
E. Suppose that a fluid twice as dense is placed into the container to the same original
height of 0.4 m and the plug is again removed. Will the new fluid land farther away from
the container, the same 0.25 m distance from the container, or closer to the container.
_____ Farther from the container
_____ Same 0.25 m from the container
_____ Closer to the container
Justify your answer.
(3 points max)
The new fluid lands the same 0.25 m away from
the container.
From part b) we derived that
vopening = 2 gy at the top of
the fluid
This equation is independent of the density of the
fluid. Therefore, the 2-D trajectory of the new fluid
will be identical to that of the original fluid.
®
1 point for stating that the fluid lands
at the same distance of 0.25 m
2 points for a complete explanation
that includes: The exit velocity of the
fluid is independent of the density of
the fluid. The trajectory motion will be
identical because the launch velocity,
acceleration due to gravity, and the
time of flight have not changed.
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Fluids
Multiple Choice
1. Shown below are five beakers that are filled with the same volume of
mercury. Blocks of different masses and sizes are floating in the liquid mercury.
The mass and volume of each block are given below. Rank in order from greatest
to least the Buoyant Force exerted on each block. If there are any situations that
have the same Buoyant Force list them as being equal.
1_____
Greatest
Mass of block
Volume of
block
1 D
2 B&C
3 A&E
2_____
A
100 g
25 cm3
B
200 g
50 cm3
3_____
C
200 g
25 cm3
4_____
D
400 g
50 cm3
5_____
Least
E
100 g
75 cm3
For floating objects the buoyant force up must equal the weight down.
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Fluids
2. Shown below are five beakers that are filled with the same volume of water. Blocks of
different masses and sizes are completely immersed in the water and have sunk to the
bottom of each beaker. The mass and volume of each block are given below. Rank in
order from greatest to least the Buoyant Force exerted on each block. If there are any
situations that have the same Buoyant Force list them as being equal.
1_____
Greatest
Mass of block
Volume of
block
1 E
2 B&D
3 A&C
2_____
A
100 g
25 cm3
B
200 g
50 cm3
3_____
C
200 g
25 cm3
4_____
D
400 g
50 cm3
5_____
Least
E
100 g
75 cm3
FBuoy = ρVg The density of the fluid and g are the same for each case.
Thus the objects with the greatest submerged volume will have the
largest FBuoy.
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Fluids
Questions 3-4
Shown below are five beakers that are filled with the same volume of water. Blocks of
different masses and sizes are completely immersed in the water and are suspended from
a string. All of the blocks would sink if not for the string holding them up. The mass and
volume of each block are given below.
Mass of block
Volume of
block
A
100 g
25 cm3
B
200 g
50 cm3
C
200 g
25 cm3
D
400 g
50 cm3
E
100 g
75 cm3
3. Rank in order from greatest to least the Buoyant Force exerted on each block. If there
are any situations that have the same Buoyant Force list them as being equal.
1_____
Greatest
1 E
2 B&D
3 A&C
2_____
3_____
4_____
5_____
Least
FBuoy = ρVg The density of the fluid and g are the same for each case.
Thus the objects with the greatest submerged volume will have the
largest FBuoy.
4. Rank in order from greatest to least the Tension in the supporting string.
1_____
Greatest
1
2
3
4
5
D
C
B
A
E
2_____
3_____
4_____
5_____
Least
The suspended block has three forces acting on it (FBuoy , weight of the
block, and tension) These three forces must cancel because the block is
in equilibrium.
FBuoy + T = mg
ρVg + T = mg
T = mblockg - ρwaterVg = g(mblock - ρwaterV)
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Fluids
Questions 5-6)
Shown below are six graduated cylinders filled with different amounts of the same fluid.
The radius of each cylinder and the height of fluid in each is given in the table below.
Radius of
Cylinder
Depth of
Fluid
A
B
C
D
E
F
0.2 m
0.2 m
0.3 m
0.1 m
0.4 m
0.4 m
0.4 m
0.25 m
0.6 m
1.0 m
0.4 m
0.1 m
5. Rank in order from greatest to least the pressure from the fluid on the bottom of each
cylinder.
1
2
3
Greatest
1
2
3
4
5
D
C
A&E
B
F
4
5
6
Least
Fluid static pressure: P = ρgh. The density and g are the same in each
case. Only the depth of the fluid makes a difference.
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Fluids
6. Rank in order from greatest to least the force from the fluid on the bottom of each
cylinder.
1
2
3
4
5
Greatest
1 E
2 C
3 A&F
4 B&D
6
Least
Force = (Pressure)(Area): F = PA = (ρgh)(πr2). The density, g, and π
are the same in each case. The cylinders with the greatest h r2 have the
greatest forces exerted on the bottom of the cylinder.
Questions 7-8)
A 40 cm tall glass is filled with water to a depth of 30 cm.
30 cm
7. What is the gauge pressure at the bottom of the glass?
A) 1000 Pa
B) 3000 Pa
C) 4000 Pa
D) 103,000 Pa
E) 104,000 Pa
Pgauge = ρgh = (1000 kg/m3)(10 m/s2)(0.3 m) = 3000 Pa
B
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Fluids
8. What is the absolute pressure at the bottom of the glass?
A) 1000 Pa
B) 3000 Pa
C) 4000 Pa
D) 103,000 Pa
E) 104,000 Pa
D
Pabsolute = P0 + Pgauge = P0 + ρgh = 100,000 Pa + (1000 kg/m3)(10
m/s2)(0.3 m) = 103,000 Pa
F
h = 70 cm
h = 70 cm
r = 10 cm2
r = 50 cm2
9. Two cylinders filled with oil and fitted with pistons are connected by a tube that allows
the oil to flow freely between them. The radius and heights of the cylinders are given. A
10 kg mass is placed on the right cylinder. What force must be applied to the left cylinder
to keep the 10 kg mass at the same height?
A) 0.4 N
B) 4.0 N
C) 100 N
D) 250 N
E) 2500 N
B
Pascals Principle: An external pressure applied to a confined fluid is
distributed throughout the fluid.
FL / AL = FR / AR
FL / πrL2 = mg / πrR2
FL = mgrL2 / rR2
FL = (10 kg)(10 m/s2)(10 cm2)2 / (50 cm2)2 = 4 N
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Fluids
10. A 20 kg cube is placed on the bottom of a large container of water and then released.
The cube has a height of 20 cm. What normal force will the bottom of the container exert
on the cube?
A) The will be no normal force because the cube will float to the surface
B) 80 N
C) 120 N
D) 200 N
E) 280 N
C
The Buoyant Force on the cube is:
FBuoy = ρVg = (1000 kg/m3)[(0.2 m)3](10 m/s2) = 80 N
The weight of the cube is: mg = (20 kg)(10 m/s2) = 200 N
The Buoyant Force is not great enough to make the cube float.
Therefore,
FBuoy + FN = mg
FN = mg - FBuoy = 120 N
Questions 11-13)
A fluid flows through a pipe of varying diameters as shown in the figure at the right.
B
C
A
11. Which of the following is true of the fluid velocity as it passed through the pipe?
A) v A = vc
B) vB > vc
C) vC > vA
D) vA < vB
E) vC < vA
E
The Continuity Equation tell us that inside of a pipe or stream tube the
volume flow rate is a constant: A1v1 = A2v2 Where the pipe is
narrowest the velocity will be the greatest: vA > vC > vB.
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Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org
Fluids
12. Which of the following is true of the pressure in the fluid as it passes through the
pipe?
A) PA = PB = PC
B) PA > PB > PC
C) PA > PC > PB
D) PB > PC > PA
E) PC > PB > PA
D
Bernoulli’s Equation tells us: P + ρgy + ½ ρv2 = constant. As the
velocity along a streamline increases the pressure must decrease to
maintain the same constant. Wherever the velocity is high the pressure
is low and when the velocity is low the pressure must increase:
PB > PC > PA
13. Which of the following is true of the volume flow rate (define the symbol, υ , for the
volume flow rate) as the fluid flows through the pipe?
A) υ A = υ B = υC
B) υ A > υ B > υC
C) υ A > υC > υ B
D) υ B > υC > υ A
E) υC > υ B > υ A
A
The Continuity Equation tell us that inside of a pipe or stream tube the
volume flow rate is a constant: A1v1 = A2v2.
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Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org
Fluids
14. A hurricane blows air at a steady 45 m/s across the top of a flat topped warehouse as
shown in the figure above. The roof has an area of 300 m2. What is net force on the roof
due to the wind? (Assume the air inside the house is stationary and the density of air is
1.29 kg/m3.)
A) The air blows parallel to the roof and does not cause a net force on the roof.
B) 1.3 × 103 N
C) 8.7 × 103 N
D) 3.9 × 105 N
E) 7.8 × 105 N
D
Bernoulli’s Equation tells us: P + ρgy + ½ ρv2 = constant.
(P + ρgy + ½ ρv2)above the roof = (P + ρgy + ½ ρv2)below the roof
Note: the difference in heights between “above” and “below”
the roof is small and so these two values cancel each other out.
In addition, the velocity of the air inside the warehouse is zero.
This leaves the following equation:
(P + ½ ρv2)above the roof = Pbelow the roof
½ ρv2above the roof = Pbelow the roof - Pabove the roof = ΔP
Fon the roof = (ΔP)A = (½ ρv2above the roof)A
F= ½ (1.29 kg/m3)(45 m/s)2 (300 m2) = 3.9 × 105 N
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Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org