Transcript phys1441-120610x

PHYS 1441 – Section 002
Lecture #23
Monday, Dec. 6, 2010
Dr. Jaehoon Yu
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Monday, Dec. 6, 2010
Similarities Between Linear and Rotational
Quantities
Conditions for Equilibrium
How to Solve Equilibrium Problems?
A Few Examples of Mechanical Equilibrium
Elastic Properties of Solids
Density and Specific Gravity
PHYS 1441-002, Fall 2010
Dr. Jaehoon Yu
1
Announcements
• The Final Exam
– Date and time: 11am – 1:30pm, Monday Dec. 13
– Place: SH103
– Comprehensive exam
• Covers from CH1.1 – what we finish Wednesday, Dec. 8
• Plus appendices A.1 – A.8
• Combination of multiple choice and free response
problems
• Bring your Planetarium extra credit sheet to the class next
Wednesday, Dec. 8, with your name clearly marked on
the sheet!
Monday, Dec. 6, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
2
Similarity Between Linear and Rotational Motions
All physical quantities in linear and rotational motions show striking similarity.
Quantities
Mass
Length of motion
Speed
Acceleration
Force
Work
Power
Momentum
Kinetic Energy
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Linear
Mass
Rotational
Moment of Inertia
M
Distance
r
t
v
a
urt
I
Angle  (Radian)
L

t


t
v

r
Force F  ma
r r
Work W  F  d
ur r
P  F v
ur
r
p  mv
Kinetic
K
1
mv 2
2
r ur
Torque   I 
Work W  
P  
ur
ur
L  I
Rotational
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
KR 
1
I 2
2
3
Conditions for Equilibrium
What do you think the term “An object is at its equilibrium” means?
The object is either at rest (Static Equilibrium) or its center of mass
is moving at a constant velocity (Dynamic Equilibrium).
When do you think an object is at its equilibrium?
ur
Translational Equilibrium: Equilibrium in linear motion  F  0
Is this it?
The above condition is sufficient for a point-like object to be at its
translational equilibrium. However for an object with size this is
not sufficient. One more condition is needed. What is it?
Let’s consider two forces equal in magnitude but in opposite direction acting
on a rigid object as shown in the figure. What do you think will happen?
F
d
d
CM
-F
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
r
The object will rotate about the CM. Since the net torque
 0
acting on the object about a rotational axis is not0.
For an object to be at its static equilibrium, the object should not
have linear or angular speed.
vCM  0   0
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
4
More on Conditions for Equilibrium
To simplify the problem, we will only deal with forces acting on x-y plane, giving torque
only along z-axis. What do you think the conditions for equilibrium be in this case?
The six possible equations from the two vector equations turns to three equations.
ur
F 0
F
F
x
y
0
0
AND
r
  0

z
0
What happens if there are many forces exerting on an object?
r’
r5 O O’
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If an object is at its translational static equilibrium, and
if the net torque acting on the object is 0 about one
axis, the net torque must be 0 about any arbitrary axis.
Why is this true?
Because the object is not moving, no matter what the
rotational axis is, there should not be any motion. It is
simply a matter of mathematical manipulation.
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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How do we solve static equilibrium problems?
1.
2.
3.
4.
5.
6.
7.
Select the object to which the equations for equilibrium are to be
applied.
Identify all the forces and draw a free-body diagram with them
indicated on it with their directions and locations properly indicated
Choose a convenient set of x and y axes and write down force
equation for each x and y component with correct signs.
Apply the equations that specify the balance of forces at equilibrium.
Set the net force in the x and y directions equal to 0.
Select the most optimal rotational axis for torque calculations 
Selecting the axis such that the torque of one or more of the
unknown forces become 0 makes the problem much easier to solve.
Write down the torque equation with proper signs.
Solve the force and torque equations for the desired unknown
quantities.
Monday, Dec. 6, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
6
Example for Mechanical Equilibrium
A uniform 40.0 N board supports the father and the daughter each weighing 800 N and
350 N, respectively, and is not moving. If the support (or fulcrum) is under the center of
gravity of the board, and the father is 1.00 m from CoG, what is the magnitude of the
normal force n exerted on the board by the support?
1m
F
MFg
x
n
MBg
Since there is no linear motion, this system
is in its translational equilibrium
D
F  0
MDg
x
F
 n M B g M F g M D g  0
n  40.0  800  350  1190N
y
Therefore the magnitude of the normal force
Determine where the child should sit to balance the system.
The net torque about the fulcrum
by the three forces are
Therefore to balance the system
the daughter must sit
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  M B g  0  n  0  M F g 1.00  M D g  x  0
x

MFg
800
1.00m 
1.00m  2.29m
MDg
350
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Example for Mech. Equilibrium Cont’d
Determine the position of the child to balance the
system for different position of axis of rotation.
Rotational axis
1m
F
MFg

x
n
x/2
D
MFg
MBg
The net torque about the axis of
rotation by all the forces are
 M B g  x / 2  M F g  1.00  x / 2  n x / 2  M D g  x / 2  0
n  MBg  MF g  MDg
  M B g  x / 2  M F g  1.00  x / 2
 M B g  M F g  M D g  x / 2  M D g  x / 2
Since the normal force is
The net torque can
be rewritten
 M F g 1.00  M D g  x  0
Therefore
x
Monday, Dec. 6, 2010
MFg
800

1.00m 
1.00m  2.29m
MDg
350
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
What do we learn?
No matter where the
rotation axis is, net effect of
the torque is identical.
8
Ex. 9.8 for Mechanical Equilibrium
A person holds a 50.0N sphere in his hand. The forearm is horizontal. The biceps
muscle is attached 3.00 cm from the joint, and the sphere is 35.0cm from the joint. Find
the upward force exerted by the biceps on the forearm and the downward force exerted
by the upper arm on the forearm and acting at the joint. Neglect the weight of forearm.
FB
Since the system is in equilibrium, from
the translational equilibrium condition
F  0
O
l
mg
 F  F  F  mg  0
F
From the rotational equilibrium condition   F  0  F  d  mg  l  0
d
x
U
y
B
U
U
B
FB  d  mg  l
mg  l 50.0  35.0

 583N
FB 
3.00
d
Force exerted by the upper arm is
FU  FB  mg  583  50.0  533N
Thus, the force exerted by
the biceps muscle is
Monday, Dec. 6, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Elastic Properties of Solids
We have been assuming that the objects do not change their
shapes when external forces are exerting on it. It this realistic?
No. In reality, the objects get deformed as external forces act on it,
though the internal forces resist the deformation as it takes place.
Deformation of solids can be understood in terms of Stress and Strain
Stress: A quantity proportional to the force causing the deformation.
Strain: Measure of the degree of deformation
It is empirically known that for small stresses, strain is proportional to stress
The constants of proportionality are called Elastic Modulus Elastic Modulus 
Three types of
Elastic Modulus
Monday, Dec. 6, 2010
1.
2.
3.
stress
strain
Young’s modulus: Measure of the elasticity in a length
Shear modulus: Measure of the elasticity in an area
Bulk modulus:
Measure of the elasticity in a volume
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Example 9 – 7
A 5.0 m long ladder leans against a wall at a point 4.0m above the ground. The ladder is
uniform and has mass 12.0kg. Assuming the wall is frictionless (but ground is not),
determine the forces exerted on the ladder by the ground and the wall.
FW
FBD
mg
FGy
O
FGx
First the translational equilibrium,
using components
 Fx FGx  FW  0
 F  mg  F
y
0
Gy
Thus, the y component of the force by the ground is
FGy  mg  12.0  9.8N  118N
The length x0 is, from Pythagorian theorem
x0  5.02  4.02  3.0m
Monday, Nov. 30, 2009
PHYS 1441-002, Fall 2009 Dr. Jaehoon
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Example 9 – 7 cont’d
From the rotational equilibrium

O
 mg x0 2  FW 4.0  0
Thus the force exerted on the ladder by the wall is
mg x0 2 118 1.5

 44 N
4.0
4.0
The x component of the force by the ground is
FW 
F
x
 FGx  FW  0
Solve for FGx
FGx  FW  44 N
Thus the force exerted on the ladder by the ground is
FG  FGx2  FGy2  442  1182  130N
The angle between the  tan 1  FGy 
1  118 
o

tan

70





ground force to the floor
F
44


 Gx 
Monday, Nov. 30, 2009
PHYS 1441-002, Fall 2009 Dr. Jaehoon
Yu
12
Ex. A Diving Board
A woman whose weight is 530 N is poised at the right
end of a diving board with length 3.90 m. The board has
negligible weight and is supported by a fulcrum 1.40 m
away from the left end. Find the forces that the bolt and
the fulcrum exert on the board.
First the torque eq.
So the force by
the fulcrum is
F2 
  F2
F2 
W
2
W
W
0
W
2
 530 N  3.90 m   1480 N
How large is the
torque by the bolt?
None Why?
Because
the lever
arm is 0.
1.40 m
Now the force eq.
F
y
  F1  F2 W  0
 F1  1480 N  530 N  0
So the force
by the bolt is
F1  950 N
Monday, Nov. 30, 2009
PHYS 1441-002, Fall 2009 Dr. Jaehoon
Yu
13
Finger Holds Water in Straw
You insert a straw of length L into a tall glass of your favorite
beverage. You place your finger over the top of the straw so that
no air can get in or out, and then lift the straw from the liquid. You
find that the straw strains the liquid such that the distance from the
bottom of your finger to the top of the liquid is h. Does the air in the
space between your finger and the top of the liquid have a pressure
P that is (a) greater than, (b) equal to, or (c) less than, the
atmospheric pressure PA outside the straw?
Less
pinA
What are the forces in this problem?
Gravitational force on the mass of the liquid
Fg  mg   A  L  h  g
Force exerted on the top surface of the liquid by inside air pressure Fin  pin A
mg
p AA
Force exerted on the bottom surface of the liquid by the outside air Fout  p A A
 pA A   g  L  h  A  pin A  0
Since it is at equilibrium Fout  Fg  Fin  0
Cancel A and
solve for pin
Monday, Dec. 6, 2010
pin  pA   g  L  h 
So pin is less than PA by g(L-h).
PHYS 1441-002, Fall 2010 Dr. Jaehoon
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Young’s Modulus
Let’s consider a long bar with cross sectional area A and initial length Li.
Li
Fex
After the stretch
F
Tensile Stress  ex
A
Young’s Modulus is defined as
Fex
Fex=Fin
A:cross sectional area
Tensile stress
Lf=Li+L
Tensile strain
Tensile Strain 
F
Y
ex
Tensile Stress
A


Tensile Strain L L
i
L
Li
Used to characterize a rod
or wire stressed under
tension or compression
What is the unit of Young’s Modulus?
Experimental
Observations
1.
2.
Force per unit area
For a fixed external force, the change in length is
proportional to the original length
The necessary force to produce the given strain is
proportional to the cross sectional area
Elastic limit: Maximum stress that can be applied to the substance
before
it becomes permanently
Monday, Nov.
30, 2009
PHYS 1441-002,deformed
Fall 2009 Dr. Jaehoon
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15
Bulk Modulus
F
Bulk Modulus characterizes the response of a substance to uniform
squeezing or reduction of pressure.
V
After the pressure change
F
F
V’
F
Normal Force
F
Volume stress
Pressure 

Surface Area the force applies
A
=pressure
If the pressure on an object changes by P=F/A, the object will
undergo a volume change V.
F
Bulk Modulus is
defined as
Because the change of volume is
reverse to change of pressure.
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P
Volume Stress  
A 
B
V
V
Volume Strain
Vi
V
i
Compressibility is the reciprocal of Bulk Modulus
PHYS 1441-002, Fall 2009 Dr. Jaehoon
Yu
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Example for Solid’s Elastic Property
A solid brass sphere is initially under normal atmospheric pressure of 1.0x105N/m2. The
sphere is lowered into the ocean to a depth at which the pressures is 2.0x107N/m2. The
volume of the sphere in air is 0.5m3. By how much its volume change once the sphere is
submerged?
Since bulk modulus is
P
B
V
Vi
The amount of volume change is
V  
PVi
B
From table 12.1, bulk modulus of brass is 6.1x1010 N/m2
The pressure change P is
P  Pf  Pi  2.0 107 1.0 105  2.0 107
Therefore the resulting
2.0 107  0.5
4
3

V

V

V




1
.
6

10
m
f
i
volume change V is
6.11010
The volume has decreased.
Monday, Nov. 30, 2009
PHYS 1441-002, Fall 2009 Dr. Jaehoon
Yu
17
Density and Specific Gravity
Density,  (rho), of an object is defined as mass per unit volume
M
 
V
Unit?
Dimension?
kg / m3
[ ML3 ]
Specific Gravity of a substance is defined as the ratio of the density
of the substance to that of water at 4.0 oC (H2O=1.00g/cm3).
 substance
SG 
 H 2O
What do you think would happen of a
substance in the water dependent on SG?
Monday, Nov. 30, 2009
Unit?
None
Dimension? None
SG  1 Sink in the water
SG  1 Float on the surface
PHYS 1441-002, Fall 2009 Dr. Jaehoon
Yu
18
Fluid and Pressure
What are the three states of matter?
Solid, Liquid and Gas
Using the time it takes for a particular substance
How do you distinguish them?
to change its shape in reaction to external forces.
A collection of molecules that are randomly arranged and loosely
What is a fluid? bound by forces between them or by an external container.
We will first learn about mechanics of fluid at rest, fluid statics.
In what ways do you think fluid exerts stress on the object submerged in it?
Fluid cannot exert shearing or tensile stress. Thus, the only force the fluid exerts
on an object immersed in it is the force perpendicular to the surface of the object.
This force by the fluid on an object usually is expressed in the form of P  F
A
the force per unit area at the given depth, the pressure, defined as
Expression of pressure for an
dF Note that pressure is a scalar quantity because it’s
P

infinitesimal area dA by the force dF is
dA the magnitude of the force on a surface area A.
What is the unit and the
Unit:N/m2
Special SI unit for
2
1
Pa

1
N
/
m
dimension of pressure?
pressure is Pascal
Dim.: [M][L-1][T-2]
Monday, Nov. 30, 2009
PHYS 1441-002, Fall 2009 Dr. Jaehoon
Yu
19
Example for Pressure
The mattress of a water bed is 2.00m long by 2.00m wide and
30.0cm deep. a) Find the weight of the water in the mattress.
The volume density of water at the normal condition (0oC and 1 atm) is
1000kg/m3. So the total mass of the water in the mattress is
m 
W
VM  1000  2.00  2.00  0.300  1.20 103 kg
Therefore the weight of the water in the mattress is
W  mg  1.20 103  9.8  1.18 104 N
b) Find the pressure exerted by the water on the floor when the bed
rests in its normal position, assuming the entire lower surface of the
mattress makes contact with the floor.
Since the surface area of the
mattress is 4.00 m2, the
pressure exerted on the floor is
Monday, Nov. 30, 2009
F mg 1.18 10 4
3




2
.
95

10
P A A
4.00
PHYS 1441-002, Fall 2009 Dr. Jaehoon
Yu
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Buoyant Forces and Archimedes’ Principle
Why is it so hard to put an inflated beach ball under water while a small
piece of steel sinks in the water easily?
The water exerts force on an object immersed in the water.
This force is called the buoyant force.
How large is the The magnitude of the buoyant force always equals the weight of the
buoyant force? fluid in the volume displaced by the submerged object.
This is called Archimedes’ principle. What does this mean?
Let‘s consider a cube whose height is h and is filled with fluid and in its
equilibrium so that its weight Mg is balanced by the buoyant force B.
pressure at the bottom of the cube
B  Fg  Mg The
is larger than the top by gh.
h
Mg B
Monday, Dec. 6, 2010
Therefore, P  B / A   gh
B  PA  ghA  Vg
B  Vg  Mg  Fg
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
Where Mg is the weight
of the fluid in the cube.
21
More Archimedes’ Principle
Let’s consider the buoyant force in two special cases.
Case 1: Totally submerged object Let’s consider an object of mass M, with density 0, is
fully immersed in the fluid with density f .
The magnitude of the buoyant force is
B   f Vg
The weight of the object is Fg  Mg   0Vg
h
Mg B
Therefore total force in the system is
What does this tell you?
Monday, Dec. 6, 2010
F  B  Fg   f
  0 Vg
The total force applies to different directions depending on the
difference of the density between the object and the fluid.
1. If the density of the object is smaller than the density of
the fluid, the buoyant force will push the object up to the
surface.
2. If the density of the object is larger than the fluid’s, the
object will sink to the bottom of the fluid.
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
22
More Archimedes’ Principle
Case 2: Floating object
h
Mg B
Let’s consider an object of mass M, with density 0, is in
static equilibrium floating on the surface of the fluid with
density f , and the volume submerged in the fluid is Vf.
The magnitude of the buoyant force is
The weight of the object is
Therefore total force of the system is
B   fVf g
Fg  Mg   0V0 g
F  B  Fg   f V f g   0V0 g
0
 f V f g   0V0 g
Vf
0

f
V0
Since the object is floating, its density is smaller than that of the
fluid.
The ratio of the densities between the fluid and the object
determines the submerged volume under the surface.
Since the system is in static equilibrium
What does this tell you?
Monday, Dec. 6, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
23
Example for Archimedes’ Principle
Archimedes was asked to determine the purity of the gold used in the crown.
The legend says that he solved this problem by weighing the crown in air and
in water. Suppose the scale read 7.84N in air and 6.86N in water. What
should he have to tell the king about the purity of the gold in the crown?
In the air the tension exerted by the scale on
the object is the weight of the crown
In the water the tension exerted

T
water
by the scale on the object is
Therefore the buoyant force B is
Tair  mg  7.84 N
mg  B  6.86 N
B  Tair  Twater  0.98N
Since the buoyant force B is
B   wVw g   wVc g  0.98N
The volume of the displaced
water by the crown is
Vc  Vw 
Therefore the density of
the crown is
c

0.98 N
0.98

 1.0 10  4 m3
 w g 1000  9.8
mc mc g 7.84
7.84



 8.3  103 kg / m 3
4
Vc Vc g Vc g 1.0  10  9.8
3kg/m
3, this
Monday,
Dec. 6,
PHYSis1441-002,
2010
Dr. Jaehoon
Since
the2010
density of pure gold
19.3x10Fall
crown is not made of pure gold.
Yu
24
Example for Buoyant Force
What fraction of an iceberg is submerged in the sea water?
Let’s assume that the total volume of the iceberg is Vi.
Then the weight of the iceberg Fgi is
Fgi  iVi g
Let’s then assume that the volume of the iceberg
submerged in the sea water is Vw. The buoyant force B
caused by the displaced water becomes
B   wVw g
Since the whole system is at its
static equilibrium, we obtain
Therefore the fraction of the
volume of the iceberg
submerged under the surface of
the sea water is
iVi g   wVw g
917kg / m3
Vw  i

 0.890

3
Vi  w 1030kg / m
About 90% of the entire iceberg is submerged in the water!!!
Monday, Dec. 6, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
25
Flow Rate and the Equation of Continuity
Study of fluid in motion: Fluid Dynamics
If the fluid is water: Water dynamics?? Hydro-dynamics
•Streamline or Laminar flow: Each particle of the fluid
Two main
types of flow follows a smooth path, a streamline
•Turbulent flow: Erratic, small, whirlpool-like circles called
eddy current or eddies which absorbs a lot of energy
Flow rate: the mass of fluid that passes the given point per unit time m / t
m1
1V1 1 A1l1


 1 A1v1
t
t
t
since the total flow must be conserved
m1 m2
1 A1v1   2 A2 v2

t
t
Equation of Continuity
Monday, Dec. 6, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
26
Example for Equation of Continuity
How large must a heating duct be if air moving at 3.0m/s through it
can replenish the air in a room of 300m3 volume every 15 minutes?
Assume the air’s density remains constant.
Using equation of continuity
1 A1v1  2 A2 v2
Since the air density is constant
A1v1  A2 v2
Now let’s imagine the room as
the large section of the duct
A2l2 / t
V2
A2 v2
300



A1 
 0.11m 2
v1
v1  t
v1
3.0  900
Monday, Dec. 6, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
27
Bernoulli’s Principle
Bernoulli’s Principle: Where the velocity of fluid is high, the
pressure is low, and where the velocity is low, the pressure is high.
Amount of the work done by the force,
F1, that exerts pressure, P1, at point 1
W1  F1l1  P1 A1l1
Amount of the work done by the force
in the other section of the fluid is
W2   P2 A2 l2
Work done by the gravitational force to
move the fluid mass, m, from y1 to y2 is
W3  mg  y2  y1 
Monday, Dec. 6, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
28
Bernoulli’s Equation cont’d
The total amount of the work done on the fluid is
W  W1 W2 W3  P1 A1l1  P2 A2 l2 mgy2  mgy1
From the work-energy principle
1 2 1
mv2  mv12  P1 A1l1  P2 A2 l2 mgy2  mgy1
2
2
Since the mass m is contained in the volume that flowed in the motion
m   A1l1   A2 l2
A1l1  A2 l2 and
Thus,
1
1
2
2
 A2 l2 v2   A1l1v1
2
2
 P1 A1l1  P2 A2 l2   A2 l2 gy2   A1l1 gy1
Monday, Dec. 6, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
29
Bernoulli’s Equation cont’d
Since
1
1
 A2 l2 v22   A1l1v12  P1 A1l1  P2 A2 l2   A2 l2 gy2   A1l1 gy1
2
2
We
obtain
1 2 1 2
 v2   v1  P1  P2   gy2   gy1
2
2
Reorganize P1 
1 2
1 2
Bernoulli’s
 v1   gy1  P2   v2   gy2 Equation
2
2
Thus, for any two
points in the flow
1 2
P1   v1   gy1  const.
2
For static fluid P2  P1   g  y1  y2   P1   gh
1
For the same heights P2  P1    v12  v22 
2
Result of Energy
conservation!
Pascal’s
Law
The pressure at the faster section of the fluid is smaller than slower section.
Monday, Dec. 6, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
30
Example for Bernoulli’s Equation
Water circulates throughout a house in a hot-water heating system. If the water is
pumped at the speed of 0.5m/s through a 4.0cm diameter pipe in the basement under
a pressure of 3.0atm, what will be the flow speed and pressure in a 2.6cm diameter
pipe on the second 5.0m above? Assume the pipes do not divide into branches.
Using the equation of continuity, flow speed on the second floor is
2
A1v1  r12 v1
 0.020 

v2 

0.5

 1.2m / s
2


A2
 r2
 0.013 
Using Bernoulli’s equation, the pressure in the pipe on the second floor is


1
P2  P1   v12  v22   g  y1  y2 
2
1
5
 3.0  10  1 103 0.52  1.22  1 103  9.8   5 
2
 2.5  105 N / m 2

Monday, Dec. 6, 2010

PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
31