Example

Report
Matter
1. Density:
m – mass
V – volume
m

V
Units:
   1kg / m
3
 water  1g / cm3  1000kg / m3
Example: Volume of 1.00 kg of iron is 128 cm3. Find density of iron.
3
m 1.00 kg  100 cm 
1.00 kg
 


3 
V 128 cm  1m 
128 cm 3
 6 cm 3 
3
3
10


7
.
8

10
kg
/
m
3 
m


m 1.00  103 g
3
3
3
 

7
.
8

g
/
cm

7
.
8

10
kg
/
m
V
128cm3
Specific gravity (SG):

SG    water at 4.0 C

1
2. Pressure
F P
Units:
F
P
A
In fluid:
F||  0
F
F
A
P  1Pa  1N / m 2
1atm  1.013 105 Pa
Example: You are walking out on a frozen lake and you begin to hear the ice
cracking beneath you. What is your best strategy for getting off the ice safely?
A) Stand absolutely still and don’t move a muscle
B) Jump up and down to lessen your contact time with the ice
C) Try to leap in one bound to the bank of the lake
D) Shuffle your feet (without lifting them) to move towards shore
E) Lie down flat on the ice and crawl toward shore
Gauge pressure = (absolute pressure - 1atm)
Pg  P 1atm
2
Fluid Mechanics
What is a fluid?
•Any substance that can flow
•Liquids or gases (there are tree common phases of mater: solid, liquid, gas)
•Fluid can not sustain a force that is tangent it its surface
•Fluids conform to the boundaries of any container in which we put them
•No crystal structure
3
A. Fluid Statics
1. Pascal’s law (principle)
F  P0 A
y
F  0 
P0  PA  P0 A  gAh  0
mg
 gAh
P  gh
P  P0  gh
h
Incompressible fluid:
  const
F  PA  ( P0  P) A
A change in the pressure applied to an enclosed incompressible fluid is
transmitted to every portion of the fluid and to the walls of its container
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Example: Three containers are filled with water to the same height and have
the same surface area at the base, but the total weight of water is different
for each. Which container has the greatest total force acting on its base?
1)
2)
3)
4)
container 1
container 2
container 3
all three are equal
The pressure at the bottom of each container depends only on the height
of water above it! This is the same for all the containers. The total force
is the product of the pressure times the area of the base, but since the
base is also the same for all containers, the total force is the same.
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2. Applications of Pascal’s law
2a. Hydraulic lift (lever)
F1
F2
PA1
PA2
F1  PA1 

F2  PA2 
W  F1x1  PA1x1  PV 

W  F2 x2  PA2 x2  PV 
F1 A1

F2 A2
W  PV
F1x1  F2 x2
Similar to solid lever
Example:
F1=10N
A1=0.01m2
A2=1m2
F2 - ?
A2
F2  F1
A1
1m 2
F2  10N 
 1000N
2
0.01m
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2b. The open tube manometer (measuring gauge pressure)
h  h2  h1
P0
P  gh1  P0  gh2
P
h
h2
Pg  P  P0  gh
h1
2c. The mercury barometer (measuring atmospheric pressure)
P=0
P0
h
P0  gh
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3. Buoyancy. Archimedes’ principle
Fb  m f g   f Vg
Fb
m g  Vg
Example: Imagine holding two identical bricks in place under water.
Brick 1 is just beneath the surface of the water, while brick 2 is held
about 2 feet down. The force needed to hold brick 2 in place is:
1) greater
2) the same
3) smaller
1
2
The force needed to hold the brick in place underwater is: W – Fb.
According to Archimedes’ Principle, Fb is equal to the weight of the fluid
displaced. Since each brick displaces the same amount of fluid, then Fb
is the same in both cases.
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3a. An object floating in equilibrium
Fb  m f g   f V f g
Fg  mg  Vg
Fb  Fg   f V f g  Vg   f V f  V


V
f
Vf
Example: An object floats in water with 3/4 of its volume submerged.
What is the ratio of the density of the object to that of water?
 Vf 3


f V 4
If the ratio of the volume of the displaced water to the volume of the
object is 3/4, the object has 3/4 the density of water.
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Example 1: A 15.0 kg solid gold statue is being raised from a sunken ship.
What is the tension in the hosting cable when he statue is at rest and
a) completely immersed; b) out of the water?
T
T = mg - Fb
Fb = mfg = ρfVg
Fb = mgρf /ρgold
T = mg(1 - ρf /ρg)
a) T = (15.0kg) (9.8 m/s2)(1 - 1.03 /19.3)=139N
mg = ρgoldVg
 gold  19.3g / cm
3
b)
air gold  1.2 103 19.3  6 105  1
T = mg= (15.0kg) (9.8 m/s2)=147N
Example 2: You place a container of sea water on a scale and note the reading
on the scale. You now suspend the statue of example 1 in the water.
Haw does the scale reading change?
Water exerts an upward buoyant force, Fb on the statue, so the statue
must exert an equal downward force on the water, making the scale
reading by Fb grater than the weight of water and container.
From example 1: Fb = mgρf /ρg= (15.0kg)( 9.8 m/s2)1.03 /19.3=7.8N
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Example: The “weight” of a chunk of metal measured in air is 100g,
and when the chunk is submerged into water the “weight” is 87.2g.
What is the density of the metal.
T1 = mg = ρVg
T1
T2
Fb
T2 = mg - Fb
T1 T 2 Fb   f Vg
mg
mg
T1
Vg



T1 T 2  f Vg  f
T1
  f
T1  T2


100g
  1.0 g / cm
 7.8 g / cm3
100g  87.2 g
3
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B. Fluid Dynamics. Bernoulli’s equation
(Ideal fluid - incompressible and has no viscosity)
y
A2
v2
A1d1  A2 d 2  V
d2
W  K U
A1
W  F1d1  F2 d 2  P1 A1d1  P2 A2 d 2  P1  P2 V
v1
d1
K  12 mv22  12 mv12  12 V (v22  v12 )
U  mg(h2  h1 )  Vg (h2  h1 )
P1  P2  12  (v22  v12 )  g (h2  h1 )
P1  gh1  12 v12  P2  gh2  12 v22
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Example:
An open water tank has a hole a distance h = 4.9 m below the water surface.
The water surface area of the tank is much bigger then the hole’s area.
What is the speed of the water emerging from the hole?
P0
P1  gh1  12 v12  P2  gh2  12 v22
v0=0
h
P0
v-?
gh  12 v2
v  2gh
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