Transcript Example
Matter 1. Density:
m V m
– mass
V
– volume Units: 1
kg
/
m
3
water
1
g
/
cm
3 1000
kg
/
m
3 Example: Volume of 1.00 kg of iron is 128 cm 3 . Find density of iron.
m V
1 .
00
kg
128
cm
3 100
cm
1
m
3 1 .
00
kg
128
cm
3 10 6
cm
3
m
3 7 .
8 10 3
kg
/
m
3
m V
1 .
00 10 3
g
128
cm
3 7 .
8
g
/
cm
3 7 .
8 10 3
kg
/
m
3
Specific gravity (SG):
SG
water
at
4 .
0
C
1
2. Pressure
F
P P
F
A
In fluid: Units: 1
Pa
1
N
/
m
2 1
atm
1 .
013 10 5
Pa F
|| 0
F
A F
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)
P g
P
1
atm
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 )
y
F
P
0
A
F
0
P
0
P
A
P
0
A
gAh
0
mg
gAh
P
gh P
P
0
gh h
Incompressible fluid:
const F
PA
(
P
0
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 4
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) container 1 2) container 2 3) container 3 4) 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)
F
1
PA
1
F
2
PA
2
F F
2 1
PA
1
PA
2
F
1
F
2
A
1
A
2
Example:
F 1 =10N A 1 =0.01m
2 A 2 =1m 2 F 2 - ?
W
W
F
1
x
1
F
2
x
2
PA
1
x
1
PA
2
x
2
P
V P
V F
2
F
1
A
2
A
1
W F
1
x
1
P
V F
2
x
2 Similar to solid lever
F
2 10
N
1
m
2 0 .
01
m
2 1000
N
6
2b. The open tube manometer (measuring gauge pressure)
h P
h
2
gh
1
h
1
P
0
gh
2 P h P 0
h
2
P g
P
P
0
gh h
1
2c. The mercury barometer (measuring atmospheric pressure)
P = 0 P
0
h P
0
gh
7
3. Buoyancy. Archimedes’ principle
F b
m f g
f Vg F b mg
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 – F b
. According to Archimedes’ Principle,
F b
is equal to the weight of the fluid displaced. Since each brick displaces the same amount of fluid, then
F b
is the same in both cases.
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3a. An object floating in equilibrium
F b F g
m f mg g
f V f
Vg g F b
F g
f V f g
Vg
f V f
V V f V
f
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?
f
V f V
3 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 F b = m f g = ρ f Vg F b = mgρ f /ρ gold T = mg - F b T = mg(1 - ρ f /ρ g )
a)
T = (15.0kg) (9.8 m/s 2 )(1 - 1.03 /19.3)=139N mg = ρ gold Vg
gold
19 .
3
g
/
cm
3 b)
air
gold
1 .
2 10 3 19 .
3 6 10 5 1
T = mg= (15.0kg) (9.8 m/s 2 )=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,
F b
on the statue, so the statue must exert an equal downward force on the water, making the scale reading by
F b
grater than the weight of water and container.
From example 1:
F b = mgρ f /ρ g = (15.0kg)( 9.8 m/s 2 )1.03 /19.3=7.8N
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Example:
The “weight” of a chunk of metal measured in air is 100
g
, and when the chunk is submerged into water the “weight” is 87.2g. What is the density of the metal.
T 1 = mg = ρVg T 2 = mg - F b T
1
T
2
F b
f Vg T 1 mg T 2 F b mg T
1
T
1
T
2
f Vg Vg
f
f T
1
T
1
T
2 1 .
0
g
/
cm
3 100
g
100
g
87 .
2
g
7 .
8
g
/
cm
3 11
y A 1 v 1
B. Fluid Dynamics. Bernoulli’s equation
(Ideal fluid - incompressible and has no viscosity
)
A 2 v 2
A
1
d
1
A
2
d
2
V
d 1 d 2
W
K
U W
F
1
d
1
F
2
d
2
P
1
A
1
d
1
P
2
A
2
d
2
P
1
P
2
V K
1 2
mv
2 2 1 2
mv
1 2 1 2
V
(
v
2 2
v
1 2 )
U
mg
(
h
2
h
1 )
Vg
(
h
2
h
1 )
P
1
P
2 1 2 (
v
2 2
v
1 2 )
g
(
h
2
h
1 )
P
1
gh
1 1 2
v
1 2
P
2
gh
2 1 2
v
2 2 12
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?
P 0 v 0 =0 h P 0 v - ?
P
1
gh
1 1 2
v
1 2
P
2
gh
2 1 2
v
2 2
gh
1 2
v
2
v
2
gh
13