Chapter 9: Fluids
•Introduction to Fluids
•Pressure
•Pascal’s Principle
•Gravity and Fluid Pressure
•Measurement of Pressure
•Archimedes’ Principle
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§9.1 Fluids
A liquid will flow to take the shape of the container that holds
it. A gas will completely fill its container.
Fluids are easily deformable by external forces.
A liquid is almost incompressible. Its volume is fixed
and is impossible to change.
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§9.2 Pressure
Pressure arises from the collisions between the particles of
a fluid with another object (container walls for example).
There is a momentum
change (impulse) that is
away from the container
walls. There must be a
force exerted on the
particle by the wall.
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By Newton’s 3rd Law, there is a force on the wall due to
the particle.
F
Pressure is defined as P = .
A
The unit of pressure is N/m2 and is called Pascals (Pa).
Note: 1 atmosphere (atm) = 101.3 kPa
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Example (text problem 9.1): Someone steps on your toe,
exerting a force of 500 N on an area of 1.0 cm2. What is the
average pressure on that area in atmospheres?
2
⎞
2⎛ 1m
−4
2
1.0 cm ⎜
⎟ = 1.0 × 10 m
⎝ 100 cm ⎠
500 N
F
Pav = =
A 1.0 ×10 -4 m 2
1 atm
⎞
6
2 ⎛ 1 Pa ⎞⎛
= 5.0 × 10 N/m ⎜
⎟
2 ⎟⎜
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⎝ 1 N/m ⎠⎝ 1.013 × 10 Pa ⎠
= 49 atm
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Pascal’s Principle
A change in pressure at any point in a confined fluid is
transmitted everywhere throughout the fluid. (This is useful
in making a hydraulic lift.)
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The applied force is
transmitted to the piston
of cross-sectional area
A2 here.
Apply a force F1 here
to a piston of crosssectional area A1.
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Mathematically,
ΔP at point 1 = ΔP at point 2
F1
F
= 2
A1 A 2
⎛ A2 ⎞
⎟⎟ F1
F2 = ⎜⎜
⎝ A1 ⎠
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Example: Assume that a force of 500 N (about 110 lbs) is
applied to the smaller piston in the previous figure. For each
case, compute the force on the larger piston if the ratio of the
piston areas (A2/A1) are 1, 10, and 100.
Using Pascal’s Principle:
A2 A1
1
10
100
F2
500 N
5000 N
50,000 N
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The work done pressing the smaller piston (#1) equals the
work done by the larger piston (#2).
F1d1 = F2 d 2
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Example: In the previous example, for the case A2/A1 = 10,
it was found that F2/F1 = 10. If the larger piston needs to
rise by 1 m, how far must the smaller piston be depressed?
Using the result on the previous slide,
F2
d1 =
d 2 = 10 m
F1
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Gravity’s Effect on Fluid Pressure
FBD for the fluid cylinder
P1A
A cylinder
of fluid
w
P2A
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Apply Newton’s 2nd Law
to the fluid cylinder:
∑F = P A− PA− w = 0
2
1
P2 A − P1 A − (ρAd )g = 0
P2 − P1 − ρgd = 0
∴ P2 − P1 = ρgd
or P2 = P1 + ρgd
If P1 (the pressure at the top of the cylinder) is known, then
the above expression can be used to find the variation of
pressure with depth in a fluid.
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If the top of the fluid column is placed at the surface of the
fluid, then P1=Patm if the container is open.
P = Patm + ρgd
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Example (text problem 9.15): At the surface of a freshwater
lake, the pressure is 105 kPa. (a) What is the pressure
increase in going 35.0 m below the surface?
P = Patm + ρgd
ΔP = P − Patm = ρgd
= (1000 kg/m )(9.8 m/s )(35 m )
3
2
= 343 kPa = 3.4 atm
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Example: The surface pressure on the planet Venus is 95
atm. How far below the surface of the ocean on Earth do
you need to be to experience the same pressure? The
density of seawater is 1025 kg/m3.
P = Patm + ρgd
95 atm = 1 atm + ρgd
ρgd = 94 atm = 9.5 ×106 N/m 2
(1025 kg/m )(9.8 m/s )d = 9.5 ×10
3
2
6
N/m 2
d = 950 m
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§9.5 Measuring Pressure
A manometer
is a U-shaped
tube that is
partially filled
with liquid.
Both ends of the
tube are open to
the atmosphere.
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A container of gas is connected to one end of the U-tube
If there is a pressure difference between the gas and the
atmosphere, a force will be exerted on the fluid in the U-tube.
This changes the equilibrium position of the fluid in the tube.
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From the figure:
At point C
Also
Pc = Patm
PB = PB'
The pressure at point B is the pressure of the gas.
PB = PB ' = PC + ρgd
PB − PC = PB − Patm = ρgd
Pgauge = ρgd
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A Barometer
The atmosphere pushes on the container of mercury which
forces mercury up the closed, inverted tube. The distance
d is called the barometric pressure.
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From the figure
and
PA = PB = Patm
PA = ρgd
Atmospheric pressure is equivalent to a column of mercury
76.0 cm tall.
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Example (text problem 9.22): An IV is connected to a
patient’s vein. The blood pressure in the vein has a gauge
pressure of 12 mm of mercury. At least how far above the
vein must the IV bag be placed in order for fluid to flow into
the vein? Assume that the density of the IV fluid is the same
as blood.
Blood:
Pgauge = ρ Hg gh1
h1 = 12 mm
IV:
The pressure is equivalent
to raising a column of
mercury 12 mm tall.
Pgauge = ρ blood gh2
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Example continued:
At a minimum, the gauge pressures must be equal. When
h2 is large enough, fluid will flow from high pressure to low
pressure.
Pgauge = ρ Hg gh1 = ρ blood gh2
ρ Hg gh1
h2 =
ρ blood g
⎛ ρ Hg ⎞
⎛ 13,600 kg/m 3 ⎞
⎟⎟h1 = ⎜⎜
⎟(12 mm )
= ⎜⎜
3 ⎟
⎝ 1060 kg/m ⎠
⎝ ρ blood ⎠
= 154 mm
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Archimedes’ Principle
F1
An FBD for an object
floating submerged in
a fluid.
w
F2
The total force on the block due to the fluid is called the
buoyant force.
FB = F2 − F1
where F2 > F1
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The magnitude of the buoyant force is:
FB = F2 − F1
= P2 A − P1 A
= (P2 − P1 )A
From before:
P2 − P1 = ρgd
The result is
FB = ρgdA = ρgV
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Archimedes’ Principle: A fluid exerts an upward buoyant
force on a submerged object equal in magnitude to the
weight of the volume of fluid displaced by the object.
FB = ρgV
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Example (text problem 9.30): A flat-bottomed barge loaded
with coal has a mass of 3.0×105 kg. The barge is 20.0 m long
and 10.0 m wide. It floats in fresh water. What is the depth
of the barge below the waterline?
∑F = F
B
FB
−w=0
FBD
for the
barge
FB = w
mw g = (ρ wVw )g = mb g
w
ρ wVw = mb
ρ w ( Ad ) = mb
mb
3.0 ×105 kg
=
= 1.5 m
d=
3
ρ w A 1000 kg/m (20.0 m *10.0 m )
(
)
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Example (text problem 9.34): A piece of metal is released
under water. The volume of the metal is 50.0 cm3 and its
specific gravity is 5.0. What is its initial acceleration? (Note:
when v=0, there is no drag force.)
FB
FBD
for the
metal
w
∑F = F
B
− w = ma
The buoyant force is the
weight of the fluid displaced
by the object
FB = ρ waterVg
⎛ ρ waterV
⎞
ρ waterVg
FB
−g =
− g = g⎜
− 1⎟
Solve for a: a =
⎜ρ V
⎟
ρ objectVobject
m
⎝ object object ⎠
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Example continued:
Since the object is completely submerged V=Vobject.
specific gravity =
ρ
ρ water
where ρwater = 1000 kg/m3 is the
density of water at 4 °C.
ρ object
= 5.0
Given specific gravity =
ρ water
⎛ ρ waterV
⎞
⎞
⎛ 1
⎞
⎛ 1
⎜
⎟
− 1⎟ = −7.8 m/s 2
a=g
−1 = g⎜
− 1⎟ = g ⎜
⎜ρ V
⎟
⎝ 5.0 ⎠
⎝ S .G. ⎠
object
object
⎝
⎠
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