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What is a fluid
A substance that can flow
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Liquids (difficult to compress)
Gases (easy to compress)
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Chapter 9
Basic properties:
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Fluids
The field of science that describes the
fluid motion is called “Fluid Mechanics”
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Density
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ρ=
∆m
∆V
Mass per unit volume.
Unit kg/m3.
The density can be different at each
point of the fluid.
An object is uniform if the density is
constant at any point of the body
A fluid is incompressible if the density
does not change as a result of an applied
pressure.
Pressure
Fluids take the form of the container.
When shear stress is applied they flow.
The molecules in fluids are in constant random
motion (Brownian motion).
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P= p=
Some examples
Interstellar space
Laboratory vacuum
Air (1 atm 20C)
Ice
Water
Seawater
Earth
Neutron star
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∆F
∆A
Force per unit area (the force is perpendicular to
the surface)
Scalar: the pressure has the same value in all
directions!
Units: 1Pa=1N/1m2.
Other popular units
1 atm = 1 bar = 1.01x105 Pa = 760 torr = 14.7 lb/in2
10-20 kg/m3
10-17 kg/m3
1.21 kg/m3
0.917x103 kg/m3
0.998x103 kg/m3
1.024x103 kg/m3
5.5x103 kg/m3
1018 kg/m3
Some Examples
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Sea level atm. pressure
Bottom of the ocean
Center of the Earth
Center of the Sun
105 Pa
1.1x108 Pa
4x1011 Pa
2x1016 Pa
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Fluid Pressure
Pascal’s Principle
Pressure arises from the collisions between the particles of
a fluid with another object (container walls for example).
A change in the pressure applied
to an enclosed incompressible
fluid is transmitted undiminished
to every portion of the fluid and
to the walls of the container.
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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.
∆p = ∆pext
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Hydraulic Lever
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An input force applied over a
distance is transformed into
bigger force over a shorter
distance
∆p =
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Gravity’s Effect on Fluid Pressure
Fi Fo
=
Ai Ao
Fo = Fi
F2 = F1 + mg
F1 = p1 A
F2 = p2 A
p2 A = p1 A + ρVg
Ao
> Fi
Ai
V = A(y1 − y 2 )
p2 = p1 + ρg(y1 − y 2 )
The work done by the input force
and the output force is the same!
V = Ai di = Ao do
W = Fo do = do Fi
Static equilibrium
If the top of the fluid column is placed at the
surface of the fluid, then P1 = Patm =P0 if the
container is open.
Ao
d
= d o Fi i = Fi di
Ai
do
p = patm + ρgh
Example (text problem 9.19): 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?
Hydrostatic pressure
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
∆P = P − Patm = ρgd
(
)(
P = Patm + ρgd
95 atm = 1 atm + ρgd
)
= 1000 kg/m 3 9.8 m/s 2 (35 m )
ρgd = 94 atm = 9.5 ×10 6 N/m 2
(1025 kg/m )(9.8 m/s )d = 9.5 ×10
= 343 kPa = 3.4 atm
3
2
6
N/m 2
d = 950 m
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Measuring pressure
Gauge pressure
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Absolute pressure = total pressure
Gauge pressure = total - atmospheric
The atmospheric pressure on the free surface
of a liquid is often neglected.
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Mercury barometer:
measures the
atmospheric pressure.
p0 = ρgh
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pgauge = ρgh
Open tube manometer:
measures the gauge
pressure of a gas.
pg = p − p0 = ρgh
Fluids at Rest: Examples
Sample problem
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• The closed U-tube shown in the figure contains a liquid of
density ρ. What is the pressure difference ∆P = PA - PB?
What gauge pressure must a machine produce
to suck mud of density 1800 kg/m3 up a tube
by a height of 1.5 m?
PA = PB + ρgh
p g = p − patm
h
Manometer
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(open) U-tube shown in the figure
contains two liquids in static equilibrium:
Water of density ρW is in the right arm, and an
oil of unknown density ρX is in the left arm.
What is the density of the unknown oil in
terms of d and L?
patm = p + ρgh
p g = − ρgh = −2.6×10 4 Pa
pa
PB = P0 + ρ x g ( L + d )
PB = P0 + ρW gL
ρX =
mud
University of Florida
Archimedes Principle-Buoyancy Force
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r
Fb
When a body is partially or fully
submerged in a fluid a buoyant force
from the surrounding fluid acts on the
body.
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The buoyancy force is upward*
It is applied at CM (displaced fluid)
The magnitude is:
Fb = F2 − F1
Fb = ρ f gh2 A − ρ f gh1 A = ρ f gV
Fb = m f g
where mf is the mass of the displaced fluid!
L
B
B
L
ρW
L+d
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Example (text problem 9.28): 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?
FBD for
the barge
Apply Newton’s 2nd Law to the barge:
∑F = F
FB
B
h1
−w=0
FB = w
h2
F2
d
PHY 2053
y
F1
h
PA = PB + ρgh
∆P = PA − PB = ρgh
p
B
A
mw g = (ρ wVw )g = mb g
ρ wVw = mb
x
w
ρ w ( Ad ) = mb
d=
mb
ρw A
=
3.0 ×105 kg
(1000 kg/m )(20.0 m *10.0 m ) = 1.5 m
3
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Example (text problem 9.40): 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.)
y
FBD for
the metal
Example continued:
Since the object is completely submerged V=Vobject.
specific gravity =
Apply Newton’s 2nd Law to the
piece of metal:
FB
∑F = F
B
where ρwater = 1000 kg/m3 is the
density of water at 4 °C.
− w = ma
x
w
The magnitude of the buoyant force equals
the weight of the fluid displaced by the metal.
FB = ρ waterVg
Solve for a:
 ρ V

F
ρ Vg
− g = g  water
− 1
a = B − g = water
 ρ objectVobject

m
ρ objectVobject


ρ
ρ water
Given
specific gravity =
ρ object
= 5 .0
ρ water
 ρ V

 1

 1

a = g  water
− 1 = g 
− 1 = g 
− 1 = −7.8 m/s 2
ρ V

 S .G. 
 5.0 
 object object 
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