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Fluids and Elasticity
Readings: Chapter 15
1
Solid, Liquid, Gas
-Solid has well-defined shape and well-defined surface.
Solid is (nearly) incompressible. Specific positions of
the molecules
-Liquid has well-defined surface (nor shape), It is
(nearly) incompressible.
- Gases are compressible. They occupy all the volume.
2
Liquid, Gas: Density
-Density is defined as a ratio of the mass of the object
and occupied volume
m

V
The mass of unit volume (1m x 1m x 1m)
Units: kg / m 3
 air
air
m

V
kg
1 3
m
 liquid
 liquid
m

V
kg
1000 3
m
V
 air   liquid
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Liquid, Gas: Pressure
The force will be distributed over the area A,
F
We can define the force per unit area pressure
F
p
A
If we place some object inside liquid (gas) then there will be normal force
acting on the object
F
The force will depend on the area. We can
define pressure. The pressure will be the same
for all orientation of the object.
F
F
F
p
A
F
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F
p
A
Liquid, Gas: Pressure
Pressure is a SCALAR. The force due to pressure
will be perpendicular to the surface.
Units: Pascal (Pa)
Pa 
N
m2
Origin of pressure:
1. Gravitational – in liquid - as gravity pull down
the liquid exerts a force on the bottom and
sides.
2. Thermal motion – in gasses – motion of atoms
(molecules) results in collision with the walls.
Atmospheric pressure – 1atm = 101300 Pa
5
Liquid: Pressure
Fnet  0
p0 A  w  pA
w  mg   Adg
w
Then
p0 A   Adg  pA
p  p0   dg
Hydrostatic pressure at depth d
6
Liquid: Pressure
Example: What is the hydrostatic pressure of water at depth 10 m, 100 m,
1000 m
p  p0   dg
  1000kg / m 3
p0  patm  105 Pa
p10  p0   dg  105  1000  10  10  2  105  2 patm
p100  p0   dg  105  1000  100  10  11  105  11 patm
p1000  p0   dg  105  1000  1000  10  101  105  101 patm
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Pressure at point A is the same as the pressure at point B – atmospheric
pressure,
p0
Pressure at point 1 is
p1  p0   hg
Pressure at point 2 is
p2  p0   hg
B
A
h
h
8
Liquid: Buoyancy force: Archimede’s principle
p2  p0   h1 g
Fnet  F1  F2
y
h1
F1
Fnet , y  F2  F1
h2
F1  p1 A  ( p0   liquid gh1 ) A
F2
p2  p0   h2 g
F2  p2 A  ( p0   liquid gh2 ) A
Fnet , y  ( p0   liquid gh2 ) A  ( p0   liquid gh1 ) A 
  liquid g( h2  h1 ) A   liquidVg
FB   liquidVg
Archimede’s principle
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Liquid: Buoyancy force: Archimede’s principle
FB   liquidVg
10
Liquid: Buoyancy force: Archimede’s principle
FB   liquidVg
w  mobject g   objectVg
w  FB
w  FB
 object   liquid
Object sinks
 object   liquid
Object floats
w  FB
 object   liquid
Equilibrium
11
Liquid: Fluid Dynamics: Equation of Continuity
Ideal-fluid model:
1. Liquid is incompressible
2. Liquid is nonviscous (no friction)
3. Flow is steady
Liquid is incompressible: Equation of Continuity
v1 A1t
A1
v1
v1 A1  v2 A2
v2 A2 t
A2
v2
The same volume of fluid crosses both planes
12
Liquid: Fluid Dynamics: Bernoulli’s Equation
No friction: conservation of energy
1 2
1 2
p1   v1   gh1  p2   v2   gh2
2
2
A1
A2
v1
v2
v1 A1  v2 A2
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Example: find v-?
A
At point A:
pA  p0
vA  0
v
hA  h
B
At point B:
pB  p0
vB  v
hB  y
1 2
1 2
pA   v A   ghA  pB   vB   ghB
2
2
1 2
v  2 g( h  y )
 gh   v   gy
2
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Example: find p at point B
B
C
v1 A1  v2 A2
1 2
p1   v1   gh1 
2
1 2
 p2   v 2   gh2
2
Water
Continuity equation:
Bernoulli’s equation:
hB  10m
hc  0
  1000kg / m 3
vC AC  vB AB
1 2
1 2
pC   vC   ghC  pB   vB   ghB
2
2
1
pB  pC   (vC2  vB2 )   ghB  102500 Pa
2
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