PHY 113 A General Physics I
9-9:50 AM MWF Olin 101
Plan for Lecture 26:
Chapter 14: The physics of fluids
1. Density and pressure
2. Variation of pressure with height
3. Buoyant forces
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The physics of fluids.
•Fluids include liquids (usually “incompressible) and gases
(highly “compressible”).
•Fluids obey Newton’s equations of motion, but because
they move within their containers, the application of Newton’s
laws to fluids introduces some new forms.
Pressure: P=force/area
1 (N/m2) = 1 Pascal
Density: r =mass/volume
1 kg/m3 = 0.001 gm/ml
Note: In this chapter Ppressure (NOT
MOMENTUM)
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Pressure
P
F
A
Note: since P exerted by a fluid
acts in all directions, it is a scalar
parameter
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Example of pressure calculation
High heels
(http://www.flickr.com/photos/moffe6/3771468287/lightbox/)
F
F
mg / 4
600 / 4 N
6
P



1
.
5

10
Pa
2
A
Aheel
0.01 0.01m
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Pressure exerted by air at sea-level
1 atm = 1.013x105 Pa
Example: What is the force exerted by 1 atm of air
pressure on a circular area of radius 0.08m?
F = PA = 1.013x105 Pa x p(0.08m)2
= 2040 N
Patm
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Density = Mass/Volume
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Relationship between density and pressure in a fluid
y
Effects of the weight of a fluid:
F ( y )  F ( y  y )  mg
F ( y ) F ( y  y ) mg


A
A
A
P ( y )  P ( y   y )  ρ g y
P ( y  y )  P ( y ) dP
lim


y

0
P(y+y)
y
dy
dP
rgy = mg/A

 ρ g
dy
P(y)
Note: In this formulation +y is
defined to be in the up
direction.
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For an “incompressible” fluid (such as mercury):
r = 13.585 x 103 kg/m3 (constant)
dP
 ρg  P  P0  ρg ( y  y0 )
dy
Example:
P0
h  y  y0 
ρg
1.013 105 Pa

13.595 103 kg/m 3  9.8 m/s 2
 0.76 m
r  13.595 x 103 kg/m3
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Barometric pressure readings
Historically, pressure was measured in terms of
inches of mercury in a barometer
dP
 ρg  P  P0  ρg ( y  y0 )
dy
P0
h  y  y0 
ρg
1.013 105 Pa

13.595 103 kg/m 3  9.8 m/s 2
 1in 
 0.76 m  0.76m  
  29.93in
 0.0254m 
r  13.595 x 103 kg/m3
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Weather report:
0.0254m
30.03in  30.03in
 0.763m
in
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Question: Consider the same setup, but
replace fluid with water (r = 1000 kg/m3).
What is h?
iclicker equation:
Will water barometer have h:
A. Greater than mercury.
B. Smaller than mercury.
C. The same as mercury.
r  1000 kg/m3
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Question: Consider the same setup, but
replace fluid with water (r = 1000 kg/m3).
What is h?
dP
 ρg  P  P0  ρg ( y  y0 )
dy
P0
h  y  y0 
ρg
1.013 105 Pa

1000 kg/m 3  9.8 m/s 2
 10.34m
r  1000 kg/m3
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r  1000 kg/m 3
h  0.5 m
iclicker question:
A 0.5 m cylinder of
water is inverted over a
piece of paper. What
will happen
A. The water will
flow out of the
cylinder and
make a mess.
B. Air pressure will
hold the water in
the cylinder.
Patm  1.013 105 Pa
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General relationship between P and r:
dP
For all fluids near Earth' s surface :
 ρg
dy
For water, mercury, etc : r  (constant)  P  P0  ρg ( y  y0 )
ρ0
For an ideal gas : ρ  P
P0
Solution :
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P( y )  P0 e
 ρ0 g 
dP


  P
dy
 P0 
r g
 0  y  y0 
P0
 P0 e
PHY 113 A Fall 2012 -- Lecture 26

y  y0
8000 m
 P0 e

y  y0
5 mi
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Approximate relation of pressure to height above sea-level
P( y )  P0 e
 P0 e

y  y0
8000 m
 P0 e

y  y0
5 mi
y-y0(mi)
Solution :
r g
 0  y  y0 
P0
P (atm)
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iclicker question:
Have you personally experienced the effects of
atmospheric pressure variations?
A. By flying in an airplane
B. By visiting a high-altitude location (such as
Denver, CO etc.)
C. By visiting a low-altitude location (such as
Death Valley, CA etc.)
D. All of the above.
E. None of the above.
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Buoyant forces in fluids
(For simplicity we will assume that the fluid is
incompressible.)
Image from the web of
Image from the web of
a glass of ice water
a floating iceberg.
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Buoyant force for fluid acting on a solid:
FB=rfluidVdisplacedg
P( y )  P( y  y )  ρ fluid gy
Buoyant force :
FB  Fbottom  Ftop
FB  P( y )  P( y  y )A  ρ fluid gyA  ρ fluid gVsubmerged

FB
A
y
rfluidVsubmergedg  rsolidVsolidg = 0
Vsubmerged
mg
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mg = 0
Vsolid

ρ solid
ρ fluid
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Summary:
Buoyant force :
Vsubmerged
Vsolid
ρ solid

ρ fluid
Some densities:
ice
fresh water
salt water
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FB  ρ fluid gVsubmerged
r = 917 kg/m3
r = 1000 kg/m3
r = 1024 kg/m3
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iclicker question:
Suppose you have a boat which floats in a fresh water lake,
with 50% of it submerged below the water. If you float
the same boat in salt water, which of the following would
be true?
A. More than 50% of the boat will be below the salt water.
B. Less than 50% of the boat will be below the salt water.
C. The submersion fraction depends upon the boat's total
mass and volume.
D. The submersion fraction depends upon the barometric
pressure.
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Archimede’s method of finding the density of the King’s
“gold” crown
Wwater
Wair
Wwater  mg  FB  r objectVobject g  r waterVobject g
Wair  mg  r objectVobject g
r object  r water
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Wair
Wair  Wwater
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Summary:
Application of Newton’s second law to fluid (near Earth’s
surface)
dP
 ρ g
dy
Incompress ible fluid : P  P0  ρg ( y  y0 )
example : ρ  1000kg/m 3 (water)
Compressib le fluid : P  P0e

ρ0 g
( y  y0 )
P0
ρ0 g
 P0  ρ 0 g ( y  y0 ) (for
( y  y0 )  1)
P0
example : ρ  1.29kg/m 3 (air)
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iclicker question:
Suppose that a caterer packed some food in an air tight
container with a flexible top at sea-level. This food was
loaded on to an airplane with a cruising altitude of ~6 mi
above the earth’s surface. Assuming that the airplane
cabin is imperfectly pressurized, what do you expect the
container to look like during the flight?
(A)
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(B)
PHY 113 A Fall 2012 -- Lecture 26
(C)
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Example:
Hydraulic press
incompressible
fluid
A1x1=A2x2
F1/A1= F2/A2
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