Transcript Lect25

About Midterm Exam 3
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When and where
Thurs April 21th , 5:45-7:00 pm
Rooms: Same as Exam I and II, See course webpage.
Your TA will give a brief review during the discussion session.
Coverage: Chapts 9 – 12 (4 chapters)
Format
Closed book, 20 multiple-choices questions (format as in practice exams)
1 page 8x11 formula sheet allowed, must be self prepared, no photo
copying/download-printing of solutions, lecture slides, etc.
Bring a calculator (but no lap-top computer). Only basic calculation functionality
can be used. Bring a 2B pencil for Scantron.
Fill in your ID and section # !
Special requests:
If different from Exam II, email me at [email protected]
One alternative exam: 3:30pm – 4:45pm, Thurs Mar. 24, Cham 5280
(as before).
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Chapter 13: Fluids
Lecture 25
Density
Pressure in fluids
Variation of pressure with depth in a fluid
Buoyancy and Archimedes’ principle
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Liquid
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Has a definite volume
No definite shape
Exist at a higher
temperature than solids
The molecules “wander”
through the liquid in a
random fashion
The intermolecular
forces are not strong
enough to keep the
molecules in a fixed
position
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Gas
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Has no definite volume
Has no definite shape
Molecules are in constant random motion
The molecules exert only weak forces on each other
Average distance between molecules is large
compared to the size of the molecules
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Density
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Density = Mass/Volume
ρ = M / V
units = kg/m3
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Pressure = Force per Unit Area
Which will hurt more?
If you are pricked by a nail
with a force equal to your weight
If your entire weight is supported
by a bed of similar nails
Both will hurt the same
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Pressure in a fluid or gas
book
v
• Impulse to book:
Fx
(or raindrops on your umbrella)
• Force is perpendicular to surface
v
Air molecule
• Force proportional to area of surface
• pressure (p)
p = Force/area [N/m2]
1 N/m2 = 1 Pascal (Pa)
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Atmospheric Pressure
Even when there is no breeze,
air molecules are continuously bombarding
everything around - results in pressure.
Normal atmospheric pressure = 1.01 x 105 Pa
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Pressure and Depth
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Examine the darker region, assumed to
be a fluid ρ
It has a cross-sectional area A
Extends to a depth h below the
surface
Three external forces act on the region
-P1A + P2A - Mg = 0
P2 = P1 + Mg/A = P1 + Mgh/V = P1 + ρgh
At the surface compared to at depth h
Po is normal atmospheric pressure
1.013 x 105 Pa = 14.7 lb/in2
P is the “absolute pressure”;
P - Po is the “gauge pressure”
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Barometer: Measure atmospheric pressure
p2 = p1 + ρgh
p1=0
patm = ρgh
Measure h, determine patm
example--Mercury
p2=patm
h
ρ = 13,600 kg/m3
patm = 1.05 x 105 Pa
ρ h = 0.757 m = 757 mm (for 1 atm)
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Question:
Is it possible to stand on the roof of a five story (50 foot) tall house
and drink, using a straw, from a glass on the ground?
CORRECT
1. No
2. Yes
Pa
P=0
h
Evacuate the straw by sucking
How high will water rise?
no more than h = Pa/ρ g (= 1.05 x 105/1000/9.8) = 33 ft
8”
nopressure
matter that
how the
hard
suck!
The
airyou
pushes
down on the liquid in
the glass is not enough to push all of the liquid up the
50ft through the straw.
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Measurement of Pressure
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Manometer
If both sides of an U-tube are open to atmosphere the levels
of the fluid are the same on both sides
If one side is connected to a “pressurized side” the level
difference between the two sides can be used to measure
pressure.
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Measuring Blood Pressure
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Blood pressure is quite high, 120/80 mm of Hg
Use higher density fluid in a manometer: Mercury
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F1
Pascal’s Principle
A1
A1
A2
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F2
The pressure applied to an enclosed fluid is transmitted undiminished to
all portions of the fluid and to the walls of its container.
This principle is used in hydraulic system
P1 = P2  (F1 / A1) = (F2 / A2)
Can be used to derive large gain by making A2 much larger than A1
» F2 = F1 (A2 / A1)
» Work done is the same: height by which the surface A2 rises is
smaller than the change in the height of surface with area A1.
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Buoyancy and Archimedes’ Principle
B = ρf V g
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King Hiero II of Syracuse’s Crown
King Hiero II asked Archimedes to check if his crown was pure gold.
Archimedes knew
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ρgold = 19.3x103 kg/m3 , ρw = 1.00x103 kg/m3
weighed:
Wc = Fg = 7.84 N in air
W’c = F’g = 6.86 N in water
Archimedes’ principle:
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B = Wc - W’c = 0.98 N = Fw = V ρw g  V g = 0.98 N/ ρw
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Thus, the crown density:
ρc = M/V = Wc / V g = (7.84 N /0.98 N) ρw = 8 kg/m3 < ρgold
King Hiero II was cheated !
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