About Midterm Exam 3

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 than@hep.wisc.edu One alternative exam: 3:30pm – 4:45pm, Thurs Mar. 24, Cham 5280

(as before).

14/18/11 Phys 201, Spring 2011

4/18/11 Phys 201, Spring 2011 2

Chapter 13: FluidsLecture 25

DensityPressure in fluidsVariation of pressure with depth in a fluidBuoyancy and Archimedes’ principle

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Liquid

Has a definite volume No definite shape Exist at a higher

temperature than solids The molecules “wander”

through the liquid in a random fashionThe intermolecular

forces are not strong enough to keep the molecules in a fixed position

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Gas 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 = Mass/Volumeρ = M / Vunits = kg/m3

Density

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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 weightIf your entire weight is supported

by a bed of similar nailsBoth will hurt the same

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Pressure in a fluid or gas

• Impulse to book:

(or raindrops on your umbrella)

• Force is perpendicular to surface

• Force proportional to area of surface

• pressure (p)

p = Force/area [N/m2]

1 N/m2 = 1 Pascal (Pa)

v

v

Fx

book

Air molecule

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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 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

hp2=pat

m

p1=0p2 = p1 + ρgh

patm = ρgh

Measure h, determine patm

example--Mercury

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? 1. No 2. Yes

The pressure that the air pushes down on the liquid in the glass is not enough to push all of the liquid up the 50ft through the straw.

CORRECT

Pah

P=0

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”

no matter how hard you suck!

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Measurement of Pressure

ManometerIf both sides of an U-tube are open to atmosphere the levels

of the fluid are the same on both sidesIf 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 Blood pressure is quite high, 120/80 mm of Hg Use higher density fluid in a manometer: Mercury

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Pascal’s Principle

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 systemP1 = 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.

A1

F1

F2

A1

A2

Buoyancy and Archimedes’ Principle

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B = ρf V g

King Hiero II of Syracuse’s Crown King Hiero II asked Archimedes to check if his crown was pure gold. Archimedes knew

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

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Archimedes’ principle:

B = Wc - W’c = 0.98 N = Fw = V ρw g V g = 0.98 N/ ρw

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 !