11/05/2013phy 113 c fall 2013 -- lecture 191 phy 113 c general physics i 11 am-12:15 pm tr olin 101...

PHY 113 C Fall 2013 -- Lecture 19 111/05/2013

PHY 113 C General Physics I11 AM-12:15 PM TR Olin 101

Plan for Lecture 19:

Chapter 14: The physics of fluids

1. Density and pressure

2. Variation of pressure with height

3. Buoyant forces


.m/s ˆ 15ˆ 10 NEv f

2212

1ff vmmK

125- E. 125 D.

ˆ225ˆ100 C. ˆ225ˆ100 B. 325 A.

?ˆ 15ˆ 10 of evaluationcorrect theisWhat 2

NENE

NE

Comment on exam

iclicker question:


Exam 2 feedback

iclicker question

For Exam 3 would you prefer:A. To have an in class exam only.B. To have both in class and take-home

components.


Webassign questions:

A driver travels northbound on a highway at a speed of 30.0 m/s. A police car, traveling southbound at a speed of 34.0 m/s, approaches with its siren producing sound at a frequency of 2500 Hz. (a) What frequency does the driver observe as the police car approaches?

(b) What frequency does the driver detect after the police car passes him?

(c) Repeat parts (a) and (b) for the case when the police car is traveling northbound.

toward

awayS

OSO vv

vvff

:Summary

http://www.webassign.net/web/Student/Assignment-Responses/last?dep=7782953


Webassign questions -- continued:

A driver travels northbound on a highway at a speed of 30.0 m/s. A police car, traveling southbound at a speed of 34.0 m/s, approaches with its siren producing sound at a frequency of 2500 Hz. (a) What frequency does the driver observe as the police car approaches?

toward

awayS

OSO vv

vvff

:SummaryPolice vs

Driver vo

http://www.webassign.net/web/Student/Assignment-Responses/last?dep=7782953


Webassign question:

A violin string has a length of 0.400 m and is tuned to concert G, with fG = 392 Hz.(a) How far from the end of the string must the violinist place her finger to play concert A, with fA = 440 Hz? cm from the bridge

(b) If this position is to remain correct to one-half the width of a finger (that is, to within 0.270 cm), what is the maximum allowable percentage change in the string tension?

a) Assuming the tension remains the same: lAfA=lGfG LAfA=LGfG


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 Ppressure (NOT MOMENTUM)


Pressure

AP

F

Note: since P exerted by a fluid acts in all directions, it is a scalar parameter


Example of pressure calculation High heels (http

://www.flickr.com/photos/moffe6/3771468287/lightbox/)

F

Pam

N

A

mg

AP

heel

62

105.101.001.0

4/6004/

F

http://www.flickr.com/photos/moffe6/3771468287/lightbox

http://www.flickr.com/photos/moffe6/3771468287/lightbox


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


Density = Mass/Volume


Relationship between density and pressure in a fluid

Effects of the weight of a fluid:

gdydP

dydP

yyPyyP

ygyyPyPA

mgA

yyFAyF

mgyyFyF

y

ρ

)()(lim

ρ)()(

)()(

)()(

0

yrgDy = mg/A

P(y+Dy)

P(y)

Note: In this formulation +y is defined to be in the up direction.


For an “incompressible” fluid (such as mercury):

r = 13.585 x 103 kg/m3 (constant)

)(ρ ρ 00 yygPPgdydP

= 13.595 r x 103 kg/m3

Example:

m 76.0

m/s 8.9kg/m 1013.595

Pa 10013.1

ρ

233

5

00

g

Pyyh


Barometric pressure readingsHistorically, pressure was measured in terms of inches of mercury in a barometer


= 13.595 r x 103 kg/m3

in93.290.0254m

1in0.76mm 76.0

m/s 8.9kg/m 1013.595

Pa 10013.1

ρ

233

5

00

g

Pyyh


Weather report:

m763.0in

0.0254min03.30in03.30


Question: Consider the same setup, but replace fluid with water (r = 1000 kg/m3). What is h?

= 1000r kg/m3

iclicker equation:Will water barometer have h:

A. Greater than mercury.B. Smaller than mercury.C. The same as mercury.


Question: Consider the same setup, but replace fluid with water (r = 1000 kg/m3). What is h?


= 1000r kg/m3

m34.10

m/s 8.9kg/m 0001Pa 10013.1

ρ

23

5

00

g

Pyyh


m 5.0

kg/m 1000 3

h

Pa 10013.1 5atmP

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.


ρ

ρ

ρ :gas idealan For 0

0

0

0

P

gP

dy

dP

PP

mi

yy

m

yyyyP

g

ePePePyP 50

800000

0000

0

)( :Solution

)(ρ (constant) :etc mercury, For water, 00 yygPP

General relationship between P and :r

gdy

dPρ :surface sEarth'near fluids allFor

PHY 113 C Fall 2013 -- Lecture 19 2111/05/2013P (atm)

y-y 0

(mi)

mi

yy

m

yyyyP

g

ePePePyP 50

800000

0000

0

)( :Solution

Approximate relation of pressure to height above sea-level


iclicker question:Have you personally experienced the effects of atmospheric pressure variations?

A. By flying in an airplaneB. 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.


Example:

Hydraulic press

incompressible fluid

A1Dx1=A2Dx2

F1/A1= F2/A2


Buoyant forces in fluids(For simplicity we will assume that the fluid is incompressible.)

Image from the web of a floating iceberg.

Image from the web of a glass of ice water


Buoyant force for fluid acting on a solid:

FB=rfluidVdisplacedg

submergedB

topbottomB

gVyAgAyyPyPF

FFF

ygyyPyP

fluidfluid

fluid

ρρ)()(

:forceBuoyant

ρ)()(

mg

FB - mg = 0

rfluidVsubmergedg - rsolidVsolidg = 0

fluid

solid

solid

submerged

V

V

ρ

ρ

A

Dy


Summary:

submergedB gVF fluidρ :forceBuoyant

fluid

solid

solid

submerged

V

V

ρ

ρ

Some densities: ice r = 917 kg/m3

fresh water r = 1000 kg/m3

salt water r = 1024 kg/m3


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.


Archimede’s method of finding the density of the King’s “gold” crown

Wwater Wair

waterair

airwaterobject

objectobjectair

objectwaterobjectobjectBwater

WWW

gVmgW

gVgVFmgW


(water) 1000kg/mρ :example

)(ρ :fluid ibleIncompress

ρ

3

00

yygPP

gdydP

Application of Newton’s second law to fluid (near Earth’s surface)

(air) kg/m29.1ρ :example

)1)(ρ

(for )(ρ

:fluid leCompressib

3

00

0000

)(ρ

0

00

0

yyP

gyygP

ePPyy

P

g

Summary:


ρ

ρ

ρ :gas" ideal"an For

d;complicate more isequation fluid, lecompressibFor

0

0

0

0

P

gP

dy

dP

PP

0000

50

800000

)( , smallFor

)(00

00

0

yygPyPyy

ePePePyP mi

yy

m

yyyyP

g

)(ρ (constant) fluid, ibleincompressFor 00 yygPP

Review of equations describing static fluids in terms of pressure P and density :r

gdy

dPρ :surface sEarth'near fluids allFor


Buoyant force

submergedB gVF fluidρ :forceBuoyant

Ftop

Fbottom

ADy

gVF

gyAFFF

submergedB

topbottomB

fluid

fluid

ρ

ρ

:fluid from cubeon forceNet


Scale reading due to buoyant force:

mg

FB

T

Bscale

B

FmgTF

mgFT

0

:reading Scale

gVmg

gVF

solid

submergedB

solid

fluid

ρ

ρ

scaleBfluid

solid

solidsubmerged

Fmg

mg

F

mg

ρ

ρ

VV

:usingdensity solid Measure

ρρ If solidfluid


Scale reading due to buoyant force:

mg

FB

T

Bscale

B

FmgTF

mgFT

0

:reading Scale

gVmg

gVF

solid

submergedB

solid

fluid

ρ

ρ

solid

submerged

fluid

solid

solidsubmerged

V

V

ρ

ρ

T

VV

:usingdensity solid Measure ;0

ρρ If solidfluid


Fluid dynamics (for incompressible fluids)

2211

2211

2211

21

fluid of massgiven aConsider

(constant)

:fluid ible"incompress"an For

vAvA

tvAtvAM

xAxAM

VVM


Approximate energy conservation in fluid dynamics Bernoulli’s equation

22

222

111

212

1

2222

1121

212

1

2222

1

121212

1

222111

fluid of piece""

on iprelationshwork -Energy

PgyvPgyv

gyvPPgyv

VM

MgyMv

VPPMgyMv

UKWUK

UKWUK

M

fffiii


Bernoulli’s equation:

22222

111

212

1 PgyvPgyv

21

2

1

2222

1

2222

11

212

1

221121

1

PPA

Av

PvPv

vAvAyy


22222

111

212

1 PgyvPgyv Bernoulli’s equation:

2

2

1

01

1122

01212

1

2222

1

1

22

AA

PPghv

vAvA

Pgyv

Pgyv


m/s 7.24

1000

10013.11818

10 :Example

2

50

1

0

01

Pv

PP

PPv

Possibilities:

Fire extinguisher P>>P0


m/s 1.3

/5.08.92

m 0.5 :Example

2

1

1

smv

h

ghv

Possibilities:

Open top: P=P0


iclicker question:Notice that if P<<P0, v1 could become very small. How might this happen?

A. Put an air-tight lid on the top.

B. Remove the lid on the top.

C. This will never happen.

2

2

1

01

1

22

AA

PPghv


Siphon

0Av

ghv

PvPghvA

2

02

21

02

21


Simplistic statement:

1222

212

1

22222

1

11212

1

PPvv

Pgyv

Pgyv

v1, P1

v2, P2


Webassign question:

The gravitational force exerted on a solid object is 5.30 N. When the object is suspended from a spring scale and submerged in water, the scale reads 3.50 N (figure). Find the density of the object.


21

1

: thatNote

2

2

yAhA

yhgg

zyhg

zggP

HgOH

Hg

HgOH

22A

M

OH

Dy

Dz


2

1

2

1

21

1

1

2

2

2

AA

h

A

Aghg

yAhA

yhgg

Hg

OH

HgOH

HgOH

m0049.02113600

1000m2.0

:2/ m, 0.2For 21

h

AA

11/05/2013phy 113 c fall 2013 -- lecture 191 phy 113 c general physics i 11 am-12:15 pm tr olin 101...

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