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 pressure2. Variation of pressure with height3. Buoyant forces
PHY 113 C Fall 2013 -- Lecture 19 211/05/2013
PHY 113 C Fall 2013 -- Lecture 19 311/05/2013
.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:
PHY 113 C Fall 2013 -- Lecture 19 411/05/2013
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.
PHY 113 C Fall 2013 -- Lecture 19 511/05/2013
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
PHY 113 C Fall 2013 -- Lecture 19 611/05/2013
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
:Summary Police vs
Driver vo
PHY 113 C Fall 2013 -- Lecture 19 711/05/2013
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
PHY 113 C Fall 2013 -- Lecture 19 811/05/2013
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 PascalDensity: r =mass/volume 1 kg/m3 = 0.001 gm/ml
Note: In this chapter Ppressure (NOT MOMENTUM)
PHY 113 C Fall 2013 -- Lecture 19 911/05/2013
Pressure
AP
F
Note: since P exerted by a fluid acts in all directions, it is a scalar parameter
PHY 113 C Fall 2013 -- Lecture 19 1011/05/2013
Example of pressure calculation High heels (http://www.flickr.com/photos/moffe6/3771468287/lightbox/)
F
Pam
NA
mgA
Pheel
62 105.1
01.001.04/6004/
F
PHY 113 C Fall 2013 -- Lecture 19 1111/05/2013
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
PHY 113 C Fall 2013 -- Lecture 19 1211/05/2013
Density = Mass/Volume
PHY 113 C Fall 2013 -- Lecture 19 1311/05/2013
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
yrgy = mg/A
P(y+y)
P(y)
Note: In this formulation +y is defined to be in the up direction.
PHY 113 C Fall 2013 -- Lecture 19 1411/05/2013
For an “incompressible” fluid (such as mercury):
r = 13.585 x 103 kg/m3 (constant)
)(ρ ρ 00 yygPPgdydP
r 13.595 x 103 kg/m3
Example:
m 76.0m/s 8.9kg/m 1013.595
Pa 10013.1
ρ
233
5
00
g
Pyyh
PHY 113 C Fall 2013 -- Lecture 19 1511/05/2013
Barometric pressure readingsHistorically, pressure was measured in terms of inches of mercury in a barometer
)(ρ ρ 00 yygPPgdydP
r 13.595 x 103 kg/m3
in93.290.0254m
1in0.76mm 76.0
m/s 8.9kg/m 1013.595Pa 10013.1
ρ
233
5
00
g
Pyyh
PHY 113 C Fall 2013 -- Lecture 19 1611/05/2013
Weather report:
m763.0in
0.0254min03.30in03.30
PHY 113 C Fall 2013 -- Lecture 19 1711/05/2013
Question: Consider the same setup, but replace fluid with water (r = 1000 kg/m3). What is h?
r 1000 kg/m3
iclicker equation:Will water barometer have h:
A. Greater than mercury.B. Smaller than mercury.C. The same as mercury.
PHY 113 C Fall 2013 -- Lecture 19 1811/05/2013
Question: Consider the same setup, but replace fluid with water (r = 1000 kg/m3). What is h?
)(ρ ρ 00 yygPPgdydP
r 1000 kg/m3
m34.10m/s 8.9kg/m 0001
Pa 10013.1ρ
23
5
00
g
Pyyh
PHY 113 C Fall 2013 -- Lecture 19 1911/05/2013
m 5.0kg/m 1000 3
hr
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.
PHY 113 C Fall 2013 -- Lecture 19 2011/05/2013
ρ ρ ρ :gas idealan For 0
0
0
0
PgP
dydP
PP
miyy
myyyy
Pg
ePePePyP 50
800000
0000
0
)( :Solution
r
)(ρ (constant) :etc mercury, For water, 00 yygPP r
General relationship between P and r:
gdydP ρ :surface sEarth'near fluids allFor
PHY 113 C Fall 2013 -- Lecture 19 2111/05/2013P (atm)
y-y 0
(mi)
miyy
myyyy
Pg
ePePePyP 50
800000
0000
0
)( :Solution
r
Approximate relation of pressure to height above sea-level
PHY 113 C Fall 2013 -- Lecture 19 2211/05/2013
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.
PHY 113 C Fall 2013 -- Lecture 19 2311/05/2013
Example:
Hydraulic press
incompressible fluid
A1x1=A2x2
F1/A1= F2/A2
PHY 113 C Fall 2013 -- Lecture 19 2411/05/2013
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
PHY 113 C Fall 2013 -- Lecture 19 2511/05/2013
Buoyant force for fluid acting on a solid: FB=rfluidVdisplacedg
submergedB
topbottomB
gVyAgAyyPyPF
FFFygyyPyP
fluidfluid
fluid
ρρ)()(
:forceBuoyant ρ)()(
mg
FB mg = 0
rfluidVsubmergedg rsolidVsolidg = 0
fluid
solid
solid
submerged
VV
ρρ
A
y
PHY 113 C Fall 2013 -- Lecture 19 2611/05/2013
Summary:
submergedB gVF fluidρ :forceBuoyant
fluid
solid
solid
submerged
VV
ρρ
Some densities: ice r = 917 kg/m3
fresh water r = 1000 kg/m3
salt water r = 1024 kg/m3
PHY 113 C Fall 2013 -- Lecture 19 2711/05/2013
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.
PHY 113 C Fall 2013 -- Lecture 19 2811/05/2013
Archimede’s method of finding the density of the King’s “gold” crown
Wwater Wair
waterair
airwaterobject
objectobjectair
objectwaterobjectobjectBwater
WWW
gVmgW
gVgVFmgW
rr
r
rr
PHY 113 C Fall 2013 -- Lecture 19 2911/05/2013
(water) 1000kg/mρ :example
)(ρ :fluid ibleIncompress
ρ
300
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
00
0
0
yyP
gyygP
ePPyy
Pg
Summary:
PHY 113 C Fall 2013 -- Lecture 19 3011/05/2013
ρ ρ ρ :gas" ideal"an For
d;complicate more isequation fluid, lecompressibFor
0
0
0
0
PgP
dydP
PP
0000
50
800000
)( , smallFor )(
0000
0
yygPyPyyePePePyP mi
yym
yyyyP
g
r
r
)(ρ (constant) fluid, ibleincompressFor 00 yygPP r
Review of equations describing static fluids in terms of pressure P and density r:
gdydP ρ :surface sEarth'near fluids allFor
PHY 113 C Fall 2013 -- Lecture 19 3111/05/2013
Buoyant force
submergedB gVF fluidρ :forceBuoyant
Ftop
Fbottom
Ay
gVF
gyAFFF
submergedB
topbottomB
fluid
fluid
ρ
ρ :fluid from cubeon forceNet
PHY 113 C Fall 2013 -- Lecture 19 3211/05/2013
Scale reading due to buoyant force:
mg
FB
T
Bscale
B
FmgTFmgFT
0:reading Scale
gVmg
gVF
solid
submergedB
solid
fluid
ρ
ρ
scaleBfluid
solid
solidsubmerged
Fmgmg
Fmg
ρρ
VV
:usingdensity solid Measure
ρρ If solidfluid
PHY 113 C Fall 2013 -- Lecture 19 3311/05/2013
Scale reading due to buoyant force:
mg
FB
T
Bscale
B
FmgTFmgFT
0:reading Scale
gVmg
gVF
solid
submergedB
solid
fluid
ρ
ρ
solid
submerged
fluid
solid
solidsubmerged
VV
ρρ
T
VV
:usingdensity solid Measure ;0
ρρ If solidfluid
PHY 113 C Fall 2013 -- Lecture 19 3411/05/2013
Fluid dynamics (for incompressible fluids)
2211
2211
2211
21
fluid of massgiven aConsider (constant)
:fluid ible"incompress"an For
vAvAtvAtvAM
xAxAMVVM
rrrr
rr
r
PHY 113 C Fall 2013 -- Lecture 19 3511/05/2013
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
VMMgyMv
VPPMgyMv
UKWUK
UKWUKM
fffiii
rrrr
rrrr
r
PHY 113 C Fall 2013 -- Lecture 19 3611/05/2013
Bernoulli’s equation:
22222
111
212
1 PgyvPgyv rrrr
21
2
1
2222
1
2222
11
212
1
221121
1
PPAAv
PvPv
vAvAyy
r
rr
PHY 113 C Fall 2013 -- Lecture 19 3711/05/2013
22222
111
212
1 PgyvPgyv rrrrBernoulli’s equation:
2
2
1
01
1122
01212
1
2222
1
1
22
AA
PPghv
vAvAPgyv
Pgyv
r
r
rr
rr
PHY 113 C Fall 2013 -- Lecture 19 3811/05/2013
m/s 7.24 1000
10013.11818
10 :Example
2
50
1
0
01
r
r
Pv
PP
PPv
Possibilities:
Fire extinguisher P>>P0
PHY 113 C Fall 2013 -- Lecture 19 3911/05/2013
m/s 1.3
/5.08.92
m 0.5 :Example2
1
1
smv
hghv
Possibilities:
Open top: P=P0
PHY 113 C Fall 2013 -- Lecture 19 4011/05/2013
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
r
r
PHY 113 C Fall 2013 -- Lecture 19 4111/05/2013
Siphon
0Av
ghv
PvPghvA
2
02
21
02
21
rrr
PHY 113 C Fall 2013 -- Lecture 19 4211/05/2013
Simplistic statement:
1222
212
1
22222
1
11212
1
PPvv
Pgyv
Pgyv
r
rr
rr
v1, P1
v2, P2
PHY 113 C Fall 2013 -- Lecture 19 4311/05/2013
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.
PHY 113 C Fall 2013 -- Lecture 19 4411/05/2013
PHY 113 C Fall 2013 -- Lecture 19 4511/05/2013
21
1
: thatNote
2
2
yAhA
yhgg
zyhg
zggP
HgOH
Hg
HgOH
rr
r
rr
22A
M
OHr
y
z
PHY 113 C Fall 2013 -- Lecture 19 4611/05/2013
2
1
2
1
21
1
1
2
2
2
AA
h
AAghg
yAhA
yhgg
Hg
OH
HgOH
HgOH
r
r
rr
rr
m0049.02113600
1000m2.0
:2/ m, 0.2For 21
h
AA