Transcript Physics 106P: Lecture 1 Notes

Physics 101: Chapter 11
Fluids


Textbook Sections 11.1-11.6
Density
Pressure
Pascal’s Principle
Textbook Sections 11.6-11.10
 Archimedes Principle & Buoyancy
 Fluids in motion: Continuity & Bernoulli’s equation
 Some problems, homework hints
UB, Phy101: Chapter 11, Pg 1
UB, Phy101: Chapter 11, Pg 2
Physics 101: Density


Density = Mass/Volume
  = M/V
 units = kg/m3
Densities of some common things (kg/m3)
 Water
1000
 ice
917 (floats on water)
 blood
1060 (sinks in water)
 lead
11,300
 Copper
8890
 Mercury 13,600
 Aluminum 2700
 Wood
550
 air
1.29
 Helium
0.18
UB, Phy101: Chapter 11, Pg 3
Qualitative Demonstration of Pressure
y
p y
average vertical force  f y 

t
 mv y 


t
UB, Phy101: Chapter 11, Pg 4
UB, Phy101: Chapter 11, Pg 5
UB, Phy101: Chapter 11, Pg 6
UB, Phy101: Chapter 11, Pg 7
Atmospheric Pressure
normal atmospheric pressure = 1.01 x 105 Pa (14.7 lb/in2)
UB, Phy101: Chapter 11, Pg 8
Chapter 11, Preflight
You buy a bag of potato chips in Urbana Illinois and forget them (un-opened)
under the seat of your car. You drive out to Denver Colorado to visit a friend
for Thanksgiving, and when you get there you discover the lost bag of chips.
The odd thing you notice right away is that the bag seems to have inflated
like a balloon (i.e. it seems much more round and bouncy than when you bought
it). How can you explain this ??
Due to the change in height, the air is much thinner in Denver. Thus, there is
less pressure on the outside of the bag of chips in Denver, so the bag seems
to inflate because the air pressure on the inside is greater than on the
outside.
PD
PU
PU
PU
In Urbana
In Denver
UB, Phy101: Chapter 11, Pg 9
UB, Phy101: Chapter 11, Pg 10
Pressure and Depth
Barometer: a way to measure atomospheric pressure
p2 = p1 + gh
p1=0
patm = gh
Measure h, determine patm
p2=patm
h
example--Mercury
 = 13,600 kg/m3
patm = 1.05 x 105 Pa
 h = 0.757 m = 757 mm = 29.80” (for 1 atm)
UB, Phy101: Chapter 11, Pg 11
Chapter 11, Preflights
Suppose you have a barometer with mercury and a barometer with
water. How does the height hwater compare with the height hmercury?
1. hwater is much larger than hmercury
2. hwater is a little larger than hmercury
3. hwater is a little smaller than hmercury
4. hwater is much smaller than hmercury
CORRECT
water is much less dense than
mercury, so the same amount of
pressure will move the water
farther up the column.
Pa  gh
p1=0
p2=patm
h
Pa
h
g
 mercury  13.6  water
so, h water  13.6 h mercury
UB, Phy101: Chapter 11, Pg 12
Chapter 11, Preflights
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
CORRECT
2. Yes
Even if a person could completely remove all of the air
from the straw, the height to which the outside air
pressure moves the water up the straw would not be high
enough for the person to drink the water.
UB, Phy101: Chapter 11, Pg 13
Chapter 11, Preflight
p=0
pa
Pa  gh
h
Pa
h
g
Evacuate the straw by sucking
How high will water rise?
no more than h = Pa/g = 10.3m = 33.8 feet
no matter how hard you suck!
UB, Phy101: Chapter 11, Pg 14
UB, Phy101: Chapter 11, Pg 15
UB, Phy101: Chapter 11, Pg 16
UB, Phy101: Chapter 11, Pg 17
UB, Phy101: Chapter 11, Pg 18
UB, Phy101: Chapter 11, Pg 19
UB, Phy101: Chapter 11, Pg 20
UB, Phy101: Chapter 11, Pg 21
Summary
• Density
• Pressure
P2 = P1 + gh
Pascal’s Principle
UB, Phy101: Chapter 11, Pg 22

Textbook Sections 11.6-11.10
 Archimedes Principle & Buoyancy
 Fluids in motion: Continuity & Bernoulli’s equation
 Some problems, homework hints
Note: Everything we do assumes fluid
is non-viscous and incompressible.
UB, Phy101: Chapter 11, Pg 23
UB, Phy101: Chapter 11, Pg 24
UB, Phy101: Chapter 11, Pg 25
Archimedes Principle (summary)


Buoyant Force (B)
weight of fluid displaced
B = fluid x Vdispl g
W = Mg = object Vobject g
object sinks if object > fluid
object floats if object < fluid
Eureka!
If object floats….
B=W
Therefore fluid g Vdispl. = object g Vobject
Therefore Vdispl./Vobject = object / fluid
UB, Phy101: Chapter 11, Pg 26
Chapter 11, Preflight
Suppose you float a large ice-cube in a glass of water, and that after
you place the ice in the glass the level of the water is at the very brim.
When the ice melts, the level of the water in the glass will:
1. Go up, causing the water to spill out of the glass.
2. Go down.
3. Stay the same.
CORRECT
B = W g
Vdisplaced
W = ice g Vice
Must be same!
 W g V
UB, Phy101: Chapter 11, Pg 27
Chapter 11, Preflight
Which weighs more:
1. A large bathtub filled to the brim with water.
2. A large bathtub filled to the brim with water with a battle-ship
floating in it.
3. They will weigh the same.
CORRECT
Tub of water + ship
Tub of water
Weight of ship = Buoyant force =
Weight of displaced water
Overflowed water
UB, Phy101: Chapter 11, Pg 28
Chapter 11: Fluid Flow (summary)

A1 P1 v1
A2 P2
v2
• Mass flow rate: Av (kg/s)
• Volume flow rate: Av (m3/s)
• Continuity: A1 v1 = A2 v2
i.e., mass flow rate the same everywhere
e.g., flow of river
For fluid flow without friction:
• Bernoulli: P1 + 1/2 v12 + gh1 = P2 + 1/2 v22 + gh2
UB, Phy101: Chapter 11, Pg 29
Chapter 11, Preflight
A stream of water gets narrower as it falls from a faucet (try it & see).
Explain this phenomenon using the equation of continuity
hmm, gravity stretches it out
The water's velocity is increasing as it flows down, so in
order to compensate for the increase in velocity, the area
must be decreased because the density*area*speed must
be conserved
A1
V1
V2
A2
First off the mass of the water is conserved, and water is a incompressible fluid.
The equation of continuity states that A1V1= A2V2. We know that the velocity or
V2 will be increasing due to gravity so in order for the equation to work the area
must decrease or get narrower.
UB, Phy101: Chapter 11, Pg 30
UB, Phy101: Chapter 11, Pg 31
Chapter 11, Preflight
A large bucket full of water has two drains. One is a hole in the side of
the bucket at the bottom, and the other is a pipe coming out of the
bucket near the top, which bent is downward such that the bottom of
this pipe even with the other hole, like in the picture below:
Though which drain is the water spraying out with the highest speed?
1. The hole
2. The pipe
3. Same
CORRECT
The speed of the water depends on the pressure at the opening through which
the water flows, which depends on the depth of the opening from the surface of
the water. Since both the hole and the opening of the downspout are at [the
same] depth below the surface, water from the two will be discharged with the
same speeds.
UB, Phy101: Chapter 11, Pg 32
UB, Phy101: Chapter 11, Pg 33
UB, Phy101: Chapter 11, Pg 34