Physics
Unit 5
 This Slideshow was developed to accompany the textbook
 OpenStax Physics
Available for free at https://openstaxcollege.org/textbooks/collegephysics
 By OpenStax College and Rice University
 2013 edition
 Some examples and diagrams are taken from the textbook.
Slides created by
Richard Wright, Andrews Academy
rwright@andrews.edu
 Phases of Matter
 Solid
 Atoms in close contact so they can’t
move much
 Set volume and shape
 Can’t compress
 Liquid
 Atoms move past each other
 Set volume
 Takes shape of container
 Hard to compress
 Gas
 Atoms far apart
 Neither set volume or shape
 Compressible
 Fluids
 Flow
 Both liquids and gases
 Density
𝜌 =
𝑚
𝑉
Where
𝜌 = density
m = mass
V = Volume
 Things with small density
float on things with more
density
 Solids most dense
 Gases least dense
See Table 11.1
 Can use density to determine unknown material
 An ornate silver crown is thought to be fake. How could we determine if
is silver without damaging the crown?
 Find its mass using a balance. (It is 1.25 kg)
 Find its volume by submerging in water and finding volume of displaces
water. (It is 1.60 × 10−4 m3)
 Find the density
 𝜌 = 7.81 × 103 kg/m3
 Table 11.1 says it is steel
 Silver’s density is 10.5 × 103 kg/m3
 Don’t be dense, you can
solve these problems
 11P1-7
 Read 11.3, 11.4
 11CQ8, 9, 11, 13, 14, 16, 17
 Answers:
 1) 1.610 cm3






2) 2.54 L
3) 2.58 g
4) 2.70 g/cm3
5) 3.99 cm
6) 0.163 m
7) 2.86 times denser
 The molecules in a fluid are free to wander around
 In their wanderings they sometimes collide with the
sides of their container (i.e. balloon)
 The more the molecules collide with the walls, the more
force is felt
𝐹
𝑃=
𝐴
P = Pressure
F = Force perpendicular to surface
A = Area of surface
 Unit: N/m2 = Pa (pascal)
1 Pa is very small so we usually use kPa or atm
 In a fluid the pressure is exerted
perpendicularly to all surfaces
 A static fluid cannot produce a
force parallel to a surface since it is
not moving parallel to surface
 You are drinking a juice box. In the process you suck all the juice
and air out of the box. The top of the box is 7.5 cm by 5 cm. If the
air pressure is 1.013 × 105 Pa, how much force is acting on the top
of the box?
 380 N = 85 lbs
 Would the force of the side of the box be more or less than the
top?
 More because more area
 The force that squashes the juice box is from the weight
of all the air above it
 Atmospheric Pressure at Sea Level
1.013 × 105 Pa = 1 atmosphere (1 atm)
 The column of static fluid
experiences several
vertical forces
 Since the fluid is not
moving, it is in
equilibrium and ∑𝐹 = 0


∑𝐹 = 𝑃2 𝐴 − 𝑃1 𝐴 − 𝑚𝑔 = 0
𝑃2 𝐴 = 𝑃1 𝐴 + 𝑚𝑔

𝜌=

𝑃2 𝐴 = 𝑃1 𝐴 + 𝜌𝑉𝑔


𝑉 = 𝐴ℎ
𝑃2 𝐴 = 𝑃1 𝐴 + 𝜌𝑔𝐴ℎ


𝑃2 = 𝑃1 + 𝜌𝑔ℎ
Or 𝑃 = 𝜌𝑔ℎ where P is the pressure due to the fluid at a depth h below the
surface
𝑚
𝑉
→ 𝑚 = 𝜌𝑉
 If the pressure is known at a depth, the pressure lower
down can be found by adding ρgh
 This assumes ρ is constant with depth
 This is a good estimate for liquids, but not for gasses
unless h is small
 Would Hoover Dam need to be just as strong if the entire
lake behind the dam was reduced to an inch of water
behind the dam, but the same depth as the lake?
 Yes, the pressure depends only on the depth
 What is the total pressure at points A and B?
 1.55x105 Pa
 Yes, there is a lot of pressure
riding on this assignment
 11P11-17, 20, 22
 Read 11.5, 11.6
 11CQ24-27
 Answers:
 11) 3.59 × 106 Pa








12) 7.80 × 104 Pa
13) 2.36 × 103 N
14) 0.760 m
15) 1.10 × 108 Pa
16) Show work
17) 30.6 m
20) 1.09 × 103 Pa
22) 24.0 N
 Pascal’s Principle
 A change in pressure applied
to an enclosed fluid is
transmitted undiminished to
all portions of the fluid and
the walls of its container.
 Basis of hydraulics
𝐹
 Since 𝑃 = , if we change the
𝐴
area, the force is changed
 How much force must be exerted at A to support the
850-kg car at B? The piston at A has a diameter of 17
mm and the piston at B a diameter of 300 mm.
 F = 26.7 N
A
B
 Gauge Pressure
Used by pressure gauges
Measures pressure relative to atmospheric pressure
 Absolute Pressure
Sum of gauge pressure and atmospheric pressure
𝑃𝑎𝑏𝑠 = 𝑃𝑔𝑎𝑢𝑔𝑒 + 𝑃𝑎𝑡𝑚
 Open-Tube Manometer
U-shaped tube with fluid in
it
One end is connected to the
container of which we want
to measure the pressure
The other end is open to the
air
𝑃2 = 𝑃𝑎𝑏𝑠
𝑃2 = 𝜌𝑔ℎ + 𝑃𝑎𝑡𝑚
𝑃2 − 𝑃𝑎𝑡𝑚 = 𝑃𝑔𝑎𝑢𝑔𝑒
 Barometer
 Used to measure air pressure
 A tube with the top sealed and filled with
mercury
 The bottom is open and sitting in a pool of
mercury
 Pressure at top = 0
 Pressure at bottom = 𝑃𝑎𝑡𝑚
 𝑃 = 𝜌𝑔ℎ
 Let me pressure you into
solving these problems
 11P24-27, 30-34
 Read 11.7
 11CQ28-31




 Answers:
 24) 2.55 × 107 Pa, 251 atm




25) 136 N
26) 5.76 × 103 N
27) 100:1, 10:1, 1:100
30) 1.60 × 104 Pa, 1.07 × 104
Pa
31) 4.08 m
32) 1.49 × 106 N
33) 159 / 119
34) 3.99 × 104 N
 Think of trying to push a beach ball under water
 The water pushes it up
 All fluids push things up because the pressure is higher at
greater depths
 The upward force is buoyant force


𝐹𝐵 = 𝑃2 𝐴 − 𝑃1 𝐴
𝐹𝐵 = 𝑃2 − 𝑃1 𝐴

𝑃2 = 𝑃1 + 𝜌𝑔ℎ  𝑃2 − 𝑃1 = 𝜌𝑔ℎ

𝐹𝐵 = 𝜌𝑔ℎ 𝐴

𝜌=


𝑉 = 𝐴ℎ
𝑚 = 𝜌ℎ𝐴
𝑚
𝑉
 𝑚 = 𝜌𝑉
 𝐹𝐵 = 𝑚𝑔 = 𝑊𝑙𝑖𝑞𝑢𝑖𝑑
 Archimedes’ Principle
Buoyant force = weight of the displaced fluid
𝐹𝐵 = 𝑊𝑓𝑙
 If buoyant force ≥ gravity, then it floats
 If buoyant force < gravity, then it sinks
 An ice cube is floating in a glass of fresh water. The cube
is 3 cm on each side. If the cube is floating so a flat face
is facing up, what is the distance between the top of the
cube and the water?
h
 0.002 m
x
 As you might have guessed
An object will float if its average density < density of
the fluid
In other words, it will float if it displaces more fluid
than its own weight
 Specific Gravity
𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 =
𝜌
𝜌𝑓𝑙
= fraction submerged
If specific gravity < 1 it floats
If specific gravity > 1 it sinks
 A man tied a bunch of helium balloons to a lawn chair and
flew to a great altitude. If a single balloon is estimated as a
sphere with a radius of 20 cm and is filled with helium, what
is the net force on one balloon?
0.3648 N
 How many balloons would be required to lift a 80 kg man
and chair?
2150 balloons
 Be buoyed up the thought of the
joy derived from solving these
problems
 11P 37-42, 44, 51
 Read 11.8, 11.9
 11CQ33-38
 Answers:
 37) 800 kg/m3
38) 815 kg/m3
39) 99.5%, 96.9%
40) 41.4 g, 41.4 cm3, 1.09 g/cm3
41) 198 g, 198 cm3, 2.73 g/cm3
42) 39.5 g, 50 cm3, 0.79 g/cm3,
ethyl alcohol
 44) 8.21 N
 51) 2920 N





 Cohesive Forces
Forces between
molecules of same type
Holding water together
in cup
 Adhesive Forces
Forces between
molecules of different
type
Water drops sticking to
grass
 Surface Tension
 Cohesive forces at surface
of a liquid cause it to have
smallest surface area
possible
 Why drops are round
 Sort of like membrane on
top of fluid
 Proportional to cohesive
force which varies by fluid
𝛾 =
𝐹
𝐿
𝛾 = surface tension
F = force
L = length on the
surface the force is
applied
 Gauge pressure inside spherical bubble
Inward surface tension causes pressure inside a
bubble to be higher than atmospheric pressure
𝑃 =
4𝛾
𝑟
 Interaction of cohesion and adhesion
 A drop of water on a surface
Cohesion wants to make a spherical drop
Adhesion wants to spread out and stick to surface
 Contact angle
Indicates which force is stronger
 Capillary Action
 Tendency of a fluid to be drawn
up a narrow, capillary, tube
 If contact angle is < 90°, then
fluid raised up tube
 As the fluid touches the sides of
tube, adhesive forces pull fluid
up, cohesive forces pull the
center up, but not as much
 The curved part is called
meniscus
 Height that capillary action
can lift fluid
ℎ =
2𝛾 cos 𝜃
𝜌𝑔𝑟
𝛾 = surface tension
𝜌 = density of fluid
𝑔 = acceleration of
gravity
𝑟 = radius of tube
 Calculate the gauge pressure inside two soap bubbles
with radii of 3 cm and 1 cm.
3 cm: 4.93 Pa
1 cm: 14.8 Pa
 That’s why it’s harder to blow up a small balloon than a
big balloon.




Stick with it, you can do well
11P54-59, 62-63, 68-69, 73-74, 77
Read 12.1, 12.2
12CQ1-5, 9-12, 14





Answers:
54) 592 N/m2
55) 0.0700 N/m, water at 40°C
56) 2.23 × 10−2 mmHg
57) 1.56 × 10−3 N
 58) 1.65 × 10−3 m, 3.71 × 10−4 m
 59) 0.566 m
 62) 14.6 N/m2, 4.46 N/m2, 7.40
N/m2, Alcohol
 63) -1.46 cm, 0 cm
 68) 479 N
 69) 5.00 × 108 Pa
 73) 3.81 × 103 N/m2, 28.7 mmHg
 74) 27.1 cm
 77) 3.98 × 106 Pa, 2.1 × 10−3 cm
 Flow Rate
𝑄 =
𝑉
𝑡
Q = Flow rate
V = Volume of fluid
t = time
 𝑄=
𝑉
𝑡
=
𝐴𝑑
𝑡
= 𝐴𝑣
 A = cross-section area
 𝑣 = average velocity of fluid
 Since flow rate is constant
for a given moving fluid
 Equation of continuity
𝜌1 𝐴1 𝑣1 = 𝜌2 𝐴2 𝑣2
 If incompressible
𝐴1 𝑣1 = 𝐴2 𝑣2
 If incompressible and
several branches
𝑛1 𝐴1 𝑣1 = 𝑛2 𝐴2 𝑣2
 Where does the water flow the fastest?
A
B
C
 A garden hose has a diameter of 2 cm and water enters it
at 0.5 m/s. You block 90% of the end of the hose with
your thumb. How fast does the water exit the hose?
 v = 5 m/s
 When a fluid goes through narrower channel, it speeds
up
 It increases kinetic energy
 𝑊𝑛𝑒𝑡 =
1
𝑚𝑣 2
2
−
1
𝑚𝑣02
2
 Net work comes from pressure pushing the fluid
 Derivation
 𝑊𝑛𝑒𝑡 = 𝐸𝑓 − 𝐸0
 𝐸 = 𝐾𝐸 + 𝑃𝐸
1
= 𝑚𝑣 2 + 𝑚𝑔ℎ
2
 𝑊𝑛𝑒𝑡 = 𝐹 ⋅ 𝑥
 𝑃=
𝐹
𝐴
→ 𝐹 = 𝑃𝐴 → 𝐹𝑥 = 𝑃𝑉
 𝑊𝑛𝑒𝑡 = 𝑃2 − 𝑃1 𝑉
𝑊𝑛𝑒𝑡 = 𝑃2 − 𝑃1 𝑉 =
1
𝑚𝑣12
2
 Divide by V and rearrange
𝜌 =
+ 𝑚𝑔ℎ1 −
1
𝑚𝑣22
2
+ 𝑚𝑔ℎ2
𝑚
𝑉
 Bernoulli’s Equation
1 2
1 2
𝑃1 + 𝜌𝑣1 + 𝜌𝑔ℎ1 = 𝑃2 + 𝜌𝑣2 + 𝜌𝑔ℎ2
2
2
 Think about driving down a road with something in your car
trunk. The object is too large to completely shut the trunk lid.
While the car is stopped, the lid quietly rests as far down as it can
go. As you drive down the road, why does the trunk open?
 The air in the trunk is still. The air above the trunk is moving.
 The air in the trunk is at a higher pressure than above the trunk.
So the trunk is pushed open.
 Demo:
 Fold paper into a tunnel and blow through it.
 What happens and why?
 Why is an aneurysm (enlarged region of a blood vessel)
bad for you?
 As the blood enters the enlarged region, it slows.
 The slower blood has a higher pressure, and pop!
 Why do all houses need a plumbing vent?
 Waste water flows through a sewer line.
 Something like a sink is connected to the line, but there is a water
trap to keep the sewer gasses from entering the house.
 The flowing water in the sewer means the air directly above the
flowing water has a lower pressure than the air above the sink.
 This pushes the water in the trap down the pipe and sewer gasses
enter the house
 How do airplane wings work (even paper airplanes)?
 The top of the wing is curved and the bottom is not. The air flows
faster over the top of the wing, than the bottom. This pushes the
wing up.
 How does a curve ball in baseball work?
 The faster you work, the less
pressure you’ll feel?
 12P2-6, 8, 11, 20-23
 Read 12.3
 12CQ18-21
 Answers:
 2) 83.3 cm3/s, 8.33 × 10−5 m3/s
 3) 27 cm/s









4) 5.03 cm3/s, 151 cm3
5) 0.75 m/s, 0.13 m/s
6) 0.139𝑣1
8) 0.166 cm/s
11) 12.6 m/s, 0.0800 m3/s
20) 33.1 m, 19.6 cm
21) 2.54 × 105 N
22) 15.3 N
23) 1.58 × 106 N/m2, 163 m
 Previous examples of Bernoulli’s Equation had
simplified conditions
 Bernoulli’s Equation work in real world
 Water circulates throughout a house in a hot-water heating
system. If the water is pumped at a speed of 0.50 m/s
through a 4.0-cm-diameter pipe in the basement under a
pressure of 3.0 atm, what will be the flow speed and pressure
in a 2.6-cm-diameter pipe on the second floor 5.0 m above?
Assume the pipes do not divide into branches.
 𝑣2 = 1.2
𝑚
𝑠
 𝑃2 = 2.5 𝑎𝑡𝑚
 The tank is open to the
atmosphere at the top.
Find an expression for the
speed of the liquid leaving
the pipe at the bottom.
 𝑣1 =
2𝑔ℎ
 Since Bernoulli’s
Equation is conservation
of energy, the water
would rise up to the same
height as the water in the
tank.
 Power in Fluid Flow
 Power is rate of work or
energy
 Bernoulli’s Equation terms
are in energy per volume
 Multiply Bernoulli’s
Equation by volume and
divide by time
Or multiply by flow rate
Q
 𝑃𝑜𝑤𝑒𝑟 = 𝑃 +
1
𝜌𝑣 2
2
 Apply yourself to these
applications
 12P25-28
 Read 12.4, 12.5
 12CQ22-24, 26-28
 Answers:
 25) 9.56 × 108 W, 2.12
 26) 71.8 m/s, 257 m/s
 27) 1.26 W
 28) 2.76 × 105 𝑃𝑎, 2.81 ×
105 𝑃𝑎
 Viscosity
 Fluid friction
Turbulent
 Laminar Flow
 Smooth flow in layers that
don’t mix
 Turbulent Flow
 Has eddies and swirls that
mix layers of fluid
Laminar
 How viscosity is
measured
Two plates with fluid
between
Top plate moved
Friction causes the
fluid to move
𝜂 =
𝐹𝐿
𝑣𝐴
 Laminar flow in tubes
 Difference in pressure causes
fluids to flow
𝑃2 − 𝑃1
𝑄=
𝑅
 Where
 Q is flow rate
 𝑃1 and 𝑃2 are pressures
 R is resistance
 Poiseuille’s law for resistance
8𝜂𝑙
𝑅= 4
𝜋𝑟
 Where
 𝜂 is viscosity
 l is length of tube
 r is radius of tube
 Since flow rate depends
on pressure
If Q is high, the
pressure must
decrease like
Bernoulli’s equation
 If R is high, the pressure
must decrease.
 In blood vessels this is
a problem with plaque
on artery walls
 How to tell if laminar or
turbulent flow
 Low speed with smooth,
streamlined object 
laminar
 High speed or rough object
 turbulent
 Reynolds number
 Below 2000  laminar
 Above 3000  turbulent
 Between 2000 and 3000
depends on conditions
2𝜌𝑣𝑟
𝑁𝑅 =
𝜂
 A hypodermic syringe is filled
with a solution whose viscosity
is 1.5 × 10−3 Pa ⋅ s. The plunger
area of the syringe is 8.0
× 10−5 m2 , and the length of the
needle is 0.025 m. The internal
radius of the needle is 4.0
× 10−4 m. The gauge pressure in
a vein is 1900 Pa (14 mmHg).
What force must be applied to
the plunger, so that 1.0
× 10−6 m3 of solution can be
injected in 3.0 s?
 𝐹 = 0.25 N
 Is the flow laminar if the
density is 1000 kg/m3?
 𝑁𝑅 = 4.44; Yes




Let the answers flow
12P30-33, 35-37, 51-54, 56
Read 12.6, 12.7
12CQ29-33





Answers:
30) 1.61 × 10−5 𝑁
31) 1.60 cm3/min
32) 0.0314 N, 0.471 mW
33) 8.7 × 10−2 𝑚𝑚3 /𝑠








35) 0.316𝑟1
36) 1.8𝑟1
37) 1.52, higher blood pressure
51) 1990
52) show work
53) 127000, 35100
54) 790000, 1690000
56) 3.16 × 10−4 𝑚3 /𝑠
 If object move in fluid (or
fluid moves past object)
 Alternate form of
Reynolds number
𝜌𝑣𝐿
′
𝑁𝑅 =
𝜂
 𝑁𝑅′ < 1  laminar flow
 1 < 𝑁𝑅′ < 10 
transition, might have
turbulent wake
 10 < 𝑁𝑅′ < 106  either
 𝑁𝑅′ > 106  turbulent
 Viscous drag
Force trying to stop
object in fluid
If 𝑁𝑅′ < 1, 𝐹𝑉 ∝ v
If 10 < 𝑁𝑅′ < 106 , 𝐹𝑉 ∝
𝑣2
 Terminal velocity
Objects falling in fluid
Viscous drag will
increase as speed
increases
Eventually drag force
will = gravity force
No acceleration
 Diffusion
Molecules in fluids move
randomly
They will travel some
distance and then hit
another molecule
Average distance
traveled
𝑥𝑟𝑚𝑠 = 2𝐷𝑡
 How long will it take hemoglobin to move 1 mm in water
by free diffusion (𝐷 = 6.9 × 10−11 𝑚2 /𝑠)?
 t = 7246 s = 2 h
 Diffusion
Goes from high
concentration to low
Faster with higher
difference of
concentrations
 Diffusion across
Membranes
Membranes are barriers
They are semipermeable
Some molecules can
cross, but others can’t
 Osmosis
 Transport of water through
membrane from differences in
concentration
 Dialysis
 Transport of other molecules
through membrane from
differences in concentration
 Reverse osmosis/reverse dialysis
Sufficient pressure is applied to reverse normal
direction of substance through membrane
Can be used to purify a fluid
Water pushed through a membrane that won’t let
salt through
 You cannot learn by
osmosis since knowledge is
not water, you must study
 12P38-39, 62-64, 66
 Answers:
 38) 𝑣 =
2𝑅 2 𝑔 𝜌2 −𝜌1
9𝜂





39) 225 mPa⋅s
62) 1.41 mm
63) 130
64) 130 s
66) 0.391 s