Transcript Unit One: AC Electronics - Helderberg Hilltowns Association

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GE253 Physics
Unit Seven:
Solids and Fluids
John Elberfeld
[email protected]
518 872 2082
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Schedule
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Unit 1 – Measurements and Problem Solving
Unit 2 – Kinematics
Unit 3 – Motion in Two Dimensions
Unit 4 – Force and Motion
Unit 5 – Work and Energy
Unit 6 – Linear Momentum and Collisions
Unit 7 – Solids and Fluids
Unit 8 – Temperature and Kinetic Theory
Unit 9 – Sound
Unit 10 – Reflection and Refraction of Light
Unit 11 – Final
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Chapter 7 Objectives
• Distinguish between stress and strain, and use
elastic moduli to compute dimensional changes.
• Explain the pressure-depth relationship and state
Pascal's principle and describe how it is used in
practical applications.
• Relate the buoyant force and Archimedes' principle
and tell whether an object will float in a fluid, on the
basis of relative densities.
• Identify the simplifications used in describing ideal
fluid flow and use the continuity equation and
Bernoulli's equation to explain common effects of
ideal fluid flow.
• Describe the source of surface tension and its
effect and discuss fluid viscosity.
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Reading Assignment
• Read and study College Physics, by
Wilson and Buffa, Chapter 7, pages
219 to 250
• Be prepared for a quiz on this
material
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Written Assignments
• Do the homework on the handout.
• You must show all your work, and carry
through the units in all calculations
• Use the proper number of significant
figures and, when reasonable, scientific
notation
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Introduction
• Until now, we have been
dealing with “ideal” objects
• Now we will look at objects that can
stretch, bend, break, flow, and
compress when forces are applied
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Introduction
• In this week, we will study real,
macroscopic objects, such as wires,
blocks, and containers of fluids.
• We will keep the concept of force and see
what happens to macroscopic objects
when forces act on them and see what
forces macroscopic objects exert.
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Stress and Strain
• The stress of taking physics may
cause a strain on your brain
• But in physics, these terms have
different and precise meanings.
• We will learn to use these terms as
we study the effects of forces on
solids and fluids – in this unit, not to
accelerate objects, but to deform
them.
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Stress and Strain
• Stress (force/area) causes strain (%
deformation of some type)
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Stress and Strain
• In physics, stress is related to the
force applied to an object
• Strain is related to how the object is
deformed as a result.
• Both also depend on the size of the
object.
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Stress
• The diagram shows a wire or rod of length
Lo and area A.
• When you apply a force F on the wire, it
stretches by a distance .
• When you apply a force F in the opposite
direction on a rod, the rod is compressed
by a distance .
• The greater the force F, the greater is the
amount of stretching or compressing.
L0
A
F
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Stress
• It is clear that a thick wire is harder to
stretch than a thin one.
• For this reason, you can define the
concept of stress as:
• Stress = F / A
• The units of stress are N/m2.
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Strain
• The longer the wire or rod, the greater is
the amount of stretching. For this reason,
you can define strain as:
• Strain = | ΔL | / L
• The vertical lines stand for absolute value,
which means the same strain results from
positive ΔL due to stretching or negative Δ
L due to compressing.
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Strain
• A strain of 0.05 on a wire means that the
wire has been stretched to a length 5%
greater than its original length.
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Young’s Modulus
• NOTE: This graph shows Young’s
Modulus: how much a wire will stretch if
you apply different forces on it.
Young’s
Modulus
Region
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Graph
• Engineers need to know the force
need to stretch material a certain
distance
• What force cause the bridge to sag
into the waves?
• Graphs show strain on the X axis and
stress on the Y axis
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Young’s Modulus
• Young’s Modulus works in the straight line
part of the graph only
• The ratio of stress to strain in a material is
known as Young’s modulus:
• Y = (F/A) / ( ΔL / L)
• Where, F is force; A is cross sectional area
perpendicular to the applied force; and L
is length of the object being stressed.
• NOTE: Young’s Modulus is an ELASTIC
Modulus because materials bounce back
to their original shape in that part of the
graph.
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Young’s Modulus
• The bigger the modulus, the stronger the
material, and the more force it takes to cause
a specific deformation.
• Because material bounce back to their
original shapes, moduli (plural) are elastic
• The units of Young’s modulus are the same
as those of stress, N/m2.
• Young’s modulus for steel is 20 x 1010 N/m2
and for bone is 1.5 x 1010 N/m2
for example.
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Practice
• The femur (upper leg bone) is the longest
and strongest bone in the body.
• Let us assume that a typical femur is
circular and has a radius of 2.0 cm.
• How much force is required to extend
the bone by 0.010%?
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Calculations
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Y = (F/A) / ( ΔL / L)
From the chart: Ybone = 1.5 x 1010N/m2
R = 2cm(1m/100cm) = .02m
A = R2 = (.02m)2 = 1.26 x 10-3m2
ΔL / L = .01% = .01/100 = 1 x 10-4
Y ( ΔL / L) = (F/A)
F = A Y ( ΔL / L)
F = 1.26 x 10-3m2 1.5 x 1010N/m2 1 x 10-4
F = 1.89 x 103 N (about 425 pounds)
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Practice
• A mass of 16 kg is
suspended from a steel
wire of 0.10-cmdiameter.
• By what percentage
does the length of the
wire increase?
• Young’s modulus for
steel is 20 x 1010 N/m2.
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Calculation
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Y = (F/A) / ( ΔL / L)
F = W = mg =16kg 9.8m/s2 = 157 N
R = .1cm(1m/100cm)/2 = .0005m
A =  R2 =  (.0005m)2 = 7.85 x 10-7m2
( ΔL / L) = (F/A) / Y = (F/  R2 ) / Y
(ΔL/L) = (157N/ 7.85 x 10-7m2) / 20x1010N/m2
(ΔL/L) = .001 = 0.1 %
NOTE: The change in length is inversely
proportional to the SQUARE of the radius!
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Shear Modulus
• To distort a rectangular solid, apply a
force on one of its faces in a direction
parallel to the face.
• Simultaneously, you must also apply a
force on the
opposite face in the
opposite direction.
• The diagram on the
screen shows how you
can distort an object by
applying force.
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Shear
• The diagram on the screen shows that the
amount of distortion can be measured by the new
angle Ø , which the faces make.
• Shearing stress is defined as F/A, where F is the
tangential force and A is the area of the surface
that the force acts on.
• Shearing strain is the angle Ø .
• Similar to linear distortions, shearing
strain is directly proportional to
shearing stress.
• The constant of proportionality is called
the shear modulus (S):
• S = (F/A) / Φ
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Shear
• Shear modulus has the same units as
Young’s modulus, N/m2.
• For many substances, shear modulus is
approximately one-third of the Young’s
modulus.
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Bulk Modulus
• You can also distort a rectangular solid by
applying forces perpendicular to its
surfaces.
• This stress causes the solid to become
smaller and is known as volume stress or
pressure.
• Volume stress is defined as F/A, where F
is the force perpendicular to the surfaces,
and A is the
area of the surface.
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Bulk Modulus
• Volume strain is the change in volume
divided by the original volume.
• Similar to shearing stress and stress,
strain is directly proportional to volume
stress.
• The constant of proportionality is called
bulk modulus (B) and is defined as:
• B = (F/A) / ( ΔV / V)
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Bulk Modulus
• Solids are usually surrounded by air that
exerts a compressing force on their
surface.
• Pressure = Force/Area
• Therefore, when you apply a force F, you
increase this force, or more precisely, the
volume stress.
• Bulk modulus is usually written as:
• B = (F/A) / ( ΔV / V)
• B = Δp / ( ΔV / V)
• Δp is the increase in pressure above
normal air pressure
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Table
1.0 x 10 9
4.5 x 109
26 x 109
2.2 x 109
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Compare Bulk Modulus
• NOTE:
• B = Δp / ( ΔV / V)
• For a gas, it takes the smallest pressure to
create a change in volume, so gasses
have the smallest bulk modulus
• Solids require a big pressure to have a
change in volume, so they have the
biggest bulk modulus
• Bigger modulus implies a stronger
material
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Practice
• By how much must you change the
pressure on a liter of water to
compress it by 0.10%?
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Calculation
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B = Δp / ( ΔV / V)
B = 2.2 x 109N/m2
( ΔV / V) = .1% (1/100%) = .001
Δp = B ( ΔV / V)
Δp = 2.2 x 109 N/m2 .001 = 2.2 x 106 N/m2
Because there is always air pressure, this
is the INCREASE in pressure to cause the
change in volume
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Pressure in a Fluid
• The molecules in a solid are tightly bound.
• In a liquid or a gas, however, the
molecules are in motion and free to move.
• The diagram shows a cubical container
filled with gas.
• The molecules of the gas are
represented by the red dots.
• The blue arrows show the
direction they are moving in.
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Pressure
• When a gas molecule with momentum p
hits the wall and bounces back, it exerts
an impulse equal to 2p on the wall.
• To calculate the force exerted by a single
molecule, you need to know how long the
collision lasted.
• The effect of a single molecule is very
small, but the effect of a room full, like in a
tornado, can be huge.
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Measuring Pressure
• The diagram shows a device
that measures the pressure of a
gas.
• Because gas molecules can
move freely, the pressure of a
gas will be the same no matter
where you place the gauge.
• Furthermore, any changes in
pressure applied to the
container will be transmitted
throughout the gas volume by
the moving molecules.
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Pascal’s Principle
• Because liquid molecules are also free to
move, the same principles also apply to
liquids and thus all fluids.
• Pascal’s principle states this effect:
• Pressure applied to an enclosed fluid is
transmitted undiminished to every point in
the fluid and to the walls of the container.
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Atmospheric Pressure
• When a gas is in a container, it exerts uniform
pressure throughout the container.
• In the case of atmosphere, however, the air stays
at the Earth’s surface because of the force of
gravity.
• Air pressure is greatest at sea-level and
decreases as the altitude increases.
• The pressure of our atmosphere at sea-level is:
• This amount of pressure is defined as 1
atmosphere (atm).
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Density
• A liquid is much more dense than a gas.
• As a result, the effect of gravity on a liquid
is much greater than that on a gas in the
same sized container on the earth.
• In ordinary sized containers, a gas exerts
uniform pressure throughout the
container it is in, but a liquid does not.
• To refresh your memory, density is the
mass of an object or system of particles
divided by the volume it occupies.
• ρ=m/V
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Depth and Pressure in a Liquid
• The diagram explains the relationship
between pressure and depth in a liquid.
• It shows a container of water with an
imaginary rectangular column of water.
• The column of water has a surface area A,
and a weight mg.
• Hence, a person holding the column of
water experiences a force mg, and a
pressure mg/A, which is the pressure of
the water at the bottom of the container.
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Demonstration
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Pressure
• You can derive the equation for pressure
using the relationship between mass and
density:
• ρ = m/V => m = ρ V
• V = A h => m = ρ Ah
• p = F/A = mg/A = ρ Ah g / A
• p=ρgh
• The total pressure of the liquid at the bottom
of the container is:
• p = p0 + ρ g h where p0 = normal air pressure
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Density
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Practice
• What is the total pressure exerted on
the back of a scuba diver in a lake at
a depth of 8.00 m?
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Total Pressure
• Pressing down on the divers back is all
the water above him, plus all the AIR
above the water.
• pwater = ρ g h = 1000kg/m3 x 9.8m/s2 x 8m
• pwater = 7.84 x 104 N/m2 = 7.84 x 104 Pa
• pAir = 1 Atmosphere = 101kPa
• Ptotal = 7.84 x 104 Pa + 101kPa
• Ptotal = 1.79 x 105 Pa
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Barometers
• A barometer is a device
that measures
atmospheric pressure.
• To make a barometer at
home, turn a glass
upside down under
water.
• When you lift it straight
up, it will remain filled
with water.
• Air pressure holds
the water up
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Barametric Pressure
• At the surface of the Earth, the mercury in a
barometer rises to a height of 760 mm.
• This means atmospheric pressure is:
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• The density of mercury is taken from the density
table shown previously.
• Here we have introduced another unit of pressure
called the torr, named after a famous scientist,
and representing the pressure corresponding to
1 mm of mercury.
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Pascal’s Principle
• The pressure of a gas in a container
is the same at every point in the gas
and on the walls of the container.
• Pressure in a liquid varies with depth
• The diagram on the screen shows a
piston that exerts a force F over an
area A on a body of water.
• This creates a pressure p that can be
experienced throughout the fluid.
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Pressure
• Total pressure is
the sum of the
pressure from
the weight of
the fluid AND
the added
pressure from
the piston
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Hydraulic Lift
• Two pistons are connected to a hydraulic lift.
• When you press the input piston, the output
piston rises.
• The pressure on the input piston is the same as
that on the output piston.
• Keep in mind that the force is given by the
pressure times the area. F = P A
• The output piston has a much larger area than the
input piston.
• Therefore, the output piston exerts a much larger
force than the input piston, and you can easily lift
an automobile. F = P A
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Lift a Car
• p1 = p2
• F1/A1 = F0/A0
• F0 = F1 (A0 /A1 )
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Practice
• The input and lift (output) pistons of a
garage lift have diameters of 10 cm and 30
cm, respectively. The lift raises a car with
a weight of 1.4 x 104 N.
• (a) What is the force on the input piston?
• (b) What is the pressure to the input
piston?
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Calculations
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p1 = p2
F1/A1 = F0/A0
R1 = 10cm (1m/100cm)/2 = .05m
R2 = 30cm (1m/100cm)/2 = .15m
A1 =  R12 = (.05m)2 = 7.85x10-3m2
A2 =  R22 = (.15m)2 = 7.07x10-2m2
F1 = F0(A1/A0)
F1 = 1.4 x 104 N (7.85x10-3m2/ 7.07x10-2m2)
F1 =1,550N
p1 = F1/A1 =1,554N / 7.85x10-3m2= 2.0x105N/m2
p2 = F2/A2 = 1.4 x 104 N / 7.07x10-2m2
p2 = 2.0x105N/m2 = checks!
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Thought Experiment
• If the output piston of a hydraulic lift
has a very large area, a two-year-old
can lift the Empire State Building.
• Does this make sense?
• Is energy conserved?
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Results
• Yes, it does make sense.
• Simple machines, such as levers,
transform small input forces into large
output forces.
• Energy is the ability to do work.
• It is the product of force and distance.
• Machines transform small forces applied
over large distances to large forces
applied over small distances.
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Practice
• A 60-kg athlete does a single-hand
handstand.
• If the area of the hand in contact with
the floor is 100 cm2, what pressure is
exerted on the floor?
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Calculation
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p = F/A
A = 100 cm2 (1m/100cm)2 = 1.00x10-2m2
p = mg/A = 60kg 9.8m/s2/ 1.00x10-2m2
p = 5.88x104N/m2
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Practice
• An oak barrel with a lid of area 0.20 m2 is filled
with water. A long, thin tube of cross-sectional
area 5.0 X 10-5 m2 is inserted into a hole at the
center of the lid, and water is poured into the
tube.
• When the water reaches 12 m high,
the barrel bursts.
• What was the weight of the water
in the tube?
• What was the pressure of the water
on the lid of the barrel?
• What was the net force on the lid
due to the water pressure?
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Calculations
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W = m g = ρ V g = ρ Ah g
A = 5.0 X 10-5 m2 , h = 12 m
W=1000kg/m3 5.0 X 10-5 m2 12 m 9.8m/s2
W = 5.88 N
p = F/A = 5.88 N / 5.0 X 10-5 m2
p = 1.18x105N/m2
OR
p =ρgh= 1000kg/m3 9.8m/s2 12m = 1.18x105N/m2
Checks!
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Calculations
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p=F/A
F=pA
F = 1.18x105N/m2 0.20 m2
F = 2.36 x 104N (More than 5000 pounds!)
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Archimedes
• Have you ever yelled, 'Eureka!' when you figured
out how to solve a challenging problem?
• Well the story goes that when Archimedes figured
out the principle named after him, he was so
excited that he not only yelled, 'Eureka,' but also
went running through town from the public bath
… naked!
• You may not get that excited over Archimedes
principle, but you will learn some useful and
interesting things about fluid buoyancy and fluid
flow in this lesson.
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Archimedes' Principle
• The upward force experienced by an
object when it is immersed in a liquid or a
gas is called buoyant force.
• Buoyant force is equal in magnitude to the
weight of the volume of fluid displaced.
• This rule is called Archimedes' principle.
• BF = mg = ρ V g
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Archimedes’ Principle
• The diagrams on the screen illustrate
Archimedes’ principle.
• Part (a) shows buoyant force as the force
that pushes up when you hold a piece of
wood under water.
• Part (b) shows
buoyant force
as the force that
reduces the
weight of an
object weighing
10 N to 8 N.
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Practice
• A spherical helium balloon has a radius of
30 cm. What is the buoyant force acting on
it in air?
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Calculations
• Find the volume if R = 30 cm = .3 m
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V = (4/3)  R3 = .113 m3
BF = mg = ρ V g = 1.29kg/m3 .113 m3 9.8m/s2
BF = 1.43 N (about 1/3 of a pound)
If the balloon and the air in the balloon has
weight less then 1.43 N, it will float: if more,
it will sink
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Practice
• The diagram shows a partially
submerged iceberg at rest.
• It is in equilibrium because its
buoyant force is equal to the
weight of the iceberg.
• Suppose a uniform solid cube of
material 10 cm on each side has
a mass of 700 g. Will the cube
float? If it will float, how much of
the cube will be submerged?
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Calculations
• Suppose the cube were completely
SUBMERGED!
• LSide = 10 cm = .1m
• V = LWH = (.1m)(.1m)(.1m) = 10-3 m3
• BF = ρ V g = 1x103Kg/m3 10-3 m3 9.8m/s2
• BF = 9.8N
• Weight of the object = mg
• W = 700 g (1kg/1000g) 9.8m/s2 = 6.86 N
• The buoyant force is greater than the weight!
• The object will be pushed up to the surface and
will rise above the surface until the weight of the
fluid displaced just balances the weight of the
object
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Floating Object
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V = LWH = (.1m)(.1m)(.1m) = 10-3 m3
W = 700 g (1kg/1000g) 9.8m/s2 = 6.86 N
BF = 6.86 N = ρ V g = 1x103Kg/m3 V 9.8m/s2
V = 6.86 N / (1x103Kg/m3 9.8m/s2 )
V = 7 x 10-4 m3
7 x 10-4 m3 is the volume of the cube below the
surface
• The remainder:
• 10-3 m3 - 7 x 10-4 m3 = 3 x 10-4 m3 is above the
surface
• It is 70% submerged
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Ideal Fluid Flow
• To mathematically describe fluid flow like
we have projectile motion, for example
can be very difficult in real cases.
• Consider the upper portion of the rising
smoke – it looks almost random!
• In order to obtain a basic description of
fluid flow, we will make some
simplifications and consider ideal fluid
flow, so the flow looks more like the lower
part of the rising smoke stream.
• The conditions for an ideal fluid are that
its flow is:
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Ideal Fluid Flow
1. Steady: The flow is not turbulent; particles
in the fluid do not collide, and their paths
do not cross. The flow looks smooth.
2. Irrotational: There is no rotational motion in small
volumes of the fluid, so there are no whirlpools. A
paddle wheel completely
embedded in the
stream does not rotate.
3. Nonviscous: Viscosity is negligible. This means
the fluid flows easily, without significant friction
or resistance to the flow.
4. Incompressible: The density is constant
throughout the fluid.
• When a fluid's flow has these four characteristics,
we say it is an ideal fluid. Unless specified
otherwise, we will consider these four conditions
to be true in the fluids we consider.
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Equation of Continuity
• When a liquid flows through a tube, the
amount of liquid entering the tube must
equal the amount of liquid coming out of
it.
• If the cross-sectional area of the tube
varies across its length, then the speed of
the liquid must vary too.
• The diagram on the screen shows what
happens to the speed of
water when the nozzle has
a smaller cross-sectional
area than the tube.
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Equation of continuity
• If the liquid is incompressible, the density of the
liquid remains the same throughout the tube. If
the liquid is incompressible, the density of the
liquid remains the same throughout the tube.
• Hence, as the above equation simplifies to:
• Av = constant
• Area x velocity = constant
• Volume/time = constant
• Equation of continuity is an equation that
describes the fact that the amount of fluid
entering a tube is equal to the amount of fluid
leaving the tube.
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Equation of Continuity
• High cholesterol in the blood can
cause fatty deposits called plaques
to form on the walls of blood vessels.
• Suppose plaque reduces the
effective radius of an artery by 25%.
• Because the area is smaller, what
must happen to the velocity?
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Bernoulli’s Equation
• Bernoulli's equation states that when a liquid is
moving fast, its pressure is reduced.
• The diagram below shows a tube of varying
cross-sectional areas.
• The pressure indicators show that the pressure is
lower in the middle region, where the smaller
cross sectional area results in the flow rate being
greater.
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Bernoulli's Equation
• To observe Bernoulli's equation, conduct the
following experiment.
• Hold two corners of a sheet of paper with both
hands.
• Rotate your hands so that the paper takes the
shape of the wing of an airplane.
• Blow over the top of the paper.
• You will observe that the sheet of paper rises.
• It lifts up because air flows on one side of the
paper, but is stationary on the other.
• This is the principle behind the lifting of air
planes: the orientation and shape of the wing of
an airplane causes the air flowing over it to flow
faster than the air under it.
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Wing Design
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Surface Tension
• When there is a boundary between a liquid and a
gas, some remarkable phenomena occur.
• The figures on the screen show two water/air
phenomena and one soap solution/air boundary
phenomena.
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Surface Tension
• 1) An insect is able to walk on water.
2) Water in air forms circular droplets.
3) Soap solutions form bubbles.
• In the phenomenon where bubbles are
made, there are two surfaces: the inside
and the outside of the bubble.
• Although the bubble is thin compared to a
piece of paper, it is very thick compared to
the size of a molecule.
• The surface in an air-liquid boundary is
only two or three molecules thick.
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Surface Tension
• The forces of attraction between the molecules of
a liquid are lower than those of a solid.
• All the same, they do exist because otherwise the
molecules of a liquid would escape into the
atmosphere.
• In a liquid, a molecule, which is at a distance from
the surface, is surrounded by other molecules.
• A molecule on the surface, however, is not
entirely surrounded by molecules.
• The molecules on the surface are held together
by a horizontal force between them, which is
called surface tension.
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Viscosity
• You must have noticed that in January,
molasses flows slower than water or
kerosene.
• This happens because of viscosity.
• Viscosity is a fluid’s internal resistance to
flow.
• It can be measured with a device called a
viscosimeter.
• There are various types of viscosimeters.
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Flow Rate
• When fluid is flowing in a pipe, as a result
of viscosity, the speed of the fluid near the
surface of the pipe is less than that at the
center.
• The flow rate is the average volume of a
fluid that flows beyond a given point
during a time interval
• The diagram on the screen shows that is
not easy to calculate the
average flow rate.
• Flow rate has the
unit of m3/s.
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Summary
• What are the elastic modulie?
• What is the relationship between
pressure and depth in a fluid?
• What is Archimedes’ principle?
• What are the essential principles of
fluid flow?
• What causes surface tension and
what is viscosity?