Physics Beyond 2000 Chapter 7 Properties of Matter.

Physics Beyond 2000

Chapter 7

Properties of Matter

States of matter

• Solid state

• Liquid state

• Gas state (will be studied in Chapter 8.)

Points of view

• Macroscopic: Discuss the relation among physics quantities.

• Microscopic: All matters consist of particles. The motions of these particles are studied. Statistics are used to study the properties.

Solids• Extension and compression (deformation) of so

lid objects

– elasticity .

• Hooke’s law for springs

– The deformation e of a spring is proportional to the force F acting on it, provided the deformation is small.

F = k.e where k is the force constant of the spring

Hooke’s law for springs

e1

e2

F1

F2

Natural length

Extension

Compression

Force F

Deformation e

extension

compression0

F = k.e

http://www.phy.ntnu.edu.tw/~hwang/springForce/springForce.html

http://webphysics.davidson.edu/Applets/animator4/demo_hook.html

http://www.phy.ntnu.edu.tw/~hwang/springForce/springForce.html

http://webphysics.davidson.edu/Applets/animator4/demo_hook.html

Hooke’s law for springs

The slope of the graph represents the stiffness of the spring.A spring with large slope is stiff.A spring with small slopeis soft.

Force F

Deformation e

extension

compression0

F = k.e

Energy stored in a spring

• It is elastic potential energy.

• It is equal to the work done W by the external force F to extend (or compress) the spring by a deformation e.

2

0 2

1keFdxU

e

e

Energy stored in a spring

• It is also given by the area under the F-e graph.

2

2

1.

2

1keeFU e

extension

F

e0

Force

Example 1

• Find the extension of a spring from energy changes.

• Find the extension of the spring by using Hooke’s law.

Example 2

• A spring-loaded rifle

Example 2

• A spring-loaded rifle

More than springs

• All solid objects follow Hooke’s law provided the deformation is not too large.

• The extension depends on– the nature of the material– the stretching force– the cross-sectional area of the sample– the original length

stress

• Stress is the force F on unit cross-sectional area A.

A

F

FA

Unit: Pa

Stress is a measure of the cause of a deformation.Note that A is the cross-sectional area of the wirebefore any stress is applied.

strain

• Strain is the extension e per unit length.

• If is the natural length of the wire,

then

e

e

F

Strain expresses the effect of the strain on the wire.

Example 3

• Find the stress and the strain of a wire.

A

F

e

stress strain

Young modulus E

• Young modulus E is the ratio of the tensile stress σ applied to a body to the tensile strain ε produced.

Ae

FE

.

.

Unit: Pa

Young modulus E

Ae

FE

.

.

The value of E is dependent on the material.

Young modulus E and force constant k

Ae

FE

.

.

and F = k.e

AE

k.

So k depends on E (the material), A (thethickness) and (the length).

Example 4

• Find the Young modulus.

Experiment to find Young modulus

Suspend two long thin wire as shown.The reference wire can compensate for the temperature effect.The vernier scale is tomeasure the extensionof the sample wire.

referencewire

sample wire

weight

levelmeter

vernierscale


Adjust the weight so that vernier scale to read zero.Measure the diameter of the sample wire andcalculate its cross-section area A.

referencewire

sample wire

weight

vernierscale

The bubble in in themiddle.

Level meter

zero

Diameter ofthe wire


Measure the length of the sample wire.

referencewire

sample wire

weight

vernierscale


referencewire

sample wire

weight

vernierscale

Add weight W to the sample wire and measure its extension e .The force on the wire isF = W = mg.

F = W = mg where m is the added mass.

More weights

More weights

The bubble movesto the left

It is because the sample wire, whichis on the right, extends.

This endis higher.

This endis lower.

Turn thisscrew (vernierscale) toraise upthe end ofthe levelmetersuspended bythe samplewire.

This end ofthe level meteris suspendedby the sample wire.

The bubble in in themiddle again.

The readingon the screwshows theextensionof the samplewire.


referencewire

sample wire

weight

vernierscale

F

Plot the graph of stress σagainst strainε.

σσ

ε

elastic limit

0


referencewire

sample wire

weight

vernierscale

F

What is the slope of thisgraph?

σ

ε

elastic limit

0

Young modulus


σ

ε

elastic limit

0

The linear portion of the graph gives Hooke’s law.The stress applied to any solid is proportional tothe strain it produces for small strain.

The stress-strain curve

permanentstrain

stress σ

strain εO

AL

B

C

D

A: proportional limitL: elastic limitB: yield pointC: breaking stressD: breaking point


permanentstrain

stress σ

strain εO

AL

B

C

D

A: proportional limit Between OA, the stressis proportional to the strain.Point A is the limit of thisproportionality.


permanentstrain

stress σ

strain εO

AL

B

C

D

L: elastic limitBetween AL, the strain can be back to zero when the stress isremoved.i.e. the wireis still elastic.Usually the elastic limit coincides withthe proportional limit.


permanentstrain

stress σ

strain εO

AL

B

C

D

B: yield pointBetween LB, the wirehas a permanent deformation whenthe stress is removed.i.e. the wire is plastic.At point B, there isa sudden increase ofstrain a small increasein stress.


permanentstrain

stress σ

strain εO

AL

B

C

D

C: breaking stressThis is the maximumstress.Beyond this point,the wire extendsand narrows quickly,causing a constrictionof the cross-sectionalarea.


permanentstrain

stress σ

strain εO

AL

B

C

D

D: breaking pointThe wire breaksat this point.This is the maximumstrain of the wire.

Example 5

• Refer to table 7.1 on p.112.

Energy stored in the extended wire

stress

strain

σ

ε

The area under the stress-strain graph = A

Fe21

2

1

where Fe is the elastic potential energy and A is the volume of the wire.

2

1

Properties of materials

• Stiffness

• Strength

• Ductility

• Toughness

Stiffness

• It indicates how the material opposes to deformation.

• Young modulus is a measure of the stiffness of a material.

• A material is stiff if its Young modulus is large.

• A material is soft if its Young modulus is small.

Strength

• It indicates how large the stress the material can stand before breaking.

• The breaking stress is a measure of the strength of the material.

• A material is strong if it needs a large stress to break it.

• A material is weak if a small stress can break it.

Ductility

• It indicates how the material can become a wire or a thin sheet.

• A ductile material enters its plastic stage with a small stress.

ε

Toughness

• A tough material is one which does not crack readily.

• The opposite is a brittle material.

• A brittle material breaks over a very short time without plastic deformation.

Graphical representation

stress σ

strain ε

stiffest

weakest

strongest

most flexible

Graphical representation

stress σ

strain ε

brittle

tough

ductile

Graphical representation for various materials

stress σ

strain ε

glass

metal

rubber

Elastic deformation and plastic deformation

• In elastic deformation,

the object will be back to its original shape when the stress is removed.

• In plastic deformation, there is a permanent strain when the stress is removed.

Plastic deformation

stress σ

0 strain εpermanentstrain

loading

unloading

elastic limit

Fatigue

• Metal fatigue is a cumulative effect causing a metal to fracture after repeated applications of stress, none of which exceeds the breaking stress.

Creep

• Creep is a gradual elongation of a metal under a constant stress which is well below its yield point.

Plastic deformation of glass

• Glass does not have any plastic deformation.

• When the applied stress is too large, the glass has brittle fracture.

Plastic deformation of rubber• Deformation of rubber would produce

internal energy.• The area in the loop represents the internal

energy produced per unit volume.

stress σ

strain ε

loading

unloading

Hysteresis loop

Model of a solid

• Microscopic point of view• A solid is made up of a large number of identical

hard spheres (molecules).• The molecules are attracted to each other by a

large force.• The molecules are packed closely in an orderly

way.• There are also repulsion to stop the molecules

penetrating into each other.

Structure of solid

• Crystalline solid: The molecules have regular arrangement. e.g. metal.

• Amorphous solid: The molecules are packed disorderly together. e.g. glass.

Elastic and plastic deformation of metal

• Metal has a structure of layers.

• Layers can slide over each other under an external force.

layerlayer


• When the force is small, the layer displaces slightly.

Force


• When the force is removed, the layer moves back to its initial position.

• The metal is elastic.


• When the force is large, the layer moves a large displacement.

Force


• When the force is removed, the layer settles down at a new position.

• The metal has a plastic deformation.

New structure Initial structure

Intermolecular forces

• The forces are basically electrostatic in nature.

• The attractive force results from the electrons of one molecule and the protons of an adjacent molecule.

• The attractive force increases as their separation decreases.



• When the molecules are too close, their outer electrons repel each other. This repulsive force prevents the molecules from penetrating each other.



• Normally the molecules in a solid have a balance of the attractive and repulsive forces.

• At the equilibrium position, the net intermolecular force on the molecule is zero.

Intermolecular separation r

• It is the separation between the centres of two adjacent molecules.

ro

ro is the equilibriumdistance.r = ro

The force on each moleculeis zero.



r > ro

The force on the moleculeis attractive.

ro

r



r < ro

The force on the moleculeis repulsive.

ro

r

Intermolecular forcesIntermolecularforce

0

repulsive

attractive

rro

ro is the equilibrium separation

The dark line is the resultant curve.

Intermolecular separation

• Suppose that a solid consists of N molecules with average separation r.

• The volume of the solid is V.

• What is the relation among these quantities?

33.N

VrVrN

Intermolecular separation

• Example 6.

• Mass = density × volume

• The separation of molecules in solid and liquid is of order 10-10 m.

Intermolecular potential energy

Intermolecularforce

0rro


Potential energy

-ε

The potential energyis zero for large separation.

The potential energy is a minimumat the equilibrium separation.


Intermolecularforce

0rro


Potential energy

-ε

When they move towardseach other from far away, the potential energy decreases because there is attractive force.The work done by external force is negative.



Intermolecularforce

0r

ro


Potential energy

-ε

When they are further towards each other after the equilibrium position, the potential energy increasesbecause there is repulsive force.The work done by external force is positive.


Force and Potential Energy

• U = potential energy

• F = external force

r

FdrU anddx

dUF

Variation of molecules• If the displacement of two neighbouring mo

lecules is small, the portion of force-separation is a straight line with negative slope.

attractive

Intermolecularforce

0

repulsiver

ro

F

r

ro

Variation of molecules

• The intermolecular force is

F = -k. Δr

where k is the force constant between molecules

and Δr is the displacement

from the equilibrium position.

F

r

roSo the molecule is insimple harmonic motion.


F

r

ro

So the molecule is insimple harmonic motion.with ω2 =

where m is the massof each molecule.

m

k


• However this is only a highly simplified model.

• Each molecule is under more than one force from neighbouring molecules.

The three phases of matter

• Solid, liquid and gas states.

• In solid and liquid states, the average separation between molecules is close to ro.

Intermolecularforce

0rro

Potential energy

-ε

The three phases of matter

• Solid, liquid and gas states.

• In gas state, the average separation between molecules is much longer than ro.

Intermolecularforce

0rro

Potential energy

-ε

Elastic interaction of molecules

• All the interactions between molecules in any state are elastic. i.e. no energy loss on collision between molecules.

Solids

Intermolecularforce

0rro

Potential energy

-ε

When energy is supplied to a solid, the moleculesvibrate with greater amplitude until melting occurs.

Solids

Intermolecularforce

0rro

Potential energy

-ε

On melting, the energy is used to break the lattice structure.

Liquids

• Molecules of liquid move underneath the surface of liquid.

• When energy is supplied to a liquid, the molecules gain kinetic energy and move faster. The temperature increases.

Liquids

• At the temperature of vaporization (boiling point), energy supplied is used to do work against the intermolecular attraction.

• The molecules gain potential energy. The state changes.

• The temperature does not change.

Gases

Intermolecularforce

0rro

Potential energy

-ε

Molecules are moving at very high speed inrandom direction.

Gases

Intermolecularforce

0rro

Potential energy

-ε

The average separation between moleculesis much longer than ro

Gases

Intermolecularforce

0rro

Potential energy

-ε

The intermolecular force is so small that it isinsignificant.

Example 7

• There are 6.02 1023 molecules for one mole of substance.

• The is the Avogadro’s number.

Example 8

• The separation between molecules depend on the volume.

Thermal expansion

Potential energy

0rro

-ε

In a solid, molecules are vibrating about their equilibrium position.

Thermal expansionSuppose a molecule is vibrating betweenpositions A and B about the equilibrium position.

Potential energy

0rro

-ε

A B

Thermal expansionNote that the maximum displacement from the equilibrium position is not the same on each side because the energy curve is not symmetrical about the equilibrium position.

Potential energy

0rro

-ε

A B

A’ B’

C’

Thermal expansionThe potential energy of the molecule variesalong the curve A’C’B’ while the molecule is oscillating along AB.

Potential energy

0rro

-ε

A B

A’ B’

C’

Thermal expansionThe centre of oscillation M is mid-wayfrom the positions A and B. So point Mis slightly away from the equilibrium position.

Potential energy

0rro

-ε

A B

A’ B’

C’

M

Thermal expansionWhen a solid is heated up, it gains more potential energy and the points A’ and B’ move up the energy curve. The amplitude of oscillation is also larger.

Potential energy

0rro

-ε

A B

A’ B’

C’

M

Thermal expansionThe molecule is vibrating with largeramplitude between new positions AB.

Potential energy

0rro

-ε

A B

A’ B’

C’

M

Thermal expansionThe centre of oscillation M , which is the mid-pointof AB, is further away from the equilibrium position.

Potential energy

0rro

-ε

A B

A’ B’

C’

M

Thermal expansionAs a result, the average separation between moleculesincreases by heating. The solid expands on heating.

Potential energy

0rro

-ε

A B

A’ B’

C’

M

Absolute zero temperatureAt absolute zero, the molecule does not vibrate. Theseparation between molecules is ro. The potential energy of the molecule is a minimum.

Potential energy

0rro

-ε C’

Young Modulus in microscopic point of view

• Consider a wire made up of layers of closely packed molecules.• When there is not any stress, the separation between two neighbouring layer is ro.

• ro is also the diameter of each molecule.

ro

wire

Young Modulus in microscopic point of vies

• The cross-sectional area of the wire is

ro

2. orNA where N is the number ofmolecules in each layer

Aarea of one molecule= 2

or

ro

ro


• When there is not an external force F, the separation between two neighbouring layer increases by r.

ro+ r

F


• The strain is

ro+ r

F

or

r


• Since the restoring force between two molecules in the neighbouring layer is directly proportional to N and r, we have F = N.k.r where k is the force constant between two molecules.

ro+ r

F


ro+ r

F

2oNrA and F = N.k.r

or

k

A

Fstress


ro+ r

F

Thus, the Young modulus is

or

kE

Example 9

• Find the force constant k between the molecules.

Density

• Definition: It is the mass of a substance per unit volume.

V

m

where m is the mass and V is the volumeUnit: kg m-3

Measure the density of liquid

• Use hydrometer

upthrust

weight

Pressure

• Definition: The pressure on a point is the force per unit area on a very small area around the point.

A

FP or

A

FP

A

0

lim

Unit: N m-2 or Pa.

Pressure in liquid

• Pressure at a point inside a liquid acts equally in all directions.

• The pressure increases with depth.

Find the pressure inside a liquid = density of the liquid

• h = depth of the point Xsurface ofliquid

X

h

Find the pressure inside a liquid

• Consider a small horizontal area A around point X.

X

h

surface ofliquid

A

Find the pressure inside a liquid• The force from the liquid on this area is the weight

W of the liquid cylinder above this area

X

h

surface ofliquid

A

W

Find the pressure inside a liquid• W = ?

X

h

surface ofliquid

A

W

W = hAg


X

h

surface ofliquid

A

W

W = hAg and P =

g hA

WP

A

W


surface ofliquid

As there is also atmospheric pressure Po on the liquidsurface, the total pressure at X is

Po

X

h

A

P

ghPP o

Example 10

• The hydraulic pressure.

Force on a block in liquid

L

h1

h2

P1

P2

A

Consider a cylinder of area A and height L ina liquid of density .


L

h1

h2

P1

P2

A

The pressure on its top area is P1 = h1g + Po

The pressure on its bottom area is P2 = h2g + Po


L

h1

h2

P1

P2

A

The pressure difference P = P2 – P1 = Lg with upward direction.

Force on a block in liquidSo there is an upward net force F = P.A = Vg where V is the volume of the cylinder.

L

h1

h2

A

F

Force on a block in liquidThis is the upthrust on the cylinder.Upthrust = Vg

h1

h2

F

V

Force on a block in liquidUpthrust = Vg Note that it is also equal to the weight of theliquid with volume V.

h1

h2

F

V

Force on a block in liquidThe conclusion: If a solid is immersed in a liquid, the upthrust on the solid is equal to the weight of liquid that the solid displaces.

h1

h2

F

V

Force on a block in liquidThe conclusion is correct for a solid in liquid and gas (fluid).

h1

h2

F

V

Archimedes’ Principle

• When an object is wholly or partially immersed in a fluid, the upthrust on the object is equal to the weight of the fluid displaced.

upthrust upthrust

Measuring upthrustspring-balance

object

liquid

compressionbalance

The reading of thespring-balance is W,which is the weightof the object.

The reading of the compressionbalance is B, which is the weight of liquidand beaker.

beaker

W

B


object

liquid

Carefully immersehalf the volume ofthe object in liquid.

What would happen tothe reading of thespring-balance and thatof the compression balance?

beaker

compressionbalance


object

liquid

The reading of the spring-balance decreases.Why?

The difference in the readingsof the spring-balance givesthe upthrust on the object.

beaker

compressionbalance


object

liquid

The reading of the compression balance increases.Why?

The difference in the readingsof the compression balance gives the upthrust on the object.

beaker

compressionbalance


object

liquid

Carefully immersethe whole object in liquid.


beaker

compressionbalance


object

liquid

Carefully placethe object on thebottom of the beaker.


beaker

compressionbalance

Law of floatation

• A floating object displaces its own weight of the fluid in which it floats.

weight upthrust

weight of the object= upthrust= weight of fluid displaced

float or sink?‘ = density of the object = density of the fluid

• If ‘ > , then the object sinks in the fluid.

• If ‘ < , then the object floats in the fluid.

density is larger than density is smaller than

Manometer• A manometer can measure the pressure difference

of fluid.• Note that the pressure on the same level in the

liquid must be the same.

liquid ofdensity

connectto the fluid X Y Same level

Manometer

liquid ofdensity

Po = atmospheric pressure

Po+P= fluid pressure

h = difference in height

X Y

Manometer

liquid ofdensity


Po+P= fluid pressure h = difference

in heightBA

The pressures at pointsA and B are equal.

Manometer

liquid ofdensity



in heightBA

The pressure at A = Po+PThe pressure at B = Po + hg

Manometer

liquid ofdensity



in heightBA

The pressure difference of the fluid P = hg

Liquid in a pipe• Consider a pipe of non-uniform cross-sectional

area with movable piston at each end.• The fluid is in static equilibrium.

Same level

X Y

hx = hY

static fluid

Liquid in a pipe

• The manometers show that the pressures at points X and Y are equal.

Same level

X Y

hx = hY

static fluid

Liquid in a pipe

• The pressures at points M and N on the pistons are also equal.

Same level

M N

hx = hY

static fluid

Liquid in a pipe

• There must be equal external pressures on the pistons to keep it in equilibrium.

PM = PN

Same level

M N

hx = hY

static fluid

PM PN

Liquid in a pipe

• As F = P.A , the external forces are different on the two ends.

FM > FN

Same level

M N

hx = hY

static fluid

FM FN

Liquid in a pipe

• Note that the net force on the liquid is still zero to keep it in equilibrium.

• There are forces towards the left from the inclined surface.

Same level

M N

hx = hY

static fluid

FM FN

Fluid Dynamics

• Fluid includes liquid and gas which can flow.

• In this section, we are going to study the force and motion of a fluid.

• Beurnoulli’s equation is the conclusion of this section.

Turbulent flow• Turbulent flow: the fluid flows in irregular

paths.

• We will not study this kind of flow.

Streamlined flow• Streamlined flow (laminar flow) : the fluid

moves in layers without fluctuation or turbulence so that successive particles passing the same point with the same velocity.

Streamlined flow• We draw streamlines to represent the

motion of the fluid particles.

Equation of continuity• Suppose that the fluid is incompressible.

That is its volume does not change. Though the shape (cross-sectional area A) may change.

Equation of continuity

• At the left end, after time t, the volume passing is A1.v1. t

• At the right end, after the same time t, the volume passing is A2.v2. t


• As the volumes are equal for an incompressible fluid,

A1.v1. t = A2.v2. t

A1.v1 = A2.v2


• Example 12

Pressure difference and work done

• Suppose that an incompressible fluid flows from position 1 to position 2 in a tube.

• Position 2 is higher than position 1.

• There is a pressure difference P at the two ends.

h2

P+P

p

x1

x2

Position 1

Position 2

A1

A2

h1


• Work done by the external forces is

(P+P).A1.x1 - P.A2.x2

P+P

P

x1

x2

Position 1

Position 2

A1

A2

h2

h1



(P+P).A1.x1 - P.A2.x2

• A1x1=A2x2=V

= volume of fluid that moves

P+P

P

x1

x2

Position 1

Position 2

A1

A2

h2

h1



(P+P).A1.x1 - P.A2.x2 = P .V• With V =

Work done = Pwhere m is the mass of the fluid

and ρis the density of the fluid

P+P

p

x1

x2

Position 1

Position 2

A1

A2

m

m h2

h1

Bernoulli’s principle

• In time t, the fluid moves x1 at position 1 and x2 at position 2.

• x1 = v1.t and

x2 = v2.tP1

P2

x1

x2

Position 1

Position 2

A1

A2

v1

v2

h2

h1


• In time t, the fluid moves x1 at position 1 and x2 at position 2.

• x1 = v1.t and

x2 = v2.t• Work done by

external pressure =

(P1-P2)P1

P2

x1

x2

Position 1

Position 2

A1

A2

v1

v2

m

h2

h1


• Work done by external pressure =

(P1-P2)

• Increase in kinetic energy =

P1

P2

x1

x2

Position 1

Position 2

A1

A2

v1

v2

m

21

22 2

1

2

1mvmv

h2

h1


• Increase in kinetic energy =

• Increase in gravitatioanl potential energy =

mgh2 – mgh1P1

P2

x1

x2

Position 1

Position 2

A1

A2

v1

v2

h2

h1

21

22 2

1

2

1mvmv


P1

P2

x1

x2

Position 1

Position 2

A1

A2

v1

v2

h2

h1

)()2

1

2

1().( 12

21

2221 mghmghmvmv

mPP

The left hand side is the work done by external pressure. It is also the energy suppliedto the fluid.The right hand side isthe increase in energy of the fluid.


P1

P2

x1

x2

Position 1

Position 2

A1

A2

v1

v2

h2

h1

2222

2111 .

2

1...

2

1.. vhgPvhgP


P1

P2

x1

x2

Position 1

Position 2

A1

A2

v1

v2

h2

h1

2

2

1vghP constant


2222

2111 2

1

2

1vghPvghP

If h1 = h2 , then

222

211 2

1

2

1vPvP

Position 1Position 2


If h1 = h2 , then

222

211 2

1

2

1vPvP


or

2

2

1vP constant


If h1 = h2 , then


2

2

1vP constant

So for horizontal flow, where the speed is high,the pressure is low.

Low speed, high pressureHigh speed, low pressure




Low speed, high pressure High speed, low pressure

h1

h2




P1 = h1..gh1

h2P2 = h2..g

Low speed, high pressure High speed, low pressure

Simple demonstration of Bernoulli’s principle

•Held two paper strips vertically with a smallgap between them.•Blow air gently into the gap.•Explain what you observe.

air

Simple demonstration of Bernoulli’s principle

air

high pressure

high pressure

In the gap, the speedof airflow is high.So the pressure is lowin the gap.The high pressureoutside pressesthe strips together.

Examples of Bernoulli’s effect

• Airfoil: the airplane is flying to the left.

Examples of Bernoulli’s effect• Airfoil: There is a pressure difference

between the top and the bottom of the wing. A net lifting force is produced.

Example 13

• Airfoil and Bernoulli’s effect

• To find the lifting force on an airplane.

Examples of Bernoulli’s effect

• Spinning ball: moving to the left and rotating clockwise.

Examples of Bernoulli’s effect• Spinning ball: the pressure difference

produces a deflection force and the ball moves along a curve.



spinning ball

not spinning



spinning ball

not spinning

Ball floating in air

air

air


air

air

weight of the ball

thrust from theair blower

force due to spinning

What is the direction of spinning of the ball?


air

air

weight of the ball

thrust from theair blower

force due to pressuredifference

It is spinning in clockwise direction.


Force due topressuredifference

Air blown out through a funnel

What would happen to the light ball?

Air blown out through a funnel

It is sucked to the top of the funnel.

weight

Force due topressure difference

Force due topressure difference

Yacht sailing

• A yacht can sail against the wind.

• Note that the sail is curved.

Yacht sailing

• A yacht can sail against the wind.

• The pressure difference produces a net force F.

• A component of F pushes the yacht forward.

Yacht sailing

• The yacht must follow a zig-zag path in order to sail against the wind.

wind

path

Jets

• When a stream of fluid is ejected rapidly out of a jet, air close to the stream would be dragged along and moves at higher speed.

• This results in a low pressure near the stream.

air

air

fluid

low pressure

Jets: Bunsen burner

• The pressure near the jet is low.

• Air outside is pulled into the bunsen burner through the air hole. gas

air

Jets: Paint sprayer

Jets: Filter pump

Jets: Carburretor

Roofs, window and door

• The pressure difference makes the door close.

fast wind,low pressure

door

Example 14

• Strong wind on top of the roof.

tile

Example 14

• Strong wind on top of the roof.

tile

fast wind on top of the tiles (outside the house)

no wind under the tiles (inside the house)

Example 14

tile

fast wind on top of the tiles (outside the house)

no wind under the tiles (inside the house)

high pressure

low pressure

A hole in a water tank

• The speed of water on the surface is almost zero.

• The speed of water at the hole is v.

v

h

Po

Po


• The water pressure on the surface is Po.

• The water pressure at the hole is Po. v

h

Po

Po


• The height of water on the surface is h

• The height of water at the hole is 0. v

h

Po

Po


• Apply Bernoulli’s equation,

v

h

Po

Po

02

10 2

0 vPghPo

ghv 2

This is the same speed of an object falling througha distance h freely.

Example 15

• A hole in a tank

Pitot tube

• Pitot tube is used to measure the speed of fluid.

h

v

statictube

total tube

h1h2

Pitot tube

• The pressure below the static tube is h1g.

• The pressure at the mouth of the total tube is h2g.

h

v

statictube

total tube

h1h2

Pitot tube

• The fluid speed below the static tube is v.• The fluid speed at the mouth of the total tube is 0.

h

v

statictube

total tube

h1h2

Pitot tube

• Apply Bernoulli’s equation,

h

v

statictube

total tube

h1h2

02

12

21 ghvgh

Pitot tube

h

v

statictube

total tube

h1h2

ghhhg

v 2)(2 12

Physics Beyond 2000 Chapter 7 Properties of Matter.

Documents

wire slide

stress strain slide

force slide

spring slide

young modulus slide

strain strain

pa slide

e f strain