Presentation on theme: "Chapter 7 Properties of Matter"— Presentation transcript:
1 Chapter 7 Properties of Matter Physics Beyond 2000Chapter 7Properties of Matter
2 States of matter Solid state Liquid state Gas state (will be studied in Chapter 8.)
3 Points of viewMacroscopic: 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.
4 Solids F = k.e where k is the force constant of the spring Extension and compression (deformation) of solid 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
5 Hooke’s law for springs Force FDeformation eextensioncompressionF = k.ee1e2F1F2Natural lengthExtensionCompression
6 Hooke’s law for springs Force FDeformation eextensioncompressionF = k.eThe slope of the graphrepresents the stiffnessof the spring.A spring with large slopeis stiff.A spring with small slopeis soft.
7 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.
8 Energy stored in a spring It is also given by the area under the F-e graph.extensionFeForce
9 Example 1 Find the extension of a spring from energy changes. Find the extension of the spring by using Hooke’s law.
12 More than springsAll solid objects follow Hooke’s law provided the deformation is not too large.The extension depends onthe nature of the materialthe stretching forcethe cross-sectional area of the samplethe original length
13 stress Stress is the force F on unit cross-sectional area A. Unit: PaStress is a measure of the cause of a deformation.Note that A is the cross-sectional area of the wirebefore any stress is applied.
14 strain Strain is the extension e per unit length. If is the natural length of the wire,theneFStrain expresses the effect of the strain on the wire.
15 Example 3Find the stress and the strain of a wire.strainstress
16 Young modulus EYoung modulus E is the ratio of the tensile stress σ applied to a body to the tensile strain ε produced.Unit: Pa
17 Young modulus EThe value of E is dependent on the material.
18 Young modulus E and force constant k F = k.eSo k depends on E (the material), A (thethickness) and (the length).
20 Experiment to find Young modulus Suspend two long thin wire as shown.The reference wire cancompensate for thetemperature effect.The vernier scale is tomeasure the extensionof the sample wire.samplewirereferencewirelevelmetervernierscaleweight
22 Experiment to find Young modulus Adjust the weight so that vernier scale to read zero.Measure the diameter of the sample wire andcalculate its cross-section area A.samplewirereferencewirevernierscaleweight
26 Experiment to find Young modulus referencewiresampleweightvernierscaleMeasure the length of the sample wire.
27 Experiment to find Young modulus Add weight W to the sample wire and measure its extension e .The force on the wire isF = W = mg.samplewirereferencewirevernierscaleF = W = mg where m is the added mass.weight
38 The stress-strain curve OALBCDA: proportional limit Between OA, the stressis proportional to the strain.Point A is the limit of thisproportionality.permanentstrain
39 The stress-strain curve OALBCDL: elastic limitBetween AL, thestrain can be back tozero when the stress isremoved.i.e. the wireis still elastic.Usually the elasticlimit coincides withthe proportional limit.permanentstrain
40 The stress-strain curve B: yield pointBetween LB, the wirehas a permanentdeformation whenthe stress is removed.i.e. the wire is plastic.At point B, there isa sudden increase ofstrain a small increasein stress.stress σstrain εOALBCDpermanentstrain
41 The stress-strain curve OALBCDC: breaking stressThis is the maximumstress.Beyond this point,the wire extendsand narrows quickly,causing a constrictionof the cross-sectionalarea.permanentstrain
42 The stress-strain curve OALBCDD: breaking pointThe wire breaksat this point.This is the maximumstrain of the wire.permanentstrain
44 Energy stored in the extended wire The area under the stress-strain graph =stresswhere Fe is the elastic potentialenergy andA is the volume of thewire.σεstrain
45 Properties of materials StiffnessStrengthDuctilityToughness
46 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.
47 StrengthIt 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.
48 DuctilityIt indicates how the material can become a wire or a thin sheet.A ductile material enters its plastic stage with a small stress.ε
49 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.
52 Graphical representation for various materials stress σstrain εglassmetalrubber
53 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.
55 FatigueMetal fatigue is a cumulative effect causing a metal to fracture after repeated applications of stress, none of which exceeds the breaking stress.
56 CreepCreep is a gradual elongation of a metal under a constant stress which is well below its yield point.
57 Plastic deformation of glass Glass does not have any plastic deformation.When the applied stress is too large, the glass has brittle fracture.
58 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 σstrain εloadingunloadingHysteresis loop
59 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.
60 Structure of solidCrystalline solid: The molecules have regular arrangement. e.g. metal.Amorphous solid: The molecules are packed disorderly together. e.g. glass.
61 Elastic and plastic deformation of metal Metal has a structure of layers.Layers can slide over each other under an external force.layerlayer
62 Elastic and plastic deformation of metal When the force is small, the layer displaces slightly.Force
63 Elastic and plastic deformation of metal When the force is removed, the layer moves back to its initial position.The metal is elastic.
64 Elastic and plastic deformation of metal When the force is large, the layer moves a large displacement.Force
65 Elastic and plastic deformation of metal When the force is removed, the layer settles down at a new position.The metal has a plastic deformation.New structureInitial structure
66 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.
67 Intermolecular forces The forces are basically electrostatic in nature.When the molecules are too close, their outer electrons repel each other. This repulsive force prevents the molecules from penetrating each other.
68 Intermolecular forces The forces are basically electrostatic in nature.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.
69 Intermolecular separation r It is the separation between the centres of two adjacent molecules.ro is the equilibriumdistance.r = roThe force on each moleculeis zero.ro
70 Intermolecular separation r It is the separation between the centres of two adjacent molecules.rorr > roThe force on the moleculeis attractive.
71 Intermolecular separation r It is the separation between the centres of two adjacent molecules.rorr < roThe force on the moleculeis repulsive.
72 Intermolecular forces ro is the equilibrium separationrepulsiverroattractiveThe dark line is theresultant curve.
73 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?
74 Intermolecular separation Example 6.Mass = density × volumeThe separation of molecules in solid and liquid is of order m.
75 Intermolecular potential energy ro is the equilibrium separationIntermolecularforcerroThe potential energyis zero for large separation.Potential energy-εThe potential energy is a minimumat the equilibrium separation.
76 Intermolecular potential energy ro is the equilibrium separationIntermolecularforcerroWhen they move towardseach other from far away, thepotential energy decreasesbecause there is attractive force.The work done by externalforce is negative.Potential energy-εThe potential energy is a minimumat the equilibrium separation.
77 Intermolecular potential energy ro is the equilibrium separationWhen they are further towardseach other after the equilibriumposition, the potential energy increasesbecause there is repulsive force.The work done by external force ispositive.IntermolecularforcerorPotential energy-εThe potential energy is a minimumat the equilibrium separation.
78 Force and Potential Energy U = potential energyF = external forceand
79 Variation of molecules If the displacement of two neighbouring molecules is small, the portion of force-separation is a straight line with negative slope.FattractiveIntermolecularforcerepulsiverrorro
80 Variation of molecules The intermolecular force isF = -k. Δrwhere k is the force constant between moleculesand Δr is the displacementfrom the equilibrium position.FrroSo the molecule is insimple harmonic motion.
81 Variation of molecules So the molecule is insimple harmonic motion.withω2 =where m is the massof each molecule.Frro
82 Variation of molecules However this is only a highly simplified model.Each molecule is under more than one force from neighbouring molecules.
83 The three phases of matter Solid, liquid and gas states.In solid and liquid states, the average separation between molecules is close to ro.IntermolecularforcerroPotential energy-ε
84 The three phases of matter Solid, liquid and gas states.In gas state, the average separation between molecules is much longer than ro.IntermolecularforcerroPotential energy-ε
85 Elastic interaction of molecules All the interactions between molecules in any state are elastic. i.e. no energy loss on collision between molecules.
86 Solids When energy is supplied to a solid, the molecules vibrate with greater amplitude until melting occurs.IntermolecularforcerroPotential energy-ε
87 Solids On melting, the energy is used to break the lattice structure. IntermolecularforcerroPotential energy-ε
88 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.
89 LiquidsAt 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.
90 Gases Molecules are moving at very high speed in random direction. -ε IntermolecularforcerroPotential energy-ε
91 Gases The average separation between molecules is much longer than ro IntermolecularforcerroPotential energy-ε
92 Gases The intermolecular force is so small that it is insignificant. rroPotential energy-ε
93 Example 7 There are 6.02 1023 molecules for one mole of substance. The is the Avogadro’s number.
94 Example 8The separation between molecules depend on the volume.
95 Thermal expansion In a solid, molecules are vibrating about their equilibrium position.Potential energyrro-ε
96 Thermal expansion Suppose a molecule is vibrating between positions A and B about the equilibriumposition.Potential energyrro-εAB
97 Thermal expansionNote that the maximum displacement from the equilibriumposition is not the same on each side because the energy curveis not symmetrical about the equilibrium position.Potential energyrro-εABA’B’C’
98 Thermal expansion The potential energy of the molecule varies along the curve A’C’B’ while the molecule isoscillating along AB.Potential energyrro-εABA’B’C’
99 Thermal expansion The centre of oscillation M is mid-way from the positions A and B. So point Mis slightly away from the equilibrium position.Potential energyrro-εABA’B’C’M
100 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 energyMABrroA’B’-εC’
101 Thermal expansion The molecule is vibrating with larger amplitude between new positions AB.Potential energyMABrroA’B’-εC’
102 Thermal expansion The centre of oscillation M , which is the mid-point of AB, is further away from the equilibrium position.Potential energyMABrroA’B’-εC’
103 Thermal expansionAs a result, the average separation between moleculesincreases by heating. The solid expands on heating.Potential energyMABrroA’B’-εC’
104 Absolute zero temperature At absolute zero, the molecule does not vibrate. Theseparation between molecules is ro. The potentialenergy of the molecule is a minimum.Potential energyrro-εC’
105 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.rowire
106 Young Modulus in microscopic point of vies The cross-sectional area of the wire iswhere N is the number ofmolecules in each layerarea of one molecule=roroA
107 Young Modulus in microscopic point of vies When there is not an external force F, the separation between two neighbouring layer increases by r.Fro+ r
108 Young Modulus in microscopic point of vies The strain isFro+ r
109 Young Modulus in microscopic point of vies 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.Fro+ r
110 Young Modulus in microscopic point of vies and F = N.k.rFro+ r
111 Young Modulus in microscopic point of vies Thus, the Young modulus isFro+ r
112 Example 9Find the force constant k between the molecules.
113 Density Definition: It is the mass of a substance per unit volume. where m is the massand V is the volumeUnit: kg m-3
114 Measure the density of liquid Use hydrometerupthrustweight
115 PressureDefinition: The pressure on a point is the force per unit area on a very small area around the point.orUnit: N m-2 or Pa.
116 Pressure in liquidPressure at a point inside a liquid acts equally in all directions.The pressure increases with depth.
117 Find the pressure inside a liquid = density of the liquidh = depth of the point Xsurface ofliquidXh
118 Find the pressure inside a liquid Consider a small horizontal area A around point X.Xhsurface ofliquidA
119 Find the pressure inside a liquid The force from the liquid on this area is the weight W of the liquid cylinder above this areaXhsurface ofliquidAW
120 Find the pressure inside a liquid W = ?W = hAgXhsurface ofliquidAW
121 Find the pressure inside a liquid W = hAg and P =Xhsurface ofliquidAW
122 Find the pressure inside a liquid As there is also atmospheric pressure Po on the liquidsurface, the total pressure at X isPoXhAPsurface ofliquid
124 Force on a block in liquid Consider a cylinder of area A and height L ina liquid of density .Lh1h2P1P2A
125 Force on a block in liquid The pressure on its top area is P1 = h1g + PoThe pressure on its bottom area is P2 = h2g + PoLh1h2P1P2A
126 Force on a block in liquid The pressure difference P = P2 – P1 = Lgwith upward direction.Lh1h2P1P2A
127 Force on a block in liquid So there is an upward net force F = P.A= Vg where V is the volume of the cylinder.Lh1h2AF
128 Force on a block in liquid This is the upthrust on the cylinder.Upthrust = Vgh1h2FV
129 Force on a block in liquid Upthrust = VgNote that it is also equal to the weight of theliquid with volume V.h1h2FV
130 Force on a block in liquid The 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.h1h2FV
131 Force on a block in liquid The conclusion is correct for a solid in liquid and gas (fluid).h1h2FV
132 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.upthrustupthrust
133 Measuring upthrust W spring-balance The reading of the spring-balance is W,which is the weightof the object.objectbeakerliquidThe reading of thecompressionbalance is B, whichis the weight of liquidand beaker.compressionbalanceB
134 Measuring upthrust spring-balance Carefully immerse half the volume of the object in liquid.objectbeakerliquidWhat would happen tothe reading of thespring-balance and thatof the compressionbalance?compressionbalance
135 Measuring upthrust spring-balance The reading of the spring- balance decreases.Why?objectbeakerliquidThe difference in the readingsof the spring-balance givesthe upthrust on the object.compressionbalance
136 Measuring upthrust spring-balance The reading of the compression balanceincreases.Why?objectbeakerliquidThe difference in the readingsof the compression balancegives the upthrust on theobject.compressionbalance
137 Measuring upthrust spring-balance Carefully immerse object the whole object in liquid.objectbeakerliquidWhat would happen tothe reading of thespring-balance and thatof the compression balance?compressionbalance
138 Measuring upthrust spring-balance Carefully place the object on the bottom of the beaker.objectbeakerliquidWhat would happen tothe reading of thespring-balance and thatof the compressionbalance?compressionbalance
139 Law of floatationA floating object displaces its own weight of the fluid in which it floats.weight of the object= upthrust= weight of fluid displacedweightupthrust
140 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 smaller than density is larger than
141 Manometer liquid of density connect to the fluid Same level X Y 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 fluidSame levelXY
143 Manometer The pressures at points A and B are equal. Po = atmospheric Po+P= fluid pressureh = differencein heightABliquid ofdensity
144 Manometer The pressure at A = Po+P The pressure at B = Po + hg liquid ofdensity Po = atmosphericpressurePo+P= fluid pressureh = differencein heightBA
145 Manometer The pressure difference of the fluid P = hg liquid ofdensity Po = atmosphericpressurePo+P= fluid pressureh = differencein heightBA
146 Liquid in a pipeConsider a pipe of non-uniform cross-sectional area with movable piston at each end.The fluid is in static equilibrium.Same levelhx = hYYXstatic fluid
147 Liquid in a pipeThe manometers show that the pressures at points X and Y are equal.Same levelhx = hYYXstatic fluid
148 Liquid in a pipeThe pressures at points M and N on the pistons are also equal.Same levelhx = hYMNstatic fluid
149 Liquid in a pipeThere must be equal external pressures on the pistons to keep it in equilibrium.PM = PNSame levelhx = hYMNPMPNstatic fluid
150 Liquid in a pipeAs F = P.A , the external forces are different on the two ends.FM > FNSame levelhx = hYMNFMFNstatic fluid
151 Liquid in a pipe Same level hx = hY M N FM FN static fluid 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 levelhx = hYMNFMFNstatic fluid
152 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.
153 Turbulent flow Turbulent flow: the fluid flows in irregular paths. We will not study this kind of flow.
154 Streamlined flowStreamlined flow (laminar flow) : the fluid moves in layers without fluctuation or turbulence so that successive particles passing the same point with the same velocity.
155 Streamlined flowWe draw streamlines to represent the motion of the fluid particles.
156 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.
157 Equation of continuity At the left end, after time t, the volume passing is A1.v1. tAt the right end, after the same time t, the volume passing is A2.v2. t
158 Equation of continuity As the volumes are equal for an incompressible fluid,A1.v1. t = A2.v2. t A1.v1 = A2.v2
160 Pressure difference and work done x2A2Suppose 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.pPosition 2x1h2A1P+PPosition 1h1
161 Pressure difference and work done x2A2Work done by the external forces is(P+P).A1.x1 - P.A2.x2PPosition 2x1h2A1P+PPosition 1h1
162 Pressure difference and work done x2A2Work done by the external forces is(P+P).A1.x1 - P.A2.x2A1x1=A2x2=V= volume of fluid that movesPPosition 2x1h2A1P+PPosition 1h1
163 Pressure difference and work done x2A2Work done by the external forces is(P+P).A1.x1 - P.A2.x2= P .VWith V =Work done = Pwhere m is the mass of the fluid and ρis the density of the fluidpPosition 2x1h2A1P+PPosition 1h1
164 Bernoulli’s principle x2A2v2In time t, the fluid moves x1 at position 1 and x2 at position 2.x1 = v1.t andx2 = v2.tP2Position 2x1h2A1P1v1Position 1h1
165 Bernoulli’s principle x2A2v2In time t, the fluid moves x1 at position 1 and x2 at position 2.x1 = v1.t andx2 = v2.tWork done by external pressure =(P1-P2)P2Position 2x1h2A1P1v1Position 1h1
166 Bernoulli’s principle x2A2v2Work done by external pressure =(P1-P2)Increase in kinetic energy =P2Position 2x1h2A1P1v1Position 1h1
167 Bernoulli’s principle x2A2v2Increase in kinetic energy =Increase in gravitatioanl potential energy =mgh2 – mgh1P2Position 2x1h2A1P1v1Position 1h1
168 Bernoulli’s principle x2A2v2P2The left hand side is the workdone by external pressure. Itis also the energy suppliedto the fluid.The right hand side isthe increase in energyof the fluid.Position 2x1h2A1P1v1Position 1h1
186 Ball floating in air force due to spinning thrust from the air blower weight of the ballairWhat is the direction of spinning of the ball?
187 Ball floating in air thrust from the force due to pressure air blower differenceairweight of the ballairIt is spinning in clockwise direction.
188 Ball floating in airForce due topressuredifference
189 Air blown out through a funnel What would happen to the light ball?
190 Air blown out through a funnel Force due topressure differenceForce due topressure differenceweightIt is sucked to the top of the funnel.
191 Yacht sailing A yacht can sail against the wind. Note that the sail is curved.
192 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.
193 Yacht sailingThe yacht must follow a zig-zag path in order to sail against the wind.windpath
194 JetsWhen 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.low pressureairfluid
195 Jets: Bunsen burner The pressure near the jet is low. Air outside is pulled into the bunsen burner through the air hole.airgas