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Physics Beyond 2000 Chapter 7 Properties of Matter.

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1 Physics Beyond 2000 Chapter 7 Properties of Matter

2 States of matter Solid state Liquid state Gas state (will be studied in Chapter 8.)

3 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.

4 Solids Extension and compression (deformation) of solid objects – elasticity. Hookes 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 Hookes law for springs e1e1 e2e2 F1F1 F2F2 Natural lengthExtension Compression Force F Deformation e extension compression 0 F = k.e

6 Hookes 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 slope is soft. Force F Deformation e extension compression 0 F = k.e

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. extension F e 0 Force

9 Example 1 Find the extension of a spring from energy changes. Find the extension of the spring by using Hookes law.

10 Example 2 A spring-loaded rifle

11 Example 2 A spring-loaded rifle

12 More than springs All solid objects follow Hookes 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

13 stress Stress is the force F on unit cross- sectional area A. F A Unit: Pa Stress is a measure of the cause of a deformation. Note that A is the cross-sectional area of the wire before any stress is applied.

14 strain Strain is the extension e per unit length. If is the natural length of the wire, then e F Strain expresses the effect of the strain on the wire.

15 Example 3 Find the stress and the strain of a wire. stress strain

16 Young modulus E Young 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 E The value of E is dependent on the material.

18 Young modulus E and force constant k and F = k.e So k depends on E (the material), A (the thickness) and (the length).

19 Example 4 Find the Young modulus.

20 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 to measure the extension of the sample wire. reference wire sample wire weight level meter vernier scale

21

22 Experiment to find Young modulus Adjust the weight so that vernier scale to read zero. Measure the diameter of the sample wire and calculate its cross- section area A. reference wire sample wire weight vernier scale

23 The bubble in in the middle. Level meter

24 zero

25 Diameter of the wire

26 Experiment to find Young modulus Measure the length of the sample wire. reference wire sample wire weight vernier scale

27 Experiment to find Young modulus reference wire sample wire weight vernier scale Add weight W to the sample wire and measure its extension e. The force on the wire is F = W = mg. F = W = mg where m is the added mass.

28 More weights

29

30 The bubble moves to the left It is because the sample wire, which is on the right, extends. This end is higher. This end is lower.

31 Turn this screw (vernier scale) to raise up the end of the level meter suspended by the sample wire. This end of the level meter is suspended by the sample wire.

32 The bubble in in the middle again.

33 The reading on the screw shows the extension of the sample wire.

34 Experiment to find Young modulus reference wire sample wire weight vernier scale F Plot the graph of stress σ against strainε. σ σ ε elastic limit 0

35 Experiment to find Young modulus reference wire sample wire weight vernier scale F What is the slope of this graph? σ ε elastic limit 0 Young modulus

36 Experiment to find Young modulus σ ε elastic limit 0 The linear portion of the graph gives Hookes law. The stress applied to any solid is proportional to the strain it produces for small strain.

37 The stress-strain curve permanent strain stress σ strain ε O A L B C D A: proportional limit L: elastic limit B: yield point C: breaking stress D: breaking point

38 The stress-strain curve permanent strain stress σ strain ε O A L B C D A: proportional limit Between OA, the stress is proportional to the strain. Point A is the limit of this proportionality.

39 The stress-strain curve permanent strain stress σ strain ε O A L B C D L: elastic limit Between AL, the strain can be back to zero when the stress is removed.i.e. the wire is still elastic. Usually the elastic limit coincides with the proportional limit.

40 The stress-strain curve permanent strain stress σ strain ε O A L B C D B: yield point Between LB, the wire has a permanent deformation when the stress is removed. i.e. the wire is plastic. At point B, there is a sudden increase of strain a small increase in stress.

41 The stress-strain curve permanent strain stress σ strain ε O A L B C D C: breaking stress This is the maximum stress. Beyond this point, the wire extends and narrows quickly, causing a constriction of the cross-sectional area.

42 The stress-strain curve permanent strain stress σ strain ε O A L B C D D: breaking point The wire breaks at this point. This is the maximum strain of the wire.

43 Example 5 Refer to table 7.1 on p.112.

44 Energy stored in the extended wire stress strain σ ε The area under the stress-strain graph = where Fe is the elastic potential energy and A is the volume of the wire.

45 Properties of materials Stiffness Strength Ductility Toughness

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 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.

48 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

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.

50 Graphical representation stress σ strain ε stiffest weakest strongest most flexible

51 Graphical representation stress σ strain ε brittle tough ductile

52 Graphical representation for various materials stress σ strain ε glass metal rubber

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.

54 Plastic deformation stress σ 0 strain ε permanent strain loading unloading elastic limit

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

56 Creep Creep 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 σ strain ε loading unloading Hysteresis 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 solid Crystalline 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. layer

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. roro r o is the equilibrium distance. r = r o The force on each molecule is zero.

70 Intermolecular separation r It is the separation between the centres of two adjacent molecules. r > r o The force on the molecule is attractive. roro r

71 Intermolecular separation r It is the separation between the centres of two adjacent molecules. r < r o The force on the molecule is repulsive. roro r

72 Intermolecular forces Intermolecular force 0 repulsive attractive r roro r o is the equilibrium separation The dark line is the resultant 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 × volume The separation of molecules in solid and liquid is of order m.

75 Intermolecular potential energy Intermolecular force 0 r roro r o is the equilibrium separation Potential energy -ε The potential energy is zero for large separation. The potential energy is a minimum at the equilibrium separation.

76 Intermolecular potential energy Intermolecular force 0 r roro r o is the equilibrium separation Potential energy -ε When they move towards each other from far away, the potential energy decreases because there is attractive force. The work done by external force is negative. The potential energy is a minimum at the equilibrium separation.

77 Intermolecular potential energy Intermolecular force 0 r roro r o is the equilibrium separation Potential energy -ε When they are further towards each other after the equilibrium position, the potential energy increases because there is repulsive force. The work done by external force is positive. The potential energy is a minimum at the equilibrium separation.

78 Force and Potential Energy U = potential energy F = external force and

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. attractive Intermolecular force 0 repulsive r roro F r roro

80 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 roro So the molecule is in simple harmonic motion.

81 Variation of molecules F r roro So the molecule is in simple harmonic motion. with ω 2 = where m is the mass of each molecule.

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 r o. Intermolecular force 0 r roro Potential 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 r o. Intermolecular force 0 r roro Potential 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 Intermolecular force 0 r roro Potential energy -ε When energy is supplied to a solid, the molecules vibrate with greater amplitude until melting occurs.

87 Solids Intermolecular force 0 r roro Potential energy -ε On melting, the energy is used to break the lattice structure.

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 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.

90 Gases Intermolecular force 0 r roro Potential energy -ε Molecules are moving at very high speed in random direction.

91 Gases Intermolecular force 0 r roro Potential energy -ε The average separation between molecules is much longer than r o

92 Gases Intermolecular force 0 r roro Potential energy -ε The intermolecular force is so small that it is insignificant.

93 Example 7 There are molecules for one mole of substance. The is the Avogadros number.

94 Example 8 The separation between molecules depend on the volume.

95 Thermal expansion Potential energy 0 r roro -ε In a solid, molecules are vibrating about their equilibrium position.

96 Thermal expansion Suppose a molecule is vibrating between positions A and B about the equilibrium position. Potential energy 0 r roro -ε AB

97 Thermal expansion Note 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 0 r roro -ε A B AB C

98 Thermal expansion The potential energy of the molecule varies along the curve ACB while the molecule is oscillating along AB. Potential energy 0 r roro -ε AB AB C

99 Thermal expansion The centre of oscillation M is mid-way from the positions A and B. So point M is slightly away from the equilibrium position. Potential energy 0 r roro -ε AB AB C M

100 Thermal expansion When 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 0 r roro -ε AB AB C M

101 Thermal expansion The molecule is vibrating with larger amplitude between new positions AB. Potential energy 0 r roro -ε AB AB C M

102 Thermal expansion The centre of oscillation M, which is the mid-point of AB, is further away from the equilibrium position. Potential energy 0 r roro -ε AB AB C M

103 Thermal expansion As a result, the average separation between molecules increases by heating. The solid expands on heating. Potential energy 0 r roro -ε AB AB C M

104 Absolute zero temperature At absolute zero, the molecule does not vibrate. The separation between molecules is r o. The potential energy of the molecule is a minimum. Potential energy 0 r roro -ε C

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 r o. r o is also the diameter of each molecule. roro wire

106 Young Modulus in microscopic point of vies The cross-sectional area of the wire is roro where N is the number of molecules in each layer A area of one molecule = roro roro

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. r o + r F

108 Young Modulus in microscopic point of vies The strain is r o + r F

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. r o + r F

110 Young Modulus in microscopic point of vies r o + r F and F = N.k. r

111 Young Modulus in microscopic point of vies r o + r F Thus, the Young modulus is

112 Example 9 Find the force constant k between the molecules.

113 Density Definition: It is the mass of a substance per unit volume. where m is the mass and V is the volume Unit: kg m -3

114 Measure the density of liquid Use hydrometer upthrust weight

115 Pressure Definition: The pressure on a point is the force per unit area on a very small area around the point. or Unit: N m -2 or Pa.

116 Pressure in liquid Pressure 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 liquid h = depth of the point X surface of liquid X h

118 Find the pressure inside a liquid Consider a small horizontal area A around point X. X h surface of liquid A

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 area X h surface of liquid A W

120 Find the pressure inside a liquid W = ? X h surface of liquid A W W = hA g

121 Find the pressure inside a liquid X h surface of liquid A W W = hA g and P =

122 Find the pressure inside a liquid surface of liquid As there is also atmospheric pressure P o on the liquid surface, the total pressure at X is PoPo X h A P

123 Example 10 The hydraulic pressure.

124 Force on a block in liquid L h1h1 h2h2 P1P1 P2P2 A Consider a cylinder of area A and height L in a liquid of density.

125 Force on a block in liquid L h1h1 h2h2 P1P1 P2P2 A The pressure on its top area is P 1 = h 1 g + P o The pressure on its bottom area is P 2 = h 2 g + P o

126 Force on a block in liquid L h1h1 h2h2 P1P1 P2P2 A The pressure difference P = P 2 – P 1 = L g with upward direction.

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. L h1h1 h2h2 A F

128 Force on a block in liquid This is the upthrust on the cylinder. Upthrust = Vg h1h1 h2h2 F V

129 Force on a block in liquid Upthrust = Vg Note that it is also equal to the weight of the liquid with volume V. h1h1 h2h2 F V

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. h1h1 h2h2 F V

131 Force on a block in liquid The conclusion is correct for a solid in liquid and gas (fluid). h1h1 h2h2 F V

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. upthrust

133 Measuring upthrust spring-balance object liquid compression balance The reading of the spring-balance is W, which is the weight of the object. The reading of the compression balance is B, which is the weight of liquid and beaker. beaker W B

134 Measuring upthrust spring-balance object liquid Carefully immerse half the volume of the object in liquid. What would happen to the reading of the spring-balance and that of the compression balance? beaker compression balance

135 Measuring upthrust spring-balance object liquid The reading of the spring- balance decreases. Why? The difference in the readings of the spring-balance gives the upthrust on the object. beaker compression balance

136 Measuring upthrust spring-balance object liquid The reading of the compression balance increases. Why? The difference in the readings of the compression balance gives the upthrust on the object. beaker compression balance

137 Measuring upthrust spring-balance object liquid Carefully immerse the whole object in liquid. What would happen to the reading of the spring-balance and that of the compression balance? beaker compression balance

138 Measuring upthrust spring-balance object liquid Carefully place the object on the bottom of the beaker. What would happen to the reading of the spring-balance and that of the compression balance? beaker compression balance

139 Law of floatation A floating object displaces its own weight of the fluid in which it floats. weightupthrust weight of the object = upthrust = weight of fluid displaced

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 larger than density is smaller than

141 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 of density connect to the fluid XY Same level

142 Manometer liquid of density P o = atmospheric pressure P o + P = fluid pressure h = difference in height X Y

143 Manometer liquid of density P o = atmospheric pressure P o + P = fluid pressure h = difference in height B A The pressures at points A and B are equal.

144 Manometer liquid of density P o = atmospheric pressure P o + P = fluid pressure h = difference in height B A The pressure at A = P o + P The pressure at B = P o + h g

145 Manometer liquid of density P o = atmospheric pressure P o + P = fluid pressure h = difference in height B A The pressure difference of the fluid P = h g

146 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 h x = h Y static fluid

147 Liquid in a pipe The manometers show that the pressures at points X and Y are equal. Same level X Y h x = h Y static fluid

148 Liquid in a pipe The pressures at points M and N on the pistons are also equal. Same level M N h x = h Y static fluid

149 Liquid in a pipe There must be equal external pressures on the pistons to keep it in equilibrium. P M = P N Same level M N h x = h Y static fluid PMPM PNPN

150 Liquid in a pipe As F = P.A, the external forces are different on the two ends. F M > F N Same level M N h x = h Y static fluid FMFM FNFN

151 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 h x = h Y static fluid FMFM FNFN

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. Beurnoullis 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 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.

155 Streamlined flow We 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 A 1.v 1. t At the right end, after the same time t, the volume passing is A 2.v 2. t

158 Equation of continuity As the volumes are equal for an incompressible fluid, A 1.v 1. t = A 2.v 2. t A 1.v 1 = A 2.v 2

159 Equation of continuity Example 12

160 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. h2h2 P+ P p x1x1 x2x2 Position 1 Position 2 A1A1 A2A2 h1h1

161 Pressure difference and work done Work done by the external forces is (P+ P).A 1.x 1 - P.A 2.x 2 P+ P P x1x1 x2x2 Position 1 Position 2 A1A1 A2A2 h2h2 h1h1

162 Pressure difference and work done Work done by the external forces is (P+ P).A 1.x 1 - P.A 2.x 2 A 1 x 1 =A 2 x 2 =V = volume of fluid that moves P+ P P x1x1 x2x2 Position 1 Position 2 A1A1 A2A2 h2h2 h1h1

163 Pressure difference and work done Work done by the external forces is (P+ P).A 1.x 1 - P.A 2.x 2 = P.V With V = Work done = P where m is the mass of the fluid and ρis the density of the fluid P+ P p x1x1 x2x2 Position 1 Position 2 A1A1 A2A2 h2h2 h1h1

164 Bernoullis principle In time t, the fluid moves x 1 at position 1 and x 2 at position 2. x 1 = v 1. t and x 2 = v 2. t P1P1 P2P2 x1x1 x2x2 Position 1 Position 2 A1A1 A2A2 v1v1 v2v2 h2h2 h1h1

165 Bernoullis principle In time t, the fluid moves x 1 at position 1 and x 2 at position 2. x 1 = v 1. t and x 2 = v 2. t Work done by external pressure = (P 1 -P 2 ) P1P1 P2P2 x1x1 x2x2 Position 1 Position 2 A1A1 A2A2 v1v1 v2v2 h2h2 h1h1

166 Bernoullis principle Work done by external pressure = (P 1 -P 2 ) Increase in kinetic energy = P1P1 P2P2 x1x1 x2x2 Position 1 Position 2 A1A1 A2A2 v1v1 v2v2 h2h2 h1h1

167 Bernoullis principle Increase in kinetic energy = Increase in gravitatioanl potential energy = mgh 2 – mgh 1 P1P1 P2P2 x1x1 x2x2 Position 1 Position 2 A1A1 A2A2 v1v1 v2v2 h2h2 h1h1

168 Bernoullis principle P1P1 P2P2 x1x1 x2x2 Position 1 Position 2 A1A1 A2A2 v1v1 v2v2 h2h2 h1h1 The left hand side is the work done by external pressure. It is also the energy supplied to the fluid. The right hand side is the increase in energy of the fluid.

169 Bernoullis principle P1P1 P2P2 x1x1 x2x2 Position 1 Position 2 A1A1 A2A2 v1v1 v2v2 h2h2 h1h1

170 Bernoullis principle P1P1 P2P2 x1x1 x2x2 Position 1 Position 2 A1A1 A2A2 v1v1 v2v2 h2h2 h1h1 constant

171 Bernoullis principle If h 1 = h 2, then Position 1 Position 2

172 Bernoullis principle If h 1 = h 2, then Position 1 Position 2 or constant

173 Bernoullis principle If h 1 = h 2, then Position 1 Position 2 constant So for horizontal flow, where the speed is high, the pressure is low. Low speed, high pressure High speed, low pressure

174 Bernoullis principle Position 1 Position 2 So for horizontal flow, where the speed is high, the pressure is low. Low speed, high pressure High speed, low pressure h1h1 h2h2

175 Bernoullis principle Position 1 Position 2 So for horizontal flow, where the speed is high, the pressure is low. P 1 = h 1..g h1h1 h2h2 P 2 = h 2..g Low speed, high pressure High speed, low pressure

176 Simple demonstration of Bernoullis principle Held two paper strips vertically with a small gap between them. Blow air gently into the gap. Explain what you observe. air

177 Simple demonstration of Bernoullis principle air high pressure In the gap, the speed of airflow is high. So the pressure is low in the gap. The high pressure outside presses the strips together.

178 Examples of Bernoullis effect Airfoil: the airplane is flying to the left.

179 Examples of Bernoullis effect Airfoil: There is a pressure difference between the top and the bottom of the wing. A net lifting force is produced.

180 Example 13 Airfoil and Bernoullis effect To find the lifting force on an airplane.

181 Examples of Bernoullis effect Spinning ball: moving to the left and rotating clockwise.

182 Examples of Bernoullis effect Spinning ball: the pressure difference produces a deflection force and the ball moves along a curve.

183 Examples of Bernoullis effect Spinning ball: the pressure difference produces a deflection force and the ball moves along a curve. spinning ball not spinning

184 Examples of Bernoullis effect Spinning ball: the pressure difference produces a deflection force and the ball moves along a curve. spinning ball not spinning

185 Ball floating in air air

186 Ball floating in air air weight of the ball thrust from the air blower force due to spinning What is the direction of spinning of the ball?

187 Ball floating in air air weight of the ball thrust from the air blower force due to pressure difference It is spinning in clockwise direction.

188 Ball floating in air Force due to pressure difference

189 Air blown out through a funnel What would happen to the light ball?

190 Air blown out through a funnel It is sucked to the top of the funnel. weight Force due to pressure difference Force due to pressure difference

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

194 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 fluid low pressure

195 Jets: Bunsen burner The pressure near the jet is low. Air outside is pulled into the bunsen burner through the air hole. gas air

196 Jets: Paint sprayer

197 Jets: Filter pump

198 Jets: Carburretor

199 Roofs, window and door The pressure difference makes the door close. fast wind, low pressure door

200 Example 14 Strong wind on top of the roof. tile

201 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)

202 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

203 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 PoPo PoPo

204 A hole in a water tank The water pressure on the surface is P o. The water pressure at the hole is P o. v h PoPo PoPo

205 A hole in a water tank The height of water on the surface is h The height of water at the hole is 0. v h PoPo PoPo

206 A hole in a water tank Apply Bernoullis equation, v h PoPo PoPo This is the same speed of an object falling through a distance h freely.

207 Example 15 A hole in a tank

208 Pitot tube Pitot tube is used to measure the speed of fluid. h v static tube total tube h1h1 h2h2

209 Pitot tube The pressure below the static tube is h 1 g. The pressure at the mouth of the total tube is h 2 g. h v static tube total tube h1h1 h2h2

210 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 static tube total tube h1h1 h2h2

211 Pitot tube Apply Bernoullis equation, h v static tube total tube h1h1 h2h2

212 Pitot tube h v static tube total tube h1h1 h2h2


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