Copyrighted material – 9781137443236 Contents Preface ix How to use this book x Acknowledgements xi Units and Measurements 1 Philosophy and practice in science 1 The S.I. system of units 1 Symbols and equations in physics 3 Making measurements in physics 4 Uncertainty and accuracy 6 Summary 8 Revision questions 8 PART 1 Mechanics 9 1 Forces in Equilibrium 11 1.1 Force as a vector 11 1.2 Turning effects 15 1.3 Stability 17 1.4 Friction 19 1.5 Equilibrium conditions 22 Summary 25 Revision questions 25 2 Dynamics 27 2.1 Speed and distance 27 2.2 Speed and velocity 29 2.3 Acceleration 30 2.4 The equations for uniform acceleration 33 2.5 Acceleration due to gravity 35 2.6 Rates of change 38 Summary 40 Revision questions 40 3 Force and Motion 41 3.1 Newton’s laws of motion 41 3.2 Force and momentum 45 3.3 Conservation of momentum 47 Summary 51 Revision questions 51 4 Energy and Power 53 4.1 Work and energy 53 4.2 Power and energy 56 4.3 Efficiency 58 4.4 Energy resources 60 Summary 64 Revision questions 64 PART 2 Properties of Materials 67 5 Matter and Molecules 69 5.1 States of matter 69 5.2 Elements and compounds 71 5.3 The Avogadro constant, N A 74 5.4 Intermolecular forces 75 Summary 77 Revision questions 78 6 Thermal Properties of Materials 79 6.1 Thermal expansion 79 6.2 Specific heat capacity 81 6.3 Specific latent heat 84 6.4 Heat transfer 86 6.5 More about thermal conduction 89 Summary 92 Revision questions 93 7 Strength of Solids 94 7.1 Measuring force 94 7.2 Force and materials 96 7.3 Stress and strain 97 7.4 Stress versus strain curves 100 7.5 Elastic energy 103 Summary 104 Revision questions 105 v Copyrighted material – 9781137443236
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Contents · 29 Simple Harmonic Motion 453 29.1 Oscillating motion 453 29.2 Sine waves 456 29.3 Forces in oscillating systems 458 29.4 Resonance 462 ... East and a force of 12.0 N.
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Copyrighted material – 9781137443236
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Contents
Preface ix
How to use this book x
Acknowledgements xi
Units and Measurements 1
Philosophy and practice in science 1The S.I. system of units 1Symbols and equations in physics 3Making measurements in physics 4Uncertainty and accuracy 6
Summary 8Revision questions 8
PART 1 Mechanics 9
1 Forces in Equilibrium 11
1.1 Force as a vector 111.2 Turning effects 151.3 Stability 171.4 Friction 191.5 Equilibrium conditions 22
Summary 25Revision questions 25
2 Dynamics 27
2.1 Speed and distance 272.2 Speed and velocity 292.3 Acceleration 302.4 The equations for uniform
acceleration 332.5 Acceleration due to gravity 352.6 Rates of change 38
Summary 40Revision questions 40
3 Force and Motion 41
3.1 Newton’s laws of motion 413.2 Force and momentum 45
3.3 Conservation of momentum 47Summary 51Revision questions 51
4 Energy and Power 53
4.1 Work and energy 534.2 Power and energy 564.3 Efficiency 584.4 Energy resources 60
Summary 64Revision questions 64
PART 2 Properties of Materials 67
5 Matter and Molecules 69
5.1 States of matter 695.2 Elements and compounds 715.3 The Avogadro constant, NA 745.4 Intermolecular forces 75
Summary 77Revision questions 78
6 Thermal Properties of Materials 79
6.1 Thermal expansion 796.2 Specific heat capacity 816.3 Specific latent heat 846.4 Heat transfer 866.5 More about thermal conduction 89
Summary 92Revision questions 93
7 Strength of Solids 94
7.1 Measuring force 947.2 Force and materials 967.3 Stress and strain 977.4 Stress versus strain curves 1007.5 Elastic energy 103
Summary 104Revision questions 105
v
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CONTENTSvi
8 Pressure 106
8.1 Pressure and force 1068.2 Hydraulics 1078.3 Pressure in a fluid at rest 1098.4 Measurement of pressure 1108.5 Flotation 112
Summary 114Revision questions 114
PART 3 Waves 117
9 Properties of Waves 119
9.1 Types of waves 1199.2 Measuring waves 1209.3 Properties of waves 1229.4 Longitudinal and transverse waves 1269.5 Polarised light 128
Summary 129Revision questions 129
10 Sound 131
10.1 The nature and properties of sound 13110.2 The human ear 13310.3 Ultrasonics 13510.4 Vibrating strings 13710.5 Acoustic resonance 139
Summary 142Revision questions 143
11 Optics 145
11.1 Interference and the nature of light 14511.2 Reflection and refraction of light 14811.3 Lenses 15111.4 Optical instruments 155
Summary 158Revision questions 158
12 Electromagnetic Waves 160
12.1 The visible spectrum 16012.2 Types of spectra 16112.3 Infra-red and beyond 16612.4 Ultraviolet radiation, X-radiation and
13.1 Static electricity 17713.2 Current and charge 18013.3 Potential difference 18513.4 Resistance 188
Summary 192Revision questions 193
14 Electric Circuits 194
14.1 Cells and batteries 19414.2 The potentiometer 19814.3 The Wheatstone bridge 20114.4 Resistivity 20414.5 Electrical measurements 205
Summary 209Revision questions 209
15 Capacitors 211
15.1 Storing charge 21115.2 Capacitor combinations 21415.3 Energy stored in a charged cell 21715.4 Capacitor design factors 21915.5 Capacitor discharge 221
Summary 224Revision questions 224
16 Electronics 226
16.1 The systems approach 22616.2 Electronic logic 22716.3 Electronics at work 23116.4 Operational amplifiers 23416.5 Multivibrators 241
Summary 242Revision questions 243
PART 5 Fields 244
17 Electric Fields 247
17.1 Electrostatic forces 24717.2 Electric field patterns 24917.3 Electric field strength 25317.4 Electric potential 258
Summary 260Revision questions 261
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CONTENTS vii
18 Magnetic Fields 262
18.1 Magnetic field patterns 26218.2 Electromagnets 26518.3 The motor effect 26618.4 Magnetic flux density 26918.5 Magnetic field formulae 27218.6 Magnetic materials 275
Summary 278Revision questions 279
19 Electromagnetic Induction 280
19.1 The dynamo effect 28019.2 Magnetic flux 28419.3 The alternating current generator 28719.4 The transformer 29019.5 Self-inductance 294
Summary 297Revision questions 297
20 Alternating Current 299
20.1 Measurement of alternating currents
and voltages 29920.2 Rectifier circuits 30220.3 Alternating current and power 30420.4 Capacitors in a.c. circuits 30720.5 Coils in a.c. circuits 31120.6 Resonant circuits 313
Summary 316Revision questions 317
PART 6 Atomic and NuclearPhysics 319
21 Electrons and Photons 321
21.1 Electron beams 32121.2 The charge of the electron 32621.3 Photoelectricity 33021.4 Electrons in the atom 33421.5 X-rays 33821.6 Wave mechanics 342
Summary 346Revision questions 347
22 Radioactivity 349
22.1 Inside the atom 34922.2 Radiation from radioactive
substances 35222.3 The range and penetrating power of
alpha, beta and gamma radiation 356
22.4 Radioactive decay 36022.5 The mathematics of radioactive
decay 363Summary 365Revision questions 366
23 Energy from the Nucleus 367
23.1 The nature of force 36723.2 Binding energy 37023.3 Nuclear power 37523.4 Probing the nucleus 380
Answers to In-text Questions andRevision Questions 505
Answers to Further Questions 534
Glossary 535
Index 543
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Forces in Equilibrium
OBJECTIVES
After working through this unit, you should be able to:
. represent a force geometrically or numerically
. combine two or more forces
. resolve a force into perpendicular components
. describe the condition for a point object to be in equilibrium
. draw a free-body diagram of the forces acting on a body
. understand and use the concept of centre of gravity
Force as a vector
Vectors and scalars
An air traffic controller needs to know the position, height, speed and direction ofmotion of an aircraft in order to monitor its flight path. Each piece of informationis important, including the direction of motion. The position and height of anaircraft gives its displacement, which is its distance and direction from the airtraffic controller. Its speed and direction of motion define its velocity. See Fig. 1.1.
Displacement and velocity are examples of vector quantities because they havea direction as well as a magnitude. Distance and speed are examples of scalarquantities because they have a magnitude only. Any vector may be represented byan arrow of length in proportion to the magnitude of the vector and in theappropriate direction.
. A vector is defined as any physical quantity that has a direction. Examplesinclude weight, force, displacement (i.e. distance in a given direction), velocity,acceleration and momentum.
. A scalar is any physical quantity that has no direction. Examples include mass,energy and speed.
1.1
UNIT
1CONTENTS
1 Force as a vector
2 Turning effects
3 Stability
4 Friction
5 Equilibriumconditions
Summary
Revision questions
11
N
SMAP
0EW
Kilometre grid
Velocity = 60 ms–1 south-eastDisplacement from 0 = 3.5 km36°E of due North (height not shown)
Fig. 1.1 Vectors and scalars
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MECHANICS12
Force as a vector
Force is measured in units called newtons (N).The weight of a mass of 1 kg isapproximately 10 N on the surface of the Earth. An object acted on by two forceswill be at rest if the two forces are equal and opposite, as in Fig. 1.2a. However, ifone of the forces is larger than the other, their combined effect is the differencebetween the two forces, as in Fig. 1.2b.
Each force in Fig. 1.2 can be represented by a vector (i.e. an arrow of lengthproportional to the force in the appropriate direction). From now on in this book, abold symbol will denote a vector. For example, in Fig. 1.2b, F1 denotes force vectorF1 and F2 denotes force vector F2.
The parallelogram of forces
Fig. 1.3 shows how to work out the effect of two forces, A and B, acting on a pointobject but not along the same line. Each force is represented by a vector, such thatthe two vectors form adjacent sides of a parallelogram. The combined effect of thetwo forces, the resultant, is the diagonal of the parallelogram between the twovectors. The resultant is therefore given by force vector B added onto force vector A.Fig. 1.4 shows how to balance out the resultant of A and B with an equal andopposite force, C.
1.1A QUESTIONS
1 A point object is acted on by a force of 5.0 N acting dueEast and a force of 12.0 N. Use the parallelogram methodto determine the magnitude and direction of theresultant force when the 12.0 N force actsa) due East,b) due West,c) due North,
d) 608 North of due East,e) 608 North of due West.
2 A point object is acted on by forces of 6.0 N and 8.0 N.Determine the magnitude of the resultant of these twoforces if the angle between them isa) 908,b) 458.
F1 = 20 N F2 = 20 N
Object
(a) F1 = F2 so object is at rest
F1 = 20 N F2 = 30 N
Object
(b) F1 < F2 so object moves towards F2. The resultant force is 10 N
Fig. 1.2 Forces in opposite directions
A = 4.0 N
A + B = 6.1
N
B = 3.0 N
Fig. 1.3 The parallelogram of forces
A
C
A + B
B0
Fig. 1.4 Equilibrium
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FORCES IN EQUILIBRIUM 13
The condition for equilibrium of a point object
In general, a point object acted on by two or more forces is in equilibrium (i.e. atrest) if the resultant of the forces is zero.
1 For equilibrium of a point object acted on only by two forces, A and B, the twoforces must be equal and opposite to each other. The resultant of the two forcesmust be zero.
Aþ B ¼ 0
2 For equilibrium of a point object acted on by three forces, A, B and C, theresultant of any two of the three forces is equal and opposite to the third force.For example, Aþ B ¼ �C. Rearranging this equation gives
Aþ Bþ C ¼ 0
The three force vectors add together to give a zero resultant if the object is inequilibrium. This can be represented as a triangle, as in Fig. 1.5.
3 In general, a point object is in equilibrium if the forces acting on it form a vectordiagram which is a closed polygon. This rule is called the closed polygon rule.
Resolving a force into perpendicular components
This is a mathematical technique which enables equilibrium situations to beanalysed without scale diagrams. Sketch diagrams are nevertheless useful to visualisesituations.
Consider a force, A, acting at point, O, of an x–y coordinate system, as in Fig. 1.6.The magnitude at A is represented by A and the direction of A is at angle � to thex-axis. This force may be resolved into two perpendicular components.
1 Ax ¼ A cos � along the x-axis.2 Ay ¼ A sin � along the y-axis.
Force A may be written in terms of its components in the form
A ¼ A cos � iþ A sin � j
where i and j are vectors of unit length along the x-axis and the y-axis respectively.Using Pythagoras’ theorem gives
A ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiAx
2 þ Ay2
� �q
and using the rules of trigonometry gives
tan � ¼ Ay
Ax
AC
A + B + C = 0
B
Fig. 1.5 Equilibrium of a pointobject
j
i
A
Ax = A cos O
Ay = A sin
A = Ax2+ Ay
2
Ay
Ax
tan =
x
y
Fig. 1.6 Resolving a force
Notes
1 Pythagoras’ theorem states that for a right-angled triangle ABC,
AB2 þ BC2 ¼ AC2.
2 The following trigonometric functions are defined from this triangle
sin � ¼ AB
AC
�¼ o
h
�cos � ¼ BC
AC
�¼ a
h
�tan � ¼ AB
BC
�¼ o
a
�
where o is the opposite side and a is the adjacent side to �, and h is the hypotenuse.aC B
A
oh
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MECHANICS14
How to calculate the resultant of two or more forces
Consider a point object, O, acted on by two or more forces, A, B, C, etc, as shown inFig. 1.7. The resultant of these forces can be calculated by following these steps:
1 Resolve each force into perpendicular components along the x-axis and they-axis.
2 For each axis, add the components, taking account of þ or � directions, to givethe component of the resultant along each axis.
3 Use Pythagoras’ equation to calculate the magnitude of the resultant, and use thetrigonometry equations to calculate the direction of the resultant.
1.1 WORKED EXAMPLE
A point object, O, is acted on by three forces, A, B and C, as shownin Fig. 1.8. Calculate (a) the magnitude and (b) the direction of theresultant of these three forces.
Solution
Resolving the three forces into componentsalong the i-axis and the j-axis gives:
A ¼ 10 cos 45iþ 10 sin 45j
B ¼ �8 sin 30iþ 8 cos 30j
C ¼ �5j
Therefore the resultant is
R ¼ Aþ Bþ C
¼ ð10 cos 45� 8 sin 30Þiþ ð10 sin 45þ 8 cos 30� 5Þjthus R ¼ 3:1iþ 1:0j
a) The magnitude of the resultant, R ¼ pð3:12 þ 1:02Þ ¼ 9:5 Nb) The angle between the resultant and the x-axis, �, is given by
tan � ¼ Ry
Rx¼ 9:0
3:1¼ 2:90
Hence � ¼ 718.
A R = A
+ B
B
O Ax
By
Ay
RxBx +x
+y
R = Rx2+ Ry
2
where Rx = Ax + Bx
and Ry = Ay + By
Fig. 1.7 Using a calculator
j
i
A
C
RB
0
A = 10
N
C = 5 N
B = 8 N
45°
30°
+x–x
+y
–y
Fig. 1.8
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FORCES IN EQUILIBRIUM 15
1.1B QUESTIONS
1 Calculate the magnitude and direction of theresultant of two forces, A and B, acting on a pointobject for each of the arrangements shown inFig. 1.9.
2 For each arrangement shown in Fig. 1.9, state themagnitude and direction of a third force, C, actingon O which would cancel out the resultant of Aand B.
Turning effects
When you need to use a spanner to unscrew a nut, the longer the handle of thespanner, the easier the task will be. The same applies to a claw hammer if ever youneed to remove a nail from a piece of wood. The longer the handle, the greater theleverage about the turning point. See Fig. 1.10.
The moment of a force about a point is defined as the magnitude of theforce the perpendicular distance from its line of action to the point (seeFig. 1.11).
The unit of a moment is the newton metre (Nm). The term torque is used todescribe the moment of a force about an axis.
For example, if a force of 40 N is applied to a spanner at a perpendicular distanceof 0.20 m from the turning point, the moment of the force about that point is equalto 8.0 Nm (¼ 40 N� 0:20 m). The same moment could be achieved by applying aforce of 80 N at a perpendicular distance of 0.1 m from the pivot (¼ 80 N� 0:1 m).Now you can see why a lesser force gives the same leverage if it acts at a greaterdistance. Prove for yourself that a force of 16 N acting at a perpendicular distance of0.5 m from the turning point gives the same moment as 40 N at 0.20 m.
The principle of momentsAny object that is not a point object can be turned when one or more forces act onit. For example, a door turns on its hinges when it is given a push. The term body isused for any object that is not a point object.
. A body acted on by only one force cannot be in equilibrium and will turn if theforce does not act through its centre of gravity.
. A body acted on by two or more forces may or may not turn, depending on theposition and direction of the forces acting on it. For example, a balanced see-saw,as shown in Fig. 1.12 carrying a child either side of the pivot will becomeunbalanced if one of the children climbs off. For balance, a child of weight 250 Nwould need to be nearer the pivot than a child of weight 300 N. The condition fora balanced see-saw is that the moment of the child on one side about the pivotshould equal the moment of the other child about the same pivot. For example, ifthe 250 N child is 1.20 m from the pivot (moment ¼ 250� 1:20 ¼ 300 Nm), theother child would need to be 1.0 m from the pivot (moment ¼ 300 N� 1:0 m)for balance.
1.2
A = 3
N
B = 2 NO
45°x
y(a)
A = 4 N
B = 2.5 N O30°
x
y(b)
Fig. 1.9
Pull onhandle
Claw hammerForceon nail
Pivot
Fig. 1.10 Applying leverage
F
Moment about P = Fd
PPull onhandle
d
Fig. 1.11 Moment = force �perpendicular distance
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MECHANICS16
The balanced see-saw is an example of a more general rule, the Principle ofMoments, which applies to any body in equilibrium.
The Principle of Moments states that for a body in equilibrium, the sum of theclockwise moments about any point is equal to the sum of the anticlockwisemoments.
1.2 WORKED EXAMPLE
A uniform beam of length 4.0 m pivoted horizontally about its centresupports a 5.0 N weight at a distance of 1.5 m from the centre, and a6.0 N weight at the other side of the centre. Calculate the distancefrom the 6.0 N weight to the centre of the beam.
Solution
Let d represent the distance from the centre to the 6.0 N weight.
Applying the principle of moments about the centre gives 6:0d ¼ 5:0� 1:5.
Hence d ¼ 5:0� 1:5
6:0¼ 1:25 m
1.2 QUESTIONS
1 The uniform beam in Fig. 1.14 is pivoted at its centre andsupports three weights, W1, W2 and W3, at distances d1,d2 and d3 as shown.
For each set of weights and distances in the table,calculate the value of the missing quantity necessary tomaintain the beam in equilibrium.
2 A simple beam balance is shown in Fig. 1.15. The scalepan is suspended from one end of a metre rule which ispivoted at its centre. The rule is balanced by adjusting theposition of a 2.0 N weight suspended from the rule at theother side of the pivot to the scale pan.
a) When the scale pan is empty, the 2.0 N weight mustbe positioned 150 mm from the centre of the rule tokeep the rule horizontal. Calculate the weight of thescale pan.
b) An object, X, of unknown weight is placed on thescale pan and the 2.0 N weight is moved to a distanceof 360 mm from the centre to keep the rulehorizontal. Calculate the weight of X.
1.0 m1.2 m
250 N 300 N
Fig. 1.12 A balanced see-saw
6.0 N 5.0 N
1.5 md
Fig. 1.13
W2
d3d2
d1
PivotW3W1
Fig. 1.14 Scale pan
2 N
Pivot
Fig. 1.15
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FORCES IN EQUILIBRIUM 17
Stability
Mass and weight
If you want to lose weight, go to the Moon where the force of gravity is one-sixth ofthe Earth’s gravity. Unfortunately, you won’t become slimmer because your masswill be unchanged.
. The mass of an object is the amount of matter it possesses. The unit of mass isthe kilogram (kg), defined by means of a standard kilogram at BIPM (BureauInternational des Poids et Mesures), Paris.
. The weight of an object is the force of gravity acting on it. The unit of weight isthe newton (N).
. The force of gravity per unit mass, g, at the surface of the Earth, is 9.8 N kg�1.The value of g changes slightly with position on the Earth’s surface, ranging from9.78 N kg�1 at the equator to 9.81 N kg�1 at the poles. The cause of this variationis due partly to the shape of the Earth not being exactly spherical and partly tothe effect of the Earth’s rotation. Strictly, this latter effect is not due to the force ofgravity but arises from the circular motion of an object at the equator. The forceof gravity on an object on the Moon, at 1.6 N kg�1, is much smaller than that onan object on the Earth because the Moon is smaller in size than the Earth.
For an object of mass m, its weight, W, can be calculated from the equationW ¼ mg
Stable and unstable equilibrium
At a bowling alley, have you noticed how easily a pin falls over? Each pin is designedto be top-heavy with a small base. Fig. 1.16 shows a bowling pin in comparison with‘Wobbly Walter’, a bottom-heavy child’s toy that rights itself when knocked over.
. The child’s toy is an example of an object in stable equilibrium – it returns toequilibrium when knocked over because its centre of gravity is always over its base.
. The standing bowling pin is an example of an object in unstable equilibrium – itmoves away from equilibrium when released from a tilted position.
. The bowling pin lying on its side is an example of an object in neutralequilibrium – it stays wherever it is moved to, neither returning to nor movingaway from where it was.
Centre of gravity
The centre of gravity of an object is the point where its weight can beconsidered to act.
. An object freely suspended from a point and then released will come to rest withits centre of gravity directly beneath the point. Fig. 1.17 shows how to find thecentre of gravity of a flat object.
. A tall object tilted too far will topple over if released. This happens if it is tiltedbeyond the point where its centre of gravity is directly above the point aboutwhich it turns, as in Fig. 1.18. If released beyond this position, the moment of theweight about the point of turning causes the object to overbalance.
1.3
A bowlingalley pin
High centreof gravity
Weight
A wobbly toy
Low centreof gravity
Weight
Fig. 1.16 Equilibrium
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See www.palgrave.com/foundations/breithaupt for aproof that the centre ofgravity of a uniform beam is atits midpoint.
MECHANICS18
. The moment of an object due to its weight, W, about any point = Wd, where d isthe perpendicular distance from the point to the vertical line through its centre ofgravity. For example, Fig. 1.19 shows a rule pivoted at one end. The rule can beheld horizontal by a vertical string attached to the other end. The moment of therule’s weight about the pivot can be proved to be equal to WL/2, where L is itslength. Therefore its centre of gravity is at a distance L/2 from the pivot.
. An object may be balanced by a single vertical force applied at its centre ofgravity. For example, it ought to be possible to support a plate on the end of avertical rod, provided the plate is placed with its centre on the end of the rod.Another example is when someone carries a ladder in a horizontal position; thetask is much easier if the ladder is supported at its centre of gravity, because theladder is balanced at this point.
1.3 WORKED EXAMPLE
A road gate consists of a 4.5 m uniform steel tube ofweight 120 N which has a 400 N counterweight fixedto one end, as shown in Fig. 1.20. It is pivoted about ahorizontal axis which passes through the beam 0.55 mfrom the centre of the counterweight.a) Calculate the position of the centre of gravity of the
road gate.b) Calculate the magnitude and direction of the force
that must be applied to the counterweight to keepthe beam horizontal.
EXERCISE
Locating the centre of gravity of a flat card
Use a plumbline to mark a vertical line through the pivot on thecard as in Fig. 1.17.Repeat for a different pivot.The centre of gravity is where the lines intersect.
Pivot P
Card freelypivoted at P
Fixed rod
Marker
Centre ofgravityis here
Plumbline
Fig. 1.17
(b)(a)
Pivot
Fig. 1.18 Stability (a) tilting without toppling,(b) on the point of toppling
Pivot P
L
W
Fig. 1.19 Centre of gravity of arule
1.7 m
Cg
(Cg = centre of gravity)
CounterweightPivot
d
0.55 m
Fig. 1.20
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FORCES IN EQUILIBRIUM 19
1.3 WORKED EXAMPLE continued
Solution
a) Let d ¼ the distance from the centre of gravity Cg of the road gate to the counterweight. The distance fromthe centre of gravity of the counterweight to the centre of gravity of the tube ¼ 2:25 m. The moment of thetube about the counterweight ¼ 120 N� 2:25 m ¼ 270 Nm anticlockwise. This is the same as the momentof the whole gate about the counterweight ¼ ð400þ 120Þd anticlockwise.
Hence 520d ¼ 270 Nm ; d ¼ 270� 520 ¼ 0:52 m.
b) Because Cg is between the counterweight and the pivot, the road gate will turn clockwise if no additionalforce is applied to it. Let F ¼ the upward vertical force on the counterweight needed to keep the road gatehorizontal.
Applying the principle of moments about the pivot, the moment of the road gate about the pivot¼ ð400þ 120Þ � 0:03 ¼ 16 Nm clockwise.
The moment of force F about the pivot ¼ 0:55 F anticlockwise.
Hence 0.55 F ¼ 30 ; : F ¼ 16� 0:55 ¼ 28 N
1.3 QUESTIONS
1 A placard consists of a uniform pole of length 4.0 m and of weight 30 Nwhich has a square board of weight 20 N fixed to one end, as shown inFig. 1.21. Calculate the distance from the other end of the pole to thecentre of gravity of the placard.
2 An advertising sign consists of a uniform board as in Fig. 1.22. The totalmass of the board is 15 kg.a) Calculate the weight of the rectangular section of the board.b) Calculate the weight of the square section of the board.c) Calculate how high above the base the centre of gravity of the whole
board is.
Friction
Friction exists between any two solid surfaces that slide over one another. Walkingand running would be very difficult without friction (e.g. on ice). See Fig. 1.23.However, friction causes undesirable wear and heating in machines and engines.Friction always acts on a surface in the opposite direction to the movement of thesurface. Oil between two solid surfaces reduces the friction because it keeps themapart and reduces the points of direct contact.
1.4
4.0 m
Fig. 1.21
0.4 m
2.0 m
0.4 m
0.2 m
Fig. 1.22
Push on
shoefrom foot
Friction on shoe
Fig. 1.23 Forces on a runningshoe
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MECHANICS20
Measuring friction
The force of friction between two flat surfaces can be measured by:
. measuring the force needed to pull a block across a fixed horizontal surface. Oneof the surfaces to be tested is fixed to the underside of the block. The othersurface is the fixed surface. If the applied force is gradually increased from zerousing a spring balance, as in Fig. 1.24, the block does not slide until the appliedforce reaches a certain value. Once the block starts to move, the applied forcebecomes slightly less. At the point of sliding, the force needed to make the blockmove is equal to the maximum possible frictional force between the two surfaces,referred to as the limiting value of the frictional force
. using an inclined plane as in Fig. 1.25. The angle of the incline is increased untilthe block on the plane is at the point of sliding down the incline. The limitingvalue of the frictional force is equal to the component of the block’s weight,W sin �, acting down the incline.
The coefficient of static friction,
The arrangement in Fig. 1.24 may be used to prove that the limiting value of thefrictional force, F, is proportional to the weight of the block. Since the weight of theblock on a horizontal surface is equal and opposite to the normal reaction, N, of thesurface on the block, the frictional force F is proportional to N. In other words, theratio F=N is a constant for the two surfaces. This ratio is referred to as thecoefficient of static friction, �.
FN
In Fig. 1.25, the normal reaction on the block is equal to W cos �, the componentof the block’s weight normal to the surface. If the plane is made steeper, the blockwill slip if its component of weight parallel to the slope, W sin �, exceeds thefrictional force. At the point of sliding, the frictional force is equal to W sin �.Therefore the coefficient of static friction is
� ¼ F
N¼ W sin �
W cos �¼ tan �, since
sin �
cos �¼ tan �.
1.4 WORKED EXAMPLE
A cupboard of weight 350 N is pushed at steady speed across a horizontal floor, by a force of 150 N appliedhorizontally near its base, as in Fig. 1.26. Calculatea) the coefficient of static friction between the two surfaces that slide over each other,b) the force needed if the weight of the cupboard is reduced to 200 N by removing some of its contents.
m
Pulled inthis direction
Pull onthe block
10N
0Frictionalforce F
Block Fig. 1.24 Measuring friction on ahorizontal surface
BlockF
Centre ofgravity Normal
reaction N
W cos
W sin
W
Fig. 1.25 Measuring frictionusing an inclined plane
Copyrighted material – 9781137443236
Copyrighted material – 9781137443236
Copyrighted material – 9781137443236
Index
A page number followed by N indicates that relevant information on that page is given only in a boxed note.
rates of changecamera 155–6capacitance 213capacitor smoothing
(a.c.) 304
capacitors 211–24in a.c. circuits 307–10,
313–16combinations 214–16design factors 219–20discharging 221–4energy stored in charged
capacitor 217–18reactance 307–11, 313–16
carsbatteries 194–5engine power and
speed 57hydraulic brakes 108seat-belt warning
light 229–30cathode rays 319–23Cavendish, Henry 441Ncells (electrical) 194–6
internal resistance 195–6,201N
standard 206Celsius temperature scale 79,
405, 417centigrade temperature
scale 417Ncentre of gravity 17–18centripetal
acceleration 433–5centripetal force 434Chadwick, James 381chain reactions (nuclear) 376changes of state 69, 425NCharles’ law 404–7chemical energy 370Chernobyl nuclear
accident 379chromatic aberration 155circuit breaker 266circuits see electric circuitscircular motion 431–9
and simple harmonicmotion 456–7
closed polygon rule 12, 21cloud chamber 356, 381
543
Copyrighted material – 9781137443236
Copyrighted material – 9781137443236
Copyrighted material – 9781137443236
INDEX544
coefficient of staticfriction 20
coefficient of thermalconductivity 89–90
coefficient of thermalexpansion 80
coherent light sources 146coils, in a.c. circuits 307,
311–18collisions 47, 48colours of light 160comets 450communications
microwaves in 168radio waves in 168–9
commutator 267–8compass 263compounds 71, 72computers, capacitors in 221concave lens 153conduction see electrical
Ffairground rides 435–7falling objects 35–7, 43,fan-out 228farad (unit) 213Faraday’s law of
electromagneticinduction 284–6
fast breeder reactor 377feedback (amplifiers) 235ferromagnetism 275–6fields see electric fields;
gravitational field strength;magnetic fields
film badges 341, 379flotation 112–14flow line 390focal length
concave lens 154convex lens 152–4
force carriers 368–9force diagrams 21, 23force–distance graphs 103forced oscillations 463forces
and acceleration 41–5centripetal 433–4frictional 19–21fundamental 367–8impulse 48measuring 95moment of 15and momentum 45–7parallelogram of 12pressure 106–7resolving into
components 13resultant 12, 13turning 15–16as vector quantities 12–13and work 53–4
Fortin barometer 111fossil fuels 60, 61four-stroke petrol engine 418free oscillations 462free-body force diagrams 21,
23frequency 120
angular 457Nmeasuring (sound
waves) 132resonant 463fundamental 137–8,
139–42frequency modulation 169friction 19–21, 58, 59fuels, supply and
demand 60–1
full-wave rectification 303–4fundamental
frequency 137–8, 139–42fuses 187fusion, latent heat of 84
GGalileo Galilei 1gamma (g) radiation 171,
353ionising power 356range and penetrating
power 356–8gas laws 402–7
ideal gas equation 407–8kinetic theory 408–11
gas thermometer 417–18gases 69, 77, 402–14Gauss’s law 257Geiger–Muller tube 355geostationary satellites 451geothermal power 63glasses 70gold leaf electroscope 178–9gradient of straight line 35grains (metals) 70, 101–3graphs 35gravitation 440–52gravitational field
strength 443–5gravitational potential
energy 54, 446–8gravity
acceleration due to 35–7,43–4, 461
force of 17, 43, 367, 441gravity wheel 436grid system 293ground state of atom 335–6