CORNELL UNIVERSITY LIBRARY
ENGINEERING
„_r All
books are subject
to recall after
Engineering Library
two weeks.
5
w
W The
Cornell University Librar y
original of this
book
is in
the Cornell University Library.
There are no known copyright
restrictions
the United States on the use of the
in
text.
http://www.archive.org/details/cu31924003964123
ELEMENTARY VECTOR ANALYSIS WITH APPLICATION TO GEOMETRY AND PHYSICS
BY
C.
E.
WEATHERBURN
M.A. (Came.), D.Sc. (Sydney) MATHEMATICS AND THEORETICAL PHYSICS, ORMOND COLLEGE UNIVERSITY OF MELBOURNE UNIVERSITY EVENING LECTURER IN PURE MATHEMATICS
LECTURER
IN
G.
LONDON BELL AND SONS, 1921
LTD.
$6-1
V/3&
jc^s
3
TO
JOHN HENRY MICHELL,
F.R.S. «
WHOSE HELP AND ADVICE
HAVE BEEN AT ALL TIMES MOST WILLINGLY
GIVEN,
AND TO WHOSE INSPIRATION THE WRITING OF THE FOLLOWING PAGES WAS LARGELY DUE, THIS BOOK
IS
GRATEFULLY ASCRIBED.
PREFACE My object in writing this
book was to provide a simple exposition and to show how it may be employed with advantage in Geometry and Mechanics. It was thought unnecessary, in the present volume, to enter upon the more advanced parts of the subject, built upon the ideas of gradient, curl and divergence. Vector algebra and the
of elementary Vector Analysis,
differentiation of vectors with respect to one scalar variable
furnish a powerful instrument even for the higher parts of
dynamics.
The work does not claim to be a complete text-book in either Geometry or Mechanics, though a good deal of ground is covered in both subjects. The use of vector analysis in the former is abundantly illustrated by the treatment of the straight hne, the plane, the sphere and the twisted curve, which are dealt with as fully as in most elementary books, and a good deal more concisely. In Mechanics I have explained and proved all the important elementary principles. The equations of equilibrium for a rigid body are deduced from the equations of motion. This
is
contrary to the ordinary practice and, of course,
recommended
for
young beginners.
for a student
is
not
who
is
certainly desirable to
show that
statics are only particular cases of the
dynamical
able to read this volume,
the principles of
But
it is
argument followed by text-books in Statics, to prove the theorems about moments, parallel forces, couples and the equilibrium of bodies, is really unnecessary. All these theorems are immediately deducible from the equations
ones,
and that the long
line of
motion of a rigid body, as shown in Chapter VIII. Another departure from the ordinary practice has been made in connection with the theory of centroids. Most students gain
of
VECTOR ANALYSIS
viii
But
their introduction to centroids through centre of gravity.
at a later stage they should learn that centre of gravity
is
only
a particular case of centroids, and that a presentation of the subject may be given which includes all cases. Arts. 9-11 were written with this object in view. It is because most students regard centre of gravity as the very essence of centroid that we continually meet such expressions as " centre of gravity of an area " or " centre of gravity of a cross section."
area and volume exist in their of
mass and weight. In treating the geometry
my object is
plane and sphere, method and notation, over other methods. The
of the straight line,
primarily to explain the vector
and not to show their superiority reader must decide for himself which tion
Centroids of
own right, and are quite independent,
is
preferable.
In connec-
with twisted curves, the use of vectors seems decidedly
In the case of the plane and the sphere
advantageous.
chiefly brevity of expression that
work
of the geometrical
gained
is
in Chapters III.
;
and IV. with the
in the sequel
is
;
and any two-dimensional
introduced merely
tion of the vector method.
will
Vectors were, hotvever, not designed
for use in elementary plane geometry
geometry
is
corre-
sponding theory given in books on analytical geometry be instructive to the reader.
it
though a comparison
by way of easy illustra-
Vector analysis
is
intended essentially
and its greatest service is. rendered in the domains of mechanics and mathematical physics. After much consideration I decided to employ the dot and for three-dimensional calculations
;
cross notation for products of vectors.
to
me the most
This has always appeared
convenient, particularly for the treatment of the
linear vector function.
During the preparation
of this book I have been greatly inH. Michell, M.A., F.R.S., who read the MS. and proof sheets, and made many valuable suggestions that have been incorporated in the work. I was allowed free use of his honour lectures in Mixed Mathematics, Part I., at Melbourne University and it was his manner of treating the subject
debted to Mr.
J.
;
me
to undertake the study of Vector Analysis. His interest and encouragement in the writing of this book
that
first
led
have been largely responsible for
its final
appearance.
My
PREFACE
ix
W.
P. Milne, the editor of this
thanks are also due to Prof.
series,
who
which
I
MS. and made many excellent suggestions was glad to adopt. Acting on his advice, I added the also read the
Historical Introduction, which will prove interesting to readers.
am
I
of
Ormond
to
some extent influenced
College, for
my
I take this opportunity of of
many
K. Picken, M.A., Master certain introductory ideas, which have
also indebted to Mr. D.
And W. Wark, B.Sc,
presentation of the subject.
thanking Mr. Ian
Ormond College, who generously undertook the task of verifying
the exercises to each chapter, and furnishing answers where necessary.
I
MacKobert,
of
also grateful to my college friend, Dr. T. M. Glasgow University, who kindly offered to revise
am
the final proofs.
have elsewhere * acknowledged my great indebtedness to E. B. Wilson's Vector Analysis, which was my early instructor in the subject and during the writing
As
to other literature, I
;
of the following pages I
unconsciously Analysis,
consulted
by
was influenced both consciously and
Professor
Wilson's
another American book,
by the
book.
has also
Coffin's
Vector
been frequently
author. .to thank the Publishers for their and the Printers for the excellence of their
In conclusion I wish unfailing courtesy,
work. C.
E.
WEATHERBURN.
Ormond College, University of Melbourne, April, 1920.
*
"
A plea for a more general use of Vector Analysis in Applied Mathematics,'
Math. Gazette, Jan. 1917.
CONTENTS Page
Preface
vii
Historical Introduction Short Courses
xvii
xxvii
CHAPTEE
I.
ADDITION AND SUBTRACTION OF VECTORS. CENTROIDS. ART. 1.
Scalar and vector quantities
2.
Length vectors
1
3.
Definitions of terms
2
1
4.
Addition and subtraction of vectors.
5.
Multiplication
6.
Resolution of a vector
7
7.
The
8
8.
Division of a line in a given ratio
9.
Centroid, or centre of
Component and Resultant
-
by a number
unit vectors
i,
j,
6
k
mean
4
9
10
position
10.
Centroids of area and volume
11.
Centre of mass
12.
Relative position.
13.
Concurrent forces.
14.
Solution of examples
12
-
12
-
Relative displacement.
Uniform
relative velo-
14
city
Vector polygon.
Lami's Theorem
15 17
18
Exercises
CHAPTER
II.
ELEMENTARY GEOMETRICAL ILLUSTRATIONS AND APPLICATIONS. 21
16.
Introductory Vector equation of a straight
17.
Bisector of the angle between two straight lines
18.
The Triangle
15.
-
line
21
23 24
VECTOR ANALYSIS
Xll
Pa ge
Akt.
25 26
The Tetrahedron
19.
20-1.
Vector equation of a plane
Linear relation independent of the origin Theorem for closed polyhedra
22.
Vector areas.
23.
30 30
-
32
Exercises
CHAPTER
III.
PRODUCTS OF TWO VECTORS.
THE PLANE AND
THE SPHERE. Scalar and Vector Products. 24.
New
25.
Scalar product of two vectors
34 34
use of terms product and multiplication
26.
The
27.
Vector product of two vectors
28.
The
29. 30.
Vector equation of a plane r. n= q Distance of a point from a plane
31.
Plane through the intersection of two planes
32.
Distance of a point from a straight line
36
distributive law -
distributive law
39 40
Geometry of the Plane.
42
-
-
44 45 46
Geometry of the Sphere. Vector equation of a sphere 34. Equation of the tangent plane at a point
33.
47
35.
Polar plane of a point
48 49
36.
Diametral plane for parallel chords Radical plane of two spheres. System of spheres with a
37-8.
radical plane
40.
Work done by a force Vector moment or torque
41.
Angular velocity
Exercises
51
-
Application 39.
51
common
of
to
Mechanics.
about a point body about a fixed axis
.
54
-
54 56
of a force
a rigid
-
-
57
CHAPTER
IV.
PRODUCTS OF THREE OR FOUR VECTORS. NON-INTERSECTING STRAIGHT LINES.
CONTENTS Art. 46.
A
47.
System
vector product of four vectors.
xill
P AGE 64
Relation between four vectors
of vectors reciprocal to a, b, c
65
Further Geometry of the Plane and Straight Line.
49.
Planes satisfying various conditions Condition of intersection of two straight lines
50.
The common perpendicular
51.
Pliicker's coordinates of a straight line
69
52.
Volume
71
53.
Two
48.
to
66 67
two non-intersecting straight
of a tetrahedron
67
lines
-
Other Applications.
54.
formulae of Spherical Trigonometry Rankine's theorem for four concurrent forces
72 73
-
Exercises
74
CHAPTBE
V.
DIFFERENTIATION AND INTEGRATION OF VECTORS. CURVATURE AND TORSION OF CURVES. Differentiation 55.
and
Integration.
Derivative of a vector with respect to a scalar variable.
Differentia-
78 80 82
tion 56.
Derivatives of products
57.
Integration of vectors and products of vectors
58.
Tangent to a curve at a given point
59.
Curvature.
60.
Binormal.
61.
Definite integral of a vector function
62.
Illustrations
Curvature and Torsion.
Principal normal.
84
Plane of curvature or osculating
85 86
plane
Torsion Definite Integrals.
63.
Line integral of a vector function
64.
Surface integral of a vector function
88 89 90
-
91
92
Exercises
CHAPTER
VI.
KINEMATICS AND DYNAMICS OF A PARTICLE. Kinematics.
Theorem of vector addition of an instant. Theorem of vector addition
65.
Velocity at an instant.
velocities
66.
Acceleration at
of accelera-
67.
Tangential and normal resolutes of acceleration
tions
-
98 99 99
VECTOK ANALYSIS
Xiv
PAQE 1"" *" 103
AM. 68.
Radial and transverse resolutes of velocity and acceleration
69.
Areal velocity about a point
70.
Motion with constant acceleration
71.
Momentum
72.
Newton's Second Law of Motion Impulse of a force. Impulsive forces Activity of a force -
Dynamics.
73. 74..
103
-
105 105
75.
The
76.
Moment
77.
Central forces
78.
Central force varying inversely as the square of the distance
79.
Planetary motion
principle of energy
106
-
momentum
of
10 *
-
or
angular
momentum about
Principle of A.M.
80.
Central force varying directly as the distance
81.
Motion of a particle on a fixed curve
a point.
-
107
-
108 108 110 112 114
-
Exercises
116
CHAPTER
VII.
DYNAMICS OF A SYSTEM OP PARTICLES AND OF A RIGID BODY. Dynamics
momentum
of a System.
86.
system Acceleration of the centre of mass Angular momentum about a point Moving origin of moments. Centre of mass as origin Equations for impulsive forces
87.
Motion of a
82.
83. 84.
85.
Linear
121
of the
122 122
-
124 125
Kinematics of a Rigid Body. rigid
of rotation
body about a
fixed point.
Instantaneous axis 126 128
-
88.
General motion of a rigid body.
89.
Simultaneous motions
90.
Angular
Screw motion
-
130
Dynamics of a Rigid Body.
momentum
of a rigid body.
Moments and products
inertia
of
92. Kinetic
energy of a rigid body
93.
Principle of energy
94. 95.
Moving axes or frame Coriolis' Theorem -
96.
Euler's dynamical equations
Exercises
-
31
.
1
.
132 134
91. Principal axes of inertia
I35 of reference
13g I37 -
138 140
CONTENTS
CHAPTER
xv
VIII.
STATICS OF A RIGID BODY. AKT. 97.
PASB 144 145
Conditions of equilibrium of a rigid body
Equivalent systems of forces Centre of gravity 100. Couples. Composition of couples 101. Poinsot's reduction of a system of forces. 98. 99.
146 147
Parallel forces.
Central axis.
wrench
148
102.
Null plane at a point
103.
Conjugate forces
104.
Principle of Virtual
105.
Equilibrium of a string under any forces
106.
Equilibrium of a wire under any forces
151
152
Work,
Exercises
Summary Answers to Exercises Index
Equivalent
-
or Virtual Velocities
-
153 154 155 156
160 179 183
HISTORICAL INTRODUCTION * The method by
of subjecting vector quantities to scalar algebra
resolution into three
sopher
Descartes
components
is
due to the French philo-
The need
(1596-1650).
of
a
calculus for
operating directly on vectors has long been recognised in 1679 Leibnitz
with
little
made an attempt
success.
;
The problem attracted the attention
subsequent writers, and in 1806 Argand showed
how
of
unreal
quantities
by coplanar
proved of considerable importance in the theory variables
;
but at the same time
it
yet entirely disappeared.
—an A
vectors of
This
has
complex
gave the unfortunate impres-
sion that the theory of real vectors
that of complex numbers
of
a geometrical
representation could be given to the complex number. representation
and
at meeting the need, but
is
necessarily dependent on
impression which has not even
little
later,
in 1826,
appeared the
by Mobius, one of the best known of Gauss's pupils. This work is a forerunner of the more general analysis of geometric forms subsequently developed by Grassmann. The Calcolo delle Equipollenze devised by Bellavitis in 1832, and subsequently enlarged, actually deals systematically with the geometric addition of vectors and the equality of Barycentrisches
Calcul
vectors.
The years 1843-44 are remarkable
in the history of
mathematics
for the almost simultaneous appearance of Hamilton's Quaternions
and the Ausdehnungslehre of Grassmann. William Rowan Hamilton was born at Dublin on 4th August, 1805. His father, Archibald Hamilton, had migrated from Scotland in his youth. * This Introduction is not essential to the argument of the book. The author hopes, however, that it will add to the value and general interest of work. the
xvm
VECTOR ANALYSIS
The son gave early evidence of genius, being a remarkable and displaying great mathematical talent. He entered Trinity College, Dublin, in 1824, where he had a brilliant and unprecedented career. His ability was so conspicuous that in 1827, while still an undergraduate, he was asked to apply for the vacant Andrews' Professorship of Astronomy in the UniHe was versity of Dublin, and was appointed to the position. conthe but not specially qualified as a practical astronomer linguist
;
appointment allowed him to advance the cause In 1835, while of Science in the way he felt best able to do so. acting as secretary to the B.A.A.S. at its meeting in Dublin, he received a knighthood and two years later the importance ditions of his
;
work was recognised by his election as President His mathematical work continued of the Royal Irish Academy. uninterrupted till his death on 2nd September, 1865, at the age of his scientific
of sixty.
It often
happens that we get our most important ideas while
not formally working at a subject, perhaps while walking in
more commonplace surroundings. From a letter of Hamilton's we learn that, on 16th October, 1843, while he was walking beside the Royal Canal on his way to preside at a meeting of the Academy, the thought flashed into his mind which gave the key to a problem that had been occupying his thoughts, and led to the birth and development of the subject of Quaternions. He announced the discovery at that meeting of the Academy, and asked permission to read a paper on quaternions at the next, which he did on 13th November. During the next few years he expanded the subject, and published his Lectures on Quaternions in 1853, the
country or by the
sea,
or even in
while the Elements of Quaternions appeared in 1866, soon after his death.
In August, 1844, appeared the
first
edition of Grassmann's
Lineale Ausdehnungslehre, a treatise of over 300 pages.
Hermann
Grassmann was born at Stettin on 15th April, 1809, died at the same place in 1877. He held the post of instructor and in mathematics and science at a gymnasium in his native town. The systems of Hamilton and Grassmann may be regarded as the parents of modern Vector Analysis. The two authors. Giinther
HISTOBICAL INTRODUCTION
xix
working independently and along different lines, each developed a wonderful analysis. The quaternion is a sort of " sum " or complex of a scalar and a vector, though originally defined as the " quotient " of two vectors. algebra of geometric forms. of vectors,
and both,
The Ausdehnungslehre is an Both systems contain an algebra
as finally developed, a theory of linear
In Hamilton's we have also the linear quater-
vector functions.
nion function, and in Grassmann's the linear function applied " Grassmann's algebra to the quantities of his algebra of points. of points
may
be regarded as the application of the methods
of multiple algebra to the notions
connected with tetrahedral
coordinates, just as his or Hamilton's algebra of vectors
may
be regarded as the application of the methods of multiple algebra to the notions connected with rectilinear coordinates." *
Each of the above systems is a remarkable and potent instrument of analysis and the devotees of each have faithfully striven to prove its power and utility in the various branches of mathematics. Among Hamilton's disciples the most noted was Prof. P. G. Tait. Peter Guthrie Tait was born at Dalkeith Scotland, on 28th April, 1831. He was educated at the Edinburgh Academy, and then for one session (1847) at the Edinburgh University. The following year he proceeded to Cambridge, ;
where he entered Peterhouse before his eighteenth birthday. He became Senior Wrangler, and in 1852 was elected Fellow and Lecturer of Peterhouse, where he remained for two years
At the end of that time he was appointed Professor Mathematics at Queen's College, Belfast. Here he was introduced to Hamilton and Quaternions, and became a staunch friend of both. In 1860 he was appointed to the Professorship longer. of
of
Natural Philosophy in the University of Edinburgh, a position
which he held
till
his
death in 1901.
His Elementary Treatise
on Quaternions was published in 1867, and a second edition in 1873.
However, neither the system
met the needs
of
Hamilton nor that
of
Grassmann
of physicists or applied mathematicians, being
too general and too complex for the requirements of ordinary * Gibbs, "
(1891).
Quaternions and the Ausdehnungslehre,'" Nature,
vol. 44, pp.
79-82
XX
VECTOR ANALYSIS
calculations.
The
quantities
mechanics and physics are
of
involved
ideas
the
in
scalar
and vector simpler than
much
those of Hamilton's theory, in which imaginaries play a large and vectors and scalars appear as degenerate quater-
part,
own
nions rather than in their
The
right.
feeling
became general
that a system was needed in which the ideas were more simply Mathematicians in various related to the facts of nature. therefore began to adapt the results of Hamilton and and Grassmann to the more elementary requirements although these investigations were carried on independently and from different points of view, the analyses arrived at are identical as regards the elements and functions introduced. It is mainly in notation and terminology that the differences lie. In Germany the starting point was the Ausdehnungslehre and among those who contributed to the formation and adoption countries
;
;
of
a
simpler
analysis
may
mentioned Foppl, Abraham,
be
Bucherer, Fischer, Ignatowsky and Gans. deserves special mention
;
In England, Heaviside
while in America, Prof.
W. Gibbs
did
much admirable work. Josiah Willard Gibbs was born at New Haven, Connecticut, on 11th February, 1839. His father, who bore the same name, was Professor of Sacred Literature in the Yale Divinity School from 1824 till 1861. The son entered Yale in 1854, and graduated
four years later after a distinguished career. his studies at
New Haven, and
He
continued
in 1863 received the degree of
Ph.D., being then appointed Tutor at Yale for a period of three
At the end
years.
of this
term he visited Europe, studying and at Berlin and Heidelberg
at Paris during the winter of 1866-67,
during the ensuing two years. in June, 1869,
and two years
later
He
returned to
was appointed
New Haven
to the Professor-
ship of Mathematical Physics at Yale, a position which he held till
his
death on 28th April, 1903.
As an
investigator in Mathe-
matical Physics, Gibbs soon gave evidence of his powers by the publication of several papers in Thermodynamics, among
which the well-known memoir On the Equilibrium of HeteroSubstances has proved of fundamental importance In the Electromagnetic Theory of to Physical Chemistry. Light also, Gibbs did much work of permanent value and
geneous
;
HISTORICAL INTRODUCTION
xxi
learned societies and Universities, both in Europe and in America, formally recognised the merits of his contributions to science.
In lecturing to students Prof. Gibbs felt the need of a simpler form of Vector Analysis than was then available. Being familiar with the work of both Hamilton and Grassmann, he was able to adapt to his requirements the best and simplest parts of both systems, thus developing an analysis which he used freely in his
In 1881 and 1884 he printed at
University teaching.
privately for the use of his pupils, a
pamphlet
New Haven,
entitled Elements
of Vector Analysis, giving a concise account of his system.
This
pamphlet was to some extent circulated also among others It was not till twenty years specially interested in the subject. later that Prof.
Gibbs reluctantly consented to the formal publica-
tion, in a fairly
he was "
complete form, of the vector analysis to which
led.
The reluctance
of
Professor Gibbs to publish his system
of vector analysis certainly did
own mind
widely employed that
it
not arise from any doubt in his
as to its utility, or the desirability of its being it
;
was not an
seemed rather to be due
more
to the feeling
original contribution to mathematics, but
was an adaptation, for special purposes, of the work of others. Of
many
portions of the
and it is rather by the
work
selection of
of the presentation that the
vector analysis.
this
But
in
is
of course necessarily true
;
methods and by systematization
author has served the cause of
the treatment of the linear vector
function and the theory of dyadics to which this leads, a distinct
advance was made which was
of
consequence not only in the
more
restricted field of vector analysis, theory of multiple algebra in general." *
Meanwhile
in
but also in the broader
England Oliver Heaviside was engaged
in
a
His work in the Electromagnetic Theory led him first to study quaternions as probably what he needed to simplify the analysis, and then to reject them as totally unsuitable. In similar task.
adapting the results of Hamilton and Tait to his own requirements he arrived at a vector algebra practically identical with that of Gibbs.
The difference of notation was
* P. xix of the Biographical
Papers.
of course to be expected,
Sketch by H« A. Bumstead in Gibbs's
Scientific
;
VECTOR ANALYSIS
xxn
Heaviside adhering partly to that employed by the quaternionists,
but introducing the admirable practice of representing
by Clarendon symbols. On receiving a copy of Gibbs's New Haven pamphlet, Heaviside expressed his warm admiration and approval, though still preferring his own notation. The work of Gibbs and Heaviside drew forth denunciations from Prof. Tait, who considered any departure from quaternionic usage in the treatment of vectors to be an enormity. " Even Prof. Gibbs," he wrote,* " must be ranked as one of the retarders vectors
quaternion progress, in virtue of his pamphlet on Vector
of
Analysis, a sort of hermaphrodite monster
compounded
notations of Hamilton and of Grassmann." well able to look after himself,
the best of the argument.
He
Prof.
of the
Gibbs was
in his reply f had a long way wisely separates the quaternionic
and
question from that of a suitable notation, and argues powerfully against the treatment of vectors
by quaternions.
The
discussion
thus begun, continued for some years with Tait and other quater-
on one side, and Gibbs and Heaviside on the The contributions of Heaviside add a human touch the controversy, and make very interesting reading even
nionists ranged
other.
to
at the present day.
"
'
Quaternion
girl to
be
'
'
was, I think, defined by an American school-
an ancient
a complete mistake.
religious ceremony.'
The ancients
This was, however,
—unlike
Prof.
Tait—knew
not and did not worship Quaternions." J "It is known that Sir W. Rowan Hamilton discovered
or
invented a remarkable system of mathematics, and that since his of
death the quaternionic mantle has adorned the shoulders Tait, who has repeatedly advocated the claims of
Prof.
Quaternions.
Prof. Tait in particular emphasises its great power, simplicity, and perfect naturalness, on the one hand and on the other tells the physicist that it is exactly what he It is also known that physicists, with great obstinacy, have been careful (generally speaking) to have nothing to do with Quaternions and, what is equally
needs for his physical purposes.
;
* Preface to the third edition of Quaternions. t Nature, vol. 43, pp. 511-13 (1891). J
Electromagnetic Theory, vol.
1, p.
136 (London, 1893).
HISTORICAL INTRODUCTION remarkable, writers
is
who take up
speaking)
(generally
the subject of Vectors are
Quaternions
possessed of the idea that
not exactly what they want, and so they go tinkering at
trying to
make
a
it
little
more
intelligible,
who would stream pure and undefiled. Now, disgust
Prof.
of
the defilers right
answer
all
?
Opinions
may
entirely right,
very
much
it,
to the
preserve the quatemionic
Tait,
is
Prof.
differ.
Tait right, or are
My own
is
that the
we put aside and look upon Quaternions
depends upon the point
practical applications to Physics,
is
xxiii
of view.
from the quatemionic point
of view,
If
then Prof. Tait
thoroughly right, and Quaternions furnishes a uniquely
simple and natural emphasis." *
way
of treating quaternions.
Observe the
"
But when Prof. Tait vaunts the perfect fitness and naturalness by the physicist in his enquiries, I think that he is quite wrong. For there are some very serious drawbacks connected with quaternions, when applied to vectors. The quaternion is regarded as a complex of scalar and vector, and as the principles are made to suit the quaternion, the vector itself becomes a degraded quaternion, and behaves as a quaternion. That is, in a given equation, one vector may be a vector, and another be a quaternion. Or the same vector in one and the same equation may be a vector in one place, and a quaternion in another. This amalgamation of the vectorial and quatemionic of quaternions for use
functions
is
very puzzling.
You
never
know how
things will
turn out." f "
However, things changed as time went on, and
after
a
period during which the diffusion of pure vectorial analysis
made much
progress, in spite of the disparagement of the Edin-
whom said some of my work was a disgrace to the Eoyal Society,' to my great delight), it was most gratifying to find that Prof. Tait softened in his harsh judgments, and came to recognise the existence of rich fields of pure vector analysis, and to tolerate the workers therein. Besides those impertinent tamperers, Tait had to stick up for burgh school of scomers (one of '
quaternionics against Cayley, for quite different reasons.
was danger
There
of a triangular duel, or perhaps quadrangular, at * Ibid. p. 301.
t
Ibid. pp. 302-3.
;
VECTOR ANALYSIS
xxiv one time, but
I
would not engage
appeased
I
in it for one.
Tait considerably (during a little correspondence we had) by disclaiming any idea of discovering a new system. I professedly derived my system from Hamilton and Tait by elimination
and simplification, but all the same claimed to have diffused a working knowledge of vectors, and to have devised a thoroughly practical system." *
Early in the present century, when the utility of Prof. Gibbs's system had been proved by twenty years' experience, he consented to its publication in a more extended form. Not having the leisure to
of his
undertake this work himself, he entrusted
it
to one
former pupils, Dr. Edwin Bidwell Wilson, then instructor
Mathematics at Yale, now Professor of Mathematics in the Dr. Wilson was allowed
in
Massachusetts Institute of Technology. free
He
scope in his presentation of the subject. f
in the
main the notation and methods
followed
but adopted
of Gibbs,
Heaviside's suggestion of Clarendon symbols for the representation of vectors.
The
success of his undertaking
is
well known.
Professor Wilson has also contributed to the vector analysis four dimensions arising in connection with the theory
of
of
Relativity. J
The present century has also witnessed the appearance of an Italian school of vector analysts, represented by Prof. R. Marcolongo, of the University of Naples, and Prof. C. Burali-Forti, the Military
of
Academy
substantially the
same
of
Turin.
Their vector algebra
pendent notation for products of vectors
and
in the
differentiation
largely influenced
the
Barycentric
an
as that of other schools, with
of
;
is
inde-
but both in
this
vectors the authors have been
by the geometric forms of Grassmann and of Mobius. For the linear vector
Calculus
function they have developed the properties of the homographs in place of the dyadic, being in this matter influenced chiefly
by the work will
of Hamilton. The simpler portions of their system be found in their Elements de Calcul Vectoriel (Paris, 1910)
* Electromagnetic Theory,
vol. 3, p.
137 (1912).
Yale University Press, 2nd ed. (1909). Wilson and Lewis, Proc. Arner. Acad, of Arts and Sciences vol
f Vector Analysis (1901). |
pp. 391-507.
' '
4-8
HQ191>' Klvl
HISTORICAL INTRODUCTION
xxv
but for a
full account the reader is referred to their larger work, Analyse Vectorielle Generate (Paris, 1912). It is apparent then that the processes of current vector analysis
have sprung from the work of Hamilton and Grassmann. The order of development of the subject has been the opposite of what one might have expected. " Suppose a sufficiently competent mathematician desired to find out from the Cartesian
mathematics what vector algebra was
and
like,
its
laws.
He
could do so by careful inspection and comparison of the Cartesian
He would find certain combinations of symbols and quantities occurring again and again, usually in systems of threes. He might introduce tentatively an abbreviated notation for these combinations. After a little practice he would perceive the laws according to which these combinations arose and how they operated. Finally, he would come to a very compact system in which vectors themselves and certain simple functions of vectors appeared, and would be delighted to find that the rules for the multiplication and general manipulation formulae.
of these vectors were, considering the complexity of the Cartesian
mathematics out
of
which he had discovered them,
incredible simplicity.
But there would be no
of
an almost
sign of a quaternion
and, for another, there would be no metaphysics or abstruse reasoning required to establish in his result, for one thing
i
;
the rules of manipulation of his vectors." *
This is the manner Analysis have originated. Vector to would expect in which one in many quarters counted parentage has But it did not and its rapidly disappearing, and the prejudice is against it. But this more and more popular. simple vector methods are becoming In advanced three-dimensional work in nearly every branch of ;
.
mathematical physics, writers are finding
it
almost indispensable.
The lack of uniformity in the notation for products of vectors was more or less inevitable, but may yet be overcome. In America the dot and cross of Prof. Gibbs are employed almost without exception. In Germany the bracket notation is the general rule. The practice in England is by no means uniform. The influence of Lorentz's work has led some writers in the Electromagnetic Theory to follow his example in the use of * Heaviside, loc.
cit.
vol. 1, p. 136.
VECTOR ANALYSIS
XXVI brackets
while others adhere to the American usage. author of these pages considers Gibbs's notation easier and ;
The
much
more elastic, especially for the treatment of linear vector functions. Gibbs's work on dyadics is the most original and important part of his theory, and will be found of great service in our second volume.
TABLE OF NOTATIONS
*
CHAPTER
I.
ADDITION AND SUBTRACTION OF VECTORS. CENTROIDS. Definitions. 1. A scalar quantity, or briefly a scalar, has magnitude, but is not related to any definite direction in space. Examples of such, are mass, volume, density, temperature, work, quantity
and
of heat, electric charge
need a unit quantity
of the
potential.
To
same type, and the
the given quantity bears to this unit, so that as
m
times the unit.
The number
V,
d, to,
v,
E,
respectively,
mechanics
etc., of density,
that
and
enter
into
it
which
ratio (m)
may
be expressed
to is called the
the quantity in terms of the chosen unit.
we
specify a scalar
measure of
It is the
measures
mass, volume, speed and energy the
equations
and
physics
of
cannot be too strongly emphasized that these symbols denote only numbers, as in ordinary algebra.
A
;
it
vector quantity, or briefly a vector, has magnitude
related to a definite direction in space ties of
law tion,
the same kind are
momentum,
force,
is
to the triangle
Displacement, velocity, accelera-
electric
of vector quantities.
and
while two vector quanti-
compounded according
of addition stated below.
examples
;
To
and magnetic
intensities
specify a vector
are
we need not
only a unit quantity of the same kind considered apart from direction,
and a number which
is
the measure of the original quantity
in terms of this unit, but also a statement of its direction. 2.
Though ordinary algebra
is
adequate for the analysis of
both scalars and vectors, when applied to the latter
it is
often
very cumbrous, necessitating the manipulation of two or three equations instead of one, and the decomposition of the vector
[CH.
VECTOR ANALYSIS
I.
Hence quantities to meet the limitations of algebraic analysis. the desirability of an analysis which will bear to vectors the same relation that algebra bears to scalars. In the latter the elements of our equations are always numbers, denoted by
In vector analysis we require, as well as these, elements involving both number and direction directed numbers so to speak. For this purpose we choose what is perhaps the various symbols.
—
simplest type of vector quantity, viz. that whose magnitude is
A
a length.
P
0,
vector of this type
determined by two points
such that the magnitude of the vector
straight line
vector
is
OP, and
quantities, of the
direction
its
usually denoted
we have
after
by OP-
that from
is
to P.
For the purposes
any such quantity can be
This
of analysis,
specified
by
a length- vector
type OP, provided their directions are the same, and
OP
is
quantity considered in terms of then,
the length of the
is
settled the units of the various types of vector
the measure of the length of
and
is
OP
also the its
measure
of the vector
appropriate unit.
Briefly
can be used to specify the vector quantity in measure
direction,
and in
way answers
this
the purpose of a directed
number.
Having then decided to use length-vectors for our directed number elements, we shall find it convenient to abbreviate the name and call them simply vectors. To prevent confusion we. shall
henceforth
of this type.
The
confine
substantive
vector
most writers seems to be
practice of
though
the
to
length-vectors
All others will be spoken of as vector quantities. in
harmony with
this,
in the majority of cases the restricted use of the term
vector is only tacitly adopted.* We allow the wider meaning, but here adopt the narrower for the sake of brevity, and to avoid
misunderstanding.
all possibility of
3.
The module
the measure of unity.
is *
The term
We
its
of a vector is the positive
length.
shall
vector
A
number which
it
unit vector
is one whose module denote vectors by Clarendon letters, f and
was invented by Hamilton and the meaning he aasig we have just adopted. Cf. Lectures on Quatemii ;
to it agrees with the one
Lecture
1, p.
15.
t For purposes of writing, Greek letters and script capitals will be convenient to denote vectors.
found
DEFINITIONS
§ § 2, 3]
modules by the corresponding
their
PQ, QR, RS may be denoted by modules * by a, b, c. Unit vectors
these
in
Thus the vectors
italics.
a, b,
c respectively,
and
their
directions
AAA
be denoted by
will
3
a,
c
b,
respectively.
Two fined
vectors
be
b are de-
a,
if they have the same direction and equal lengths and this is denoted symbolically by a = b, which
to
equal
;
A
the two relations
A
a=b and
one whose module
is
a
A
= 6.
therefore equivalent to
zero vector, or null vector, is
All zero vectors are to be regarded
zero.
Like vectors are vectors with
as equal, irrespective of direction.
the same direction.
is
The vector which has the same module as is denned as the negative of a, and
a but the opposite direction is
denoted by - a. Thus, while the value of a vector depends on
'direction,
it
localised in
is
any
its
length and
independent of position, the vector not being
A
definite line.
single vector
cannot therefore
completely represent the effect of a localised vector quantity,
:
such as a force acting on a rigid body. -the line of action of the force
'two vectors are necessary for
When
I
i'as
force
:senting
and
;
velocity, they will always be length-vectors reprein
measure and
our equations will be
the
equation F = ma, used
direction.
of the
All the vectors entering
same kind.
for the
Newton's second law of motion,
i
be shown later that
its specification.
vectors F, v are used to specify vector quantities such
and them
-into
tof
This effect depends on
it will
m
For instance,
in
mathematical expression of is
the measure of the mass
the particle, and F, a vectors representing respectively the
force acting *
on the particle and the consequent acceleration.
The notation mod a or a |
|
is
also used for the
module
of a.
[CH.
VECTOR ANALYSIS
I.
Addition and Subtraction of Vectors. 4.
The manner
of
which the vector quantities
in
and physics are compounded is expressed by which may be stated as follows
of addition,
mechanics
the triangle law
:
If three points 0, P,
OR is
then the vector
R
are chosen so that
called the (vector)
sum
OP =
3,
and PR=b,
or resultant of a
and
b.
R
Denoting this resultant by
c,
we
write
c=a + b, borrowing the sign + from algebra, and using the term vectoi addition for the process
by which the resultant c is obtained from the components a and b. The above definition is not an arbitrary mathematical assumption.
way tities
is an expression of the which the vector quanof physics and mechanics
It
in
are compounded.
that
the
sum
of
We two
see
also
vectors
OR
a=OP and b = OQ is the vectoi determined by the diagonal of the parallelogram of which
OP
and OQ
are sides.
Pfi-£>=b,sothat
For
a+b . Op +p-^_
^
Thus the triangle law of addition is identical with the parallelogram law involved in the so-called " parallelogram of forces." Further, since
Q R = OP = a
it
follows that
b+& = OQ+QR = OR, showing that
b + a=a+b
=r
(say).
ADDITION AND SUBTRACTION
4]
Again,
we may add
to this another vector c
= RS,
obtaining
->
the result
0£=r + c = (a + b)+c
i
\
=e + (a + b).
But a glance
at the figure shows that this vector
is
also
OS = OP+P# = a + (b+c) = (b+c)+a, and the argument obviously holds for any number of vectors. Hence the Theorem. The commutative and associative laivs hold for the A& The sum is independent of the addition of any number of vectors. order and the grouping of the terms.
two
b
ref
ther<
We
have already stated that -b is to be understood as the which has the same length as b, but the opposite direction. The subtraction of b from a is to be understood as the addition We denote this by )f - b to a. rector
u
a-b = a + (-b),
VECTOR ANALYSIS
6
borrowing the - sign from algebra.
from
a, reverse the direction
Multiplication
number, *
its length.
a by
m
;
and
the vector in the
This
a
is
-ma
m
times
direction
either
and
of
by a negative and multiply by m.
= - a,
the above
may the
it is
any
positive
real
m
as the result of multiplying
to multiply a vector
b
is
In agreement with the preceding Art.,
From
b
the result of dividing a by m.
in the opposite direction to a,
its
I-
in the figure,
direction as a, but of
same
may be regarded
similarly
-
QO = RP=\),
and
by a number.
ma means
times
If,
a-b=6p+PR=OR. // m
then
CH
to subtract the vector
Thus
of b and add.
OP=QR=& 5.
[
clear that,
be expressed as
real
if
is
then the vector length.
Thus,
number -m,
reverse
its
a and b are
like vectors,
a multiple of the other.
number - being the
Thus
ratio of the length of b to that ,'
of a. tli
In particular
if
A
en
a
is
the unit vector in the direction a a
of
a,
a=oa and ma=m(aa) =(wa)a. Q'
The general laws
of association
and distribution
multipliers hold as in ordinary algebra.
numbers, positive or negative, it argument that m ( wa ) = ( mw j a = n
real
and
also that *
for
scalar
m
and n are any follows from the above
(
If
m&^
(m + n) a = ma + wa.
Imaginary and complex numbers are excluded from our discussion.
COMPONENTS OF A VECTOR
§ § 5, 6]
m (a + b) = ma + mb
Lastly, the formula
proved geometrically.
easily
is
—>
OQ=a+b.
But
that OP'
so
and
m
times
:OP = OQ':OQ=m,
if
OP=a OP
and PQ=b, then
and
then P'Q'
OQ
is
respectively
parallel to
PQ
Thus P'Q' = wb, showing that
in length.
it
m(a,
+ b)=OQ' =OP' +P'Q' =ma,+mb. Components
6.
For
P', Q' are points in
if
7
of a Vector.
Three or more vectors are said to be coplanar when a plane
can be drawn parallel to
all of
them
;
otherwise they are non-
coplanar.
Any
vector
can be ex-
r
sum
pressed as the
of three
any three
others, parallel to
Let
non-coplanar vectors.
given
three
directions.
non-coplanar
With any point
OP =r,
as origin take
OP
a,
be unit vectors in the
C
b,
and on
diagonal construct
as
parallelepiped with edges
OB, OC
parallel to
Then
respectively.
a, if
a
OA, c
b,
x,
y,
Fig.
z
are the measures of the lengths of
as the
sum
Thus
edges, r
is
expressible
-»--»•-»-»--»•
-* t
its
7.
= OA+AF + FP=OA+OB + OC = xa,+yb+zc.
the resultant of the three vectors x&, yb,
zc,
which
are called the components of r in the given directions.*
This
r is
resolution of r
unique, because only one parallelepiped can
is
OP
be constructed on given directions. *
x if
The numbers
as diagonal with edges parallel to the
Hence,
x, y, z
may
if
two vectors are equal the components
be either positive or negative.
be positive if the component of the opposite direction.
will
OA
For instance,
has the same direction as a
;
negative
;
.
[CH.
VECTOR ANALYSIS
one are equal to those of the other, each to each. Conversely, if two vectors have equal components they must be equal. Given several vectors vv r2 r3 ... they may each be resolved into components in the given directions, and expressed in Jh^iorin of the
,
r1
Their
sum
is
(x x a
,
=x1 a + «/1b + 21c,
then
+ yja + ZjC) + (a;2a + y 2b + z2c) + = {x1 +x 2 +x3 + ...)a + (y1 + y 2 + ...)b + (z 1 +2 2 + ...)C,
showing that vectors
. .
may
be compounded by adding their
like
components. 7.
The
unit vectors
resolution of vectors
is
i,
j,
k.
The most important
case
of
that in which the three directions are
^
Fig. s.
The right-handed system of directions is found most convenient OY and OZ are in the plane of the paper, and OX perpendicular to it pointing toward the reader. To an observer at the origin 0, right-handed rotations about the axes OX, OY, OZ are from Y to Z, Z to X and X to Y respectively. The unit vectors parallel to these axes are denoted by i, j, k and if x, y, z are the lengths of OA, OB, OC respectively measured in these directions, mutually perpendicular.
OX, OY, OZ
represented in the figure
;
the vector
OP
is
r
= xi + yj + zk.
UNIT VECTORS
§ § 7, 8] If a,
cos '
!
"
!
are the angles which
y
f3,
(3,
cos
makes with the
axes, cos a,
are called the direction cosines of the line
y
'
x A
so that
r
Thus the
OP
9'
=r
cos a,
=r
y
cos
= r cos
2
/3,
OP
;
and
y,
1
= -r = (cosa)i + (cos,8)j +(cosy)k.
coefficients of
i,
j,
k
the
in
rectangular resolution
of a unit vector are the direction cosines of that vector relative
Rectangular components are generally
to the rectangular axes.
termed
resolutes or resolved farts.
The numbers
x, y, z are called
relative to the axes
the coordinates of the point
OX, OY, OZ.
It
OP 2 = OA 2 +AF 2 + FP 2
that that
is
giving the distance of
P
obvious from the figure
is
,
r2
=x 2 +y 2 +z 2
P
from the origin in terms
,
of its coordi-
nates.
where the axes are oblique, x, y, z are still called the but the expression for r coordinates relative to those axes in terms of these involves also the mutual inclinations of the In Art.
6.
;
axes.
(Cf.
Exercise
(4),
Art. 26.)
Cartesian analysis deals with vectors and vector quantities
by
resolving
them
into
rectangular
as far as possible,
analysis,
we
components.
In vector
treat the quantities
without
resolution.
Centroids. Definition.
P
of a point
vector
OP
is
used
to specify it
the position
is called the position
P for the origin 0.
vector of 8.
When a
relative to another point 0,
To find
the point
which divides
the join of tioo points in
a
given ratio.
Let A, relative to
dividing
B
be the two points and
a,
b their position vectors
Then AB =b -a and the ratio m n it follows that
an origin 0.
AB
in
;
:
ifl = ^_(b_a). m+n
if
R
is
the point
[OH.
VECTOR ANALYSIS
10
The
position vector of T
R
I.
therefore
is
= OR = OA+AR =a +
m+n (b-a)
wa + mb
9. Definitions.*
to
an origin
Given n points whose position vectors
relative
G
is
are a, b,
c,
...
Off is called the centroid or
If p,
q, r,
...
are
n
vectoris
the point
,
whose position vector
= -(a+b+c + ...) n
mean
centre of
position of the given 'points
real numbers, the point
+ gb + + ^_ yap+q+r + rc
G
whose position
...
...
is called the centroid of the q, r, ...
p,
given points with associated numbers
f
respectively.
The centroid divides the line
of
B
two points A,
AB in the ratio
q
:
with associated numbers
p.
For
p, q
in this case
p+q which proves the statement. *
Regarding this treatment of the theory of centroids see remarks in
the
Preface. f The term number."
strength
may
be found more
convenient
than "associated
.
,
CENTKOIDS
§ 9]
11
Theorem.
The centroid is independent of the origin of vectors. Let 0' be a point whose position vector relative to is 1. If 0' is taken as origin, the position vectors of the points A, B, C, ... are a -1, b -1, 0-1, ... and the centroid is a point G' such that ;
y(a-l)+g(b-l) + ... p+q+r + ...
^7g,
_pa,
+ qb+rc + ... p + q+r
.
=oe-i=o 5. 7
Hence the points G, centroid
is
G' coincide,
independent
and the position found
for the
of the origin of vectors.
a-1
FIG. 10.
Theorem.
If
H
is the centroid
with associated numbers p, .of points
A', B', C",
...
q, r,
with
...
of a system of points A, B, C, ,
and H'
associated
that of
...
a second system
numbers p'
,
q', r' ,
...
then the centroid of all the points is the centroid of the two points
H and H'
with associated numbers
(p+q+r + ...) and (p' +q' +r' + ya+gb+fC + - =gg off= zp p + q+r + ...
For
and
?'*'?** + & + OH' f p +q +r + ...
similarly J
...).
- =¥*' Ip
Hence the centroid of H and H' with associated numbers Ip and l,p' respectively is a point G such that
-
a'
(l )6H + (2p')dH' = Z£a + 2y u _ P Xp Xp'
Xp + Zp'
Hence G
is also the centroid of the
+
combined system
of points.
VECTOR ANALYSIS
12
[
CH
-
x-
The theorem has been stated for only two sub-systems of points with centroids H and H'. But the same argument applies, and the theorem is true, for any number of sub-systems. In calculating the centroid of the combined system of points, each sub-system may be replaced by a single point (its centroid) with associated number ~Ep for that sub-system. Suppose the surface of any figure 10. Centroid of Area. (not necessarily plane) to be divided into a large number n of small elements. Consider one point in each element, and with these n points associate numbers proportional to the areas of the elements. Such a system of points with associated numbers has a centroid G. Now let the number n increase indefinitely, in such a
a point.
way
that each element of the surface converges to
Then the
limiting position of
G
is
called the centroid
of area of the figure.
Suppose any
Centroid of Volume. into a large
number n
solid figure to
be divided
Consider one point
of small elements.
each element, 'and with these w points associate numbers,
in
Such a system and the limiting position of G as n
proportional to the volumes of the elements.
G
of points has a centroid
;
tends to infinity and each element converges to a point,
is
called
the centroid of volume of the figure.
The centroids
and volume
of area
of the simpler figures of
plane and solid geometry are easily found
by
considerations of
symmetry, without the need of introducing vectors. Students, of geometry and mechanics become familiar with these results, at an early stage, and we shall not here enter upon their proofs.
The determination
of centroids
by integration
of vectors will be
referred to in Art. 62. 11. Definition. set
respectively
r l5 r 2 , r 3 ,
...
associated
numbers
The
*
1
2,
,
2
the
is
m m m 1;
of mass, or centre of mass, of a
m m m 3
,
...
,
9
,
situated at the points*
...
centroid
of
these
points
_
=
wtiri+OTgf>+
m
1
with
.
centre of mass (cm.) of the system of particles
the point
is r
The centroid
of particles of masses
..
+ m 2 +...
j
is
therefore
= 2mr
2m'
When the origin of vectors is understood, the point whose may be conveniently referred to as the point r.
position vector
CENTROIDS
§§10,11] It will be
shown
13
in Art. 99 that this is the point
through which
passes the line of action of the resultant of any system of parallel forces acting
on the
particles, the forces being proportional to
the masses of the particles. as defined above
system
is
into
n
Also,
one at the
of
is
the
cm.
each sub-system and of mass
equal to the total mass of that sub-system. the second theorem of Art.
independent
whole system
of the
cm.
is
the system of particles be divided
if
n sub-systems, the cm. particles,
then that the s.m.
Being a centroid, the cm.
of particles.
of the origin of vectors.
of
It will follow
identical with the centre of gravity of the
This follows from
9.
From the above formula for the cm. of a system of particles we may easily deduce the usual scalar equations. Take a set of axes
Let
through the origin parallel to the unit vectors
x, y, z
be the coordinates of a particle of mass m, and
those of the
cm.
Then the r
a, b, c. x, y, z
position vectors of these points are
=xa,+yb+zc
i=xa+yb+zc.
and The formula
for the
cm.
is
_
.
therefore
xa,+yb+zc= "
—2m
2m(cca+wb+zc) =-^
-•
The equal vectors represented by the two members of this equation must have equal components. Hence, equating the coefficients of like unit vectors, we have 1/mx
2m
_. '
2m?/
2m
_
2mz
2m
'"
'
and these formulae are true whether the axes
are rectangular
or oblique.
The cm.
of a continuous distribution of matter,
surface or a volume distribution,
is
the distribution be divided into a large
Take n
points, one
situated in
whether a
defined as follows.
Let
number n of small elements.
each element, and with these
numbers proportional to the masses of the elements. This system of n points with associated numbers has a centroid G. Now let n tend to infinity in such a way that each of the elements The limiting position of G is of mass converges to a particle. distribution. continuous the of mass of called the centre
associate
[CH.
VECTOR ANALYSIS
14
I.
uniform, the cm. coincides with the centroid of the geometrical figure occupied
When
by the
the density of the distribution
is
distribution.
Elementary Physical Applications.
The displacement
12.
placement from
of a point
a vector quantity, a dis-
is
A to B being specified in magnitude and direction And two
by the vector AB.
successive
compounded
displacements
are
according
the
to
law of addition
vector
displacement from A to
for
;
a
B followed
by another from B to C brings the moving point to the same position as a single displacement from
toC. The
Illi. 11.
point
which
may
be in motion,
And
vector PQ. point of
is
PtoO; The
Q
to another
sum
one
of
P, both
a vector quantity specified
clearly the relative position of
the vector for
is
position
relative
Q
of the relative positions of
by
A
of
the
to a third
Q
P
to
and
OQ = OP+PQ.
relative displacement of
Q
to
P
during any interval
is
the
change in relative position during
that
then P,
Q
the
of
interval.
If
are the positions
points
at
the
be-
ginning, and P', Q' at the
end is
the
of
change
interval,
the
in relative position
determined by the vector
difference
PQ' -PQ.
This
vector specifies the relative
displacement
during
the PlG
interval.
The
relative velocity of
-
Q
of its position relative to P,
12
-
with respect to
P is the rate
and
uniform,
therefore,
if
is
of
change
determined
PHYSICAL APPLICATIONS
12, 13]
§§
by the
relative
velocity
is
displacement
one
in
15 This
second.
relative to another point
may
(which
relative
Q and
the vector difference of the velocities of
P
be regarded as fixed).
For, supposing the velocities uniform, let the points be displaced
one second from P,
in
Q
The vector specifying
to P', Q'.
their
which represents the change in their
relative velocity is that
—
>
relative position in unit time,
->•
P'Q' - PQ, which
i.e.
may
also
be
expressed
oq -op'
-
(6q-6p)=(oq' -oq)- (op'-6p)
=W'-PP'=v-u, where
u
v,
are the velocity vectors for
Hence the
relative to 0.
velocity of
difference of the velocities of
The case
Q
Q and P
of variable velocities
Q and P
relative to
respectively
P is
the vector
relative to 0.
and
relative acceleration will
be considered in Chapter VI., after differentiation of vectors has
been dealt with. 13. Concurrent forces.
and may be represented
force has
magnitude and
by a
direction,
vector.
But a
has also a definite line of
force
and
action, is
A
in these respects
altered
upon a body
its effect
if
this fine of action is
changed, even though the direction
may
not vary.
We
shall for the
present confine our attention to forces
whose
lines
of
action are
concurrent, as for instance
when
on a single Now, experiment shows particle. that the joint action of two concurrent forces has the same dynamical effect as that of a single force which is equal to all
the
forces
their vector
And
if
act
sum, and acts through their point
of concurrence.
there are several forces acting on a body, represented
Fv F 2 concurrent at a point P, the single force represented by
the vectors
,
.
.
R=F
.
1
respectively,
+ F 2 +F 3 +
.
.
and with
.=2F,
by
lines of action
;
VECTOR ANALYSIS
16
[
CH L -
and acting through the same point P, is dynamically equivalent to the system of forces, and is called their resultant. The vector R is determined by the vector polygon, that is a polygon the lengths and directions of whose sides are those of
F1; F 2 F 3
the vectors
,
be closed, nor If
will it
,
This polygon will not in general be plane unless the forces are coplanar. .
.
.
.
AB is the first vector and BE the last, then AE is the resultant R = 2F. When
the vector
sum
of all the forces
zero,
is
they are together
equivalent to zero force, and are said to be in equilibrium
the particle or body on which they act is said to be in equilibrium under the action of the forces. In this case the vector polygon is closed. And since the resultant vanishes, the sums of the components of the several forces in any three non-coplanar Conversely, directions must vanish separately. (Cf. Art. 6.) if the sum of the components of the or,
forces vanishes
non-coplanar
ponents
each
for
three
of
the com-
direction^,
of the resultant are zero, and
This then
the resultant vanishes. the necessary
and
is
sufficient condition
for equilibrium of the forces. If Fig. 14.
tin
acting at a point
three Jforces
are in equilibrium, the closed vector
polygon
is
a triangle.
and the length
The vectors F 1; F 2 F 3
are then coplanar,
,
of each is proportional to the sine of the angle
between the other two.
This
is
Lami's Theorem,
viz.
If
:
three
concurrent forces are in equilibrium they are coplanar, and each is
proportional
to the
sine of the angle between the other two.
In the case of n concurrent forces,
let
to an origin 0.
A Av A lt
F1; F 2
the points whose position vectors are
.
,
Then the vector representing the
R = 2F = dJ + 672 +
.
.
.
3,
.
.
,
.
.
F„
A n be
,
relative
resultant
is
.+~OA n
= n OG, .
where
G
is
the centroid of the points
forces are in equilibrium
if
G
A A x,
coincides with 0.
2,
.
.
.
,
An
.
The
.
PHYSICAL APPLICATIONS
§§ 13, 14] 14.
Examples.
(1)
A man
an hour finds that the wind from the North. On doubling his speed he finds come from N.E. Find the velocity of the wind.
travelling East at 8 miles
seems
to bloiv directly
that
appears
it
Let
17
to
j represent velocities of 8 miles an hour toward E. and N. respectively. Then the original velocity of the man is i. Let that of the wind be xi + y}. Then the velocity of the wind relative to the
man
i,
is
,
.
(xi
But
this is
When him
to
from the N., and
man
the
is
is
,
therefore parallel to
.
-
j
Hence x = 1
.
.,
from N.E., and
is
.
+ yj)-2i. ,
this is
.
doubles his speed the velocity of the wind relative (aa
But
.,
+ yj)-i.
therefore parallel to -(i
+ j).
Hence
y = x-2= -1.
Thus the velocity of the wind an hour from N.W.
is
i
- j, which is equivalent
If two concurrent forces are represented by n
(2)
respectively, their resultant is given
sothat
By
by (m
+ n) OR,
.
to 8
m OB R divides AB
OA
ivhere
V% miles
and
.
n.AR = m.RB.
reference to Fig. 9
it will
be seen that
OA = OR+RA,
OB = OR + RB.
and
Hence the resultant
of the forces
n
(m + n)OR + (n
.
OA
.
RA +m
And
the last part of this expression the ratio n.
m
(3)
circle
—>
.
,
.
.
Pn
are
n
equal parts. -> —>
by
AP AP 2
the
plane of the
lt
is
.
m OB is .
RB).
zero because
R
divides
AB in
:
Pv P2 into
and
,
.
.
.
,
AP n
n points dividing the circumference of a Find the resultant of forces represented ivhere
A
is
any
point, not necessarily in
circle.
_^_
the preceding Art. the resultant is represented by n AG, P n And by where G is the centroid of the points P v P 2 symmetry G coincides with the centre of the circle.
By
.
,
,
-
(4) A particle is acted on by a number of centres of force, some of which attract and some repel, the force in each case varying as the distance,
W.V.A.
e
VECTOR ANALYSIS
18
[
CH
-
and the intensities for different centres being different. Show that the resultant passes through a fixed point for all positions of the particle. those of 0.,, Let be the position of the particle, and O
P
r,
the centres of force. The forces on the particle due to the different centres ar« represented by —y ->
Ml where ni>
/"2>
•
and negative
•
•
for
.
the
p, lt fx 2 ,
2
P0 2
.
,
etc.,
->
_^
= 0*i + /*« + centroid of the points O v 3 2 respectively. And this is a
Ml .P0 1 + ^ 2 is
fj.
are constants, positive for the centres that attract The resultant force on the those that repel.
^
particle is
where 67 numbers
PO v
.PO a +
.
.
.
,
.
•
.
,
.
-)PG, .
with associated
fixed point indepen-
dent of the position of P.
EXERCISES ON CHAPTER 1. 6i
—
2j
I.
Find the sum of the vectors 3i + 7j - 4k, i - 5j - 8k and + 12k. Also calculate the module and direction cosines of
y
each. 2.
If
the position vectors of
respectively, find 3. If
PQ
P
and
and determine
Q
its
are
i
+ 3 j - 7k and
5i
- 2j + 4k
direction cosines.
the vertices of a triangle are the points
a 1i + a 2 ]+a 3 k,
& 1 iH-& 2
j+& 3k
what are the vectors determined by
and
c-,i
its sides
?
+ c 2j + c 3k,
—*
Fiatl the lengths
of
these vectors. 4.
The position vectors a, b,
2a
+ 3b
->-->->Express AC, DB,
BC
of the four points
and- a
- 2b
A, B, G,
respectively.
->
and
CA
in terms of a
and
D are
\
—y
b.
5. If a, b are the vectors determined by two adjacent sides of a regular hexagon, what are the vectors determined by the other sides taken in order ?
6. A point describes a circle uniformly in the i, plane taking j 12 seconds to complete one revolution. If its initial position vector relative to the centre is i, and the rotation is from i to find the j,
position vectors at the end of of 1J and 4| seconds.
1,
3, 5,
7 seconds
;
also at the end
7. In the previous exercise find the velocity vectors of the moving point at the end of 1|, 3 and 7 seconds.
y
>
EXERCISES ON CHAPTER
I.]
The
19
I.
is represented by that of the water relative to the earth by i-3j. What is the velocity of the boat relative to the earth if i and j represent velocities of one mile an hour E. and N. respectively 1
8.
velocity of a boat relative to the water
+ 4j, and
3i
9. Two particles are moving with the same speed v ft. /sec, one along a fixed diameter of a circle and the other round its circumference. Find the velocity of the first relative to the second when the radius to the latter makes an angle 6 with the direction of motion of the former, 6 increasing. [Assume that variable velocities obey the same law of composition as uniform velocities. Cf. Art. -65.]
10. Two particles, instantaneously at A and B respectively, 15 feet apart, are moving with uniform velocities, the former toward at 3| ft. /sec. B at 5 ft./sec, and the latter perpendicular to Find their relative velocity, their shortest distance apart, and the
AB
instant
when they
are nearest.
Find the sum
of the three vectors determined by the diagonals adjacent faces of a cube passing through the same corner, the vectors being directed from that corner.
11.
of three
12.
3
2,
A
lb.
particle at the corner of a
cube which meet at the particle. 13.
cube
is
acted on by forces 1, faces of the
wt. respectively along the diagonals of the
Find their resultant.
Find the horizontal force and the force inclined at 60° to the whose resultant is a vertical force P lb. wt.
vertical,
14. If the resultant of two forces is equal in magnitude to one of the components, and perpendicular to it in direction, find the other
component. 15.
Two
forces act at the corner — —*
by
represented
Show mid
and CD. P,
Q
AB
are the
and
AD
;
AC,
ABCD,
of a quadrilateral
—>
and two at C represented by
that their resultant points of
A
by iPQ, where
represented
is
OB
BD respectively.
16. Find the cm. of particles of masses 1, 2, 3, 4, 5, 6, 7, 8 grams respectively, placed at the corners of a unit cube, the first four at of one face, and the last four at their projecthe corners A, B, C,
D
C, D'
tions A', B', 17. n]
;
respectively on the opposite face.
Find the centroid
k, 2k,
.
.
.
,
of the
3w points
i,
2i, 3i,
.
.
.
,
n\
;
j
,
2j
,
3j
,
.
.
,
rik.
- 18. Five forces act at one vertex
A
of a regular
hexagon in the
directions of the other vertices, and proportional to the distances Find their resultant. of those vertices from A.
VECTOR ANALYSIS
20 19.
ABC,
If
sum
sum
O'O
of O'A, O'B,
21.
D, E,
F
O'O.
.
;
>
of AO', O'B,
and __^
O'C
OF
A,B,C.
equivalent to the system represented
is
acting at the
same
point.
B
A,
respectively, find
AB
produced such that AC = 3AB that of a point C in = 2BA. of a point in BA produced such that
;
and that
BD
D
C
A, B,
through C.
P
are fixed points and
resultant of forces at
ABC
J-
of concurrent forces repre-
22. If a, b are the position vectors of
23.
00'
___^
show that the sum
any point 0, the system
OA, OB, OC
by OD, OE,
2
is
is
points of the sides of the triangle
mid
are the
that, for
sented by
-
AD is a diameter of the circumcircle.
AD, where
Show
OA, OB, OC
of the vectors
20. In the previous exercise is
CH
triangle the circumcentre and 0' the orthocentre of a-* _*.-»-»-
is
prove that the
that the
[
P
represented
Find the locus
a variable point such that the —> —>
by
PA
and
PB
always passes
of P.
a triangle and P any point in BC. If PQ is the —> —> resultant of AP, PB, PC, the locus of Q is a straight line parallel 24.
is
—>
to
BC. 25.
forces
Prove that the magnitude
P P 2 P3 lt
.
,
.
.
is
of the resultant of
R 2 = SP 2 + 22Pr P 26. Forces P,
Q
P
s ).
28. Particles of equal
A
line
exists
one,
linear relation.
a regular polygon of
AB
on ad infinitum
:
is
n
cm. from B
2) of
the corners
of
Find their cm.
bisected in
and
is
mass are placed at (n -
sides.
particles
Px PX B ,
in
P 2 P%B ,
whose masses are m,
in
P3
,
and
-m — m
so
etc.
P v P 2 P 3 etc. Prove that the distance of equal to one-third of the distance from A to B.
are placed at the points their
cos (P r
OB OC
OA
29.
s
and have a resultant R. If any transaction at A, B, C respectively, show that
Between any four non-coplanar vectors there
and only one,
of
act at
versal cuts their lines of
27.
any number
given by
,
,
ch.
15, 16]
ii. ; § §
CHAPTER
II.
ELEMENTARY GEOMETRICAL ILLUSTRATIONS AND APPLICATIONS. we
15. In the present chapter of a straight line,
shall consider the vector equation
one form of vector equation for a plane, and
the quantity called " vector area." of these, some
In connection with the
first
examples are taken from elementary plane geometry.
This, however,
is
merely for the purpose
of illustrating the vector
method, and not with the intention of recommending the use of vectors in elementary geometry.
Vectors were designed essenti-
ally for three-dimensional cal-
culations
and
;
author
the
explicitly disowns any attempt
to
recommend
their
use in
elementary
two-dimensional
problems,
especially
of
geometry. 16.
equation of a
Vector
straight
line.
To
find the
Fig. 15.
vector equation of the straight line through If
P
is
a given point
A
parallel to a given vector b.
a point on this straight line the vector
AP
is
parallel
where t is some real number to b, and A, and negative for points side of one points on positive for point. Thus, if a is the point to from on the other, varying is
therefore equal to
position vector of A, that of
P
fb,
is
i=OP=OA+AP = a + ib 21
•(1)
a
a
[CH.
VECTOR ANALYSIS
22
And
any point on the given vector represented by this equation
straight line has a position
since
some value
for
Cor.
t
The vector equation
parallel to b
the point a +tb
lies
of the straight line
is
of
t,
we speak is
also
on the given
line.
It
straight line.
of (1) as the vector equation of the
clear that for all values of
through the origin
= ib
r
II.
(2)
[If (x, y, z) and (%, a 2 a 3 ) are the coordinates of P and A respectively referred to rectangular axes through the origin 0, ,
while b
= bj +b 2) +6 3k, the above equation may be written xi + yj + «k = a x i + a 2j + a 3 k + i(6 xi + b 2\ + b 3 k).
Then, equating coefficients of this equation,
we deduce the x-
like vectors
_y — a 2 _z -
±
b2
!>!
on opposite
sides of
relations s
bs
which are the ordinary equations of coordinate geometry for the straight line through the point (a v a 2 « 3 with direction ,
)
cosines proportional to (b 1; b 2 b 3 ).] ,
To
find the vector equation of the straight line passing through
the points
A
and B, whose position vectors are a and
b,
we
p
O Fig. 16.
observe that the point
A
AB =b - a
so that the straight line
;
parallel to b
- a.
or
Its vector
r
= a + z(b-a)
r
= (l-*)a + rt>
[If (a v
a 2 a 3 ) and
we find
as above that the equation (3)
,
(b u b 2 , b 3 )
x ~
&i-«i
equation
is is
one through therefore
(3)
are the coordinates of is
A
and B,
equivalent to the relations
= y-« 2 = z-a 3 ~ b 2 -a 2 b 3 -a 3
'
§ §
EQUATION OF STRAIGHT LINE
16, 17]
which are the Cartesian equations the points
A
(3)
of the straight line
through
and B.]
P
The three points A, B, equation
23
are collinear
and
;
connecting their position vectors
is
if
the linear
written
{l-t)a,+tb-v = Q, with
all
of the
the terms on one side, the algebraic vectors
This
zero.
is
sum
of the coefficients
the necessary and sufficient
is
That it is a any point P
condition that three points should be collinear.
necessary condition has just been proved
A
collinear with
some value
of
t.
and
B
It
also sufficient
is
for
;
has a position vector given by ;
between
satisfied in a linear relation
for, r,
(3) for
assuming the condition
a and
b,
we may make
the coefficient of r unity and write the relation
r=sb + (l showing that
P is
17. Bisector
of
-s)a,
a point on the straight line
between two straight
angle
the
find the equation of the bisector of the angle lines
OA
and OB,
take the point
AB.
A parallel to the unit vectors a
as origin,
and
let
P
To
lines.
between the straight A
and b respectively, be any point on the bisector.
Fig. 17.
Then,
OPN
if
PN is
and
NOP
drawn
AO cutting OB in N, the angles ON =NP But these are parallel that ON = ib and NP = where is
parallel to
are equal, and
and a respectively so some real number. The position vector to b
:
r
= <(a + b).
ta,,
of
P is therefore
t
>
VECTOR ANALYSIS
24 This as
[
the required equation of the bisector, the value of
is
P moves The
along the
bisector
OP'
-
-
varying
line.
of the
supplementary angle
bisector of the angle between straight lines
are those of a
t
CH n
and - b
and
;
equation
its
is
B'OA
whose
is
the
directions
therefore
A
A
r=i(a-b). 18.
The The
(i)
the side
Triangle.
internal bisector of the angle AC. in the ratio
AB
BC
A
of a triangle
ABC
divides
:
—
Let
AB = c
and
bisector of the angle
AC = b. Then, A is the line
.
with
/eb
+ 6c
V
be
A
as origin, the internal
-t
where, according to our usual notation, b is the module of corresponding unit vector. Giving t the value bc/(b + c)
b,
and b
we
the
see that
the bisector passes through the point (cb + 6c)/(6 + c), which is the centroid of the points B and C with associated numbers b and c respectively, and therefore lies in BC, dividing it in the ratio c 6 or AC. Hence the theorem. Similarly the external bisector of the angle A is the straight line :
AB
:
T
C = t(c-b) = t( -* \C
= t (bc-cb\ be
which passes through the point
(6c
/'
-
cb)/(6
- c)
GEOMETRICAL APPLICATIONS
§§18,19] This point b
is
the centroid of B and C with associated numbers * AC. BC externally in the ratio
AB
and -c, and divides (ii)
The
25
:
internal bisectors of the angles of a triangle are concurrent.
P at which the internal bisector of the angle A cuts the centroid of B and C with associated numbers b and c. The centroid of the three points A, B, C with associated numbers a. a, b, c therefore lies on AP and divides it in the ratio (b + c) But by symmetry this point must also lie on the bisectors of the Hence the bisectors are concurrent at this angles at B and C. The point
BC
is
:
centroid.
Similarly
it
may
be shown that the internal bisector of
A
and
the external bisectors of B and C are concurrent at the centroid of the points
A, B, C with associa, -b, -c respec-
ated numbers tively. (iii)
The medians of a
triangle
are concurrent.
Let D, E,
F
be the mid points
of the sides of the triangle.
D
Then
the centroid of B and C and the centroid of the three points A, B, C therefore lies on and divides it in the ratio 2:1. And by symmetry this point also lies on the other medians, and is a common point of trisection. It is well known that this point r is also the centroid of area of the is
;
AD
-
triangle
19. (i)
ABC.
The Tetrahedron. The joins of the mid points of
opposite edges of a tetrahedron intersect and bisect each other.
The mid point of DA is the cenD and A. The mid point of BC is the centroid of B and C. Hence the point of bisection of the line joining these mid points is the troid of
centroid of the four points A, B, C, D. And by symmetry this centroid lines joining the
mid
is also the point of bisection of the If other pairs of opposite sides. the points of
The formula of Art. 9, defining the position of the centroid, whether the associated numbers are positive or negative. *
is
to apply
VECTOR ANALYSIS
26
[CH.
relative to D,
C respectively J(a + b + c).
c are the position vectors of A, B,
a, b,
common
this
point of intersection
is
(ii) The lines joining the vertices of a tetrahedron of area of the opposite faces are concurrent.
N
II.
to
the centroids
ABC
is the centroid of the The centroid of area of the face three points A, B, C. Hence the centroid of the four points A, B, C, D lies is
on
AN
and divides
And by symmetry
3:1.
in the ratio
it
this
also a point of quadrisection of the lines joining the other vertices
to the centroids of area of the opposite faces.
known
It is well
of
volume
that this point of intersection
is
also the centroid
of the tetrahedron.
The
Otherwise.
two theorems
last
may
be proved by means
Relative to D as origin the mid points of and BC are Ja and -J(b + c) and the equation of the straight line passing through these points is of the vector equations of the straight lines, thus
:
DA
sa
=2
r
;
+(1 -
,
s)
Similarly the straight line through the
+ -^— (b
c)
mid
DB and CA
points of
is
These straight lines will intersect if real values of s, t can be found which give identical values for r. This requires the coefficients of like vectors to be equal in the two expressions i.e. ;
s
which are
satisfied
= l-t; by
s
l-s = t;
= t = \.
l-s = l-t,
Hence the
lines intersect at the
point J(a + b + c). Again the centroid of area of is the point J (a line joining to this point has the equation
ABC
+ b + c)
;
and
the
D
r
Also the centroid of area of joining this to
£
J)
AC
is
the point +(a
+ c)
;
and the
line
is
r
These two
= s(a + b + c).
lines intersect
the point J(a + b + c). second theorem follows.
is
= ib + l_
?±c
(
at the point
From
.
for
which
s
= t = \\
that
the symmetry of this result the
20. Vector equation of a plane. (1)
To
find the vector equation of the plane through the origin
parallel to a If
and
b.
P is any point on the plane its position vector OP is coplanai
EQUATION OF PLANE
$§19,20] with a and
b,
•27
and may therefore be resolved and expressed in the form
into
components
parallel to these
r
= sa + i!b,
(1)
Fig. 21.
where plane. s
and
numbers which vary as the point P moves over the Any point on the plane is given by (1) for some values of and for all values of these variables the point sa + tb lies
s, t
t
;
are
We may therefore
on the given plane.
speak
of (1) as the vector
equation of the given plane. (ii)
To find
parallel to a
the vector equation of the
and
plane through the point
b.
connecting them Pig. 22.
Let c be the position vector of C, and the given plane.
may
The vector
CP
is
cop 1
'
-^d
•
therefore be written
'
-*
LP=s&A case. as in the previous r
(1)
r tb'
a,
..on will
a variable number, positive e opposite side of the origin
is
^ .„
(2)' *
'
,
,
,
be found
m Art. .
.
,
,„ 43.
VECTOR ANALYSIS
28
[CH.
i=OP=OC+CP
Then
= c+sa + ib This
is
II.
(2)
numbers
the required equation to the plane, the
varying as
P
moves over the
can be represented by
And
plane.
no point
off
s,
(
the plane
(2).
find the equation of the plane through the three points
To A, B,
C whose
position vectors are
^4.B=b-a
a, b, c,
we observe
that
AC=e-a.;
and
Fig. 23.
so that the plane "*3S
lme
equation
is
one through
is
A
parallel to b
c
- a.
= a+s(b-a) + £(c-a) r = (l-s-f)a + sb + *c
r
j
Also the -1
ot.
at j ons (i^ (2)
and
(3)
(3)
involve each two variable numbers
analogous to the equations of the lines considered
°
'Hs form of the equation for a plane
These two is
- a and
therefore
lines iiior is it the
the point J(a
+ b -t?r the
most convenient.
scalar product of
is not, however,
Another
will be
two vectors has
been
second theorem follows. 20. Vector equation of •&> (1)
To find
parallel to a If
the vector
and
b.
Pm
Fig. 23 are coplanar
equg their position vectors
is
;
and
if
the
written
+ sb + tc - r = 0,
P is any point on the plane iu algebraic sum of the coefficients
)
GEOMETRICAL APPLICATIONS
I§ 20, 21] the vectors
29
and sufficient condition That it is necessary has just been proved. It is also sufficient for, assuming the condition satisfied in a linear relation between r, a, b and c, by making the coefficient of r unity, and denoting those of b and c by s and t, we obtain the relation in the form of
is
This
zero.
is
the necessary
that four points should be coplanar,*
;
r=sb+ic + (l -s-t)&,
P is a point in the plane ABC.
showing that 21.
As an example
of the use of this
of a plane consider the following
form
of the vector equation
:
within a tetrahedron ABOD is joined to the vertices, If any point and AO, BO, CO, DO are produced to cut the opposite faces in P, Q, R, S respectively, then
With
S-jp=l.
D be a, b, c, d be expressed»in terms
as origin let the position vectors of A, B, C,
Any one
respectively.
of these vectors
may
Fig. 24.
•of
the other three, so that there may be written
is
a linear relation connecting
them
which
la
+ mb + nc+pA =
The vector equation of the plane through B, C,
while the line
OP
is
r = sb + te + (l-s-«)d, - Ma, where u is a variable number, = r
for points of the line *
An
(1
D is
which
lie
on the opposite
(2)
positive
side of the origin
equivalent condition will be found in Art. 43.
[CH.
VECTOK ANALYSIS
30 to A.
OP
In virtue of the relation
as
write the equation, of
we may
(1)
II.
u
= y (mb + wc + pi)
r
From
by
these,
;
addition,
the ratio
at a point for which (2) and (3) b, c, d are non-coplanar vectors,
BCD
This line intersects the plane give identical values for r this requires
(3)
and since
we have u = l/(m + n + p), showing
Qp
u
'that
l
AP 1+u
l
+ m + n+p
The other ratios may be written down by cyclic permutation of the symbols, and their sum is obviously equal to unity. 22. Linear relation independent oi the origin.
The necessary and
condition that a linear relation,
sufficient
connecting the position vectors of
any number of fixed
be independent of the origin, is that the algebraic
sum
points, should
of the
coefficients
is zero.
Let a 1; a 2
A A2 t,
,
.
.
.
.
.
,
An
,
a„ be the position vectors of the fixed points
,
.
relative to
an
origin 0,
and
let
the linear relation
be written fc 1
a1
+ ^2a 2+
•
•
•
+&„a„ =
(1)
Let 0'
be another point whose position vector
Then
0'
and also,
if
is
taken as origin of vectors, the point
(Fig.
1
Am
in order that the relation (1) should hold for the
is
10).
a m -l;
new
origin
we must have fc
1 (a 1
-l)+A 2 (a 2 -l)+
which, in virtue of
(1),
The condition (2) holds,
is
+ k2 +
1,
+& K (a„-l)=0,
.
.
.
.
.
+k n =0
therefore necessary.
each vector in
constant vector
.
reduces to
k1
if
is
(1)
may
and the equation
(2)
It is also sufficient
for
;
be diminished by the same
will still
be true.
Hence
the
relation will hold with 0' as origin. 23. Vector areas.
Consider the type of vector quantity whose
Such a quantity is associated with each plane figure, the magnitude being the area of the figure, and the magnitude
is
an
area.
direction that of the
normal to the plane
of the figure.
This
,
VECTOR AREAS
22, 23]
§ §
vector area therefore specifies of the
But
plane figure.
31
both the area and the orientation
might be either of two some convention is necessary. itself, and can be regarded as
as the direction
opposite directions along the normal,
The area
clearly has
no sign in
positive or negative only with reference to the direction in
boundary
the
of
the
figure
which
is
described, or the side of the plane
from which
it is
viewed.
Consider the area of the figure
bounded by LMN, which
the is
closed
curve
regarded as being
traced out in the direction of the
The normal vector PP'
arrows.
bears to this direction of rotation
the same relation as the transp
lation to the direction of rotation
a
of
area of
right-handed
LMN
is
screw.
fig. 25.
The
regarded as positive relative to the direction
PP'.
With
convention a vector area
this
may
be represented by a
vector normal to the plane of the figure, in the direction relative to which
it is
positive,
The sum
the area.
and with module equal to the measure and b
of two vector areas represented by a
of is
defined to be the vector area represented by a +b. It
is
a well-known geometrical result that the area A' of the
orthogonal projection of a plane figure of area A, upon another plane inclined at an angle 6 to
A
it, is
cos
6.
If
then
a, a'
the vectors representing the vector areas A, A' respectively,
are a' is
the resolute of a in the direction normal to the plane of projection for a'
=a
cos
6.
If there are several figures of areas
planes inclined at angles
the
sum
.
.
.
.
,
respectively to a given plane,
of the areas of their projections
A and, as a vector area, of a, b,
6,
.
.
.
cos 6 is
+B
;
A, B, ... in
cos
represented
on this plane .
.
is
.
by the sum
of the resolutes
perpendicular to the plane of projection. Thus the
sum of the areas of the projections the sum of the vector areas.
is
equal to the projection of
[CH.
VECTOR ANALYSIS
32
the areas are those of the faces of a solid figure they are regarded as positive relative to the outward drawn normals.
When
be represented by vectors
Hence the vector areas
of the faces will
in the directions of the
normals drawn outward.
The sum of
Theorem. polyhedron
the vector areas of the faces of
a
closed
is zero.
Consider the areas of the projections of the faces on any
fixed
and some negative for some of the faces have their outward normals directed from the plane, and others toward the plane. But the sum of the areas of the projections of the former is equal in magnitude to that of the latter for each sum is equal to the area enclosed by the projection of the (skew) polygon composed of those edges of the polyhedron which separate the former faces from the latter. Hence the
Some
plane.
of these are positive
:
;
sum
of the areas of the projections of all the faces is zero
therefore the projection of the
And
sum
of the vector areas
sum
of the
a parallelogram bisect each other. the diagonals of a quadrilateral bisect each other
it is a
since this
vector areas
is
is
true for any plane of projection, the
identically zero.
EXERCISES ON CHAPTER Prove the following by
The diagonals
1.
versely,
if
and
;
is zero.
vector
methods
II.
:
of
Con-
parallelogram.
The
2.
join of the
to the third side, 3.
mid
and
mid points
of
two
sides of a triangle
is parallel
of half its length.
The four diagonals of a parallelepiped, and the joins of the points of opposite edges, are concurrent at a common point
of bisection. 4. In a skew quadrilateral the joins of the mid points of opposite edges bisect each other. Also, if the mid points of the sides be joined in order, the figure so formed is a parallelogram. 5.
Show that the sum
medians 6.
of the three vectors
of a triangle directed
from the vertices
is
The three points whose position vectors are
determined by
the
zero. a,
b and 3a - 2b
are
collinear. 7.
A BCD
is
of its diagonals.
a parallelogram and the point of intersection Show that for any origin (not necessarily in the
EXERCISES ON CHAPTER
II.]
33
II.
plane of the figure) the sum of the position vectors of the vertices is equal to four times that of 0. 8.
What
points
i
the vector equation of the straight line through the
is
- 2j + k and 3k - 2j
Find where this
and
points 4j 9.
2i
No two
?
line cuts the plane
through the origin and the
+ k.
lines,
joining points in
two non-coplanar straight
lines,
can be parallel. 10. If
gram
N are the mid points of the sides AB, CD of a paralleloDM and BN cut the diagonal AG at its points of which are also points of trisection of DM and BN respec-
M,
ABCD,
trisecti on,
tively.
11. Three concurrent straight lines
to D, E,
F
respectively.
AB and DE, BC and 12.
intersection are real 13.
EF, OA and
Prove the converse
The internal
and
OA, OB, 00
Prove that the points
FD are collinear.
of the last exercise
collinear,
are produced
of intersection of
then
—that
if
DA, EB, FG
bisector of one angle of a triangle,
the points of
are concurrent.
and the external
bisectors of the other two, are concurrent. 14.
The mid points
of the six edges of a
cube which do not meet a
particular diagonal are coplanar. 15. The six planes which contain one edge and bisect the opposite edge of a tetrahedron meet in a point. 16. If
two
parallel planes are cut
by a
third plane the lines of
^intersection are parallel. 17. If a tetrahedron is cut
edges the section
is
by a plane
parallel to
two opposite
a parallelogram.
18. If a straight line is drawn parallel to the base of a triangle, the line joining the vertex to the intersection of the diagonals of the trapezium so formed bisects the base of the triangle. 19. The straight lines through the mid points of three coplanar edges of a tetrahedron, each parallel to the line joining a fixed point to the mid point of the opposite edge, are concurrent at a point P, such that OP is bisected by the centroid (of volume) of the tetra-
hedron. 20. Using the vector equation of a straight line, show that the mid points of the diagonals of a complete quadrilateral are collinear and also establish the harmonic property of the figure. ;
[CH. HI.
CHAPTER
III.
PRODUCTS OF TWO VECTORS. THE PLANE
AND THE SPHERE. 24.
From
the nature of a vector, as a quantity involving
direction as well as magnitude, it is impossible to say a priori what the product of two vectors ought to be. But by examining the ways in which two vector quantities enter into combination in Physics and Mechanics we are led to define two distinct kinds of products, the one a number and the other a vector. Each is jointly proportional to the modules of the and each follows the distributive law, that the product of a with b + c is equal to the sum of the products of a with b and of a with c. We borrow from algebra the term multiplication for the process by which a product is formed from two vectors and these are called the factors of the product.
of these products
two vectors
;
;
The
Scalar or Dot Product.
25. Scalar quantities are of frequent occurrence,
each upon two vector quantities in such a proportional to their magnitudes
mutual
inclination.
An example
it
to the cosine of their
such
convenient to adopt the following
Definition.
The
scalar product of two vectors
directions are inclined at and is written *
an angle
a*b *
which depend
as to be jointly
is the work done by a body acted upon. We there-
of
force during a displacement of the fore find
and
way
An
= a&
cos
6, is
the real
a and b, number ab
= b*a.
alternative notation for the scalar product
34
is
(ab).
whose cos
6,
i§
THE SCALAR OR DOT PRODUCT
24, 25]
The order
may
of the factors
value of the product.
be reversed without altering the
Further, b cos 6
acute or obtuse.
is
the other in If
a,
positive or negative
Hence a #b
two numbers, which measure the length
and the length
the measure of the
is
length of the resolute of b in the direction of
according as 6
35
of
is
the product of
one of the vectors
of the resolute of
its direction.
two vectors
b are perpen-
a,
dicular, cos (9=0, and their scalar product is zero. Hence the condition of perpendicularity of two finite vectors is expressed by
a-b If the vectors
direction, cos
= 0. have the same
= 1, and a*b=a&.
If their directions
scalar
are opposite, cos 6
= - 1, and
product of any two unit vectors
is
a-b
=
-ab.
The
equal to the cosine of
the angle between their directions. When the factors are equal vectors their scalar product a-a is
called the square of a,
and a
2
is
written a 2
= a*a=a
.
Thus
2 ,
the square of a vector being thus equal to the square of
The square of any unit vector
is
unity.
its
module.
In particular,
i2=]2=k 2 =l, but since these vectors are mutually perpendicular
H=j-k=k*i = 0. be constantly employed. the scalar product If either factor is multiplied by a number, multiplied by that number. For
These relations
is
will
(na)'b
=nab
cos 6 =a*(«b).
may occur as the Since the scalar product is a number, it s(a«b)c is a vector numerical coefficient of a vector. Thus a-bc with module a-bc. The combination a-bc-d simply the product of the two numbers a-b
in the direction of c of four vectors
and
c-d.
is
[CH.
VECTOR ANALYSIS
36
an angle
If r is inclined at
part of r in the direction of a
to a, the resolute or resolved
is
ar cos r cos
Hence
if
cb 21
a
(p
n a
a*r =
two components
r is resolved into
a-r
and
r,
these components
a,
a*r
and
a
a.
in the plane of a
one parallel and the other perpendicular to are
III.
r
respectively.
Similarly,
if
any vector
to the unit vectors respectively,
and
i,
j,
26. Distributive
Law.
components
k, these components are
= iTi + j'rj It
is
parallel
j-rj,
k*rk
+k*rk.
easy to show that the distributive
of multiplication holds for scalar
a*(b
i'ri,
Thus
sum.
r is their
r
law
r is resolved into
products
;
that
is,
+ c) = a*b + a*c. B
Let OA, OB,
OB,
BC
on OA.
algebraic
sum
BC
be equal to
b,
Project
c respectively.
of the projection of
OB
of the lengths of the projections of
Therefore
OC
is
the
and BC.
->
a-(b
Repeated application
+ c)=a-(OC)=a.CW = a{OM + MN)=a =a*b + a'C. of
result
this
product of two sums of vectors algebra.
a,
Then the length
may
.
shows
OM +a MN .
that
Thus (a
+b +
.
.
.)*(I
+m +
.
.
the
scalar
be expanded as in ordinary
.)=a'l+a*m + + b-1 + b*m +
.
.
.
.
.
.
,
§
THE SCALAR OR DOT PRODUCT
26]
In particular, while
As another
(a
+ b) 2 =a 2 +2a'b+b 2
(a
+ b)«(a-b)=a 2 -b 2
37
,
-
useful particular case, suppose that
are expressed in terms of the unit vectors
i,
two vectors
j,
b
a,
Then, with
k.
the usual notation,
= (%i + a 2j + a 3k)'(6 1i + b$ + 6 3k) = a 1b1 + a ib i +a 3b 3
a*b
,
two perpendicular vectors is zero. And because the direction cosines I, m, n of a are aja, aja, a3 ja, and those V, m', n' of b are bjb, b 2 /b, b 3 jb, it follows that since the scalar product of
cos 6
where 8
= IV + mm' + nn'
the angle between the directions of a and
is
b.
Examples. of masses
Particles
(1)
A, B, C, point P,
.
m AP x
Take G as
respectively,
.
.
2
m± m m ,
G
is
3,
.
.
are placed at the points
.
cm.
their
Prove that for any
+ mJBP* + ... = mJJF + m 2 BG2 + ... + (2m)PG72
origin,
those of A, B,
.
.
and .
let s
be the position vector of P, and
-
r ]; r 2
,
.
.
.
Then
respectively.
2m 1r = 0.
_>
1
AP = s-r
Also
2,
and
1
,
so that
2m JP2 =Sm 1
1
(s-r 1 ) 2
(,
= Ew^s - 2s*(2m 1r 1 + Sto,^ 2 = (I,m1 )PG2 + I,m AG2 2
)
.
1
tetrahedron, if two pairs of opposite edges are perpendicular, and the sum of the pair are also perpendicular to each other squares on two opposite edges is the same for each pair.
In a
(2)
the third
;
Using the notation of Fig. 20 we have
CB = b-c.
Hence
if
BD is
perpendicular to CA,
b-(c-a) that
is
And
similarly,
b*c if
DA
is
AB = h-a,, AC = c-a
= 0, = a-b.
perpendicular to BC,
a-(b-c)=0, that
is
a*b=c*a.
and
cos
U
THE VECTOR OR CROSS PRODUCT
26, 27]
§ §
and
if
given s
P'
the point
is
the angle
x', y', z'
by j cos
d>
=
r .r
—
39
OP
between
and OP'
is
,
;
,
rr
whose value
down.
easily written
is
The Vector or Cross Product. 27. Vector quantities are of frequent occurrence,
each upon two other vector quantities in such a
which depend
way
as to be
magnitudes and to the sine
jointly proportional to their
of their
mutual inclination, and to have a direction perpendicular to each of
We are therefore led to
them.
The
Definition.
directions are inclined at
ah sin
is
and whose
9,
perpendicular
to both
positive relative to to b.
adopt the following
vector product of two vectors
an angle
direction
a and
a and
6, is the vector
b,
whose
whose module
is
b, being
a rotation from a
(Arts. 7, 23.)
We write
*
it
a*b, so that
rr
a«b=a& sin#n, where n
is
a unit vector perpendi-
\6
cular to the plane of a, b, having
a
same direction as the translation of a right-handed screw due to a rotation from a to b. From this it opposite direction to a
a
The order
b=
Fig. 28.
follows that b a has the length, so that
-b*a.
of the factors in a vector
product
is
not commutative
;
for a reversal of the order alters the sign of the product.
OAPB
OA, OB have The area of and directions of whose boundary the figure is ab sin 6, and the vector area OAPB, = This a*b. is described in this sense, is represented by ab sin#n The vector simple geometrical relation will be found useful. area OBPA is of course represented by ba. For two parallel vectors sin 6 is zero, and their vector product Consider the parallelogram
a and b
the lengths
The
vanishes. *
An
relation a*b
=
is
whose
sides
respectively.
thus the condition of parallelism
alternative notation for the vector product
is
[ab].
VECTOR ANALYSIS
40 of
two
=
In particular, r*r
vectors.
finite
is
[CH.
III.
for
all
true
vectors.
however, a and b are perpendicular, a*b is a vector whose module is ab, and whose direction is such that a, b, a*b form a If,
right-handed system of mutually perpendicular vectors. If a, b are both unit vectors the module of ab is the sine of
For the particular unit vectors
their angle of inclination.
we have
i
i=jxj
j,
k
=k= -j jk=i= -kj,
while
i*j
k-i
These relations
will
i,
=j =
-i'k.
be constantly employed.
If either factor is multiplied
multiplied
by that number. (ma)*b
28.
i,
=k*k = 0,
The
Distributive
by a number,
=mab sin0n = a*(mb). Law. We shall now show
distributive law holds for vector products also
a*(b
Suppose
first
their product
is
For
that
;
that
the
is
+ c)=a*b + a*c.
that the vectors are not coplanar, and consider
the triangular prism whose three parallel edges have the length
and direction of a, and whose parallel ends PQR, P'Q'R' are triangular with
PQ=h, QR = c, and fore
there-
PR=b+c. The
sum
of the vector areas of the
faces
of
hedron
this closed polyis
zero
(Art. 23).
But these are represented by the outward drawn normal vectors |c*b and ib*c for the triangular
Pig. 29.
ends,
and
a*(b+c) for the other faces. Of these five two are equal in length but opposite in direction.
b*a, c*a,
vectors the
first
Hence the sum
of the other three
a
must vanish
(b+c)+ba+c*a=0,
identically
;
that
is
.
§
THE VECTOR OR CROSS PRODUCT
28]
which
41
equivalent to
is
a(b + c)=a*b+a*c. term we may
On changing the
sign of each (b
also write the relation
+ c)*a=b«a + c*a.
This proves the distributive law for non-coplanar vectors. needless to
is
remark that the order
maintained in each term to a If
change of
The
sum
the
PQR, P'Q'R'
triangles
may
be regarded as a plane
are congruent, and therefore
of the areas of the parallelograms
equal to that of
PRR'P'
Hence the
.
Kepeated application
this
of
product of two sums of vectors
shows that the vector
be expanded as in ordinary
algebra, provided the order of the factors
is
maintained in each
Thus (a
+b +
.
.
.)"(I
+m+
.
.
.)
= a*l + a*m + + b l+b*m + + .
.
terms of their rectangular components.
.
.
.
.
.
.
.
As a useful particular case consider two vectors in
is
a.
result
may
PQQ'P', Q'QRR'
relation
b*a + ca = (b+c)
term.
must be
equivalent
is
sign.
the vectors are coplanar Fig. 29
figure.
of the factors
change of order
for a
;
It
a,
b expressed
Then, with the usual
notation,
a^b=(a 1i+a 2j + a 3k)-(b 1i + b 2i +6 3k) = (a 26 3 - a 3b 2 )\ + {a 3b x - aj> 3 )\ + (a^ - a^k, in virtue of the relations result
is
very important.
proved in the preceding Art. It
may
This
be written in the determinantal
form
a*b= a 1
a2
a3
b1
b2
b3
i
J
k
Hence, on squaring both 6. 2 2 above equation and dividing by a 6 we find the sine of the angle between two vectors a and b,
This vector has a module ab sin
members for
of the
sm
w
~~
,
ks+cv^W+V+V)
;
[CH.
VECTOR ANALYSIS
42 If
n' are the direction cosines of
m, n and V m',
I,
,
III.
a and b respec-
tively, this is equivalent to 2
+ (nl' -n'l) 2 + (lm' -I'm) worth noticing that if b = C + na, where n is any real number, sin 2 6
It is
= (win'
then
.
a b=a*(c + na)=a*c.
Conversely,
if
a-b=ac
does not follow that b=c, but that b
it
may
or
may
Let p be the length of the
perA
from c by some vector parallel to
differs
a,
which
not be zero.
Geometry of the Plane. 29. Equation of a plane.
ON
pendicular
to the given plane, and n
from the origin
to N. the unit vector normal to the plane, having the direction -> A Then ON =pn. If r is the position vector of any point P on the
A plane,
rn
to p.
Thus the equation
is
the projection of
OP
on ON, and
is
therefore equal
A
T'ii=p
by the
is satisfied
and
may If
n
clearly
position vector of every point on the plane
no point
off
is
any
the plane satisfies this relation.
plane, the equation
may
it
be written
vn=np = q which we
We
as the equation of the given plane. A vector parallel to n, and therefore normal to the
therefore speak of
shall take as the
(say),
standard form for the equation
(1)
of a
§
GEOMETRY OF THE PLANE
29]
plane.
If x, y, z
the equation (1)
is
is
I,
referred to
rect-
m, n are the direction cosines of n,
equivalent to Ix
which
P
are the coordinates of
angular axes through 0, and
43
+my + nz=p,
the form used in coordinate geometry.
Conversely, every equation of the form (1) represents a plane.
For
r 1; r 2 are
if
and m, ml any
two position vectors
satisfying the equation,
(m,T 1 +m'T 2 )/(m+m') That is to say, if P 1; P 2 are two points on the surface represented by the equation, any point in the straight line PiP 2 hes on the surface, which must therefore be a plane.
also satisfies
numbers, the vector
real
it.
Consider the plane through the point d and perpendicular If r is
any point on
the vector r - d
it,
and therefore perpendicular to
is
to n.
parallel to the plane,
Thus
n.
(r-d)-n=0 r*n=dTi
or
This
is
of
the form
considered.
to the plane
The length is
and
(1),
(2)
the equation of
is
of the perpendicular
the plane
from the origin
It is equal to the projection of
d-n/n.
OD
on
D being the point whose position vector
the normal to the plane, is d.
The angle between two planes whose equations are r-n
= g,
r-n'=g', is
between their normals.
equal to the angle
n*n' so that the required angle
= nn'
cos
is
— nn r
nn '
a 6
To i'n
find the intercepts
= q, we observe that
a-axis the point
x xi
on
= cos -1i
'
the coordinate axes
xx
if
lies
But
6,
is
made
bjr the plane
the length of the intercept on the
on the plane, and therefore
x^n = q
showing that
x l = ~-
rn
Similarly, the lengths of the intercepts
«i 31
= i.-
yn
and
on the other axes are
z,= r^-k*n
;
VECTOR. ANALYSIS
44
30. Distance of a point
length
p
[CH.
III.
from a plane. We have seen that the ON from the origin to the plane
of the perpendicular
vn = q
is q\n. Suppose we require the perpendicular distance any other point P' from the plane. Consider the parallel plane of to this through P'. The length p' of the perpendicular from plane is r'-n/w, where r' is the position vector of P'. Therefore
the length of the perpendicular from P' to the given plane
p-pr r The vector
P'M
,
=±rt-r'-n n
determined by this perpendicular (g
~
is
r
'
is
n) n.
n2
In deducing the above formula we have considered the distance
from the plane as positive when measured in the direction The expression found will therefore from to N. be positive for points on the same side of the plane as the origin of P'
of n, that is
:
negative for points on the opposite side.
Fig. 31.
To
Fig. 32.
find the distance of P' from the plane measured in the direction
drawn through P' cut the Then if d is the length of P'H, the position vector + db and since this point lies on the given plane,
of the unit vector b, let a parallel to b
plane in H. of
H is
r'
;
(i'
giving
This
is
+dbyn=q,
d=the required distance.
-r'-n
b*n
— GEOMETRY OF THE PLANE
§ § 30, 31]
The equations of
45
the planes bisecting the angles between two given
planes
r-n
= g,
T'ri=q',
are obtained from the fact that a point on either bisector
equidistant from the distance as positive
two
Since
planes.
is
we consider perpendicular
when measured
in the direction from the on the plane bisecting the angle the perpendicular distances will have the
origin to the plane, for points in
which the origin
lies,
same sign but for points on the other bisector, opposite signs. Hence if n, n' are directed from the origin toward the plane in each case, for any point r' on the bisector of the angle in ;
which
lies
we must have q
And
since this relation
- r'-n _q' - r'-n' n n'
by every point on the
is satisfied
the equation of this plane
r.(n-n')
=
'
2-C n
n
Similarly the equation of the other bisector
Wn+n
)
=
'
^
31. Planes
*-
n
+-, n
is
•
through the intersection of two planes.
two given planes
r-(n
Consider
m
r-n^,]
[
r-n' =
Then the equation
bisector,
is
'
- An') =q - Xq' n - An'.
(2)
And
any vector r Hence satisfies each of the equations (1), it also satisfies this. any point on the line of intersection of the given planes lies on the plane represented by (2), which is therefore the equation of a plane through this line. And by giving a suitable real value to A we can make it represent any plane through the intersection for the normal to the plane, viz. n - An', can be given any direction For instance the plane may be made in the plane of n and n'. to pass through any point d by choosing A so that is
that of a plane perpendicular to
if
;
d-(n-An')=<2-A
is
A =5
d-n
—-q
t
, •
VECTOR ANALYSIS
4C>
[CH.
of the line of intersection of the planes (1) This line is perpendicular to both n
The equation
found as follows.
and
may
be
and
n',
The perpendicular ON from n and n', so that
therefore parallel to n*n'.
is
the origin to the line
III.
parallel to the plane of
is
ON = ln + l'ri, where
this vector
must
{ln
1
The line n n'.
y
s
t
qn
-q
Its
n*n
w a n' a -(n'n') 2
is
equation
r
'
is
therefore
r
= Zn + Z'n'
a variable number, and
from a
32. Distance of a point
on both
planes,
I
and V are determined
as
q n* - <m n w 2w' 2 -(n'n') 2 -
=
of intersection of the planes
to
where
=
lies
(1), i.e.
+ l'riyn = q, + l'riyri = q'.
these two equations the values of i
N
since
both the equations
satisfy
(ln
From
But
are real numbers.
I'
I,
a line through
is
N parallel
+ta*n',
I,
V have the values found above.
To
line.
distance
of
find the perpendicular
a
point P'
from
the
straight line
r=a + «b, where b is a unit vector. Let P', A be the points whose position vectors are
r',
a respectively,
and P'N the perpendicular from to the given line. Then
P
P'l=a-r', and
P'N =P'A*-NA 2 = (a-r') 2 -{b-(a-r')} 2 2
Fig. 33.
This equation gives the length /As a vector
-^
p
-s-
of
is
(1)
-*-
P'N = P'A-NA = (a-r')-b»(a-r')b, and p
.
the perpendicular P'N.
(2)
the module of this vector.
Other theorems relating to the geometry of the plane and the straight line will be considered in the next chapter, after products of three vectors have been dealt with.
GEOMETRY OF THE SPHERE
§§32,33]
Geometry ){$&. Equation of a sphere.
and radius and
r,
P
If
a.
Consider the sphere of centre
c are the position vectors of P,
an origin 0, the vector
CP
(r
we put k = c - a we may 2
2
r
This equation
C
C
of the sphere,
respectively relative to
=r - c has a length equal to the radius,
and therefore If
of the Sphere.
any point on the surface
is
47
2
-c
2 )
= a2
.
write this relation
-2r-c +
fc
=
(1)
by the position vector of every point the sphere, and by no other. It will therefore
satisfied
is
on the surface of
be called the equation of the sphere relative to the origin 0.
The
may
first
member
of (1), regarded as a function of the vector r,
for convenience be denoted by F(i)
written briefly F(t)
and the equation
;
= 0.
Fig. 35.
Fig. 34.
Consider the points of intersection of the surface with the straight line through the point
The equation
of this line
D
parallel to the unit vector b.
is
r=d+rt>,
where d
points of eliminate
(2)
The values of r for the If then we intersection satisfy both (1) and (2). r from these equations we find, for the values of t
is
the position vector of D.
corresponding to the points of intersection, the quadratic equation 2 t
the coefficient of
+ 2b-(d -c)t + (d 2 -2c-d +k) =0, t
2
being unity, since b
equation has two roots,
t-,,
t
2
which are
{b-(d-c)}
2
is
real
>P(d),
a unit vector. if
(3)
The
VECTOR ANALYSIS
48
[CH.
Corresponding
the meaning of which will be apparent presently.
two roots there are two points of The product the absolute term of (3), i.e.
to these
DP=t^ and DQ=t 2
that to
.
III.
intersection, P, Q, such of these roots is equal
DP .DQ = F(d). independent of
This
is
lines
through D.
straight line
b,
and
therefore the
is
same
for all straight
T, the
Q tend to coincidence at becomes a tangent, and we have DT* = DP. DQ = F(A) If
the points P,
(4)
Thus F(A) measures the square on the tangent from D to the surface of the sphere, and all such tangents have the same length. In particular, F(0)=k is the square on the tangent from the origin. If is within the sphere h is negative, and the tangents from are imaginary. The tangents from D are generators having
of a cone, called the tangent cone,
its
vertex at
D
and
enveloping the sphere.
\
34.
Equation of the tangent plane at a point.
on the surface (3) is zero.
of the sphere, F(A)
If
=0, and one root
In order that the line
(2)
D
is
a point
of the equation
should touch the surface,
the other root also must be zero, for a tangent line intersects
the surface in two coincident points. are zero,
then both roots
of (3)
b-(d-c)=0,
showing that the tangent If r is
If
line is perpendicular to the radius
any point on the tangent line,
CD.
this condition is equivalent to
(r-d)-(d-c)=0 This to
equation represents
CD
plane,
a
plane
(5)
through
D
perpendicular
showing that all tangent lines through D he on which is called the tangent plane at D. Adding ;
zero quantity F(A) to the first
equation
which we
r-d -c-(r
member + d)
+/t
of (5),
we may
this
the
write the
=0,
(6)
shall take as the
standard form of the equation the tangent plane at the point d.
We
of
can now interpret the above condition for reality of the For b-(d-c) is the projection of CD along b, B (3).
roots of
being any point, and F{&)
on the tangent DT.
is
Hence
the square of the projection of
CD
for reality of the roots the angle
of
GEOMETRY OF THE SPHERE
34, 35]
inclination of the line (2) with
tangent with DC.
That the tangent cone from D.
is
Since a tangent plane
is
DC must
be
less
to say, the line (2)
49
than that of the
must
lie
within
perpendicular to the radius to the
point of contact, the square of the perpendicular from the centre
C
to a tangent plane
that the
plane
vn = q
must be equal to a 2
should touch the sphere
q-cn --c
Further,
if
2
Hence
.
the condition
(I) is that
-k.
•(?)
two spheres cut each other
at right angles, the
tangent plane to either at a point of intersection passes through
Hence the square on the
the centre of the other.
their centres is equal to the If
sum
line joining
on their
of the squares
radii.
then the equations of the two spheres are r2 r2
-2r-c +
fc
= 0,]
-2r-c'+&'=0,J
the condition that they should cut orthogonally
(c-c') 2
that
is
=a 2 +a' 2 = c 2 -k+c' 2 -k',
2c-c'-(& + £')=0
is
35. Polar plane of a point.
to the sphere
whose equation
(8)
The tangent plane is
(1)
is
at the point d
represented
by
(6).
If
plane passes through the
this
point h, satisfy (6)
h must
the vector ;
that
is
h-d-c-(n + d)+& = 0. Further, every point such as d,
tangent plane
the
passes through
at
which
h, satisfies this
and therefore lies on the plane whose equation is relation,
r-h-c-(r '
This
is
+ h)+& =
Fig. 36.
(9)
called the polar plane of the point h.
Since the terms
involving r may be written r-(h - c), the plane is perpendicular to the straight line joining the point h to the centre of the sphere.
VECTOR ANALYSIS
50
If this straight line cuts
perpendicular
-fc-ch + c(c+h)
is
a
by Art.
30,
2
CH'
V(h-c)*
H
III.
the polar plane at N, the length of the
CN from C to the polar plane is, CN=-
where
[CH.
the point whose position vector
is
This shows
h.
N and H The polar plane of H cuts the sphere in a circle, and the line joining H to any point on this circle is a tangent line. All such tangent lines together form the tangent cone from H to the sphere. Theorem 1. If the polar plane of the point H passes through G, Let h, g be the position the polar plane of G passes through H. vectors of the two points. Then if g lies on the polar plane that is of h it must satisfy (9) are inverse points with respect to the sphere.
that
;
g*h-c*(g+h)+A = 0, and the symmetry plane of
shows that h
lies
on the polar
drawn from a point
to intersect
of this relation
g.
Theorem
2.
Any
straight line
harmonically by the surface and the polar plane of 0. Take the point as origin, and let (1) be the equation of the
a sphere
is cut
Then, by
surface.
(9),
the polar plane of
=
r-c
The equation vector b
is
r
of the straight line
= tb
;
and the value
(10)
ft
through tx
Similarly the values
t2
and
t
3
parallel to the unit
for the point
R
at which
for the points P,
Q
at which
of
this line cuts the polar plane is given
tjrc
is
t
by
= k.
of
t
the same line cuts the sphere are the roots of the equation t
„ Hence
2
-2tb-c + k=0.
1 1 fcj+ 2 _^ = 2b-c -+_= __^ = _. i
ts
But tv t 2 t 3 measure the lengths of OR, OP, OQ and therefore , , „ = + OP OQ OR' ,
which proves the theorem.
respectively;
§ §
X
GEOMETRY OF THE SPHERE
35, 36, 37]
Diametral plane.
36.
Consider again the points of intersection
with the straight
of the sphere (1)
51
line
r=d + fb, through the point d parallel to
be equal and opposite
will
(2)
The
b.
roots of the equation (3)
if
b-(d-c)=0, and d
will
then be the mid point of the chord joining the points This equation shows that
of intersection.
bisecting chords parallel to b, lie
all
points such as d,
on the plane whose equation
is
b'(r-c)=0. This plane, which passes through the centre of the sphere and
perpendicular to the chords of the
it
bisects,
sphere for chords parallel to
Further,
if
condition for
is
b.
any point on the straight line (2), the above equal and opposite roots of the quadratic in t shows r is
ttat
(r-d)-(d-c)=0,
which
is
the equation of a plane through the point d perpendicular
to the line joining d to the centre of the sphere. of
is
called the diametral plane
the sphere, which are bisected
through
D perpendicular to
37. Radical plane of
the point d to the
by the point D,
Thus
all
chords
in the plane
lie
CD.
two spheres.
If
the tangents drawn from
two spheres
= r 2 -2r-c+& = 0, ] F(r) = r 2 -2r-c'+£'=o} that is length, F(A) = F'(A) .F(r)
are equal in
;
2d«(c-c') =*-*'.
Thus the point d
lies
on the plane whose equation
is
2r-(c -c') = k -k'
This
is
called the radical plane of the
two
If
the spheres intersect,
radical plane.
it
all
For any value
the radical plane.
is
perpendi-
;
to the
both spheres also satisfies F(v) of
It
and the tangents two spheres are equal in length. the points of intersection lie on the
cular to the straight line joining their centres
$ drawn from any point on
(11)
spheres.
of r satisfying the equations of
-F(r)=0, which
is
the equation
VECTOR ANALYSIS
52
[CH.
III.
We shall now consider a system of spheres with a common radical Let be the origin, and A, B the two points a, -a
plane.
respectively on the surface
= a 2 Consider the line AB, and which
r2
.
of
the sphere whose equation is lie on the straight
spheres whose centres
The
cut the above sphere orthogonally.
is a point ma, where m is a real number, and the equation of the sphere is then
centre of any one of these positive or negative,
r2
But by have
cuts the sphere r 2
(8), if this
k=a 2
.
=a 2
Hence the required system
by the equation
m
-2mi-a + & = 0.
r
2
we must
orthogonally,
of spheres is represented
-2mra+a 2 = 0,
(12)
denoting a real number, positive or negative.
The
radius of
a sphere of the system depends on the value of m, the square of the radius
being c
2
-k=m
2
a2
-a 2 = (m 2 - l)a 2
.
Thus the spheres increase in size as their centres get further from the origin. Since the square of the radius must be positive, the value of m cannot lie between ± 1. For these limiting values of
m the radius
spheres being
of the sphere is zero,
A
limiting points of
The values
These points are therefore called the system of spheres.
radical plane of the ffl]
and
m
the centres of the vanishing
and B.
2 is :
two spheres corresponding
the
to the
m virtue of (11), 2r (m a-m a) = 0, ,
1
that
which is
2
r«a
is is
= 0,
the plane through the origin perpendicular to
the same for
all
pairs of spheres.
Thus the system
AB,
and
of spheres
R
§ §
GEOMETRY OF THE SPHERE
37, 38]
common
has a
radical plane.
the spheres for the values
R
2 x
-
2 2
m
1
If
and
C x and C 2
m
a relation showing
is
x
are the centres of
and
R
their radii,
2
= (m x 2 - l)a 2 - (m 2 2 -l)a 2 = mfa? -m 2 2 a 2 = OC\ 2 -OC 2 2
which
R
and
2,
53
how
,
the radical plane divides the
two spheres. The limiting points A, B are inverse points with respect to any sphere of the system. For if C is the centre ma, and R the radius of any sphere, join of the centres of
AC BC = (ma -a)-(wa + a) = (m
- l)a 2 = R 2
2
.
which proves the statement. either limiting point,
,
It follows that the polar plane of
with respect to any sphere of the system,
and
passes through the other limiting point,
is
perpendicular
to the line joining them.
X
38. If
i^(r)=0 and F'(r)=0 are the equations of two spheres,
then
F(r)-X*"(r)=0
also represents
on division by
(13)
a sphere, \ being an arbitrary number.
(1
-A)
this equation takes the
For,
form
^( T^M-r^)=° c
which represents a sphere whose centre
is
<
the point
(c
13 '>
- Xc')/(1 - X)
line through the centres of the two given spheres. shown that any two spheres given by (13') for different X have a radical plane
on the straight It
is
easily
values of
2r-(c-c')
which
=&-&',
(14)
the same for any two spheres, and is the radical plane two given spheres. Hence the equation (13), for real
is
of the
values of X, represents a system of spheres with a
common
radical
plane, perpendicular to the line of centres. If
the two spheres
plane
and
all
J(r)=0 and jF'(r)=0
intersect, the radical
spheres of the system (13) pass through their circle For any value of r satisfying the equations of
of intersection.
both spheres clearly satisfies both (13) and
(14).
VECTOR ANALYSIS
54
[CH.
III.
Application to Mechanics.
Work done by a
39.
work when the
force.
particle
is
A
force acting
on a particle does
displaced in a direction which
is
not
The work done is a scalar quantity proportional to the force and the resolved part of the
perpendicular to the force. jointly
We
displacement in the direction of the force.
choose the unit
quantity of work as that done when a particle, acted on by unit force, is displaced unit distance in the direction of the force.
Hence,
F, d are vectors representing the force
if
ment respectively, work done is The work done
is
inclined at
w
an angle
id cos,
zero only
,
,
.
.
.
,
Fn
.
separate forces total
work done
and the
displace-
the measure of the
= „, F*d.
fl ft
when d
is
perpendicular to F.
acted on by several forces Then during a displacement d of the particle the do quantities of work F/d, F 2 -d, F K *d. The
Suppose next that the particle
F2 F 2
6,
is
.
is
.
.
,
«
]>>d=d-2F=d-R, i
and is therefore the same as by its resultant R. Note.
A
if
the system of forces were replaced
force represented by the vector
referred to as a force F.
No
F may
misunderstanding
be conveniently
is possible,
for our
Clarendon symbols always denote Similarly we may
length-vectors.
speak of a displacement velocity v,
d, or
a
as we have already
done of a point r. 40. Vector moment or torque of a force. (or
The vector moment moment) of a
briefly the
force
F about
a point
is
a
vector quantity related to an axis through
perpendicular
the plane containing
and
to
the
Fig. 38.
line of action of the force.
magnitude
is
cular distance
jointly proportional to the force
ON upon its line of
action.
Its
and the perpendi-
The moment
or torque
APPLICATION TO MECHANICS
§§39, 40]
may
the force about
of
therefore be represented
perpendicular to the plane of r and F, where vector relative to
any point
of
And with the
force.
55
P
choice of unit
r is
by a vector the position
on the
line of action of the
moment
as that of unit force
localised in a line unit distance
the
from 0, the vector representing For the direction of this positive relative to the rotation about the axis through
moment
vector
is
of
F about
in the sense indicated
module
of the vector
is
r*F.
by the
direction of the force
;
and the
is
F. OP.
sinOPN = F. ON
as required.
there are several forces
If
same point
P
this resultant
F1; F 2 F 3 acting through the R = 2F. The moment of .
,
about
.
.)
.
.
.
is
.
is
r-R=r*(F1 +F 2 + =rxF 1 + rxF2 + and
.
they have a resultant
therefore equal to the vector
sum
.
,
of the
moments
of the
Thus, if the system of forces through P is replaced resultant, the moment about any point remains unchanged.
separate forces. by
its
Express the vectors F, ponents,as
r in
terms of their rectangular com-
F=Xi+Fj r
Then the moment
+ Zk,
= xi + yj + zk.
of the force
F about
is
M = (yZ-zY)i+(zX-xZ)i+(xY-yX)k. In this expression the coefficients of
moments
of the force
i,
j,
k are the ordinary
about the coordinate axes.
system of rectangular axes through
may
And
scalar as the
be chosen so that one
them has any assigned direction, it follows that the ordinary moment of a force F about any straight line through is the resolved part along this line of the vector moment of F about 0. With the above notation the scalar moments about the coordinate axes are M*i, M-j and M-k respectively. of
If
there are several concurrent forces
moment the sum of
that the scalar is
equal to
about that axis.
of the resultant
the scalar
it
follows from the above
about any axis through
moments
of the several forces
VECTOR ANALYSIS
56
[CH.
III.
In agreement with the above we adopt the following
The moment about
Definition.
the origin of the vector
v
localised
in a line through the -point r is the vector r*v. 41.
Angular velocity of a rigid body about a fixed axis. Conmotion of a rigid body rotating about a fixed axis ON. at
sider the
the rate of
shown in Art. 87 that, body is fixed, the instantaneous motion of the body is
w radians per second. if
It will be
of the
one point
one of rotation about such an axis through 0, every point on the axis being instantaneously at
rest.
For the
we take the rotation as given about the axis ON. The angular velocity of the body is uniquely specified by a vector A whose module is w and
present
whose direction is parallel to the axis, and in the positive sense relative to FlG 39 -
the rotation.
-
be any point on the fixed
Let
axis,
P
a point fixed in the
and PN perpendiThen the particle at P is moving Its velocity path, with centre N and radius p =PN. perpendicular to the plane OPN and of magnitude
body, r the position vector of
P
relative to 0,
cular to the axis of rotation. in a circular is
therefore
A
po}=rw
sin
PON.
Such a velocity
is
A*r, the sense of this vector being the
represented by the vector
same
as that of the velocity.
P is
In other words, the velocity of the particle at
v=A*r. Examples. (1)
A
i
+ 3j - k.
an angular velocity of 4 radians - k passing through the point particle at the point 4i - 2j +k.
rigid body is spinning with
per second about
Find
an axis
parallel to 3j
the velocity
of the
Relative to the given point on the axis of rotation the position vector of the 1particle is r
= 3i-5j+2k.
The vector specifying the angular
A= 4(3j -k)
velocity
is
4
W^TVio
(3]
- k)
-
APPLICATION TO MECHANICS
41]
§
Hence the velocity
of the particle
57
given by
is
A*r = -^-(3j-k)x(3i-5j+2k)
(i-3j-9kj.
VlO This represents a speed of of the vector A*r.
Laws
(2)
iV^ = 12
approximately, in the direction
Let n, a, b, c be of Reflection and Refraction of Light. normal to the surface of separation of two
unit vectors, the first
media, and the others in the directions of the incident, reflected and refracted rays the laws of reflection and refraction are specified ;
y
= b n, = ot'cxn ua>
and
/
respectively,
where
/j.,
fx
/
are the indices of refraction for the two
media.
For the
first
equation makes the incident and reflected rays
coplanar with the normal, as well as the angles of incidence and reflection
equal.
The second makes the incident and refracted and also gives
rays coplanar with the normal,
yusin
where
i
and
r are
i
= jx
sinr,
the angles of incidence and refraction respectively.
EXERCISES ON CHAPTER Give vectorial solutions of the following 1.
The perpendiculars
let fall
HI.
:
from the vertices
of
a triangle to
the opposite sides are concurrent. 2.
The perpendicular
bisectors
of
the sides of a triangle are
concurrent. 3.
The vector 'area
a, b, c is
of of
|(b*c
of the triangle
whose vertices are the points
+ c*a + a*b).
4. The area of the triangle formed by joining the mid point one of the non-parallel sides of a trapezium to the extremities the opposite side is half that of the trapezium. 5.
What
is
the unit vector perpendicular to each of the vectors Calculate the sine of the angle between 1
— ] + k and 3i + 4:j - k these two vectors. 2i
6.
by
3i
Find the torque about the point i + 2j - k acting through the point 2i - j + 3k.
+k
of a force represented
VECTOR ANALYSIS
58 7.
Show that
[CH.
twice the vector area of a closed plane polygon is sum of the torques about any point of forces
equivectorial * with the
represented by the sides of the polygon taken in order. 8. Find the equation of the straight line through the point equally inclined to the vectors a, b, c. 9.
If
d,
is equally inclined to three coplanar straight perpendicular to their plane.
a straight line
lines, it is
10. If a point
equidistant from the vertices of a right-angled mid point of the hypotenuse is perpendicular
is
triangle, its join to the
to the plane of the triangle.
ON
is drawn to a perpendicular 11. From an external point to a straight line PQ in the plane. Prove a plane, and another that is perpendicular to PQ.
OM
MN
12.
Find the two vectors which are equally inclined to
perpendicular to each other and to
i
k,
and
are
+ j + k.
13. The sum of the squares on the edges of any tetrahedron is equal to four times the sum of the squares on the joins of the mid points of opposite edges.
14. Prove that in any parallelogram the sum of the squares on the diagonals is twice the sum of the squares on two adjacent sides the difference of the' squares on the diagonals is four times the rectangle contained by either of these sides and the projection of and the difference of the squares on two adjacent the other upon it sides is equal to the rectangle contained by either diagonal and the projection of the other upon it. ;
;
15. Show that, in a regular tetrahedron, the perpendiculars from the vertices to the opposite faces meet those faces at their centroids. Find the angle between two faces, and the angle between a face and an edge which cuts it. 16. Find the equation of the plane through the point 2i + 3j - k perpendicular to the vector 3i - 4j + 7k. 17. Find the equation of the plane passing through the and the line of intersection of T'& = p and fb = q.
origin,
18. Show that the points i-j+3k and 3(i + j+k) are equidistant from the plane r(5i + 2j - 7k) + 9 = 0, and on opposite sides of it. 19.
and
Find the equation + 4j-2k) = 2.
of the line of intersection of r(3i
- j +k) = 1,
r(i
20. Determine the plane through the point i + 2j - k, which is perpendicular to the line of intersection of the planes in the previous exercise. *
i.e.
specified
by the same vector.
EXERCISES ON CHAPTER
III.]
a,
59
III.
21. Find the perpendicular distance of a corner of a unit cube from diagonal not passing through it. 22.
Show
that the
sum
of the reciprocals of the squares of the
made by a fixed plane systems of rectangular axes with a given origin.
intercepts on rectangular axes all
is
the same for
23. Find the coordinates of the centre of the sphere inscribed in the tetrahedron bounded by the planes r*i
24.
r(7j
r*j
= 0,
r*k
= 0,
and
r*(i
+ j + k) = a.
that the planes r(2i + 5j +3k)=0, r*(i have a common line of intersection.
j
+4k)
= 2, and
-5k) + 4 =
25. is
Show
= 0,
The line of intersection of r-(i + 2j + 3k) = and r-(3i + 2j + k) = _1 i and k, and makes an angle £sec 3 with j.
equally inclined to 26.
Find the locus
27.
forces 28.
which moves so that the difference from two given points is constant.
of a point
of the squares of its distances
Find the locus of a point about which two given coplanar have equal moments. Three forces P, 2P,
3P
act along the sides
an equilateral triangle ABC. Find the point in which its line of action cuts BC. spectively of
A
AB, BC, CA
their resultant,
re-
and
*
on by constant forces 4i + j -3k and 3i + j -k, 29. is displaced from the point i + 2j + 3k to the point 5i + 4j + k. Find the total work done by the forces. 30.
particle, acted
Show
at a point 31.
that any diameter on the surface.
The equation
of a sphere
subtends a right angle
which has the points (r-g) - (r-h) = 0.
of a sphere
extremities of a diameter
is
g,
h
as the
32. The locus of a point, the sum of the squares on whose distances from n given points is constant, is a sphere.
33. The locus of a point which moves so that its distances from two fixed points are in a constant ratio n 1 is a sphere. Also show that all such spheres, for different values of n, have a common :
radical plane. 34.
in P.
From a In
point
O
a straight line
is
drawn
to
meet a fixed sphere Find
OP a point Q is taken so that OP OQ is a fixed ratio. :
the' locus of Q.
35.
Find the inverse of a sphere with respect to a point (ii) on its surface.
(i)
outside
the sphere, 36.
The distances
of
two points from the centre
of a given sphere
are proportional to the distances of the points each
plane of the other.
from the polar
VECTOB ANALYSIS
60 37.
[CH.
III.
The sphere which cuts F(i)=0 and J"(r)=0 orthogonally - \F'(r) = orthogonally.
also cuts F(r)
from any point on the surface of a sphere, straight lines to the extremities of any diameter of a concentric sphere, the sum of the squares on these lines is constant. 38.
are
If,
drawn
39.
From
the relations
ch.
iv., § § 42, 43]
CHAPTEE
IV.
PRODUCTS OF THREE OR FOUR VECTORS. NON-INTERSECTING STRAIGHT LINES. Products of Three Vectors. 42. Since the cross
vector,
product
we may form with
product a*(b*c)
be
of the vectors b, c
is itself
a
and a third vector a both the scalar and the vector product a*(b*c). The former is it
a number and the latter a
Such
vector.
are
triple
frequent
of
products
occurrence,
/
and we shall find it useful to examine their properties. 43.
Scalar
triple
/
/
\
\
/
\
\
product,
Consider the paral-
a*(b*c).
whose concurrent OA, OB, OC have the lengths and directions of the
lelepiped
edges
vectors a, b,
c
respectively.
which we may denote by n, is perpendicular OBDC, and its module n is the measure of the area that face. If 6 is the angle between the directions of n and a,
Then the vector
b*C,
to the face of
the triple product
where
The
V
triple
is
a
.
(bx(j)
=
m cog Q = y^
the measure of the volume of the parallelepiped.
product
is
positive
if
9
is
acute, that
is if a, b,
c
form
a right-handed system of vectors.
The same reasoning shows that each of the products b*(c*a) and c(a*b) is equal to V, and therefore to the original product. The cyclic order a, b, c is maintained in each of these. If, 61
VECTOR ANALYSIS
C2 however, that order
changed, the sign of the product
is
For since b*C = -C b
it
= -(c*b)-a= = b*(c*a) = (c>a)-b = -(ac)*b = =c(a"b) = (a^b)«c= -(b-a)*c =
factors,
These
-b-(a*c)
-c(b*a).
product depends on the cyclic order of the
of the
[abc],
their cyclic order.
Then
which indicates the three factors and
= - [acb].
[abc] If the three vectors are
zero.
-a«(c*b)
is
above product by
is
changed.
independent of the position of the dot and cross. It is usual to denote the be interchanged at pleasure.
but
may
is
follows that
F=a-(b*c) = (b*c)-a
Thus the value
[CH. IV.
For
be
product vanishes.
coplanar their scalar triple product
then perpendicular to
is
Thus the vanishing
that the vectors should be coplanar.
parallel this condition
are equal the product
is satisfied. is
If
a,
and
of [abc]
two
is
their scalar
the condition
of the vectors are
In particular,
if
two
of
them
zero.
Expressing the vectors in terms of their rectangular components, a =aj.
+ a 2\ +a 3 k, and
[abc]
is
the well
we have -
=
This
so on,
= a^btfz - 6 3 c2) + a 2 (b s c1 - bjC3 ) + a 3 (b 1 c 2 •Vi)
%
a2
&i
K h
cx
c2
known
a3
c3
expression for the volume of a parallele-
More generally, m, n we write
piped with one corner at the origin. of three non-coplanar vectors
1,
if
in terms
a=a l+a 2m+a 3n, 1
and so on,
it is
easilv
shown that
[abc]
=
[lmn].
h K The product
[ijk],
b
of three rectangular unit vectors, is obviously
equal to unity. 44. Vector triple of a
and be,
viz.
product.
Consider next the cross product
P = a<(b*c).
§
PRODUCTS OF THREE VECTORS
44]
This
a vector perpendicular to both a and
is
normal to the plane of b and It is therefore expressible in
so that
c,
63
But b*c
b>
P must he
is
in this plane.
terms of b and c in the form
P = /!b+mc. To find the actual expression for P consider unit vectors j and k, the first parallel to b and the second perpendicular to it in the plane b, c. Then we may put b=6j,
c=c 2j +c3k. In terms of
j,
k and the other unit vector
system, the remaining vector a
a
Then b*C=&c3i, and the ax(b^c)
may
i
of the right-handed
be written
= a x i + ay + a 3k.
triple
product
= %&c 3j -a 2bc 3 = (a t c 2 + a 3 c 3 )M - a 2b(c 2 + c 3k) ls.
]
•
This
=a-cb-a*bc
the required expression for
is
(1)
P
in
terms
of
b and
c.
Similarly the triple product
(b*c)*a
=
-a«(b
=a*bc-a-cb It
be noticed that the expansions
will
written
down by the same
factor outside the bracket of the
Bach and the
rule. ;
(2) (1)
and
(2) are both
scalar product involves the first is
the scalar product
extremes.
In a vector triple product the position of the brackets cannot be changed without altering the value of the product.
For
a vector expressible in terms of a and b
one
(a*b)*c
is
expressible in terms of b
and
c.
The products
;
a> (b*c) is
in general therefore
represent different vectors. If a
a and
vector r r,
is
resolved into two components in the plane of
one parallel to a and the other perpendicular to
former component
is
-y a, and a*r
a-
therefore the latter
_ a*ar - a*ra _ a> (r *a) a2
a*
it,
the
VECTOR ANALYSIS
64
[OH. IV.
Products of Four Vectors.
A
45.
The products already
scalar product of four vectors.
considered are usually sufficient for practical applications. But we occasionally meet with products of four vectors of the following types.
This
Consider the scalar product of a b and C*d. easily expressible in
vectors.
terms
of the scalar
is
a number
products of the individual
For, in virtue of the fact that in a scalar triple product
may
the dot and cross
be interchanged, we
may
write
(a*b)-(c*d)=a-b*(c*d)
= a*(b*dc - b*cd) = a-cb-d - a db*c. -
Writing this result in the form of a determinant,
=
(a*b)-(c-d)
46.
A
a«c
a*d
b*c
b'd
product of four vectors.*
vector
vector product of a^b and c*d.
This
is
to c
and
Similarly
it is co-
It
d.
To express the product it
b.
must therefore be parallel to the line of d. parallel to a and b with another parallel a plane
and
intersection of
Consider next the
a vector at right angles
to a*b, and therefore coplanar with a and
planar with c
we have
(a
as the vector triple product of (a 'b)"(C'd)
Similarly, regarding
(a
it
b and m, where
b,
regard
m =C*d.
Then
= (a*b)*m=a'mb -b*ma = [acd]b-[bcdja
write
b)^(cd)=n(c d) =ivdc-n-cd = Labd]c-[abc]d
(2)
the
a, b, c, d, viz.
[bcd]a-[acd]b + [abd]c-[abc]d = *
d,
it
Equating these two expressions we have a relation between four vectors
(1)
as the vector triple product of n, c and
may
where n =a>b, we
a,
and
The bracket notation
for the
(a[bc])
([ab][cd])
above products
= a»(b
= (axb)Kcxd)
;
and
(3)
and four vectors
of three
= ax(bxc) [[ab][cdj] = (axb)x(cxd).
[a[bc]]
;
is
.
PRODUCTS OF FOUR VECTORS
§§45,46,47] Writing
we may express any vector
r instead of d,
three other vectors a, b, c in the r
which is
is
except
"~
a', b', c'
when the denominator
[abc] vanishes, that
system of vectors to
a, b, c.
In terms of the
denned by the equations
we may write the
=
C
;
[abc]'
[abc]
relation (4) above,
= ra'a + r-b'b + r*c'c.
r a',
'
coplanar.
a_ [abc]
The vectors
{
"label
a, b, c are
47. Reciprocal
vectors
terms of
r in
form
_ [rbc]a + [rca]b + [rab]c
valid except
when
65
b',
c'
are called the reciprocal system to a, b,
C,
which are assumed non-coplanar so that the denominator [abc]
The reason for the name a . a = b«b' = cc' = 1
does not vanish. relations
They
product of any
also possess the property that the scalar
other pair of vectors, one from each system, a«b' =[aca]/[abc] since, in
two
the numerator,
are equal.
terms of
a', b',
reciprocal
;
c,
then
a, b, c is
r
may
if
a',
b', c'
the reciprocal
be expressed in
form
in the r
The system
=0,
Hence any vector
c'.
For instance,
above relations shows that
of the
a', b', c'
zero.
factors of the scalar triple product
the reciprocal system to a, b,
system to
is
= 0.
Similarly c*b'
The symmetry is
in the obvious
lies
'
= r'aa'+r«bb'+r'cc'.
i, j, k is easily seen to be and we have already proved for this case the
of unit vectors
its
own
relation
r=rii+r«jj +r*kk.
The
scalar triple product [abc],
vectors,
is
formed from three non-coplanar [a'b'c'] formed from the
the reciprocal of the product
reciprocal system.
For
be [a'b'c']
c a
[b*c, c*a, a*b]
a*b
=
_[abc] [abc] [abc]J
Now, by of the
Art. 46, (b*c)*(C'a) =[abc]c,
above expression
is
~
[abc] 3
and therefore the numerator Hence the result.
equal to [abc] 2
.
;
VECTOR ANALYSIS
66
[CH. IV.
Further Geometry of the Plane and Straight Line. 48. Planes satisfying various
how
We
conditions.
the triple products considered above
may
shall
now
Let
the vector equation of a plane subject to certain conditions. us examine the following typical cases
:
Plane through three given points A, B, C.
(i)
see
be used in forming
Let
a, b, C
be
the position vectors of the points relative to an assigned origin 0, and r that of a variable point P on the plane. Since P, A, B, C all
b-C
he on the plane, the vectors r-a, a-b,
and
their scalar triple product
is
are coplanar,
Hence
zero.
(r-a)-(a-b)*(b-c)=0.
we expand
and neglect the triple products vector occurs twice, the equation becomes If
this,
r-(b*c
Thus the plane
is
in which any
+ c*a + a*b) = [abc].
perpendicular to the vector
n = b*c + c*a + a*b, which represents twice the vector area
n is the module of n, the length p from the origin to the plane is
of
the triangle
ABC.
of the perpendicular
If
ON
[abc]
r A
~*
n
1
ON =pn = -^ [abc](b*c + c*a + a-b).
while
Plane through a given point parallel
(ii)
Let a be the given point, and
lines.
to the given lines.
Then b*c
is
b,
C
By
Art. 29 this
(r-a)«b*c that
is
(iii)
first line
of the plane through
is
= 0,
straight line
and
parallel to another..
be represented by r
while the second
parallel
c=[abc].
r-b
Plane containing a given
Let the
two given straight
two vectors
perpendicular to the plane
and we have only to write down the equation a perpendicular to b-C
to
is
= a + tb,
parallel to c.
Then the plane
in question
THE PLANE AND STRAIGHT LINE
§§ 48, 49, 50]
contains the point a, is
by the
therefore,
and
is
parallel to b
c.
Its equation
last case,
r-b*c (iv)
and
= [abc].
Plane through two given points and parallel
Let
straight line.
parallel to
a and
therefore
to
a given
b be the two given points, and c a vector
a,
the given straight line.
passes through
67
parallel to b
is
The required plane then - a and c. Its equation is
- a )« c = [a, b - a,
r .(b
c]
= [abc]. (v)
Plane containing a given straight line and a given point.
Let
r=a + ib be
and
is
the given straight line and c the given point. Then the plane in question passes through the two points a, C parallel to b.
Hence by
(iv) its
equation
is
r-(a-c)*b=[abc]. 49. Condition of intersection of
two
straight lines.
Let the
equations of the given straight lines be
r=a + ib, r so that
they pass through the points
respectively.
If
|
= a'+sb'J a, a'
they intersect, their
and are
common
parallel to b, b'
plane must be
-b
FIG. 41.
parallel to
product
is
b, b', a -a', whose scalar Hence the required condition is
each of the vectors therefore zero.
[b, b',
50. lines.
triple
a-a'] = 0.
The common perpendicular to two non-intersecting Let the equations of the two straight lines be
r=a + «b, r=a'+sb'.
straight
VECTOR ANALYSIS
68
[CH. IV. •
Then the vector n=b*b' are the points a, a'
perpendicular
is
perpendicular to both lines, and
is
common
therefore parallel to their
respectively, the length
In finding
'-
n
is
this value of
from P' to P.
b',
a -a'].
p we have assumed
that the direction
This will be the case
when the moment
about the
-^^
^~"^\A
vector
/ -^^T -^
is
1
/
/
/
/
—
n.
n-(a-a')
—i p r =
= -[b,
-
p
A, A'
If
common
of this
equal to the length of the projection of A' A on
Hence
of b*b'
perpendicular P'P.
^~^-^^
-»-b
A'P'
of the
in
AP
positive or right-handed
the
D
T7
A
line
localised
relative to b'.
„
I
___-L
b
In
this case
moment about
A'P' to b.
AP
....
the vector is
b'
of
localised in>
positive
We may
relative
adopt
this
convention for the sign of the perpendicular distance, so that p In so doing we attach is positive when the moment is positive.
two lines the sense of the vectors b and The condition found above for the intersection to the
is,
b' respectively.
of the
two
lines
of course, equivalent to the vanishing of p. If r is
any point on the common perpendicular to the two
the vectors r-a, b and b*b' are coplanar.
and
Hence the
b*b' are coplanar.
the given lines at right angles planes [
r
_a,
[r-a',
is
lines,
Similarly r-a',
V
straight line which cuts both
the line of intersection of the
b,
b*b']=0,
b',
b*b']=0.
The equations of two non-intersecting straight lines can be put form by choosing as origin the middle point of their common perpendicular. The equations of the lines are then
in a convenient
where
r
= c+ib,
c
= \P'P = \pa. = \p~ =
2^ i
[b
r= -c+sb',
)
b',a-a'](b«b').
§ §
THE PLANE AND STRAIGHT LINE
50, 51]
The vector
c,
being perpendicular to both
equations
69 satisfies
lines,
the
c-b=c-b'=0.
Example.
When a
ray of light
from a plane mirror, the shortest and any straight line on the mirror reflected ray and the same straight line.
is reflected
distance between the incident ray is
equal to that between the
Let the unit vector in the direction of the incident ray be expressed in the form a - b, where a is parallel to the mirror and b perpendicular Then the unit vector in the direction of the reflected ray is to it. a + b, because reflection at the mirror reverses the direction of the normal component, but leaves the other unchanged. Hence if the point of incidence is taken as origin the incident and reflected rays are
= «(a-b) r = Z(a + b),
r
and respectively.
Any
(2)
straight line in the plane of the mirror
represented by
r
(3) is
may
= c + *d,
where c and d are perpendicular to
W and
(1)
be (3)
The
b.
shortest distance between
[ a-b,d, -c] '
|(a-b)*d|
But
a, d, c are coplanar, so that the numerator is equal to [bdc], which merely changes sign when b is replaced by -b. And since a*d is perpendicular to b>
V(a»
which remains unchanged
when
+ (bxd) 2
b
is
,
replaced
by -b.
Hence the
result.
51. Pliicker's coordinates of
a straight
vector parallel to a given line, origin of the vector
and
m
Let d be a unit
line.
the
d localised in the given
moment about
line
;
the
so that
m=r*d, where
r is
the position vector of any point on the
the position of the line
is
line.
uniquely specified by d and m.
direction of the line is that of d.
And
since
m
is
Then The
perpendicular
and the given line, the direction of m determines a plane in which the straight line must lie, and the magnitude of m determines its distance from the origin. to the plane containing the origin
Thus,
The
if
d,
m
are given the position of the straight line
six quantities
known
is
given.
as Pliicker's coordinates of the line
VECTOR ANALYSIS
70 the
are
resolutes
(scalar)
of
m
d,
jCH. IV.
along
rectangular
axes
two
lines,
through 0.
Let
and
m
d,
and'd',
Pliicker's coordinates of
m' be the
the position vectors of two points P, P' one on each line. Then the moment about P' of the unit vector d localised in the first line is (r - r')* d,
r, r'
,
and the moment of d about the second is therefore, by Art. 40, -r>d-d' = (r*d)-d' + (r'*d')«d = nvd' + m'*d.
(r
This result
symmetrical, and represents
is
moment about
the
either
It
the mutual moment of the two
m
senses the
Fig. 43.
same
where 6
is
lines is
P=
M
is
lines,
called
with
as d, d' respectively.
the angle of inclination of the two
common
Then, by Art. 50, the length of the
where
is
Since d, d' are unit vectors, the module
d>
two
a unit
of
line
vector localised in the other.
m
line
their
ra a> * *n [d,d',r-r']
perpendicular to the
jw sin 6'
sin
mutual moment. bv ,,
of the lines is given °
of
lines.
Hence the mutual moment n
M=j}sm6.
Examples. (1)
Find
planes i'a
The
Pliicker's
= q and
lli'
coordinates for the line of intersection of
line of intersection is perpendicular to ,
where
2V
= mod
planes, so that
Then
n>
the
= q'.
If
b
and
b*n =q'.
m = bxd = bx(n*n')/iV jy
(b*n'n
q'n
n', so that
n>
any point on the
is
b*n = 2
=
both n and
- qri
- b-nn')
line,
b
lies
on both
THE PLANE AND STRAIGHT LINE
§§51, 52]
Find
(2)
to
If
d>
plane through the line
the equation of the
Since the required plane
b
is
is
parallel to d
and
may
is
parallel to the
Hence
[rdc] = nrc.
be written
the required equation of the plane.
Find
(3)
parallel
= 0,
(r-b)-(d*c)
is
is
= m,
and, r being a current point on the plane, r - b plane, and therefore perpendicular to the normal.
This
m parallel to c.
normal
c, its
a point on the given line b>
which
d,
71
plane through the point a and the line
the equation of the
m.
d,
a point on the given line b*d = m. The plane is parallel Hence if r is d, and its normal is parallel to (b-a)xd. a current point on the plane (r - a) is also parallel to the plane, and If
to
b
is
b-a and
therefore
which
(r-a)-(b-a)*d = 0,
may
r*m -
be written
= a*m.
[rad]
s\52. Volume of a tetrahedron. let
points
a,
b,
c
With one vertex
C
be the
respectively.
Then and
the other vertices A, B,
the vector area of
OBC
is
as origin,
|b*c,
the volume of the tetrahedron
is
F = ia-(ibxc) = i[abc]. Suppose we require the length
p common perpendicular to the The directions two edges AB, 00. of
4he
of these lines are
b-a and
c,
those of the vectors
while
points, one on each
a,
line.
of inclination,
c
are
two
Hence, by Art. 50, [b
- a,
the relation
where edges.
M
is
6
is
their angle
a - c]
c,
V= AB.OO. The numerator
if
sin
6'
of this expression reduces to [abc] or
6F.
Hence
y = *AB 00 p sin d = \AB.OO.M, .
the mutual
moment
.
.
of the straight lines along the
VECTOR ANALYSIS
72
[CH. IV.
of a tetrahedron whose vertices are the points i[a - d, b - d, c - dj, which reduces to
The volume
Cor. a, b, c,
d
is
i{[abc] - [abd]
+ [acd] - [bed]}.
Spherical Trigonometry. 53.
Two fundamental
Let A, B,
formulae.
C
be points on the
surface of a sphere of unit radius, whose position vectors relative
are the unit vectors
to the centre
points are joined
by
a spherical triangle. angles
BOO, COA,
The angle and AOC
A ;
arcs of great
The
AOB
1,
m, n
sides a,
b,
If these
respectively.
the figure so formed
circles,
which these arcs subtend at the
of the triangle
is
centre.
the angle between the planes
and the other angles are
is
c of this triangle are the
AOB
similarly interpreted.
Consider the expansion (lxm)'(l*n)
= Mrn-n - l*nl*m Since the vectors are
as applied to the spherical triangle.
lxm
is
nrn=cosa, n*l=cos
all
and l*m=cos c. Further, a vector whose module is sine, and whose direction is perpendicular to the plane AOB drawn
unit vectors,
b,
Similarly lxn has a module
inward.
and a direction perpendicular to the plane AOC drawn outward. Hence sin b,
(l"m)'(lxn)
=sin b
sin c cos
The above expansion then
A.
gives the
relation Fia 4o
'
This
is
A = cos a - cos b cos
sin b sin c cos
'
c.
one of the fundamental formulae of spherical trigonometry
and two
similar ones
of the sides
and
may
angles.
be written
They
down by
;
cyclic permutation
are usually put in the form
cos a
= cos
b cos c
cos b
= cos
c cos
cos c
= cos a
+ sin
a + sin
cos b
b sin c cos c sin
+ sin a
A,
a cos B,
sin b cos C.
Another formula for the spherical triangle from the equality (lltm)x(lxn)
=M
L
may
be deduced
.
'
SPHERICAL TRIGONOMETRY
53, 54]
§ §
73
In having the meanings already stated, their cross a vector perpendicular to the normals to the planes and AOC, and therefore in the direction of 1 while the
For, l*m and
product
AOB
is
;
module of the vector is sin c sin b sin A. The above equation shows that sm c sin 6 sin .4 1 = [lmn]I, .
that
sin b sin c sin
is
Cyclic permutation of the sides sin c sin
a sin
have the same value.
B
sin a sin b sin
C
this it follows that
A
B
sin
sin a
which
and angles shows that
and
From sin
A = [lmn].
sin
sin b
C
sin c
another fundamental formula.
is
Concurrent Forces. 54.
Four forces in equilibrium.
If
point are in equilibrium, their vector
four forces acting at a
sum
zero.
is
Let
be unit vectors in the directions of the forces, and
sl
.
.
+ F 2b + F 3 c + J 4d =
we multiply throughout
If
a, b, c,
lt
,
d
Ft
Then
the measures of the forces.
Fx
F
scalarly
by
(1)
c*d,
two
terms
of the
disappear containing triple products with a repeated factor, and
we
find
F, [acd] +
F
2 [bcA]
Similarly,
on forming the scalar product
with a*c,
we have
From
i? 2 [bac]
= 0. of the first
member
and a third equation which may be derived same way, we find p _ pt F p2 these
_
i
[bed]
Thus each force
is
theorem
~ [abd]
'
~~
[abc]
proportional to the scalar triple product of unit
of the parallelepiped is
in the
s
[cad]
and therefore
to the
determined by those vectors.
This
vectors in the directions of the other three,
volume
of (1)
+ i^4 [acd] = 0.
usually attributed to Eankine.
Lami's theorem for the equilibrium of three forces, already considered in Art. 13, if
may
be proved by a similar method.
three forces are in equilibrium,
i^a + i^b + J^O.
For
VECTOR ANALYSIS
74
On
by
cross multiplication
and
we
find
F^a, + F^c = 0.
similarly
From
a,
[CH.
the last two equations
follows that
it
be
ab
c*a
Since their cross products are parallel, the vectors
and each force
coplanar,
is
a,
b, C are
proportional to the sine of the angle
between the other two.
EXERCISES ON CHAPTER 1.
Show
2.
Prove the relation
a-<(b*c)+b*(c*a)+cx(a*b)
that
IV.
= 0.
a*{b*(c*d)} = b'daxc-b*ca*d,
and hence expand 3.
Show that
and express the
a*(bx{c*(dxe)}).
[axb bxc cxa]
result
by means
= [abc] 2
,
of determinants.
Prove that
4.
[Imn] [abc]
=
I-b
l*a
m*a n*a
and give
is
6.
Cartesian equivalent.
of the plane which contains the line r = ta, perpendicular to the plane containing r = wb and r = vc.
Find the equation
5.
and
its
Find the equation of the plane containing the two
lines
= a + sb, ,
r 7.
What
is
r
=a
8.
r
,
+vo.
?
Show that the plane containing the two - a = sa '
parallel
.,
the equation of the plane containing the line r - a = tb>
and perpendicular to the plane rc = q
and
/
is
represented by [raa']
Give a geometrical interpretation.
= 0.
straight lines r
- a = ia'
.
EXERCISES ON CHAPTER
IV.]
The
9.
75
IV.
shortest distances between a diagonal of a rectangular
parallelepiped,
whose
sides are a,
b, c,
and the edges not meeting it, are
ca
ab
Vc2 + a*'
Va?+b 2
be
'
Vtf + c2
'
The shortest distance between two opposite edges
10.
tetrahedron
is
of a regular equal to half the diagonal of the square described on
an edge.
r
11. Find the shortest distance between the straight lines r = (k and - a = sb, and determine the equation of the line which cuts both at
right angles.
Show that the perpendicular
12. line
whose
Find the equation
13.
Find the straight
to the plane 15.
moves
A
ra = 0, and
m is mod (m + d>
of the straight line
intersecting both the lines r 14.
distance of the point a from the
Pliicker's coordinates are d,
- a = sb and
line
r
through the point
intersects the line
straight line intersects
is
parallel
two non-coplanar straight lines, and Prove that the locus of a point
parallel to a fixed plane.
16. The locus of the middle points of all by two fixed non-intersecting straight lines
common 17.
which
r-a' = *b.
which divides the intercept in a constant ratio
planes
c,
c,
is
a straight
line.
straight lines terminated is
a plane bisecting their
perpendicular at right angles.
Find the locus of a point which
rn 1 = q1
,
i'n 2 = q 2
,
T-a s
= q3
is
equidistant from the three
.
Show that the six planes, each passing through one edge tetrahedron and bisecting the opposite edge, meet in a point.
18. of a
The six planes bisecting the angles between consecutive faces tetrahedron meet in a point.
19. of a
Prove the formulae
20.
[a*b, c*d, e*f]
= [abd] [cef] - [abc] [def] = [abe] [fed] - [abf] [ecd] = [cda] [bef] - fcdbl [aef].
Show that the volume
21.
of the tetrahedron
bounded by the
four planes
r(mj+wk) = 0, r(wk + Zi)=0, is
3
2p /3hnn.
T'(li
+ mj)=0
and
v(li
+ mj +rik)=p
[CH.
VECTOR ANALYSIS
76
22. Prove that the four points and 4(-i+j+k) are coplanar.
4i
+ 5j + k,
(j+k),
3i
+ 9j+4k
23. If a straight line is drawn in each face of any trihedral angle, through the vertex and perpendicular to the third edge, the three lines thus
24.
drawn are coplanar.
Prove the formula (b*c)-(a*d)
and use
it
to
sin (A
25.
The
+ (c*a)-(b*d) + (a*b)*(c*d) =0,
show that
scalar
+ B)
sin
(A - B)
moment about
= sin2 A - sin 2 B = J (cos 25 -cos 24).
the line
CD,
by the localised vector AB, is 6V/1, where V the volume of the tetrahedron ABCD.
I
is
of a force represented
the length of
CD
and
mutual moment of the opposite edges same for each pair, prove that the product of their lengths is the same for each pair.
26. If in a tetrahedron the is
the
also
27. From the Cor. to Art. 52 show that the volume of a tetrahedron is given, in terms of the coordinates of its vertices. by the determinant
a1
-d 1
TV.] 32. If the
EXERCISES ON CHAPTER
mutual direction cosines of two sets by the accompanying table,
unit vectors are given
show that the sum of the squares of the terms in any row or column is equal to unity that the sum of the products of corresponding terms ;
any two rows or any two columns is zero and that the determinant of nine terms has the
in
value unity.
77
IV.
;
of
rectangular
[CH. V.
CHAPTER V DIFFERENTIATION AND INTEGRATION OF VECTORS. CURVATURE AND TORSION OF CURVES. 55.
ation
Our object in the present chapter is to explain the differentiand integration of vectors with respect to a scalar variable.
Only cases
of
one independent variable
differentiation
being
left
will
be considered, partial
Within the
volume.
for the second
limits of a single chapter it is impossible to
examine
fully the
points that arise in connection with continuity and the existence of a limit. is
We
therefore lay no claim to rigour of the kind that
necessary in a treatise on the infinitesimal calculus.
object
is
Derivative of a vector. Suppose that a vector r and single-valued function of a scalar variable t.
value of
Our
to explain rather than to prove.
t
corresponds only one value of r
;
origin 0,
a continuous
Then
and as
tinuously, so does
a fixed
is
t
to each
varies con-
Relative to
r.
let
P
be the
point whose position vector
is r.
Then if t varies continuously, P moves along a continuous curve in space.
Let
OP
be the value of
r corre-
Fig. 46.
sponding to the value scalar variable.
An
increment
8t in
t
of the
the latter will produce an
Thus the value t + dt of the scalar + dv of the vector, and this is the another point P' on the curve. The increment
increment dv in the former. corresponds to the value position vector of dv
is
r
equal to the vector PP'.
The quotient 78
j-, of 6t'
the vector
<5r
.
§
DIFFERENTIATION
55]
by the number <5r
dt, is itself
moves up
the point P'
a vector.
If
As
be small.
will also in general
79
now dt is a dt
and the chord PP' The hmiting
to coincidence with P,
P
to coincidence with the tangent at
small increment,
tends to the value zero,
to the curve.
value of the quotient j-, as dt tends to zero,
is
a vector whose
which
is
the direction of
direction
the limiting direction of
is
the tangent at P.
the derivative or differential dt
exists, is called
respect to
t,
and
is
denoted by dt
of r
dt
^o
dt
also in general a function of
is
possesses a derivative is
which
d 2t denoted by -^.
is
the third derivative of
A
Similarly the derivative of this r,
and
is
denoted by
case of special importance
and
particle
Then
itself
is
r,
called
dh ,-^.
that in which
is
r the position vector of a
to the origin 0.
and
t,
called the second derivative of
.
variable,
it
of determining the derivative is called differentiation
The derivative
and
_ st
when
coefficient of r with
Thus
--,-.
_
dt
The process
dt,
This hmiting value of the quotient,
moving
t
the time
is
particle
P relative
dt represents the displacement of the
during the interval
-5-
The hmiting value
velocity during that interval. velocity as dt tends to zero
and therefore
dt,
is
the average
of this
average
the instantaneous velocity of the
particle.
Hence the vector v representing the instantaneous
velocity of
P
is
v
= fa
.
dt
This vector
is,
of course, in the direction of the tangent to the
path of the particle.
Similarly,
if
dv
is
the increment in the <5v
velocity vector v during the interval
dt,
the quotient
the average acceleration during that interval.
-j-
represents
The instantaneous
acceleration of the particle is the limiting value of this average
acceleration as dt tends to zero.
Thus the vector
_dv _d*t & ~dt~dt* represents the instantaneous acceleration of the
moving
particle.
'
VECTOR ANALYSIS
80
[CH. V.
The derivative of any constant vector c is zero for the increment dt produces no change in c. 'The derivative of the sum r + s of two vectors r and S, which ;
are both functions of
For
if
di
and
increment
<5s
dt,
equal to the
is
t,
sum
of their derivatives.
are the increments in these vectors due to the
d(r+s)=(T + dr +S + ds) -
(r
+ s)
= di + ds, and therefore the quotient (5(r
+ s)_<5r (5s + ~di di'
dt
Taking limiting values of both sides as d dt
The argument
is
\_di
.
(1JrS '
=
di
+
dt
tends to zero,
we have
ds di'
sum
obviously true for the
of
any number
of
vectors.
Suppose that r is a continuous function of a scalar variable s, and 5 a continuous function of another, t. Then an increment dt in
the last produces increments
tend to zero with
dt.
The
<5r,
ds in the others,
which both
relation <5r
_
<5r
ds
di^dsdi is
an algebraical identity, the number placed after the vector
having the same meaning as when placed in front. limiting values of both sides as dt tends to zero
formula
Taking
we have
the
drjids dt
ds dt
as in algebraic calculus. 56. Derivatives of products. of vectors
is
found in the same
The derivative
way
sum of the quantities got by and leaving the others unchanged.
being equal to the single factor
Take
for instance the
of
any product
as for an algebraic product, differentiating a
product ut of a scalar u and a vector r, t. If du and (Sr are their increments
both functions of the variable
due to the increment
dt,
d (ut)
the increment in the product
= (u + du) (r + (5r - wr = dur+ u dv + du dr. )
is
.
DERIVATIVES OF PRODUCTS
§ 56]
Dividing throughout by
we have
dt,
V
du
<5(wr)
du
di
r+u
-ar
.
.
{uT)=
du
dt
From is
this
expressed as the
sum
sides,
I+U
dt
and the preceding
„
5 + **«
and on taking limiting values on both d
81
di
,,
(1)
dt
if any vector r components by the formula
results it follows that,
of rectangular
x=xi+y\ + zk, then, since
i, j,
k
are constant vectors, the derivative of r
di
=
dx
+
1
di
di
dy
.
.
+
1
dt
is
dz, dt
'
and similarly for the second and higher derivatives. The same argument as above shows that the derivatives of the scalar
and vector products vs and r*s are given by the formulae d
^ d
t
and
,
di
,
r,s
,
di
.
{T * s)=
xS
dt it
ds
,„.
>=r s+r ^ + Ix
dt
(2)
ds
.„.
(3) '
dt>
being understood that, in the last formula, the order of the
any term must not be changed unless the same time. To prove (2) we have 3 (r*s) = (r + <5r)«(s + <Ss) - r-s
factors in
at the
sign
is
changed
= drs + r-<5s + <5r*<5s. Dividing throughout by dt
and proceeding
to the limit,
we obtain
the required result. If in (2)
we take
s
equal to
d
,
.
{l ' l)
=
dt
But
if
r
is
the module of
derivative of r 2
is
2r
-=-.
we
r,
r,
find the useful formula
dr 2
„
!i=
2l
di
'W
the product
Thus
at
dv
dr
dt
dt
In the case of a vector a of constant length,
a2 W.V.A.
=a 2 =a F
constant,
r*r
= r 2 =r 2 and ,
the
VECTOR ANALYSIS
82
and therefore
= 0,
a* -t-
showing that the derivative
whose position vector
is
a,
and the vector
perpendicular to
-j- is
a then da,
radius
[CH. V.
lies
on the surface
a.
A
point
of a sphere of
parallel to the tangent plane at the
-r, is
point. If in (3)
we
replace s
by
dv -57,
we
find the formula
d( Tx dv\
d 2v dtJ~*'dt 2
dt\ since the cross product of
-r*
dv
two equal vectors
-5-
zero.
is
This
result is frequently useful.
Triple products are differentiated on the
above.
Thus,
triple products
if
p, q, r
are functions of
t,
and vector
m dt
djt
[Mt]:
I
principle as
have derivatives given by v
It
and
same
their scalar
px(axr)
Tt
v
dt.
.
= g*(<^) +P*(§*r) + p*(V|
the order of the factors being maintained in each term of the
second formula, and the cyclic order in each term of the
first.
The formulae are easily proved by a double application of (2) and (3), or from first principles as above. 57. Integration. Having given one vector r, the process of finding another vector F whose derivative with respect to t is equal to r, is called integration. Thus integration is the reverse process to differentiation. The vector F is called the integral of r with respect to t, and is written '-Jr*. The symbol integration,
The
is J
called the integral sign,
and the function
relation just written
is
t
is
be integrated
r to
the variable is
equivalent to
dF
dt~
T
'
so that the derivative of the integral
is
of
the integrand.
equal to the integrand.
:
;
INTEGRATION
§ 57]
The
integral
83
F is indefinite to the extent of an additive arbitrary c. For if the derivative of F is equal to r, so is of F + c. For this reason F is called the indefinite
constant vector the derivative
The
is termed the constant of In the application of integration to a definite problem, the value of c will, as a rule, be determined from some
integral.
arbitrary constant C
integration.
or geometrical condition to be satisfied.
initial
From the
results of the preceding Art.
the values of the following integrals, of integration,
and
which
write
down
illustrate the process
subsequently be found useful
will
di\
ds
7
* +S '*r =r S+C
r,
we may
-
'
di
f
\2v-=-dt=V'V+c=v 2 + c, d 2 i ^ dv dv dt= dl'dfi dt'd-t
f„ dv
2 }
f
cZ
2
r
and
a
if
is
dr r\
dx
+c
>
>
A
r
,
a constant vector
dl
^
f \a,*j7dt=a.xi
The constant of integration In the
{dt)
di
,,
rde dt=t *cTt +c 1
/dv\ 2
+c=
first
+ c.
is
of the
same nature
three of the above results
in the last three
it is
as the integrand.
therefore a scalar
a vector.
As a further example in integration, suppose we are given the relation
from which write
down
^2 r
dv it is
dv
with 2
-j-.
Now we
the integral of the second member, since r
known function side
required to find the value of
-=-,
of
t.
But on
we have 2
is
not a
forming the scalar product of each
the equation
d 2v dl'dT>
n dv
cannot
= - 2nl dv -dt>
.
VECTOR ANALYSIS
84
which can be integrated as
it
CH. V.
stands, giving
(|)'=c-**r», di
from which the value of -^
is
found in terms
of r.
at
Curvature and Torsion of a Curve. 58.
Let
Tangent at a given point.
A
length of the arc from a fixed point variable point P.
Then the
s be (the measure of) on a given curve up to
the
the
position vector r of P, relative to
a fixed origin 0, scalar variable
a function of the
is
We
s.
shall consider
only curves for which r and
its first
two derivatives with respect
to s are
Then,
continuous. points
r
r,
if
P, P' are the
+ <5r on the
curve, corre-
sponding to the values
s
and s+ds
Fig. 47.
respectively, the vector
The quotient
di/ds
a vector in
is
the same
direction as di
the chord
And
PP' and the
Thus
;
<5r.
and of
further, since the ratio of the lengths
of
arc
PP' tends
moves up to the module of dtjds is
to unity as P'
coincidence with P, the limiting value of unity.
is
becomes that
in the limit, as ds tends to zero, this direction
the tangent at P.
PP'
dt
t=
<5r
Lt
:
(say)
t
..(1)
ds is
a unit vector in the direction of the tangent to the curve at P.
When no
misunderstanding
is
possible
we
shall refer to it briefly
as the unit tangent. If x, y, z are
the Cartesian coordinates of
P
referred to
angular axes through 0,
r=d+2/j +zk and
,
_ dv
dx
ds
ds
.
dy ds
.
dz ds
'
(IT
the direction cosines of
t
being therefore
-=-
ds
,
(IfJ
fl"*
ds
-^ ds
~,
rect-
CURVATURE AND TORSION OF A CURVE
§ § 58, 59]
85
The vector equation of the tangent line at P may be easily For the position vector R of any point on the tangent is given by = r + wt, written down.
R
where u
is
a variable number, positive or negative.
This
is
the
equation of the tangent.
any point
is
of the curve
the arc-rate of rotation
Thus,
of the tangent.
measure
circular
The curvature
Principal normal.
59. Curvature.
at
the tangents at
if
36
of the angle
P
and
P',
the
is
between
-5-
the
is
OS
average curvature of the arc PP'. Fia. 48.
The limiting value of this as 8s tends to zero
is
the curvature at the point P.
Thus
by*.
K=
d6
Though
t is
^o
a unit vector,
os
and
t
+ <5t
it is
a function of
Then TT'
HT'.
is
Let
s,
its
direction
t
be
its
value at P', equal respectively to
its
it
d6
changing from point to point of the curve. at P,
denote
shall
=-7-. as
-5-
-Lit Ss
We
value
HT
and
THT
is equal to the angle and the angle P and P', that is 89. The quotient the same direction as <5t, and in the limit as (St,
between the tangents at dt/ds is a vector in <5s-M), this
HT Lt
perpendicular to the tangent at P.
is
Since
a unit vector, the module of the limiting value of dtfds
is
-5-
direction
= k.
Hence the
is
relation
os <5t dt T -,-= Lt ^as 5^0 os ,
where n in the
is
= «m,
,
|
(2)
a unit vector perpendicular to the tangent at P, and
plane of the tangents at
P
and a consecutive point P'
This plane, passing through three consecutive points at P,
may
be called the plane of curvature or the local plane of the curve at P. It is also commonly called the osculating plane.
The unit vectors plane
is
t,
n
are perpendicular to each other,
the plane of curvature.
The
straight line
and
their
through
P
VECTOR ANALYSIS
86
n
parallel to
where
R = v+un,
R is
the position vector of any point on
s is a positive quantity
tangent
The curvature
it.
but the direction of rotation of the
;
we may
given by the vector n, which
is
(principal) normal.
In virtue of
be written
(1),
the unit
call
may
the equation (2)
also
^2 r
Squaring both
sides,
curvature Kt
we have a well-known formula
for
the
2 _(§*)*/&*}* (&y\ + f^Y _ +
W/
R is any point in the plane and n are coplanar. Hence [R-r,
and this is the equation The normal plane at to the tangent. Hence
\ds 2 )
\ds»)
\d8*)'
of curvature, the vectors
If t
Its equation
called the principal normal at P.
is
clearly
is
[CH. V.
t,
R-r,
n]=0,
of the osculating plane.
P
the plane through
is
equation
its
P
perpendicular
is
(R-r)-t=0.
The
circle
of curvature at
P
is
the circle passing through three
points on the curve ultimately coincident at P. called the radius of curvature,
This circle clearly leave
it
curvature p lies
lies
an exercise
as
is
and
its
Its radius
= -.
P
;
and we
show that the
radius of
in the osculating plane at
for the student to
given by p
is
centre the centre of curvature.
The centre
of curvature
C
then
on the principal normal, so that
PC = P n=-n. K
60. Binormal.
Torsion.
The
straight line through
dicular to the plane of curvature parallel to the vector b
system
= t*n, and
is t,
n,
perpenIt
is
b form a right-handed
mutually perpendicular unit vectors.
of
P
called the binormal.
We may
speak
of b as the unit binormal.
Since b
that
-j-
is
is
a vector of constant length,
perpendicular to
b.
it
follows, as in Art. 56,
Further, by differentiating the
CURVATURE AND TORSION OF A CURVE
§ GO]
= 0, we
relation t'b
87
find «:n'b
+ 1* t- = 0. as
The
first
term
is
n
zero because
equation then shows that
db -r- is
is
perpendicular to
perpendicular to
t.
b,
But
and the it is
also
/b
Fig. 49.
We
perpendicular to b, and must therefore be parallel to n.
may then
write
db
= -An.
.(3)
ds
And
just as in formula (2) the scalar k
of turning of
the unit vector
t,
so here
of turning of the unit vector b.
binormal
is
measures the arc-rate
X measures the arc-rate
This rate of turning of the
called the torsion of the curve at the point P.
It
is,
•of
course, the arc-rate of rotation of the plane of curvature, since
b
is
perpendicular to
this
indicates that the torsion of the
vector
binormal as t.
A
s
is
The negative
plane.
sign in
(3)
regarded as positive when the rotation
increases
is
right-handed relative to the
glance at the figure shows that in this case
db ,
has
the opposite direction to n.
Having found the derivatives For
of
and
t
b,
we can deduce that
of n.
-r= T b4 = b* Un) ds ds
- An*t
= -a + Xb The value
of the torsion
(4)
may now
derivatives of r with respect to
s.
be found in terms
of the
Since the second derivative
VECTOR ANALYSIS
88
Forming then the
of r is *n, the third is -t-(kii).
product of the triple
first
[CH. V.
three derivatives of
scalar triple
and neglecting those
r,
products which contain a repeated factor, we have ~dv d2 n d 3 T~\
_[
Kn K
_ds ds 2 ds 3 }-^'
da
+
ds
'
dx
n
ds
.
-[t,*n,*(Xb-*t)]
= AK Hence the value
of the torsion
2
is
[tnb]=X/c
by
>
(4),
2-
given by
dt d2 v d 3 f
1
ds ds 2 ds 3 .
The equation of the binomial
is
R = r + wb. Or, since b
= t>
this
may
R=T+V u, v
put in the form
also be
dT d 2! ds ds 2
being variable numbers.
Line and Surface Integrals.
Definite Integrals. 61. Definite integral of
for values of
t
dependence we
ranging from a to
may
Let
a vector function.
vector function of the scalar variable
finite
t,
To
b.
f
be a given
and continuous
indicate the functional
use the ordinary notation
i(t)
for the vector.
Let the range b - a be divided into a number of sub-ranges, which correspond to increments dt x 6t 2 Also dt„ of the variable t. ,
let
l{t x )
,
.
.
.
,
be one of the values assumed by
in the first sub-range,
f
i(t 2 )
one of the values assumed in the second, and so on.
the
sum
It
S = & 1 f(< ] ) +
dt 2 f(t 2 )
Consider
+ ...+ 8t nl(t n ).
can be shown that, with the ordinary functions occurring
practical
applications,
indefinitely,
and each
if
the
of the
number
of
increments
in
sub-ranges increases
dt
tends to zero,
this
sum S tends to a definite finite limit which is independent of the mode of subdivision of the range b - a. This limiting value of S is
equal to the difference of the values of the indefinite integral
DEFINITE INTEGRALS
§§ 61, 62] the function
F(f) of
that
is
This
is
of the variable
t
;
LtS=P(6)-F(o). called the definite integral of the function
a and
limits
and a
for the values b
i(t)
80
b,
and
is
t(t)
between the
denoted by i(t)dt.
i
The
result
may
be put in the form I
= Lt ^i(t).dt=F(b)-F(a),
f (t)dt.
f(i), in the second expression, is one of the values taken by f in the sub-range corresponding to the increment dt, the summation taking in all increments of t from a to b, and each
where
increments tending to zero as a limit.
of these
By way
of illustration consider the displacement of a moving with a variable velocity v, which is a function Let the interval from t to t x be divided of the time variable t. During one of into a large number of infinitesimal intervals. 62.
point P,
these,
whose duration is
where
v(<) is
dt,
the displacement of the point
a value assumed
by v during
displacement during the interval from
displacements
the
of
limiting value
we
during
when each
the
t
to
t
x
is
dt
dt
.
v(t),
total
the vector
sum
sub-intervals.
of the quantities
is
The
this interval.
Taking the
tends to zero,
find for the total displacement of the point
Lt Similarly,
if
y
v(«)
.
St
= \\(t)dt.
the moving point has a variable acceleration
a(<),
the increment in the velocity during the short interval dt dt
to
.
t
a(£) x
;
and the
total increase in velocity during the interval
is ta
.
is
Lt Va(*).<5«=
The position
of the centre of
mass
*{t)dt.
of a
body may be found by
imagining the body divided up, in some convenient manner,
number of small when the number tends
and proceeding to the and each element of the element of volume
into a large
portions,
limit
to infinity,
volume converges to a point.
If
dv
is
round the point P, whose position vector relative to a given origin
VECTOR ANALYSIS
90
[CH. V.
and the density at this point, the mass within the element Then, since the cm. of a system of particles of volume is dv. is given by the formula f = v mr/2m) is r,
/j.
/j. .
the
cm.
of the infinite
number
ceeding to the limit as above,
is
of particles, arrived at
Lt Xru Cu.dv dv .
r
where
M
is
=Ltv^6v~ =
If
Mr
,
dv
'
volume occupied by the body. Consider a
63.* Tangential line integral of a vector function.
given curve, and a vector function
respectively.
If
t
F*t is the
measure
tangent.
The
is
for
F
which
s
it.
Let A,
B
be
has the values a and b
the unit tangent at a point of the curve,
F
in the direction of the
definite integral of this quantity with respect to b,
along the curve from \"
is
of the length s of the arc
of the resolute of
between the limits a and
It
F
curve measured from a fixed point on
two points on the curve,
vector
pro-
the mass of the whole body, and the range of integra-
tion includes the whole
of this
by
given by
F-t
s,
is
called the line integral of the
A
to B.
We
write
it
ds=W£F-tds.
also frequently written
F'eZr
= Lt][)Fv5r,
i:
where A,
B
are the end points of the arc of integration,
the infinitesimal vector ds
.
t
and
<5r is
parallel to the tangent at the point
considered.
F represents the force acting on a which moves along the curve from A to B, F*dr measures the work done by the force during the infinitesimal displacement If,
for instance, the vector
particle
(5r and the definite integral from A to B represents the total work done by the force during the displacement from A to B. ;
If
F
represents the value of the electric (or magnetic) intensity
at P, the line integral represents the
moves from A between those two
B
(or pole) as it
to
potential
points.
;
work done on unit charge that
is,
the difference
In hydrodynamics,
if
of
v
LINE AND SURFACE INTEGRALS
§§ 63, 64]
91
represents the velocity of a particle of the fluid, the line integral
Ivt ds= \vdr taken round any closed curve drawn in the circulation
fluid is called
the
round the curve.
Normal surface integral of a vector function. Consider a curved surface, and a vector F varying from point to point of the surface, and possessing at each point a definite value. Let n be a unit vector parallel to the normal at the point P of the surface, 64.*
drawn outwards if the surface is closed, or always toward the same side if it is not closed. Then F*n is the resolved part of F along the normal. If the surface is divided up into a large number of small elements, and dA is the area of the element round P, the
sum
extended to
S = 2FmdA all
the elements of the surface tends to a definite
when each number tends to
hmiting value
of the quantities
and their
infinity.
the surface integral surface.
We write
the function
of
8A tends
This hmiting value
F
to zero, is
called
over the given curved
it
(F-n
with the surface integral of the scalar function F*n
and, after the formation of this scalar product,
is
;
a matter of
ordinary calculus.
The vector dA n represents the vector area of the element It is often written <5A, and the above equation .
of the surface.
put in the form v
If,
„
f
|F-dA=LtZF*3A.
for instance, the vector
F
represents the value of the electric
or magnetic induction at the point P, the surface integral gives
the value of the total normal induction over the surface. if
the surface
is
drawn
in the region occupied
by a
liquid,
Or
whose
from point to point, the surface integral of the vector v gives the rate at which liquid is flowing across the surface, in units of volume per unit time. Some worked examples will be found among the Note. velocity v varies
following exercises.
VECTOR ANALYSIS
92
EXERCISES ON CHAPTER
[CH.
V.
the following expressions, in which r is a function module, and the other quantities are constants
1. Differentiate
of
t,
r its
:
dr i.
r 2r
+ a rb. #
\
EXERCISES ON CHAPTER
V.]
By means
9.
A
d
of the relation t
= -j-(rr),
A
93
V.
prove that for any curve
dr ds
-m
Hence, using the value p = of the length of the perpendicular from to the tangent, prove that for a plane curve
dp _
dr
ds
ds'
and therefore that
k
=-
—
•
r dr
The
10.
circular helix.
right circular cylinder,
angle.
This is a curve drawn on the surface of a and cutting the generators at a constant
— -a
Let a be the radius of the cylinder,
the angle at
A
which the curve cuts the generators, and k the unit vector in the Then, with a fixed point on the axis as origin, the position vector of any point on the helix may be expressed in the form
direction of the axis of the cylinder.
r
= acos#i + asin#
j
+«#tancck.
Differentiation with respect to s gives ,
t
Since this
is
d6,
dt
= -r = a-y
a unit vector
(
n
.
-
*n = ^p
is
l
+ cosftn-] + tana tk).
square
its
a J sec J a
Thus the derivative of 6
.
sine?
is
-r-
\ds)
(j^j
=
1.
Further, to find k
constant.
= -a
unity, showing that
(cos#
l
we have
+ sin0 j),
which shows that the principal normal is always perpendicular to k, and therefore to the axis of the cylinder. On squaring both sides of the last equation,
we
find
'-*©* that
To
k
is
find the torsion,
d3i 173 ds 3
Wh6nCe
dh
d3 r
^S"3
= -cos 2 a. a
we have /d6\>,
= \-rJ
\dsj
=a29
a a (asin0i-acos0j),
fd6\\ k
UJ
.
.
s
-
VECTOR ANALYSIS
94 *2 X
Therefore
=
£
^^ =
-
Substituting the value of
we
*,
A=-
a cos
sin
is
also a circular helix
^
of curvature of a circular
find the condition that
and
;
-
a.
Prove that the locus of the centre
11.
CH
find
a
helix
a 3 ten a
[
it
may
be traced on the same cylinder. The centre of curvature is the point
n
c
= r + -. K
Substitution of the values of these quantities gives the result. 12.
Prove that the curvature of the curve defined by i
= 2a
cos# a(a 2
is
equal to
i
+ 2a
sin#
13.
Find
«-,
2
+ b8 2k
2 )*
+ &2 + 62
2(a 2 + & 2
j
f
and the equations to the principal normal and the
plane of curvature, for the curve r
= Aa
cos 3
i
+ ia
sin 3
+ 3c
j
cos
20
k.
14. In the curve r
= a (St -t3 )i + 3at2 + a(3t + ]
show that
,
t
3
)k,
=X = _L_.
15. If the curve r is
= acos
6i
+l
sin 6]
+f(8)k
a plane curve, determine the form oif(6). 16.
Prove that
for
any curve
—•—-=- K \. as
as
and the binormal at a point of a curve respectively with a fixed direction, show that
17. If the tangent
angles 6,
18.
the
A
curve
same angle
is
j3
sin
6 dd
sin
drawn on
make
k
dcp~
A
a right circular cone, always inclined at
to the axis.
Prove that
X=K
COt
(j.
EXERCISES ON CHAPTER
v.
V.
95
19. A circle of radius a is drawn on a sheet of paper, which is then folded to form a cylinder of radius b. Show that for the new . curve i i a*
where
s is
a
b*
the length of the arc measured from a certain point.
20. If k is the curvature of a curve, then that of its projection on a plane inclined at an angle (3 to the plane of curvature is K cos /3 if the plane is parallel to the tangent, and K sec 2 /3 if it is parallel to the principal normal. 21. Prove that the circular helix and torsion are both constant. d 2i
From
the relation
— (fe
constant, that
2
=/cn
dh ds a
the only curve whose curvature
follows
by
differentiation, since k is
= /c(-/ct + Xb), 3
and therefore and
it
is
ds 2 ds a
/c
b
+
/c
2
At,
this is a constant vector, for its derivative is easily
Denote
vanish.
it
by
d.
shown to
Then t-a=* 2 \
and
b'd = k 3 ,
while
rrd=0.
Thus the inclination of the tangent to d is constant, the principal normal is perpendicular to d, and the plane of curvature has a constant inclination to a plane perpendicular to d. From this last fact and the constancy of k it follows, by the last exercise, that the curvature of the projection of the curve on a plane perpendicular to d is constant. The theorem is then obvious. 22. Radius of spherical curvature. through four points on a curve c r ultimately coincident with the point P is called the osculating sphere to the curve at that
point.
Its centre
called the centre
The sphere which
and radius are and radius of
Relative to be the position
spherical curvature.
an origin
let c
vector of the centre curvature,
C of spherical
and r that
of the point ->
Then the vector PC = c-r determines the radius of spherical P.
j- I&
_
50 _
passes
VECTOR ANALYSIS
96
[CH.
be denoted by R and its module by R. Then, and curve have four points in common, the first Hence, by three derivatives of c and R 2 with respect to s vanish. differentiating 2 = 2 (c r) R Let
curvature.
it
since the sphere
(c-r) - t =
we have
(1)
Another differentiation gives (c
and a third leads to
From dh ds 3
and
,
(1) is
and
(3) it
<
r)
-r
c
*^2=1
(2)
'
>'S=o.
R
follows that
(3)
di
perpendicular to both
is
-=-
and
ds
therefore of the form di d s T ds ds 3
To (2),
find the value of k multiply scalarly
we
by
dh -j-j.
Then, in virtue
of
ds
find
~dr
dh dh~
_ds ds 2 ds 3 _
d 3 r dr ds 3 ds
R=
so that
.ds ds 2
This
is
the vector PC.
It
may k
2
.(i)
dr d 2 r d 3 i
also
X
ds 3
.
be put in the form
ds
1
1
k
A ds\K/
d_fl\ h
(5)
This gives the well-known formula
R2 23.
Find the radius
24.
Show
that, for
~K 2
+
of spherical
.(6)
X 2 \ds k) curvature for the circular
any curve, 1
(dh\ 2 ds 3 )
= 1+A 2 R 2
.
helix.
EXERCISES ON CHAPTER
v.]
25.
Find the radius of spherical curvature
V.
for the curve in
97 Exer-
cise 14.
26.
A
variable point
P
relative to a fixed origin 0.
traced out
27.
by
Show
OP as P
on a given curve has a position vector
r
Show
that the vector area of the surface moves along the curve from A to B is given by rB rB r*tds = |l T*di. J
that, for a closed surface,
\ndS = Q, where n surface.
the unit normal, and dS the area ^of an element of the (The vector area of a closed surface is zero.)
is
[CH. VI.
CHAPTER
VI.
KINEMATICS AND DYNAMICS OF A PARTICLE. 1.
65. Velocity at
an
Kinematics.
P
Let
instant.
be a moving point, and
position vector relative to another point 0, which
moving
of as either
may
The velocity
or stationary.
r its
be thought
P
of
relative
the rate of change of P's position relative to 0, and
is
therefore represented in measure
of
change of
and direction by the rate the increment in r during an interval
dt
to
is
r.
If
<5r
is
di
seconds, the quotient j-
.
is
the average rate of change of r during
and therefore represents the average velocity of P during the interval St. The limiting value of the quotient as dt tends to zero is a vector which represents the
this interval,
relative to
velocity of
P
relative to
at that instant.
may
we
Briefly then,
say that dv
Lt %'
•(1)
dt
the instantaneous velocity of
is
P
relative to 0.
Suppose that there are n points P1? P n and that rm m_ 1 is the
P2
,
.
.
.
,
,
_
position vector of the
the
(m-l) th
.
m
th
Then that
relative to of the
n th
relative to the first is *n,
And
since, as
we have
position vector of this
is
1
~"n,n-l"'"ffl-i
l
jI
_2
+
.
•
•
+r
2i i-
just seen, the derivative of the relative
the relative velocity vector, differentiation
equation gives
+ v n—\,n98
•+V 2
(2)
§ § 65, 66,
That
is
KINEMATICS
67]
to say, the velocity of the
nth
99
relative to the first
equal
is
sum of the velocities of the nth relative to the (n - l) th relative to the (n-2f h and so on as far as that
to the vector
,
the (w-l) th of the
,
second relative to the
This
first.
is
the theorem of vector
addition of velocities.
The case
of three points
may
the theorem
P
to
2
is
is
one of frequent occurrence, and
then be stated
:
the velocity of
P3
relative
P3
equal to the vector difference of the velocities of
relative to
P
and
t
P
2
relative to
in Art. 13 for the case of
an
66. Acceleration at
Pv
This theorem was considered
uniform velocities only. instant.
The
relative to another, 0, is the rate of relative to 0.
If
v
is
acceleration of a point
change of the velocity
position vector r
P P
the vector representing this velocity, then
the relative acceleration.
j- represents
of
= OP, we may
In terms
of their relative
write
_d\_dH
&
~dt~dt 2
for the acceleration. If
there are
n points it follows from the above that relative compounded by vector addition. For differen-
accelerations are
tiation of (2) gives "«, 1
= ^n, n-1 "*"^ri-l,n-2 +
•
•
•
+&2,
l>
where the suffixes have the same meaning as above.
This
is
the
theorem of vector addition of accelerations. 67. Tangential
and normal resolutes of acceleration.
the motion of a particle its
P
along a fixed curve (Fig. 47).
position vector relative to a fixed origin 0.
seen, the velocity of
dv
v = T-.
P
at
any instant
But, by Art. 55, this
V= ds
where v=-j-
may
is
Consider
Let
r
be
Then, as already
given by the vector
also be written
di ds
dsdt=
.,.
,
vt
(1) >
.
is
the speed of the particle along the path, and
the unit tangent to the curve at that point.
t is
This equation
merely expresses that the velocity has the direction of the tangent
and the magnitude
of the speed.
VECTOR ANALYSIS
100
[CH. VI.
The vector representing the acceleration at the instant considered
is
^v
or
a,
d
dv
dt ds
dt
ds dt
= ^-t+KV 2n,
(2)
at
where k
the curvature and
is
n
the unit principal normal.
formula shows that the acceleration of the particle to the plane of curvature.
by
dv -=-,
and
The former
its
is
Its tangential resolute is
the rate of increase of the speed and
The
at
P
If
66
is
Denoting
this
by
is
zero.
independent
the rate of rotation of the
we may
w,
fa
V
For uniform motion
.
depends on the curvature
latter
-j- is
dt,
of the acceleration
resolute
by kv 2
the angle turned through by the tangent
during the interval
tangent.
is
This
parallel
measured
resolute parallel to the principal normal
of the shape of the curve.
and the speed.
is
write the normal resolute
dQ
d&dt
s=va,n
in a circular
-
path of radius a the tangential
Thus the acceleration
is
always normal to the
path and equal to v 2 /a. Note.
Differentiations with respect to the time variable
frequently denoted tiated.
by placing dots over the quantity
Thus V
and
_dv
fc V
~dt'
~df
t
are
differen-
^w 2
v=
r
so on.
and transverse resolutes of velocity and acceleration. Suppose now that the path of the moving particle P is a plane curve, and that it is required to find the resolutes of the velocity 68. Radial
and acceleration parallel and perpendicular to r = OP. Let 66 be (the circular measure of) the angle turned through by OP during the interval anti-clockwise.
6t,
reckoned positive when the rotation
Then g>=tt
,
the angular velocity of
P
is
the rate of turning of OP, or A A about 0. Let r, s be unit vectors is
KINEMATICS
§§ 67, 68] parallel of
and perpendicular to
r,
Then the points whose
&).
101
the latter in the positive direction
A position vectors are r
in circles of unit radius with equal speeds w.
The
A
and
s
move
velocity of the
A
former
is
A of -r.
If
in the direction of
s,
and that of the latter in the direction
.
Therefore
now
r is
di
ds
the module of
r,
by
the velocity of the point dt
d
"irdt
,
P is
given
A ,
{rt)
= rl+rcoS,
(1)
showing that the radial and transverse resolutes of the velocity dr are
-57
dt
,
and
dd
,.
,
r -3- respectively. dt
iv
VECTOR ANALYSIS
102
[CH. VI.
Exercises. (1)
A
particle
about 0.
P
™.
prove that
By
formula
tion of
w PO,
moves in a plane with constant angular velocity
the rate of increase of its acceleration is parallel to
If
(2) of
the above Art., since
P is
w
a=
- rw 2
(r
)
T
is
constant the accelera-
A
A
+ 2rwS. A
Since the rate of increase of this is parallel to This is easily shown to be is zero.
r,
A the s-component of a
- ru> 3 = 0,
3rd)
r = |no 2
so that as required. (2)
A
particle
P
is
moving on
the surface of
a body
B with
relative to the body, while the latter is rotating relative to
objects with
an angular
u> about a fixed axis parallel surrounding objects.
velocity
P relative
the velocity of
to
velocity v surrounding A to a.
Find
on the fixed axis as origin, let r be the position With any point vector of the particle P. Then the velocity of the point Q of the is body which is instantaneously coincident with
P
A
u = a>a*r, A since &>a is the angular velocity of the body.
velocity of
P
velocity of
Q
is ;
the vector that is
sum
the resultant
Q and
the
A v
69. Areal Velocity.
And
of its velocity relative to
+
Considering again the general case of
when its path is not necessarily a plane curve, we may form the moment about of the velocity vector v regarded as localised in a line through P. As seen in Art. 40 this moment motion
is
of P,
the vector
rv
perpendicular to the plane of r and
the unit vector in this direction, and dicular
ON
to the tangent at P, r>
v.
If
k
is
the length of the perpen-
p we may
write
=pvk.
But, using the value of v found in Art. 68, the particle moving instantaneously in the plane
OPN, we have A
r« v
A
= r* (fr + rws) = r 2 wk,
KINEMATICS
§ § 69, 70, 71]
where
co is
for the
OP
the rate of turning of
moment, we
103 Equating these two values
find the relation
w=pv
r2
The
areal velocity of the point
tion of vector area by- the line
vector Jr*v. is 6tv,
P
(1)
about
OP.
This
the rate of descrip-
is is
represented
by the
and the vector area swept out by
rate of description of vector area of the areal velocity will
OP
is
|r*(&v).
therefore |r*v.
is
The
The measure
be denoted by \h, so that
Ji=pv = r 2w 70.
(2)
Motion with constant acceleration.
This
considerable importance, illustrated approximately of a projectile
under gravity.
is
a
case
of
by the motion
Let d be a unit vector in the
whose measure Then
direction of the constant acceleration r
P
For, in a short interval dt the displacement of
is g,
and
the position vector of the moving particle P.
d 2i
dv
Tt=d¥ Integration with respect to
t
,
=ga
-
gives
v=^id + v where v
is
,
the constant of integration, obviously representing
the velocity at the instant
t
= 0.
Integrating again,
we
find for
the position vector of P, r
= !^ 2 d + Zv
,
where we have made the constant of integration zero by choosing as origin the initial position of P.
locus of
P is sum
a parabola whose axis
This equation shows that the is
in the direction of d.
two vectors in fixed proportional to t 2 and t respectively. r is
the
of
2.
71.
Momentum.
Dynamics
directions, with
For
modules
of a Particle.
The momentum
of a
moving
particle
is
a
vector quantity, jointly proportional to the mass of the particle
and to its velocity, and having the same direction as the velocity. The unit of momentum is chosen as that of a particle of unit mass moving with unit velocity. Hence, if m is the measure of the
VECTOR ANALYSIS
104 mass, and
M vectors representing
v,
[OH. VI.
the velocity and
momentum
of the particle respectively,
M = mv. The
momentum
rate of change of the
dM_
dv
dm
dt
dt
dt
as
If,
is
usually the case, the
mass
dM and
Newton's Second
72.
which
is
also a
is
constant,
dv
Law
of Motion.
acceleration.
According to this law,
the foundation of the ordinary theory of dynamics,
is
the force acting on a particle of
of the particle
same direction as the
therefore in the
is
of the particle
by
vector quantity, represented
is
proportional to the rate of change
momentum produced by it, and has the same direction.
Hence,
with the choice of unit force as that which produces unit acceleration in a particle of unit mass,
we have the
relation
F = |(mv), where F this
is
may
When m
the vector representing the force.
is
constant
be written simply
F = ma, showing that the acceleration produced in the motion of a This equation
it.
particle
mass has the same direction as the force producing
of constant
called
is
the equation of motion for the
particle.
Suppose that the particle
F2
,
.
tions
.
,
.
F„, which,
a^,
particle
is
a2
,
.
.
.
,
if
a'
a„ respectively.
Their joint effect on the
SF
which produces
given by n
n
i
i
Thus the actual acceleration
a'
the accelerations a x
an
is
,
acting separately, would produce accelera-
the same as that of a single force
an acceleration
each force
acted on by several forces F x
is
,
a2
,
.
.
.
,
is .
equal to the vector
That
is
sum
of
to say, the effect of
uninfluenced by the action of the others.
Each
§§
DYNAMICS OF A PARTICLE
72, 73, 74]
produces the same effect as
if it
105
were the only force acting on
the particle.
Let
be the position vector of the particle relative to a fixed
r
point 0, and suppose that referred to rectangular axes through 0. r=cci+2/j
+ zk,
F=Zi + Yj+Zk. Then the equation
This
motion
of
ma = SF may
be written
equivalent to the three scalar equations
is
mai
= HX,
mi/
= 2 Y,
= 2Z,
m'z
which are the ordinary Cartesian equations of motion for the particle.
73.
The impulse of a force F acting on a any interval of time, is the change of momentum
Impulse of a force.
particle during,
produced by quantity.
during that interval.
it
If,
during the
interval,
v to v x the impulse of the force I
When
the force
instant
t
variable,
is
is
It
is
given by
= to(v 1 -v
).
and acts from the instant
Fdt=m\ a^=m(v1 -v
When
J
to the
(
),
to
and therefore represents the impulse
62,
the force
simply
t
v the definite integral J*o
by Art.
therefore a vector
the velocity changes from
-t
)F.
of the force.
constant the value of the definite integral
is
Briefly
we say
that the vector I
is
is
the time-
integral of the vector F.
An
impulsive force, also frequently called an impulse,
large force acting for a of the force
The
very short time.
the position of the particle
is
is
practically unchanged.
effect of the impulsive force is represented completely
change of
momentum
impulsive force
by
its
produced by
We
by the an
therefore specify
impulse in the original sense.
74. Activity of a force.
work given in Art.
it.
a very
Hence during the action
39, the
In accordance with the definition of
work done by a
force
F
acting on a
VECTOR ANALYSIS
106
[CH. VI.
during a small displacement dr of the particle,
particle,
sured by
F*<5r.
is
mea-
If this displacement takes place during an interval <5r
dt,
the average rate of working during that interval
and proceeding
to the limit as dt->0,
we
F*-j-
is
;
find for the instantaneous
rate of working, or the activity of the force,
=F-v. F-f at
The work done during a short
interval dt
therefore
dt
;
and the
to the instant
t
total £•,
is
«,
LtZF-v<5«
The
75.
F-v
is
work done by the force from the instant
particle
is
=
F-v*.
The
principle of energy.
kinetic energy of a
a scalar quantity jointly proportional to
and to the square
of its speed.
The unit
moving mass
its
of kinetic energy
is
taken as twice that of a particle of unit mass moving with unit
Hence the
speed.
with a velocity v
kinetic energy of a particle of
is
is
m
moving
T = \m,v* = Jmv 2
If the velocity is variable,
resultant
mass
owing to the action
of forces
F, the rate of increase of the kinetic energy
whose
is
dT dv -—=MV-f-= ma*v = F-v, dt
and
is
dt
therefore equal to the activity of the resultant force F.
In other words, the rate of increase of the kinetic energy is equal to the rate at which the force F is doing work on the particle. It follows that, during
any
kinetic energy of the particle interval
by the resultant
true whether
F
is
finite interval, is
the increase in the
equal to the work done during that
force acting
on the
constant or variable, and
is
particle.
This
is
called the principle
of energy for the particle.
we begin by
defining the kinetic energy of the particle can do in virtue of its velocity, we may find its value thus. Suppose that at the given instant t the particle has a velocity v and that it does work against a variable force F [If
as the
work
till it is
it
finally
brought to rest at the instant t v Then at any it is doing work against the force
instant the rate at which
-
DYNAMICS OF A PARTICLE
§ § 75, 76]
F
dv
is
- F*v = - m -v- v •
and therefore the
;
the force before the particle
m dv •vdt =
-
dt
This then
work done against
total
brought to rest
is
107
is
- Jm[v 2 ]J' = %mv
2
the value of the kinetic energy due to the velocity V
is
F from
Suppose that, owing to the action of the force instant
to the instant
t
tum The
i-
e
t,
the velocity of the particle changes
;
change of momen-
of the force is the
i=«(v,-v
-
).
increase in the kinetic energy during this interval iwiVj 2
that
t
The impulse
from v to v v
is,
initial
76.
- §tov
2
= ^m(v 1 - v )-(v 1 + v = JI-(v1 +v ),
half the scalar product of the impulse
and
and the sum
Moment
momentum.
of
Let
r
be the position vector
particle relative to a fixed point 0.
is
its
momentum mv.
Regarding the
localised in a straight line
through the
moment about
H =r
This moment of
The rate
of the
final velocities.
moving v = r and
angular
is
)
the
its
.]
the
,
momentum
x
dt
d
.
as
for its
about
is
also called
0.
of increase of this angular
-T-
momentum we have
mv
of the particle
momentum about dH.
particle,
of
Its velocity
momentum
is
dv
.
=^r (rxmv)=vxmv+rxm-— dt
dt
=rxF, where
F=mv
is
the resultant force acting on the particle.
the rate of increase of the
about
is
the -particle.
equal * to the
This
In particular,
if
is the
angular
momentum (A.M.)
moment about
of the resultant force on
principle of angular
momentum.
the resultant force has zero
moment about
the a.m. of the particle about that point remains constant. is
Thus
of the particle
0,
This
the principle of the conservation of A.M. for the particle.
* The term equivectorial may be used of vector quantities which have the same measure and direction, and are therefore represented by equal vectors. But such quantities are commonly said to be equal.
VECTOR ANALYSIS
108 77. Central forces.
by a
for instance, the particle
If,
is
acted on
toward the point 0, the moment of and therefore the a.m. of the particle
force always directed
the force about
about that point invariable,
is is
zero,
Assuming the mass, of the
constant.
we have the
result that r>
pendicular to the plane of r and v. is
[CH. VI.
invariable
;
that
is
is
particle
a constant vector per-
Hence the plane of r and v moves in a plane curve, The path of the particle is
to say, the particle
whose plane contains the point 0. called its orbit.
Such a
always toward a fixed point,
force, directed
a central force, and the fixed point
is
is
called
the centre of force.
78. Central force varying inversely as the square of the distance.
To
by a particle acted on by a central toward 0, varying inversely as the square of the distance of the particle from 0. Let r be the position vector of the find the path described
force
and ^m/r 2 the measure of the force on the The constant is called the intensity offarce, representing the force per unit mass at unit distance from 0. The acceleration of the particle has the direction of - r, and is thus specified A by the relation //2 r particle relative to 0, particle.
jj.
%=-> ,,
w
Further, considering the area! velocity of the particle about 0,
we have
(Art. 69)
r>
= Ak =r 2o>k,
where k is a unit vector perpendicular to the plane of the orbit, and w the rate of turning of r. From these equations it follows that
ld 2r
,„..,
;p*( Ak )
A
=
r
-wrxk = ft,s,
-yA r%<>k)=
A where s is the unit vector -r>s = t- ; and since, in the _
case of a central force, h
is
constant,
1 /u.
we may
write the last
^
equation 7
dt \dt
dt'
which on integration gives immediately 1
A
A
-v*Ak=r + ea,
(2)
5
§ § 77,
DYNAMICS OF A PABTICLE
78]
109
A
where ea
is
now
is
If
6
the constant vector of integration, whose module is e. A the variable angle between r and a, the scalar product A
ra = / cos
8.
Hence, on multiplying the last equation scalarly by the second
member r(l
first,
we
r,
and writing
find
= -r-vx^k = -r>
+ecos
0)
= hk'hk = h* — = ,.(say). .
(
/t
ft
Hence the equation
of the orbit of the particle I ;
is
= l+ecos0,
(3)
r
representing a conic whose eccentricity
rectum
is
l=h 2/fi, and one
The orbit will be an as e
is less
whose
of
ellipse,
is
e,
whose semi-latus
foci is at the centre of force.
a parabola or an hyperbola according Squaring the
than, equal to or greater than unity.
A value of ea given by
(2),
wefind ,
e2
and noticing that v
And it
since
follows that e
the particle
is
2
<, =
or
perpendicular to k,
. 2h* , rvxk + 1.
—
= A*»« 2 ft
A r*v*k
is
ft
= r* vk/r = hjr,
> 1 according as v
projected with a speed
V
<, = or
2
>—
.
If
then
at a distance c from the
centre of force, the orbit will be an ellipse, a parabola or an
V2
hyperbola according as
is less
than, equal to or greater than
In any case the value of the eccentricity
2m/c.
A2 jX
The trated
first
case
is
/F 2 _22 V
is
an
ellipse
illus-
The
with the sun at
This fact, as well as the constancy of the areal
of a planet's orbit
discovered
(4)
of the planets relative to the sun.
velocity of a planet about the sun, size
+l
one of considerable importance, being
by the motion its foci.
given by
c
ft
orbit of a planet relative to the sun
one of
^
is
and
its
and a
relation
between the
period of revolution, were
first
by the astronomer Kepler, from whose observations
VECTOR ANALYSIS
HO
inverse square law of gravitational attrac-
Newton deduced the tion
[CH. VI.
between any two bodies.
Exercise 18 at the end of this
(Cf.
chapter.)
motion
The
a, b
be the semi-axes of
rate of description of area
to the centre of force
is
of the
of a particle in
centre of force at one focus,
Let
importance attaching an elliptic orbit under a we shall examine it a little closer. the ellipse, whose area is then irdb.
On account
79. Planetary motion.
to the case of the
A/2.
by the line joining the particle Hence the periodic time, or the
time of one complete revolution of the particle, _,
irab
„
,
T =^rr = ^ab^r\
/
is
P = —27rv=a* —
.
4.
u.
= b 2ja.
Hence
(5)
moving and with the
since the semi-latus
rectum
in different ellipses
about the same centre
same intensity
of force, the squares of their periodic times
(ft)
is I
for particles
of force,
major axes
are proportional to the cubes of the
of their orbits,
This relation was observed by Kepler in the case of the planets.
To
find the speed v of the particle at
multiply
(1)
scalarly
by
di 2-r.
d 2T _
di
2= Jt'df
On
any point
of its path,
Then 2fi
dx_
~^ T 'Jt~ M
2fxdr
"r*
dt
integration, therefore,
©-< To
C
find the constant
of integration, consider the speed v 1 at
the end of the minor axis.
tangent at this point
For
is b,
The perpendicular distance
this point the last equation
so that
Another expression
is a.
becomes o2
a
C — - n\a, and
to the
while the distance r from the focus
b-
a
the speed at any point /2
l\
\r
a.
for the speed
is
is
given
sometimes
by
useful.
are the perpendicular distances from the centre of force
If ]h
and
f
the
DYNAMICS OF A PARTICLE
§79]
111
other focus respectively, to the tangent to the ellipse at P, the
speed at
Pis w
and
is
_*_*£.'_^ p~pp'~ b 2
thus proportional to
Given the
may be found
is
'
p'.
initial position of
projection, the orbit
*
'
the particle and
determined.
velocity of
its
For the semi-major axis
at once
from the equation
(6),
using the initial values of r
and
The other on the line
v.
focus 0'
is
through the point
P
equally
projection,
PO
clined with
of in-
to the
direction of projection
and since
the point 0'
The centre bisects
;
OP + PO' = 2a, is
known.
of the ellipse
and
00',
the FIG. 53.
major axis
lies
along
it.
The value h =vp is found from initial values, and the semi-minor from the relation h 2 =/ub 2/a.
axis
Examples. (1)
A
focus 0.
particle
Show
P
describes
an
ellipse
that its velocity at
any
under a central force instant
may
two components of constant magnitude, perpendicular axis
at
and
OP
to
the
be resolved into to
the
major
respectively.
Let O'N be the perpendicular from the other focus to the tangent P Then, by formula (7) above, the velocity v of the particle is.
proportional to
duced to meet
O'N and
OP
at right angles to
it.
O iN = iOV = U
and both which is otherwise obvious from equation (2)
A
f article of mass
centre offorce, the blow.
it
if
O'N
is
pro-
m
Hence
the- result,
(2) of Art. 78.
is moving in an ellipse with a focus as end of the minor axis, in its motion from the receives a blow which changes its orbit to a circle. Find
At
centre of force.
But
in Q,
the
VECTOR ANALYSIS
112
[CH. VI.
Let be the. focus, C the centre of the ellipse, and B' the end of the minor axis. Take i, j unit vectors in the directions OC and B'G respectively. Then, in virtue of formula (6), the original velocity at
Fis
In the
,-
circular orbit the velocity
V
and the speed
a2
a or
F = V//u/«,
must be perpendicular
to
OB'
such that '
so that the speed is unaltered.
Now,
since
OB' = aei-bj, the unit vector at right angles to v 2 in the circular orbit, is
the direction of the velocity
this, in
-(6i + oej),
v 2 =a/ — (-i + ej
so that
Hence the blow
is
given by
m(v 2 -v 1)=«y^cj The module
of this is
— a
and
its
direction
-^-
i
y/2fi(a-b),
makes an angle tan
-1
\
v
r
a-o
with GO.
80.* Central force varying directly as the distance.
on the of force and force
acceleration
particle
is
r
toward
be
m,p.r,
where
/x is
Let the
the intensity
the distance of the particle from 0.
Then
the
-pi, and the equation of motion for the particle
^=-^ dx
Multiplying scalarly by 2 -j dr
r
(^
we have d 2t _
is
dt
Jt'dt^-'^'jl'
DYNAMICS OF A PAKTICLE
§§79,80]
113
which, on integration, gives
p
If
is
the perpendicular distance of
from the tangent at
r to
the orbit, this equation gives v2
1
But the
p,
r
equation of an ellipse referred to
p an
a, b
ellipse
a + b -r
~
2
its
centre
is
2
o*P
are the semi-axes.
whose centre
2
2
1
where
1
'
Hence the
orbit of the particle
is
0, while
is
h
and
= VJi
C=n{a we
Substituting these values
ab
.
2
+b*).
find for the speed at P,
2
v =fi(a 2 =/j,
+b 2 -r 2
OB 2
.
)
,
v
FIG. 54.
where
OD
is
the semi-diameter conjugate to OP, and therefore
parallel to the
tangent at P.
Thus
\=Vi*.6b is
the velocity at the point P.
F VECTOR ANALYSIS
114
The
periodic time
ellipse at the rate ^
[CH. VI.
the time to describe the whole area of the
is
This
.
is
2irah
T
h
2-n-ab _ 2tt ~Vji.ab~V/*'
The period T thus depends only on the intensity /u of the force, and not on the size of the orbit. If r is expressed in terms of two constant unit vectors a, b in the plane of the orbit, as A
A
r=xa+yb, the equation of motion
(1)
becomes
A i'-a
and
is
A
A
+ yb =
-fj,(xa
A
+ yb),
equivalent to the two scalar equations
y=-[iy,
x=-fix;
which represent simple harmonic variations of
The
period 2-ir/VJi.
velocity of
P in A
v
the ellipse A
is
common
motion with the same period and the
into elliptic
centre.
(Cf. also
Exercise 3 at end of chapter.)
81.* Motion of a particle on a fixed curve. particle
period and
but in different directions and with different phases,
compound same
y of common
= r=«a + 2/b.
Thus two simple harmonic motions with a centre,
x,
moving along a
fixed wire.
fixed curve, e.g. a
Let v denote
its
Consider next a
bead
sliding along a
speed at any point and
t,
n, b the
unit vectors in the directions of the tangent, principal normal
and binomial at that point. Given the external force F acting on the particle, it may be resolved into components in these directions,
F = F1t +
n+F b,
2
3
while the action of the curve on the particle the
sum
of
two
forces
;
may
be specified by
one
R = R n + J?3b 2
perpendicular to the curve, and the other a frictional force parallel to the tangent,
former, where
/j,
is
and equal
in
magnitude to p times the
the coefficient of friction.
The
frictional force
is
DYNAMICS OF A PARTICLE
80, 81]
§.§
thus
-fj,Rt,
having the opposite direction to the velocity v = vt.
The acceleration
of the particle, as
dv
shown
in Art. 67,
The equation
b.
of
ma.=F+R-fiRt, = (FJ + F 2n + F3b) + (R2n + R3b)
m(vt + Kv n) 2
is
is
,
,
having zero resolute in the direction of motion for the particle is therefore
that
115
-fiVR 2 2 + R3 H. This
is
equivalent to the three scalar equations
mv = F 1 -/LiVR 2 i + R 3 2 m,KV i = F 2 + R 2 = F3 + R 3 Thus the resolutes
(1) (2) (3)
and the reaction
of the external force
of the
curve in the direction of b are equal and opposite.
For a smooth curve /j,=0 and the equation (1) If
Fx
is
given as a function of
dv
dv ds
dt
ds dt
s,
viz.
_ V dv "
jl ds
F^s),
is
dv simply m-j-
we may
= F1
.
write
d
= ds j-Mv2
)'
and the equation becomes
mjs (^)=F
1
(s),
which, on integration, gives the formula
M« v being the speed at A.
2
2 -«o )=£^i(s)*>
This result
is
simply that the increase
in the kinetic energy of the particle is equal to the
work done
by the external force on the particle. Having determined v we find R 2 from the equation (2), while R a is known from (3).
When
the external force is at each point parallel to the plane of
curvature,
or smooth.
F
3
=0, and therefore
The equations
of
R3 =0 whether the curve
[
mKv
rough
motion are then
\hn^ = F 2
is
1
-iAR2
= F2 + R2
,
(2)
VECTOR ANALYSIS
116 Elimination of
[CH.
R 2 gives dv 2
|m -j- + m/iiKV 2 = F t + fiF2 which
,
a linear differential equation of the
is
whose solution
mined v 2 we
found in the usual way.
is
find
R 2 from
first
order in v 2
,
Then, having deter-
(2).
Further worked examples will be found among the
Note.
following exercises.
EXERCISES ON CHAPTER
VI.
1. Three particles A, B, C at the points r lt r 2 r 3 are moving with velocities v v v 2 v 3 respectively. Find the rate of change of the vector area of the triangle ABC. ,
,
D
at the points r 1( r 2 r 3 r 4 are moving 2. Four particles A, B, C, with velocities v 1( v 2 v 3 v 4 respectively. Find the rate of change of the volume of the tetrahedron ABCD. ,
,
,
,
3. If a, b are constant vectors and t the time variable, show that a particle whose position vector at any instant is
r = cosnt a
+ sirrnt
b
moving in an ellipse whose centre is the origin and that the motion is that due to a central force varying as the distance.
is
;
4.
Show
when a
that,
the action of gravity,
its
moves on a smooth curve under speeds u, v at any two points P, Q are
particle
v 2 = u2 + 2gh,
connected by
where h 5.
is
the vertical depth of
Hodograph.
Q
below P.
The hodograph of a moving point P is the Q whose position vector at any instant is equal
locus of another point
the velocity vector of P. Suppose, for instance, that P is moving with constant acceleration
(or proportional) to
gi (Art.
70).
Then
its
velocity at
v=v
Hence the
position vector of
Q
is
any instant
is
+ tgH. equal (or proportional) to
R = v + tg&. Thus the locus
a straight line parallel to d. And since t uniformly along this straight line. The velocity of Q is equal (or proportional) to R=^d that is to the acceleration of P. This is always true. For, since R = cv, it follows that R = cv = ca, where a is the acceleration of P. of
Q
increases uniformly,
is
Q moves
;
,
EXERCISES ON CHAPTER
VI.]
the motion of
If
and
its
P
velocity at
that due to a central force varying as the moves in an ellipse whose centre is at 0,
is
then (Art. 80)
distance,
P
is v = V/u. OD, where OD is the semiThus, for the point Q on the hodograph, R
any instant
diameter conjugate to OP. is
proportional to
described in the
OD, showing same
that the hodograph
periodic time.
proportional to the velocity of Q,
A
6.
A
bead
The
is
a similar ellipse
acceleration of
and therefore
particle describes a circle
the hodograph
117
VI.
to
P
is
PO.
with uniform speed.
Show that
a circle described uniformly.
is
down
the circumference of a smooth vertical Show that the a equation of the hodograph in polar coordinates is r = c sin ^. 7.
circle,
slides
starting
from rest at the highest point.
// a particle describes a conic under a central force to the focus, hodograph is a circle. From equation (2) of Art. 78, on transposing and squaring we find 8.
the
A r2
W
he* — a*vxk + e
M
h
= —s2 v2 - 2
2 ,
2
that
v2
is
,M = 0. -2v-(^k*a) + (e2 -l)^
Hence the hodograph
whose centre
is
the circle
R
-2R-c + (e2 - l)p = 0,
2
eu.
is
the point c = -^
* If
k>
and the hodograph passes through the
the conic
is
a parabola, e =
1
origin.
Show 9. A particle describes an ellipse under a force to the centre. that its angular velocity about a focus varies inversely as its distance from that focus and that the sum of the reciprocals of its angular ;
velocities
about the two
foci is constant.
Show that the path of a point P, whose velocity is the sum two components of constant magnitude u, v, the first in a fixed direction and the second perpendicular to the line joining P to a 10.
of
fixed point S, is a conic with focus at
A
S and
eccentricity -.
an ellipse with a focus as the centre of that the speed at the end of the minor axis is a mean proportional between the speeds at the ends of any diameter. 11.
force.
particle describes
Show
VECTOK ANALYSIS
118
[OH.
12. In the previous exercise, show that the angular velocity of the particle about the other focus varies inversely as the square of the normal.
A
describing an ellipse under a central force to the reaches the end of the minor axis the intensity of force is diminished by one-third. Find the position and size of the new orbit, and show that the join of its centre and focus is bisected 13.
particle
When
focus.
by the minor
is
it
axis of the original orbit.
in the previous exercise, the change at the end of the minor an alteration of the law of force to variation directly as the distance, the magnitude of the force at that point remaining the same, show that the periodic time is unaltered, and that the sum of 14.
axis
If,
is
the new axes is to their difference as the distance between the foci.
A
sum
of the old axes to the
an ellipse with one focus S as the centre the particle is at P the centre of force is suddenly removed to the other focus S'. If k, k are the curvatures at P of the old and new orbits, show that 15.
particle describes
of force.
When
k:k' = S'P 2 :SP 2
-
Show
that the rate of rotation of the direction of motion of a particle in an ellipse under a force to the focus, is a maximum or minimum when the particle is furthest from that focus, according as the eccentricity is greater or less than i. 16.
17. In elliptic
arrives at
motion under a force to the focus, when the
speed being unaltered. ellipse
particle
P the direction of motion is turned through a right angle, the Show that
the particle will then describe an
whose eccentricity varies as the distance
18. Central force
of
any function of the distance.
P from the
centre.
Suppose a
particle
A
- Ft per unit mass, where F is a r from the centre of force. Then the equation
moving under a central function of the distance of motion
force
is
J2_
a
-=-Ft dt 2
and forming the
On
scalar product of each side with 2 n di
d 2i
A dt
nT,dr
dt
dt 2
dt
dt
integration this becomes
dJ= c -Yit dt v2
= C-2\ Fdr.
di -=-
we
find
'
EXERCISES ON CHAPTER
VI.]
VI.
119
This equation gives the value of the speed at any distance. Writing h 2 /p 2 instead of v 2 and differentiating the last equation with respect ,
to
we
r,
find
7
,
,
3
p dr This
the p,
equation of the orbit when the function F is given. Or, given the orbit, this equation tells the law of force to the pole (origin) under which the orbit can be described. For instance, the p, r equation of an ellipse referred to a focus is is
r differential
s
JL-lf? 2 2
V
p
""
p The law
elliptic orbit ,,
7, j
v_
3
)
\r
Hence, for a particle describing an to the focus,
I a)
®P
" a
dr
b2
under a central force
-,
1 .
r2
'
of force is therefore that of the inverse square.
19. Find the law of force to the centre of an ellipse under which the ellipse will be described. This is the reverse of Art. 80. For an ellipse, with origin at the
centre,
a2
l 2
a2b 2
p _
,
+ &2 _ r2
2
h dp
h2
*-?£-*»''
whence
varying directly as the distance. 20. Show that the law of force to the pole under which the equiangular spiral p = r sin a. can be described is that of the inverse cube. Also that the speed varies inversely as r.
21.
Prove that the hodograph of the motion is also an equiangular spiral.
the previous
in
exercise
22. Show that the force under which a particle describes a circle, with the centre of force on the circumference, is J = n/r 5 and that the speed varies inversely as r 2 /
,
.
23.
Show
exercise 24.
A
is
that the hodograph of the motion
particle
is
and when it arrives at a in the focus moving toward the vertex, the centre of ceases to act for a certain interval T. Prove that when the operates again the new orbit will be an ellipse, parabola or (fj)
;
distance r from the focus, force
the previous
describing a parabolic orbit (latus rectum 4a)
about a centre of force
force
in
a parabola.
hyperbola according as
T<,
=, >2r\/(r-a)/2/j..
VECTOR ANALYSIS
120
[CH. VI.
A
smooth wire in the form of a circular helix has its axis and a bead slides down it under gravity. Determine the and show that the speed at a depth z, and the action on the wire time to fall this depth is cosec a.V2z/g. 25.
vertical,
;
Show that
26.
the hodograph of the motion in the previous on the surface of a right circular cone
exercise is a curve described
of semi-vertical angle «
- «-.
27. A particle P moves on a smooth helix (a, a) under the action on the axis equal to /j.m OP, only of a central force to a fixed point where in is the mass of the particle. Show that the action on the curve cannot vanish unless the maximum speed of the particle is .
rjx
a /\Z
.
sec
a..
28. If the
hodograph be a circle described with constant angular on its circumference, show that the path of
velocity about a point
the particle
P
is
a cycloid.
a point on the tangent at a variable point Q to a fixed radius a ; QP is of length r, and makes an angle 6 with a fixed tangent. Show that the resolutes of the acceleration of P 29.
is
circle of
along and perpendicular to
QP
f-rd 2 + a6
are
and
-
4
(r
2
d)
+ ad2
,
r at
A particle of unit mass is placed in a smooth tube in the form an equiangular spiral of angle a, and starts from rest at a distance 2d under a force /x/r 2 to the pole. Show that it will reach the pole 30.
of
in a time ir sec 31.
by a
A
move on a circular wire, is acted on on the circumference varying inversely the distance. Prove that the action on the
particle, constrained to
central force to a point
as the fifth
wire
ocVd3///.
is
power
of
constant in magnitude.
ch. vii., § 82]
CHAPTER
VII.
DYNAMICS OF A SYSTEM OF PARTICLES AND OF A RIGID BODY. 82. Linear
Momentum.
Consider a system of moving particles,
independent of each other or under any sort of mutual action. Their relative positions may be changing, or they may either
be rigidly attached to one another as in the case of a rigid body.
Let
m be the mass of any one of the particles, r its position vector
and v=r its velocity. Then the momentum of the particle is mv. We define the linear momentum of the system as the vector sum of the linear momenta relative to a fixed origin 0,
linear
of the separate
particles.
vector
It is therefore represented
by the
M = Smr = atg 2mr. -Y-
But the centre given by where
M = Zm
of
mass
of the
system has a position vector
r
Mi = 2mi, is
the mass of the whole system.
Hence d)
«-s<*r>-*ar**. v denoting the velocity of the centre of mass.
Thus
momentum
-particle
of the system is equal to that of
a single
the linear
of mass
equal to the total mass of the system, moving with the velocity of
cm. The rate given by the
of increase of the linear
m
momentum
is
system
is
d
w=Jt (Mv)=m,
where a
of the
the acceleration of the 121
cm.
(2)
VECTOB ANALYSIS
122 83. Equation of
force acting
the particle
on the
motion of the
cm.
[CH. VII.
F
Let
be the resultant
Then the equation
particle to.
of
motion
for
j
is
F=^(tov). Taking the vector sum
of the resultant forces
we have
Now
j
the vector
sum 2F
on
the particles,
all
»j
includes both forces whose origin
is
external to the system, and forces of internal action between the
But
separate particles of the system.
Law
with Newton's Third
between the particles forces, the vector
of
if
we assume,
in accordance
Motion, that the internal action
represented by pairs of equal and opposite
is
sum
2F we
In calculating
of these is zero.
need therefore only consider the external forces acting on the
The equation
particles of the system. vector
sum
(3)
then shows that
the
of the external forces acting on the system is equal to the
rate of increase of its linear
The equation may
also
momentum.
be written
2F = |(Mv) = Ma, so that the
cm. has
(4)
the same acceleration as a particle of mass
equal to the total mass of the system, acted on by a force equal to the vector 84.
of all the external forces.
momentum,
angular Art. 76)
by
sum
Angular Momentum. is
The moment
of
represented by the vector TOr*v, and
defined as the vector
about 0.
sum
momentum,
the particle to about the origin
The angular momentum
toT'V.
particles
of
of the
it
rate of change
system about
momenta of by H, we have
of the angular
Representing
its
its
rate of change
(5)
is
dH.
_
d
,
.
_
„
^=2r^(TOv)-r*F,
where
F
is
is
the separate
H = Xr*TOV, and
or (cf.
the resultant force acting on the particle at
(6)
r.
The
product r*F represents the (vector) moment or torque of the force F about 0. In the summation the internal actions may
DYNAMICS OF A SYSTEM OF PARTICLES
83, 84]
§§
123
be neglected, since each pair of equal forces has zero
moment about
and opposite colli near The equation (6) then states
0.
A.M. of the system about a fixed sum of the torques about of all the
that the rate of increase of the point
is
equal to the vector
external forces acting on the system.
The angular momentum
the system about a fixed line A in the direction of the unit vector c, is the (scalar)
through
of
resolute in this direction of the a.m. H. It is therefore given A A dH d A by c*H, and its rate of change by t^(c'H)=c*-^-. Further, by .
Art. 40, the
sum
forces acting
of the
moments about
on the system
this line of all the external
is
A
A
A JVt
Zc-r*F = c-2r*F=c-^. at
Hence the
rate of increase of the a.m. of the system about a fixed
equal to the
line is
about that
sum
of the
moments
of the external forces
line.
In calculating the a.m. of the system about any point 0,
it is
frequently convenient to use the velocity of the centre of mass G,
and the velocity of the particles relative to G. position vector and velocity of the cm., and r', particle
m relative to r
v are the
If
r,
v'
those of the
G,
= r + r'
Hence the a.m. about
and
v=v+v'.
is
H = 2mr* v = 2m(r + r>(v + v') = 2wifxv + 2mr'*v' + (2mr')*v+rx2mv'.
Now
the last two terms vanish
and therefore
also
its
about of G,
to
for
2mr'
is
constantly zero,
The equation may then
h =r*Mv + 2r'*mv'.
be written
Thus the a.m.
;
derivative Zwv'.
of the
system about
of a particle of
mass
and the a.m. about
G
is
the vector
sum
of the a.m.
M at G moving with the velocity
of the
system in
its
motion relative
6?.
Example.
The motion of a body is given by the velocity v of its cm. and the A.M. Find the A.M. about a straight line through the that point.
H about point
A
parallel to the unit vector b.
VECTOR ANALYSIS
124 Let
M
[CH. VII.
A A
be the mass of the body and a the position vector of
Then -a
cm.
relative to the
and by the theorem
is
cm.
that of the
relative to
proved the a.m. about the point
just
A
;
is
H + (-a)*Mv. Therefore the (scalar) a.m. about an axis through
A
parallel to b is
b'(H - Ma- v) = b'H + M[abv]. 85.
Moving
Instead of the fixed point 0, suppose we choose as origin of moments
moments.
origin of
moving point
a
whose variable is s. Then
0',
position vector relative to if r,
r'
are the position vectors of the
m
particle tively,
relative to
r=s+r'.
0, 0' respec-
Let H, H' represent
the angular momenta of the system about 0, 0' respectively. Then
FlO. 55.
H = Srxmv = 2(s + r')*mv = s*2wv + Zr'xmv = s*M+H', where tion
M
we
the linear
is
momentum
find for the rate of
_dW_ ds ~ dt + dt
dt
By
of the system.
differentia-
change of H,
dFT ''
M + s dM dt :
+ v' xM+s*2F,
dt
in
which
v'
=
But the
s is the velocity of 0' relative to 0.
rate of
increase of the a.m. about
the torque of the external forces about 0. torque,
and
L' the torque about 0'
,
If
L
is
equal to
represents this
we have
L = 2r*F = 2(s + r')xF = sx2F + L'. Equating this value
of
L
to the value found above for -j-,
dK' ~dT
which
=L
v'xM,
we
find
•(7)
is the relation between the rate of change of the a.m. about a moving point, and the torque of the external forces about
that point.
DYNAMICS OF A SYSTEM OF PARTICLES
§§ 85, 86]
A
case of particular interest
point 0'
is
the
cm.
so that the last
For then
v'
= v and
M =Mv,
Hence with the cm. as
(7) vanishes.
moments, whether moving or at
origin of is
that for which the moving
is
of the system.
term in
125
rest,
the relation -=-
=L
always true. In calculating
H
we may
the cm.,
and
v'
cm.
treat the
consider only velocities relative to
that of the particle
as a fixed point, and
For
it.
if
v
the velocity of
is
m relative to
it,
H = 2r'*m(v + v') But the
first
term
is zero,
= (Zmr')xv + Zr'*mv'. since 2mr' = 0. Therefore
H = 2r'*mv', and depends only on the motion
of the
system relative to the cm.
86. Equations for impulsive forces.
on by a
set of impulsive forces.
Suppose the system acted
When
there are any connections
between the particles of the system, these forces cause
impulsive
action
will in general
But we assume,
between them.
in
accordance with Newton's Third Law, that such action between
two
particles consists of a pair of equal
forces along the line joining
of the particle
m just
after
and
and opposite impulsive
Let v and v be the
them.
resultant impulsive force I on this particle of
momentum
it
is
equal to the change
produces in the particle (Art. 73)
I=wi(v-v
where M,
2I = 2m(v-v
M
are the linear
:
or
).
Taking the vector sum of the impulsive forces on
we have
velocities
Then the
just before the blow.
)=M -M
momenta
all
the particles, (8)
,
of the
system just after
and just before the impulsive forces act. In forming the sum SI we may neglect the internal impulses which consist of equal and opposite pairs. The equation (8) then shows that the vector
sum
of the external impulsive forces acting on the system
is
equal
momentum produced in the system. Similarly, taking moments about a fixed point 0, relative which the position vector of the particle m is r, we have
to the increase of linear
2rxI
where
H
and
H
= 2rxw(v-v
)=H-H
0!
are the a.m. of the system about
to
(9)
before and
VECTOR ANALYSIS
126
[CH. VII.
In the summation on the left-hand side we mayneglect the internal impulses for the same reason as before, thus after the blow.
sum
obtaining the result that the vector external impulsive forces about crease in
A.M. produced about
of the moments of the
a fixed point
is
equal to the in-
that point.
Kinematics of a Rigid Body. 87.
Any
Motion about a fixed point.
body with one point
fixed
is
equivalent to
displacement of a rigid
a rotation about a
definite
axis through 0.
To prove having
about is
its
its
consider a unit sphere fixed in the
this,
centre at 0.
centre
;
As the body moves,
and the motion
of
body and
this sphere turns
any point
P
of the
body
determined by the motion of that point of the spherical surface
which
lies in
the line OP.
We may
thus think of the body as a
sphere of unit radius, and consider the motion of points on the
Suppose that during any displacement moves to A' and the point B to B' Join A, A' and also B, B' by arcs of great circles, and bisect these arcs at right angles by other arcs of great circles, DC and EC, intersecting at C. Then clearly the arcs AC, A'C are equal, and also the arcs BC, B'C. But since the particles of a rigid body remain at the same distance apart, the arcs AB and A'B' are equal. Hence the spherical triangles ACB and A'CB' are congruent. Thus the portion of the spherical surface which originally occupied the position ACB, occupies the position A'CB' after the displacement. The point C of the body occupies its surface of this sphere.
the point
A
.
§
KINEMATICS OF A RIGID BODY
87]
127
and the same is therefore true of all points on OC. The displacement is therefore equivalent to a rotation about the axis OC through an angle original position,
the line
ACA' = BCB'. Suppose now that the displacement
is one which takes place owing to a continuous motion of the body, the circular measure of the small angle AC A' of
in a short interval 5t
and that S9
is
rotation about interval
is -j-
The average angular
OC.
radians per unit time about OC.
w
to zero, the limiting value
OC
If
now
dt
tends
of this average angular velocity
called the instantaneous angular velocity, of
velocity during the
and the
is
limiting position
In the case of a body
the instantaneous axis of rotation.
turning about a fixed point 0, the instantaneous angular velocity
may be completely and whose direction
by a vector A, whose module
specified is
parallel to the instantaneous axis
is
w,
and
in
the positive sense relative to the rotation.
The instantaneous velocity now be found as in Art. 41.
any
of
particle of the
body may
the position vector of the
If r is
particle relative to 0, its instantaneous velocity is perpendicular
to the plane of r
The
and A.
in a circle whose centre
particle
instantaneous axis (Fig. 39)
;
and
w. PN=u,r Hence the instantaneous velocity
for the
moment
N of the particle on the
speed
its
sin
moving
is
the projection
is
is
therefore
PON.
of the particle
is
v = A>
A If
F
rigid body is turning about a fixed point
of the
cm.
sum
of the external forces, relative to 0, show that the force
is the vector
given by
with angular velocity A.
and
R
r the position vector
of constraint at
^a
F + R = M-r-xr + .MAx(Axr). at
The velocity
of the
cm. _
a=
A*r, and
is
d dt
..
its
rfA -
_.
{A * T) = .
.
I+Ax
-dt*
dk .. -. = ^-*r + A*(A*r), at
acceleration therefore
dr
Tt
is
;
VECTOR ANALYSIS
128 since
-=-
is
the velocity of the
cm.
the forces on the body to Ma,
(cf.
[ch. VII.
Equating the vector sum of Art. 83)
we have the
all
required
result.
motion of a
88. General
point of the body
is
rigid body.
Suppose now that no
Then the most general displacement
fixed.
body is equivalent to a translation in which all particles have the same displacement, together with a rotation about some definite axis. For the body may be translated without into its final position rotation so as to bring any one point and then the whole body may be brought into its final position by rotation about some axis through 0. of the
Consider the small displacement that takes place in a short interval St owing to the finite velocity of the body.
This
is
equivalent to a translation ds of every particle, such that a certain point
is
brought into
its final position,
together with a A
rotation 68 about an axis through
parallel to
some unit vector
<5s
Then
-^ represents the average velocity of
terval,
and
68* -^-
a
during the
a.
in-
the average angular velocity of the body ds
about 0.
Taking limiting values as St^-0, we have v = -*- for the
instantaneous velocity of 0, and
taneous angular velocity about 0.
(68 *\
A=Lt(^-a)
.
for the instan-
During the interval
St the
displacement of the point P, whose position vector is r relative A to 0, is Ss + S8a,*T, the first term being due to the translation,
and the second to the rotation
SO.
The average (5s
point during the interval values
we have
is
68 A
therefore ^- +-j-
a>
for the instantaneous velocity of
velocity of this
Taking limiting
P
V=v+A*r The of
first
term
P relative
is
(10)
the velocity of 0, the second
is
the velocity
to 0.
Consider another point 0' whose position vector relative to is s,
and whose velocity
v' is therefore
v'=v+A><s.
given by
KINEMATICS OF A RIGID BODY
§88]
The position vector
P
of
relative to 0'
P given by
the velocity of
(10)
may
is r'
129
=r -s
(Fig. 55),
and
be expressed,
V = (v+A*s)+A*(r-s) = v' This
P
same form as
of the
is
+A*r'. (10),
showing that the velocity of
A of the body The value A of the angular velocity is thus the same for all origins, and is a property of the body as a whole. The formula (10) will be found very useful for writing down the velocity of any point of the body. relative to 0' is that 0'
about
.
If
'.
the point 0'
of the
due to an angular velocity
is
such that
V
angular velocity,
Tnatls
is
its
velocity
parallel to A,
is
parallel to the axis
and therefore
A>=v'
= 0.
A*(v+A*s)=0,
A*v+A'sA-A2 s=0, from which
follows that
it
A*v
+wA,
where,
by
substitution,
the point A> v/A 2
The
any point on
parallel to
motion
A.
of the
found that the value
it is
of 0' is a straight line parallel to
Thus the locus of
.(11)
this
of
is
arbitrary.
velocity line
AA
is
The instantaneous body is therefore
equivalent
to
about this
line,
a
motion
screw
which
called
is
Axv
Every of point on the axis is moving along the axis, while the body is turning round it with an angular the axis
u
A passing through
the
screw.
2
A
velocity A. Fig. 57.
From
(10)
V-A=vA, and all
it
follows
this expression has therefore the
points of the body.
and
will
that
It
is
be denoted by T.
same value
called an invariant of the
This invariant property
that the resolute of the velocity in the direction of for all points.
We have already seen that A
being the same for
all origins.
2
A is
is itself
is
for
motion simply
the same
an invariant,
VECTOR ANALYSIS
130
The
motion
pitch of the screw
the advance along the axis Since, for a point 0' on the
is
per radian of rotation of the body. axis of the screw, v' has the
given
The
same
[OH. VII.
direction as A, the pitch
by
A
and
is
positive
the screw
is
right-handed.
that
;
screw
is
is, if
is
T
. „
pitch, as thus defined,
sense
p
if
It
v' is
have the same negative
if
the
left-handed.
A body is said to possess several when the velocity of each particle is the
89. Simultaneous motions.
motions simultaneously, vector
sum
separately.
of the velocities it If,
would have due to each motion
for instance, the body possesses simultaneous
A
angular velocities
1
and
A
2
about a fixed point 0, the particle
whose position vector relative to
is r
has a velocity
A *r+A 2 *r = (A +A 2 )xr. 1
1
But
due to a single angular Hence simultaneous angular velocities
this is the velocity of the particle
velocity
A x +A
2
about 0.
about a fixed point are compounded by the law of vector addition. The argument is clearly true for any number of angular velocities.
Suppose that the body has simultaneous angular
A =
a> 2
about parallel axes through the points
Si
A = wjH
and
1
2
Then the
velocity of the particle at r
velocities
due to each
;
that
velocities
A
A
is
a
and
S 2 respectively.
the vector
sum
of the
is
wA + "*« + A 2 *(r -s 2 ) = (A, + A 2 )*(r ),
A,-(r - Sl )
showing that the motion
is
equivalent to an angular velocity
Aj + A 2 about a parallel axis through the point (co^!
which divides the
+
ft>
2 S 2 )/(ft>l
line joining Si
however, the angular velocities
sum
A
+ w 2 ),
and s 2 in the ratio &> 2 m v If, 1 and A 2 are equal and opposite, :
and the velocity of the particle at r is simply is the same for all particles. Hence two simultaneous equal and opposite angular velocities about parallel their
A 2 x(s 1 -s
is
2 ).
zero,
This
axes are equivalent to a velocity of translation of the body as a whole equal to
A2
><(s 1
-s 2 ).
This
is
perpendicular to the plane
KINEMATICS OF A RIGID BODY
§ § 89, 90]
containing both axes, and of module pw, where cular distance
between the axes and
p
131
is
the perpendi-
the angular speed about
to
These results are analogous to those for the resultant
either.
of a pair of parallel forces.
Any two
simultaneous motions of a rigid body
compounded. Vj
and v 2
A1
A2
and
vector relative to
combined respectively with angular The particle whose position
+A
about 0.
is r
has a velocity equal to the vector
due to each
of the velocities (v 1
1
that
;
xr)+(v 2 + A 2 *r)
+ v2
of the point
about
it.
The
sum
is
= (v 1 +v 2 )+(A 1
showing that the resultant motion Vj
similarly
Let the two motions be equivalent to velocities
of the point
velocities
may be
+A
2 )*r,
equivalent to a velocity
is
combined with an angular velocity Aj + A 8
first of
the two simultaneous motions v 1;
equivalent to a screw whose invariants are
and whose pitch p 1 is their ratio. Similarly invariants of the combined motion are
Ax
is
r i =V1 'A 1 and A^, The for T 2 and p 2 .
r = (v x + v 2 )«(A 1 + A 2 = I\ + T 2 + Vj-Aa + v 2 «A, )
(A x + A 2 ) 2
and and
its
pitch
p
is
of a Rigid Body.
Angular momentum.
moving about a fixed point 0. of
Consider Its
position vector of the particle is
the case of a body
first
motion at any instant consists
an angular velocity A about this point.
particle
,
the ratio of these invariants.
Dynamics 90.
= A x 2 + A 2 2 + 2A 1 -A 2
If r is
the instantaneous
m relative to 0, the velocity of this
A*r, and the a.m. of the body about
is
H = 2r*mv = Zr<m(A*r) = 2mr 2A-Smr-Ar
(12)
Eeferred to rectangular coordinate axes through the point the directions of
i, j,
k, let
the vectors
r,
A
in
have the values
i=xi+yj +zk, (o 2 3 + ^sk,
A = fc^i +
so that &>!, 6D 2 w 3 are the angular velocities (or speeds) of the body about the coordinate axes, and x, y, z the coordinates of ,
VECTOR ANALYSIS
132
m
the particle of
A
r,
in (12),
Substituting these values
referred to these axes.
we
[CH. VII.
find
K = 1m{(y
2
+z 2 )w 1 -xyco 2 -xzw 3}i
+ Sw{(z 2 + a5 2 )o) 2 - yz«>3 - yxwi}j + 2,m{(x 2 +y 2 )w 3 - zxm x - zyw 2}k. 2
2 The values of the sums I,m{y 2 +z 2 ), ?,m(z 2 +x ), 2m(a; +y ) and Hmyz, ~Lmzx, "Zmxy, which are denoted by A, B, C and D, E, F respectively, depend on the distribution of mass relative to the coordinate axes, and are constant only if these axes are The quantities determined by A, B, C are fixed in the body. called the moments of inertia of the body about the axes of x, y, z 2
respectively.
F
Those determined by D, E,
are the products of
and x, x and y respectivelyEach moment of inertia involves only one axis. For A = Y.mp 2 where p is the perpendicular distance of the particle m from the inertia relative to the axes of y
and
z,
z
,
a;-axis.
In terms of these moments and products of inertia the a.m. of the
body about
may
be written
H=h i+h£ +h k, li
x
h2 Ji
=Aw
(13)
3
1
where
1
-
Fw 2 - Ew z
,~\
= Boo 2 -D«o 3 -Fcc v
3 =(7ft) 3
(14)
Y
Dw 2 .'
— Ewi —
The quantities h 1; h 2 hs are the angular momenta ,
of the
body
about the coordinate axes.
Suppose next that no point of
the
body
the position vector of the r', v'
cm. and v=r
be the position vector and velocity
to the
cm.
is fixed,
With
require the a.m. about a given point 0.
and that we
as origin let r be
Also
its velocity.
of the particle
Then, by Art. 84, the a.m. of the body about
H=r*Mv + 2r'xmv' The second term
is
the a.m. of the
by
(13)
if
is
(15)
body about the cm.,
as though that point were at rest.
let
m relative
Its value
we use moments and products cm.
is
calculated
therefore given
of inertia relative to
axes through the
91. Principal axes ol inertia.
each point of the rigid body there set of
We is
shall
one,
and
now show
that for
in general only one,
mutually perpendicular axes relative to which the products
DYNAMICS OF A RIGID BODY
§ 91]
of inertia vanish.
133
These are called the principal axes of inertia
at that point.
Let
be the given point of the body, and let us endeavour such that an angular velocity A about gives an a.m. about in the same direction as A. For such
to find an axis through it
H
an axis we must have
Aw x - Fw 2 - Ew s _ Bw 2 - Dw 3 - Fw! — _ Cw 3 -Etox Dw 2 »1
«>2
-
—A
,
.
(say),
<»3
clearing fractions,
or,
(A -X)^- Fw - Ecc =0,'\ FU)1 +{B -X)u>i- Do>3=0,[ - EMl - Doo 2 + (C - X) w 3 =0 J 2
3
-
Eliminating
w2
co 1;
we have
co 3 ,
,
A-X
-F -E B-X -D -D C-X
-F -E This
is
With
=0.
a cubic equation in X, and therefore has one real rootXj.
X any two
this value of
w1
ratios
(i)
:
o> 2
:
(i)
Let this be chosen as the
the required property.
an angular velocity
Hence by
of the equations
determine the
giving the direction of an axis which possesses
co 3 ,
u> 1
about
it
gives an a.m. in the
x-axis.
same
Then
direction.
(14)
0=h 2 = -Fw 1 and Q=h 3 = -Ew v showing that
E = F = 0.
Thus both the products
of inertia involv-
ing the a;-axis vanish.
Try now to
an axis perpendicular to
find
possessing the required property that for
H
this one,
an angular velocity about such an axis w 1 = the condition for parallelism of A and H
E=F =
Bw 2 - Dw 3 = Cw 3 - Du> 2
.
(B -/j,)co 2 -Dw 3 =0,\ - D<0 2 + (C - fi) w 3 = 0. J
Hence Eliminating » 2
:
w 3 we ,
2
fi
With
,
;
is
w3
w2
which
and also Then and since
parallel to A.
is
is
/x
find
-fi(B+C)+(BC-D*)=0,
a quadratic in
a value of
(")
pi
whose roots are
easily
shown
to be real.
equal to one of these roots either equation
(ii)
VECTOR ANALYSIS
134
determines the ratio w 2
co 3 ,
:
[OH. VII.
which gives an axis having the
Let this be taken as the axis of y. Then an angular velocity w 2 round it produces an a.m. in the same direction. Hence D = 0. Since then all the products of inertia vanish, the z-axis is also a principal axis. Thus there are three required property.
mutually perpendicular principal axes at the given point 0. There in
X
is
is
in general only one such set
for the original equation
;
a cubic, showing that for each point there are only three
axes possessing the required property.
The moments
of inertia
about the principal axes are called the principal moments of inertia at that point.
For the motion
of a
body about a
fixed point 0,
if i, j,
k have
the directions of the principal axes at that point, and A, B, are the principal
moments
of inertia, the
C
formula (13) for the
becomes simply
a.m. about
H=Aw The
92. Kinetic Energy. is
1i
+ Bw 2j +Cw 3k
(16)
kinetic energy of a system of particles
defined as the
sum
of the kinetic energies
of the separate particles.
Consider a rigid body moving about a fixed with an angular velocity A whose
point
module
is
w.
the particle
If r is
m
the position vector of
relative to 0, its velocity
A*r, and the kinetic energy of the body
T = J2m(A*r) 2 If Fia
p
is
particle 5g
which
is
(17)
the perpendicular distance of the
m from
the instantaneous axis, and
/ the moment of inertia that axis, I
is
is
of the
= ~£mp 2 and the equation (17) may be T = \-ZmpW = lI w \ ,
body about
written (18)
analogous to the formula §mv 2 for the kinetic energy
of
a particle. It is also
worth noticing that if H is the a.m. of the body about the
fixed point 0,
|A-H = iA-2mr*(A*r)
= |2m(A*r)-(A*r) = |2mv 2 = T, a simple relation between the a.m. and the kinetic energy.
(19)
DYNAMICS OF A RIGID BODY
§§ 92, 93]
Introducing
through
again
same
the
as in Art. 90, let
I,
rectangular
135 axes
coordinate
m, n be the direction cosines of
the instantaneous axis of rotation, so that
A = («(Zi+mj +wk). Then the expression
may be
body
(17) for the kinetic energy of the
written
r = |Sm{A 2r 2 -(AT) 2} = %I,m{(l 2 +m 2 +n 2 )(x 2 +y 2 +z 2 - (lx + my + nz) 2} co 2 = 1{A1 2 + Bm 2 + Cn 2 - 2Dmn - 2Enl - 2Flm}w 2 )
-
Comparing
this
I
2
with the value \Iw found above, we have
=Al 2 + Bm 2 + Cn 2 - 2Dmn - 2Enl - 2Flm,
a formula which gives the value of the
any axis through a point 0,
moment
of inertia
about
in terms of its direction cosines relative
to a set of rectangular axes through 0,
and the moments and
products of inertia relative to these axes.
Suppose now that the body has no point velocity of its
cm., and
relative to the
cm.
The
and the kinetic energy
Let v be the
fixed.
the position vector of the particle
r'
velocity of this particle
of the
body
is
m
then v + A>
is
T = p>ra(v+A*r') 2 = ^Smv 2 + J2m(A*r') 2 +v«AxSmr'. The
last
term vanishes because 2mr'=0.
T = iMv 2 + |Smv' 2 showing that
Also
A>
m relative to the cm.
to the velocity v' of the particle
(20)
,
can be expressed as the
the kinetic energy
equal
Thus
sum of
two parts, one of which represents the kinetic energy of translation of the body as a whole with a velocity equal to that of the cm., while the other represents the kinetic energy
of rotation about the
cm.
regarded as a fixed point. 93. Principle of Energy.
It
was shown
in Art. 75 that the
rate of increase of the kinetic energy of a particle activity of the resultant force for all the particles of the
the kinetic energy of the of all the forces
on
on the
body,
we have
body equal
all its
particle.
equal to the
the rate of increase of
sum But the
to the
particles.
is
Hence, summing of the activities
internal action
VECTOR ANALYSIS
136
[OH. VII.
between two particles consists of a pair of equal and opposite
And
along the line joining them.
forces
between two particles of a
two particles Hence the activity representing their mutual action is zero. velocities of the
joining them.
line
the distance
since
body remains unaltered, the have equal resolutes along the
rigid
of
the pair of forces
The same
every pair of forces in the internal action.
We
is
true for
need therefore
only consider the activity of the external forces on the body
and
it
follows
from the above that
energy of a rigid body
Hence the of
time
is
;
the rate of increase of the kinetic
equal to the activity of the external forces.
any finite interval work done on the body by external
increase in the kinetic energy during
equal to the total
is
forces during that interval.
Moving axes or frame of reference. We have spoken of whose length and direction remain unchanged. But its direction must be expressed relative to some frame of reference and this frame will be in motion relative to many other frames that might be used for reference. Thus 94.
a constant vector as one
;
directions that are constant relative to the former are variable
to
relative
For practical
the latter.
frame, and both think and speak of
We know,
however, that
it is
it
purposes we
terrestrial
choose the earth and bodies rigidly attached to
as our standard
it
as a fixed frame of reference.
moving
relatively to similar frames
belonging to the sun and other heavenly bodies.
St and S 2
Consider two frames of reference,
which we may think of as fixed, while the latter to
It will
it.
be
sufficient to consider a
with angular velocity is
A
about a point
common fixed point for both
thus a
is
fixed in
frames.
view
of the
frame
changes from
OP
to the frame
Sv
frame
of
Sv
Sv
If in
relatively
Let
r
Draw
of
S2
This point
be a variable
some moving
required to determine the relation between
change relative to the two frames. of
moving
motion of rotation
vector, say the position vector relative to It
is
the former of
,
its
point.
rates of
a figure from the point
a short interval
dt
the vector
r
OR, then PR is the increment (<5r)j relative Now, during this interval the point of the
to
S 2 that was initially at P has moved to Q, where PQ = &A*r,
so that
QR
is
the increment (dt) 2 of r relative to the frame
S2
MOVING AXES
§§94,95]
—^
"Rnf
—
—^
137
?-
PR=PQ+QR,
and therefore Dividing by interval
;
(dr^
dt,
we
= dtA.*! + ((5r) 2
.
rind the average rate of change during that
and proceeding to the
we
limit as dt tends to zero,
obtain the required relation dv\
.
.(21)
\dt)
the suffix denoting the space relative
which the rate
to
of
change
is
con-
sidered.
A
In this formula
Sv one.
denotes the
S2
angular velocity of
to
relative
But the formula is a reciprocal °/ For we may regard »S 2 as
fixed, the
Sx
angular velocity of
lative to it being A'
= - A, and
the above formula gives
QAiX +A as
we should
reference
All motion
expect.
may
FIG. 59.
re-
'*
v>
relative.
is
Any frame
be regarded as fixed, and the motion
expressed relative to
of
of
any other
it.
In the equation
95.* Coriolis' theorem.
(21), first
take r as the
moving point P relative to 0. Then its relative to the two frames are connected by v x and v 2
position vector of a velocities
/dr\
(dt
'Hit
=
=
v 2 +A*r
(21) to the vector
Next apply the formula acceleration of
* + A*r
(22)
dt.
P relative to *S
X
v 1; and obtain the
as
dtJi
-(j
[v 2
+A*r]) + A*(v 2 +A*r) 2
t
dA\ dtJ z
But
dA\ dt)i
dk dt
'
dik + A*A
.
v+A
>
dA\
-('
it) 2
+ Axv 2 + Ax(A*r).
VECTOR ANALYSIS
138
and may therefore be written simply is
[CH. VII.
dh
The above value
-=-.
for a x
therefore
a1
which
is
dA
= a 2 + 2Axv 2 + ^->
a theorem due to Coriolis.
upon the motion
of
P
relative to
>S 2
the acceleration of a point fixed in
The formula
cident with P.
is
first
(23)
two terms depend
while the other two give
;
S 2 and
instantaneously coin-
a reciprocal one
and 2 may be interchanged, provided the sign at the same time. 1
;
for the suffixes of
A
changed
is
a rigid body have already seen that, if H is the a.m. of the body about 0, and L the torque of the external forces about the same point,
dynamical
96. Euler's
Consider
equations.
We
turning about a fixed point 0.
da
T
This equation was formed expressing the motion of the body
some frame $j independent
relative to
A=«) 1i+&> 2j+
about 0. It
Such a frame
k
by body at 0, which are taken
inertia of the
the (moving) frame parallel to
<S 2 .
is
specified
We now
these principal axes,
S2
vectors relative to the frame
by
3
make use
of a
moving
fixed in the body, for expressing the changes of
»S 2
motion.
is,
<*>
however, frequently convenient to
is,
frame
the body, and regarded
of
Relative to this frame the body has an angular velocity
as fixed.
-
its
the principal axes of as coordinate axes for
i, j, k as remaining and therefore constant unit The a.m. of the body about
consider
H = 4 Wli + 5«ji+Ca a k,
(16),
C
where A, B.
are the principal
moments
the torque of the external forces about
of inertia at
;
and
is
L=L i+i j+i k, 1
2
where
Lx L 2 L3
are their (scalar)
axes.
Applying
(21) to the vector
,
,
da\
3
moments about the H, we have
/da\
„
TT
principal
EULBR'S DYNAMICAL EQUATIONS
§ 96]
This vector equation
is
139
equivalent to Euler's scalar equations.
For on substituting the above values of L, H and A, remembering that A, B, G and i, j, k are constant relative to S 2 we fi Q(l an equation which is equivalent to the three scalar equations ,
A-tt -{B- C)a> 2(0 3 =L lt dt
B -j£ - (0 - A)w C —j-
— (A —
3 co 1
= L,
.(25)
B)co 1 w 2 = L
These are Euler's dynamical equations for the motion of a rigid
body about a fixed point, referred to axes fixed in the body, and coinciding with the principal axes of inertia at the fixed point.
Examples. (1)
Prove
that,
in the above problem, if
the body,
T
is the kinetic
energy of
ij,
^=A L '
-
On forming
the scalar product of both members term vanishes, showing that
last
•(f), 1
=
-c
du>i
,
_
with A, the
dtJi
doo„ 2
dt
of (24)
_
dt
duio 3
dt
^.H:
A-L^f)^®* dT - ld {AB) (A .m-Tt' -2dt by
(19).
torque (2)
L
If,
The expression A*L
therefore measures the activity of the
acting on the body. in the same problem, the kinetic energy is proportional to and L momentum, prove that the plane of
H
the square of the angular is
perpendicular
to that
of
H and A.
The kinetic energy T = kH?, where k is constant. The vector and H*A is normal to that is normal to the plane of H and L These are perpendicular if of H and A.
H*L
;
(H*L)-(H*A)
= 0.
.
VECTOR ANALYSIS
140
But the
first
member when expanded
[CH.
is
H L-A-H-AL-H = H2 ^-2T^-H at 2
at
= H-*
obviously zero.
is
Hence the
result.
EXERCISES ON CHAPTER
VII.
1. A homogeneous sphere rolls without slipping on a fixed rough plane under the action of forces whose resultant passes through the centre of the sphere. Show that the motion of the sphere is the same as if the plane were smooth, and all the forces were reduced to five-sevenths of their former value. 2. A cube is rotating with angular velocity w about a diagonal, when suddenly one of its edges which does not meet the diagonal becomes fixed. Show that the ensuing angular velocity about this
ed g e 3.
is
^ V$-
An
inelastic cube, sliding
down
a plane inclined at
a. to
horizontal, strikes symmetrically a small fixed nail with speed
the
V
tumbles over the nail and goes on sliding down the plane, show that the value of V 2 is not less than 16ga(\/2 - cos a. - sin cx.)/3, where 2a is the length of an edge of the cube. If it
4. A rod moves with its extremities on two intersecting lines. Find the direction' of motion of any point. If two lines do not intersect but are at right angles, examine whether the motion can be represented by an angular velocity only.
is
5. Prove that a straight line through the cm. G of a body, which a principal axis of inertia at G, is a principal axis at any point.
6. A rigid body, hinged at to a fixed point, is set rotating about a principal axis at 0. If it is acted on by no forces but those at the hinge, show that it will continue to rotate with constant angular velocity about the same axis.
If
hinge
the cm. of the body, show that there is zero action on the and hence that a principal axis at the cm. is an axis of
is ;
free rotation. 7. A fly-wheel of mass M, concentrated at the rim of radius a, rotating with angular velocity w about a fixed axis through its centre inclined at an angle 6 to the axis of the wheel. Show that the constraint due to the fixed axis of rotation is equivalent to a couple
is
EXERCISES ON CHAPTER
"VII.]
whose scalar moment
is
\Mu?a^
sin
d cos
141
VII.
and whose plane contains
6,
the axis of the wheel and the axis of rotation. 8.
Determine the screw motion which is equivalent to the two A and v', A' whose axes are given and show that its
screws v, invariant
;
T
is
+ y>(A + K)+AA M> >
{y
M
is the mutual moment of the two axes. Also prove that the axis of the resultant screw intersects at right angles the common
where
perpendicular to the axes of the two given screws. 9. Prove that the points, whose position vectors relative to a given origin are equal to the velocity vectors of the particles of a rigid body, all lie in a plane.
that any motion of a rigid body may be represented by velocities A and A', about axes one of which may be chosen arbitrarily. Also that the common perpendicular to the two axes intersects perpendicularly the axis of the resultant screw. (The axes of A, A' are called conjugate axes.) 10.
Show
two angular
11. If
one conjugate axis of an instantaneous motion
is
perpen-
dicular to the axis of the screw, the other meets this axis
:
and
conversely. 12.
A body possesses simultaneously twelve equal angular velocities
about axes forming the edges of a cube, those about parallel axes being in the same sense. Prove that the axis of the resultant motion is
a diagonal of the cube.
A
body possesses simultaneous angular velocities about the skew polygon taken in order, their magnitudes being proportional to the lengths of the corresponding sides. Show that every point of the body has the same velocity. 13.
sides of a
14. If four simultaneous angular velocities are equivalent to zero of any two is equal Also that the invariant of any three
motion in the body, show that the invariant to that of the other two. is zero.
15. The coordinates of a moving point P are x, y relative to rectangular axes traced on a plane lamina which is rotating about the origin, in its own plane, with variable angular velocity u>. Prove that the acceleration of P has resolutes, in the directions occupied
by the moving axes at any
instant, given
x-yw- xu? - 2wy 16.
A
particle
B
own
is
and
y
by
+ xw- yu? + Iwx.
moving on a smooth plane curve which
is
plane with constant angular velocity u> about a If P, Q are the tangential and normal resolutes of fixed origin 0. the force on the particle, R the reaction of the curve, and (j> the rotating in its
1
VECTOR ANALYSIS
142 angle which of
motion
OB
may
makes with the tangent, show that the equations be put in the form p,
:>>-«4i: m (kv2 + 2o>v + s
[CH.
a>
2
r sin
= R+Q,
being the length of the arc measured from a fixed point on the
curve. 17. A circular wire is constrained to turn round a vertical tangent with a uniform angular velocity w. A smooth heavy bead, starting from the highest point without any velocity relative to the wire, descends under the action of gravity. Find the velocity and the
reaction in
A
any
position.
smooth
about its axis, with uniform angular velocity oo. Find the motion of a particle descending on it under the action of gravity. 18.
which
helical wire is constrained to turn
is vertical,
19. A heavy particle is moving in a smooth surface of revolution whose axis Oz is vertical, and vertex downward. If z, r, are the <j>
cylindrical coordinates of the particle, s the length of the meridian
from the vertex, R the reaction of the surface and the equations of motion in the form
arc
S
2
-
,
wr
ds
in-
-=- (tor 2 )
=
prove
= - m9 dz
Is) dz
2
,
dr\
u>
ds'
R-mg dr Is
,
= 0.
r at
20.
The position of a moving point
is
given in spherical polar
Find the resolutes of its acceleration in the radial direction, perpendicular to the meridian plane and in the meridian plane. coordinates
21.
r,
6,
.
Hence show that the equations of motion of a heavy particle, by a light inextensible string of length I, are
tied to a fixed point
m(W 2 + 10
sin 2 00 2 )
- 1 cos 6 sin
^|jsin 6 dt
B
2
=T -mg cos 9, = - g sin
9,
Ti)=O. 9
sin
22. A solid cubical body is in motion, under no external forces, about a fixed corner. If w v w 2 u> 3 are the angular velocities about the three edges meeting at the fixed point, prove that coj + aia + cua and &) 1 2 + co 2 2 + co 3 2 are both constant. ,
EXERCISES ON CHAPTER
VII.]
A
body turning about a
143
VII.
is acted on by forces about an axis perpendicular to A. Show that the angular velocity cannot be uniform unless two of the
23.
fixed point
which, tend to produce rotation
moments
principal
A
of inertia at the fixed points are equal.
acted on by a force P per unit mass, varying relative to a fixed origin 0, r is the position vector of the point P where the density is /m, v the velocity of the cm., and H the a.m. about 0, show that 24.
rigid
body
is
from point to point.
If,
^= where dv
is
J
M r*F
the element of volume at P.
A
symmetrical body, such as a top, is rotating about a fixed point on the axis of symmetry and the external forces have zero moment about that axis (L 3 = 0). The principal moments of inertia at are A, A, C. Show from Euler's dynamical equations that w 3 remains constant. Then, by considering the velocity v and the acceleration a of the point P on the axis of symmetry whose 25.
;
position vector is
k
relative to 0,
Aa, = L 2i
Hence prove that
P
show that
- Zj + Cft) 3k> 22 )k.
moves
as a particle of
mass
A
smoothly con-
strained to the surface of a unit sphere (centre 0) and acted on by tangential forces CVo 3k*v and P, the latter having the same torque
about
as the external forces.
26. If, in the previous exercise, the only external force is the weight of the body, and the cm. is at a distance h from 0, show that for a steady precession of the axis at an angle 9 to the vertical, the angular velocity Q of precession is given by
Q = (Cw s - VCW 3 - lAMgh cos where
M
is
the mass of the body.
0)/2A cos
6,
[CH. VIII.
CHAPTER
VIII.
STATICS OF A EIGID BODY. The equations
97. Conditions of equilibrium.
of
motion
for
a rigid body, as found in the previous chapter, are equivalent to
f -2F, at
§-** where the
M
cm.
is
momentum of the ZF the vector sum
the linear
(Art. 85),
and ZrxF the torque
of these forces
body,
H
about the cm.
remains at rest under the action of the forces,
permanently
its a.m.
about
of the external forces,
M
If the
and
body
H
are
Thus
zero.
2F=0
2r*F =
and
(1)
are necessary conditions of equilibrium for the body.
But they are
also sufficient conditions, provided the
initially at rest.
For;
they are
if
and therefore remain permanently
H = Awj + £
and
the principal axes at the v,
« l5
ft>
2>
w 3 must
all
fore in equuibrium.
cm.
zero.
23
body
is
M and H are constant,
satisfied,
But
+ Cco 3k, Hence
being chosen for reference.
remain equal to
The conditions
zero, (1)
and the body
is
there-
are therefore sufficient
conditions of equilibrium for the body.
Suppose we take moments about any other point position vector relative to the
external forces about
P
cm.
is r'.
is
E(r-r')xF=SrxF-r' x 2F, 144
P
Then the torque
whose of the
which is
is
BODY
STATICS OF A RIGID
§ § 97, 98]
zero in virtue of
satisfied for
any
is
Thus the condition of zero torque if SF=0 and the torque
(1).
Conversely,
point.
about any one point
145
zero, it is zero
about
all
points.
conditions (1) therefore, the point chosen as origin of
In the
moments
may
be any whatever. The conditions of equilibrium just found are equivalent to six scalar conditions. For if, choosing rectangular axes through the origin of moments, we write
F=Xi + Fj+Zk, v
the equations (1)
= xi+yj
+zk,
become 2(Xi + Yj+Zk)=0,
l{yZ -zY)i+2(zX -xZ)] + I,(xY -yX)k=0,
and
which are equivalent to the six scalar conditions
SX=0,
2Y=0,
2Z=0,1
Z(yZ-zY)=0, (2)
S(zX-zZ)=0, •2(xY-yX)=0, that
is,
the
sum
of the resolved parts of the forces
for each of the coordinate axes,
forces
and the moment
must vanish
of the external
about each of them must also vanish.
no need to choose rectangular axes as we have done, And, with regard to moments about a line, it is worth noticing that if 2F = the moment is the same for all parallel lines. For the There
is
but these are generally more convenient than oblique.
A
moment about
a line parallel to
whose position vector
a,
drawn through the point
P
is r', is
a-2(r - r')*F
= a«Sr*F - aV*2F
= a«Sr>F, which
is
independent of A
r',
and therefore the same
for all axes
parallel to a.
98. Equivalent
systems of forces.
equilibrium found in the previous Art.
From it
the conditions
of forces are statically equivalent in their
of
two systems action on a rigid body
follows that
VECTOR ANALYSIS
146 their vector
if
For
point.
if
sums
are equal
[CH. VIII.
about any and then both
also their torques
and
one of the systems
is
reversed,
act together on a rigid body, the combined system of forces
the conditions
satisfies
and the body
(1),
determined by
vector
its
sum SF, and
Thus
in equilibrium.
is
body
the statical effect of a system of forces on a
is
completely
torque Sr>
its
specified point.
The point
of application of a force
may
to any other point in its line of action either of the
above quantities.
;
This
therefore be shifted
for this does not alter is
the principle of the
transmissibility of forces.
Further, the effect of a system of forces on a rigid body unaltered
by introducing a
pair of equal
For the vector sum
the same line of action.
and
zero,
also its torque
99. Parallel
forces
parallel r 1; r 2
,
.
.
which
.
,
p 2a,
.
r„ respectively,
.
.
,
p n&
acting
of such a pair
Given a system
If there
is
separate forces, that is to (Pi+p 2 + + Pn)&must have the same torque about the origin. through the point r we must have -
r*
Hence where the value of
t
is
-
(Sy)a
= Srx^a,
r*a
= ^r— x a. 2p
f
of
through the points such a force
must, by the preceding Art., be equal to the vector
or
is
required to find a single force
it is
their statical equivalent.
is
is
forces with
about any point.
Centre of gravity.
forces. p^a,,
and opposite
And
sum
further,
then
If
it
of the
it
it
acts
Ear
= ^J- + ta, 2p Thus the
arbitrary.
single force,
which
equivalent to the system of parallel forces, acts through the point J.pvl'Zp, its line of action being parallel to the others, is
and is
magnitude equal to the sum
of their magnitudes. This independent of the direction of the parallel forces, and the centroid of the points r^ r 2 i n with associated numbers its
point
Pi,
is
,
1*2,
•
,
.
.
.
,
Vn-
m m
m
In particular, if there are particles of masses X) 2 , ... respectively at the above points, their weights constitute a system of parallel forces which may be represented by &• a, a, ...
m
1
m
2
m
and the
BODY
STATICS OF A RIGID
§§99,100]
single force
which
is
147
equivalent to these has a line of
action passing through the point
r
whatever the direction of
= Emr
2m
This point
a.
and
gravity of the system of particles,
called the centre of
is is
identical with their
centre of mass, as defined in Art. 11.
sum Epa
of the parallel forces is zero, "Lp vanishes.
If "Lpi is also zero
they have zero torque about the origin and
If
the
But
are in equilibrium.
not, their resultant
if
acting at an infinite distance from the origin.
a
of particles
cm. given by
Sm
a zero force
This case will
Here we observe that
be considered in the following Art.
system
is
for a
cannot vanish, so that they always have
the above formula.
Law
100. Couples.
Consider
composition.
of
parallel forces, equal in
a
pair
of
magnitude but opposite in direction and
Such a pair of forces and the plane containing the
localised in different lines. is
called a couple,
two
lines
of
action
The vector sum is
is
the plane of the couple.
of the forces is zero,
no single force which
is
and there
statically equivalent
to the couple.
Let
F and -F
be the two
forces,
action passing through the points position vectors relative to r 2 respectively.
about
with lines of
P and Q,
an origin
Then the torque
of the couple
is
r 1 xF
+ r2 x(-F)=(r1 -r2
x )
F.
But ix - r2 is the vector QP, and the torque is fore independent of the point 0.
It
the
scalar
moment,
is
thereFig. 60.
may therefore
be called simply the torque of the couple. plane of the couple
pendicular to the called
whose and
are i x
Its direction ;
and
its
is
per-
magnitude,
measured by Fp, where p
is
the perpendicular distance between the lines of action of the forces.
from Art. 98 that two couples are statically equivalent For SF is zero for both couples, they have the same torque.
It follows
if
;
VECTOR ANALYSIS
148
[CH. VIII.
and 2r*F is the same for each, since their torques are equal. Thus for two couples to be equivalent their planes must be each
parallel, for
The
torque.
is
perpendicular to the vector representing the
direction of the forces in the plane
but the scalar
moment Fp must be
and the sense
of the torque also the same.
of a couple acting
by
on a
body
rigid
is
immaterial
the same for both couples,
The
statical effect
thus completely determined
is
torque.
its
Suppose a body acted on by several couples whose torques are
Lv L 2
,
.
.
,
.
forces the vector
L„ respectively. Then for the whole system of sum is zero, and the torque about any point is
L = L 1 +L 2 + The whole system torque This
is
L
is
is
is
.
+L„.
therefore equivalent to a single couple whose
the vector
what
.
.
sum
of the torques of the separate couples.
meant by saying that
couples are compounded
by vector addition of their torques. 101. Poinsot's reduction of a system of forces.
be acted on by forces
P1 F2 ,
,
.
.
.
Fn
,
Let the body
with lines of action passing
through the points whose position vectors relative to an origin are r 1; r2 ,
.
.
.
,
The
i n respectively. is
statical effect of the
unaltered by introducing at
and opposite forces
F1; F 2
,
forces .
.
.
±F
F„ at
,
1;
system
pairs of equal
±F 2 ..., +F„. have a resultant
The
,
R = 2F through that point
;
(1)
and the remaining
forces
on
the body constitute couples whose torques are r i xF i> r 2 xF 2;
,
r„^F n respectively.
These are
equivalent to a single couple whose torque
G=2r*F Thus the .
a single force
.
sum torque forces
is
original
R
system of forces
is
(2) is
equivalent to
through 0, equal to the vector with a couple whose
of the forces, together
equal to the vector
sum
of the torques of the separate
about 0.
The force R is the same for all origins but not so the couple G. For if we take as origin a point 0' whose position vector relative ;
to
BODY
STATICS OF A RIGID
§ 101]
(Fig. 55),
is s
through
we
0', together
149
find the system equivalent to a force
with a couple whose torque
R
is
G'=S(r-s)-F = Er«F-sxSF
=G-s*R This relation
R
at 0'
is,
(3)
of course, obvious
from the fact that a force together with a
equivalent to an equal force at
is
couple of torque s*R. Since
R
For by
We
is
the same for
all
,
.
2
is
an invariant of the R and G.
the scalar product of
.
.
-|
denote this second invariant by T.
fact that the scalar
about
is
R G =R G _ [RsR = R G
(3)
shall
R
all origins,
Another invariant
system.
moment
of the
is
the same
These invariants are analogous
R.
lines parallel to
It expresses the
system of forces
to those of Art. 88.
We to R.
naturally enquire If 0' is
if
there
a point for which
is
G
is
parallel
such a point, the vector product RxG' must vanish,
and therefore
R*(G-s*R)=0, 2 s + R'sR=0,
R *G-R from which
it
follows that
R-G
_,
where, by substitution, it is found that t is arbitrary. the locus of points 0' possessing the required property straight line through
A
corresponds to
R
R*G/R
and v to
central axis of the system
2
parallel to
R
and the system
;
with a couple whose plane
is
and couple constitute what
is
of the force is the axis of the
Fig. 57,
This straight line
G).
been proved equivalent to a force
(cf.
is
Thus is
the
where
called the
of forces has thus
R along the central axis, together it.
Such a force
The
line of action
The torque
of the couple
perpendicular to called a wrench.
wrench.
same for all points on the central axis. For if in (3) It is called s is increased by tR the value of G' is unaltered. The pitch p of the wrench is the ratio of the principal torque. the parallel vectors G' and R. Thus is
clearly the
G'
P ~R
G'-R
r
R
R
2
2
'
VECTOR ANALYSIS
150
and The
[CH. VIII.
two invariants of the system. positive if G' and R have the same
therefore the ratio of the
is
pitch, as thus denned, is
direction,
wrench
i.e.
the wrench
if
couple only,
a single force
R R
is
right-handed
;
negative
if
the
the system of forces reduces to a If it reduces to is zero and the pitch is infinite. at 0', the couple G' is zero and the pitch vanishes.
left-handed.
is
If
Examples. (1)
Forces of magnitudes
edges of
mo, nc act along three non-intersecting Prove b, c respectively.
la,
a parallelepiped whose
lengths are a,
T of the system is (mn + nl + lm)V, where V is the volume of the figure. Let a, b, C be the vectors determined by the edges of the paralleleand piped, all directed from one corner, which we take as origin let forces la, mb, nc act respectively through the origin and the that the invariant
;
points e and (a
Then the vector sum
+ b).
R= and the torque about the
+ mb + nc,
la.
origin
of the forces is
is
G = c*»ib + (a + b)*nc. Hence the invariant
R*G = (la, + mb + wc) - (wa*c + wb*c + mc>b)
= (lm + mn + nl) [abc], which proves the result. (2)
Two
forces
F 1 and F 2
Prove
act along non-intersecting lines.
common perpendicular to the two ratio F 2 *(F 1 + F 2 F^Fx + Fa). Also principal couple is F 1 F 2 M/ |F X + F 2
that their central axis intersects the lines,
and divides
that the scalar
where
M
is the
Let PP' is
it
in
the
)
moment of the mutual moment of the
(Fig. 42)
:
|
lines of action of the tivo forces.
common perpendicular. The central axis sum F! + F 2 = R say, and is therefore at right
be the
parallel to the vector
angles to PP', since each force is so. Further, each of these forces has zero moment about PP' and therefore the equivalent wrench ;
G has
moment.
central axis is perpendicular to PP' has zero resolved part along it. Hence the central axis cuts PP', for otherwise the moment of the force R about this line would not vanish. Take the point of intersection of these lines as origin, and let
R,
so that
zero
But the
;
G
OP = mk and OP' = - nk, k being perpendicular to Fj and the torque of these forces about is mkxFj - nk»F 2
,
F2
.
Then
F
and since
BODY
STATICS OF A RIGID
§§ 101, 102]
151
this is parallel to R,
{kx^Fj Expanding
by Art.
this
-nF^(F
44,
we
1
+ F 2 ) =0.
find
»F 2' (Fj + F 2 )k -
>riF 1 '(W
+
1
which gives the first result. Finally equate the values of the invariant
Then
2 )k
= 0, and
for the origins
R-G = (m + w)k=
P'.
)
= (m + »)kxF 1 F 2 = MF X F 2 ,
,
where
M
is
the mutual
moment
are parallel,
^^ 1*^ G--
which
is
Hence, since
of the lines.
R and G
2
the required result.
Two
wrenches (F lf pj? t ) and (F 2 , pj? 2 ) acting on a body have moment is M, and whose shortest distance apart is 2h. Show that the central axis of their resultant wrench intersects perpendicularly the common perpendicular to their axes, at a distance from its middle point equal to (3)
axes whose mutual
2h*(F 1 *-F 2*)-MF 1 F 2 (p 1
-p 2
)
2£(F 1 + F 2 ) 2
may
be shown exactly as in the previous example that the common perpendicular PP' (Fig. 42) at right angles. With the point of intersection as origin, let It
central axis intersects the
OP = (h-x)k The torque about
and
of the original
OP'= -{h + x)k. system
is
G = {h- a^Fi - {h + x)k"F 2 +p 1F 1 +p^ 2 And
.
since this is parallel to the central axis
G*(F 1 + F 2 )=0, which gives
On
a;(F 1
+ F 2 2 k = MF 1 2 -F 22 )k-(j3 1 -y 2 )F 1 «F 2
.
)
forming the scalar product of both sides with 2Ak the result
follows.
102. Null plane at a point.
Let the system
of forces acting
on the body be equivalent to a force R O and a couple Then the scalar moment of the system about any of torque G. through
straight line through lines lie in the
perpendicular to
plane through
G
is
zero.
All such
perpendicular to G, which
is
.
VECTOR ANALYSIS
152
[CH. VIII.
called the null plane at the point 0, while
is
called the null
point of the plane. If
the point 0'
null line
Hence
that
;
it lies
in the null plane at 0, the line
lies
00'
Thus, if the null plane
in the null plane at 0'.
a
is
moment about
to say the system has zero
is
it.
at
passes through 0', the null plane at 0' passes through 0.
To to
find the equation of the null plane at the point
we
as origin,
relative
s,
have only to observe that the couple for that
P ointis
G'=G-s*R,
and the equation
through
of the plane
perpendicular to G'
s
is
(r-s)-(G-s*R) =
r«(G-s*R)=S'G.
or
the equation of the null plane at
This
is
in r
and
s
so that
;
the null plane at
if
null plane at r passes through
of one
may
symmetrical r,
the
system of forces acting on a body
two forces, of which the line of action
statically equivalent to
is
is
passes through
s.
Any
103. Conjugate forces.
It
s.
s
Such a pair
be chosen arbitrarily.
of forces are called
conjugate forces of the system.
Take a
straight line through
direction, say that of a.
to a force
R
through
components F 1
=#
Let the system and a couple G.
A 1
an arbitrary point
F 3 =R-F 1
a and
parallel to the plane of the couple G.
0=F that
2
-G
is
of forces
We
in
be equivalent
can resolve
in such a
any
way
R
that
into
F2
is
This only requires
= (R--F 1 a)-G,
F-y
= -^
—
a«G
Eeplace
R
by
F2
F x and F 2
these two forces
transform the couple
G
to consist of a force
The system
in the null plane at 0.
A at
parallel to a
We
can then
and another
then equivalent to Fj
A
and F 2
forces are uniquely
=R -i^a
determined
;
Thus,
G = r>F 2 where ,
in the null plane at 0.
and the
also determined, for the torque of
equal to G.
is
at 0.
- F 2 at
line of action of
this force r is
point on this line of action, referred to
about
Both
F2
is
must be
the position vector of any as origin.
STATICS OF A RIGID
§§ 103, 104]
BODY
153
The axis of F 2 may also be expressed as the line of intersection two planes. For since this axis lies in the null plane at r any point on it must satisfy the equation
of
=
r«G
But
it
that
on the null plane
also lies
is,
of
(1)
any point
a
in the plane
aa,
on the axis
of Fj
;
a
r-(G-aaxR)=aa'G.
By
we
subtraction
find that
r(R*a) =a«G is
(2)
a plane through the line of intersection of the other two.
the axis of
F2
of action of lines
is
the line of intersection of
the conjugate forces
(1)
F 1 and F 2
and
lines
are called conjugate
or conjugate axes.
104. Principle of Virtual
Work
with any finite motion consistent with the
started
// a
or Virtual Velocities.
system of bodies, in equilibrium under any set of forces,
its
Thus
The
(2).
is
supposed
connections of
work of the forces on the system is zero. forces on the system are such that, however the
parts, the initial rate of
Conversely, if the
system is set in motion, the initial rate of work of the forces then the system is in equilibrium
We
shall
under
prove the principle for a single rigid body.
any point
position vectors,
as origin of
is zero,
the forces.
and
let
Choose
the initial
motion of the body be equivalent to a velocity v of the particle
and an angular velocity A about it. is v+A*r. If F is the particle, the initial rate of working of F is at of
the particle at r
Then the
velocity
force acting at this
F*v+F»A*r. Considering
the forces on the body,
all
initial activity of
we have
for the total
the forces
v-2F + A-2r*F. of equilibrium of the
And, in virtue of the conditions
body, this
expression vanishes whatever the values of v and A.
Conversely,
body must be only (A
2F = 0.
if
this activity
in equilibrium.
zero for
is
all initial
motions, the
For, choosing one of translation
= 0), we must have v2F
zero for all values of v.
Similarly choosing one of rotation only about
Hence
we
VECTOR ANALYSIS
154
[CH. VIII.
A*2r*F zero for all values of A. Therefore 2r*F must also be zero, and the conditions of equilibrium are satisfied. Such velocities as we have imagined given to the body are and the work done by the forces owing called virtual velocities The above principle is to these velocities is called virtual work.
find
;
frequently stated differently. displacement of the
infinitesimal interval dt,
The
virtual
it.
is,
to the first order,
in the
owing to the virtual velocity imparted
SW of the forces during this displacement
work
to
Consider the infinitesimal virtual
body from the equilibrium position
vttt
SW=^-dt. at
And,
if
the body
is
in equilibrium, this vanishes not
on account
of
the convergence of dt to the limit zero, but because the coefficient
dW —j-
.
is zero,
generally calculated, not in terms of 88,
8<j>,
The value
representing the initial activity.
etc., of
St,
of
8W is
but of the small increments
the coordinates expressing the position of the body.
Equilibrium of Strings and Wires. 105.* String under any forces. string
and
Let
measured from a fixed point
ds the length of the
string to be acted
A
s
be the length
element PP' (Fig
on by a force
F
of the
up to a variable point P, 47).
Suppose the
per unit length, varying from
Then F 8s is the force on the element PP' due But if T and T + 8T are the values of the tensions at P and P' respectively, and t and t + 8t the unit tangents at those points, the element PP' is also acted on by the forces - Tt and (T + <3T)(t+^t). For equilibrium of the element the vector sum of these forces must be zero that is point to point.
to external action.
;
8Tt + T8t + 8T8t + Fds = 0. Dividing throughout by
we have
ds,
1T
and taking limiting values as
^t + T~ +F = as ds which, by Art 59,
is
8s->0,
7.
t
0,
equivalent to
dT
^-t + ?Vn + F = 0,
(1)
EQUILIBRIUM OF STRINGS AND WIRES
§§105,106] where k
i.e.
is
P
the curvature at
The
point.
force
F must
155
and n the unit normal at that t and n,
therefore be in the plane of
the plane of curvature.
If
we
write
F=F t+Fp, 1
we may replace the
single vector equation (1)
equations
7
m
ds
by the two
scalar
>
1 ']-
(2)
These are the equations of equilibrium for the string. 106.* Wire or thin rod under any forces. In the case of a wire or rod the stress across any section is not as a rule tangential, nor can
it
be represented by a single
force.
In general
consists
it
and a couple due to bending and twisting. Let S and L represent the force and couple respectively at the point P, acting on the end of the portion AP. Then the element PP' (Fig. 47), of length 8s, is acted on by forces -S at P and S + <5S at P', and by bending couples of torque - L and L + <5L and in addition there is a force F 8s and a couple of torque G ds due to external action on the element. For the equilibrium of the rigid element PP' the vector sum of the forces must be zero, and the torque about any point (say P) also zero. Hence the equations of a force
;
<5S+Fc5s=0 and
<5L
where
<5r
is
+ G.<5s + <5r*(S + <5S) = 0,
the vector PP'.
proceeding to the limit, in the
form
Dividing throughout by
we have r
jC
^+t*S + G=0 These may,
by
if
desired, be replaced
resolving
along
suitable
varying from point to point,
t,
n, b,
if
we put
and
o
f +F =°
got
8s,
the equations of equilibrium
by
six Cartesian equations,
The rectangular set most convenient. Thus,
axes. is
(2)
S-SJ+Sp+SJ,,
VECTOR ANALYSIS
156
and
so on,
and remember that di
db
=Kn
ds
we
[CH.
find in place of (1)
and
da.
.
dT
>
-Xn
= Xb - /ct,
Ts'
'
the six equations
(2)
.(3)
dS, ds
+ \S 2 + F 3 = 0,
dLi — kL 2 + G 1 =0,
and
ds
dL 2 ds
dL
(*)
+Xi 2 + S 2 + (? 3 = 0.
,
For a ^awe system
+ K L 1 -\L t - S 3 + G 2 = 0,
\=0=S 3 = F 3
and the axes
,
of the couples
L and G are in the directions of b, so that L L 2 G G 2 If then L 3 = L and G 3 = G the equations reduce to 1}
d
-§-«s 2+Fl dS,2
+kS 1
ds
dL ds
addition the system
If in
L
is
zero.
+ S9
is
2
all
vanish.
^ =0,
.(5)
+G = 0.
perfectly flexible, the
bending moment
so that, if there is no external S2 + G = normal resolute S 2 must vanish, making the stress
This makes
couple G, the
+F
lt
,
;
purely tangential, as found in the previous Art.
EXERCISES ON CHAPTER 1.
A
body
is
VIII.
under four forces acting along the Prove that the forces are proportional
in equilibrium
sides of a cyclic quadrilateral.
to the lengths of the opposite sides. 2.
If four forces
acting along the sides of a cyclic quadrilateral
are inversely proportional to the lengths of those sides,
show
that
EXEKCISES ON CHAPTER
VIII.]
157
VIII.
their resultant acts along the line joining the intersections of opposite sides.
Forces act along the sides of a quadrilateral proportional Show that, if the body = and q r are the ratios in which the diagonals divide each other. 3.
to p, q, r, s times the lengths of those sides. is in equilibrium, pr qs, and the ratios p q
:
:
Four forces act along the
4.
skew quadrilateral repre-
sides of a
by a.AB, b. BC, c CD, d DA respectively. Show that they cannot be in equilibrium. If a=b=c=d they are equivalent to a couple whose plane is parallel to the diagonals AC, BD. But if ac=bd they have a resultant whose line of action intersects the sented
.
.
diagonals. 5. Forces act along the sides of a skew polygon taken in order, proportional to the lengths of the sides along which they act. Show that they are equivalent to a couple. 6. Show that a body cannot be in equilibrium under six forces acting along the edges of a tetrahedron. 7.
Six equal forces act along the edges of a regular tetrahedron Prove that in the directions AB, BC, CA, DA, DB, DC.
A BCD
their central axis is the perpendicular 8. If
D to the face ABC.
from
exercise the tetrahedron
in the previous
and the forces are proportional to the lengths
is
not regular,
of the edges along
which they act, their central axis is parallel to the line joining D to the centroid G of the face ABC and if
Forces act at the vertices of a tetrahedron outward, perpenand proportional to their areas. Prove that the body on which they act is in equilibrium. 9.
dicular to the opposite faces
Twelve equal forces act along the edges of a cube, the parallel having the same sense. Prove that their central axis is a diagonal. If, instead of forces, there are twelve equal couples whose planes are parallel to the faces of the cube, show that their central 10.
forces
axis is parallel to a diagonal. 11.
points
Show
that, for
on the central
12. Forces
the vectors
any system
of forces, the couple
G
least for
is
axis.
through the points
A XA{, A 2A 2 A
',
.
.
.
,
Av A2
A nA n
'
,
.
.
.
,
A„
are represented by
respectively.
If
G
A
is
the
centroid of the n points m and G' that of the n points m', show that the central axis of the forces is parallel to GG' If the lines of action of the forces intersect any plane perpendicular to the central ,
VECTOR ANALYSIS
158
B B
B
[ch.
axis in n show that the central axis meets this plane v 2 in the centroid of these points with associated numbers proportional ,
,
.
,
to the resolutes of the forces in the direction of the central axis.
middle points of the sides of a plane polygon, and perpendicularly to the sides. If they are proportional to the lengths of the sides, and act either all inward or all outward, prove that the body is in equilibrium. 13. Forces act at the
in the plane of the figure
14. Forces act at the centroids of the faces of a closed polyhedron, proportional to the areas of the faces. If they are normal to the faces, and either all inward or all outward, prove that the system is
in equilibrium. 15.
Extend the previous exercise to the case of a closed curved by increasing the number of faces indefinitely.
surface,
16. If four forces are in equilibrium,
any two
show that the invariant T
equal to that of the other two. invariant of any three is zero. of
is
17. If the origin is taken
on the axis
Also that the same
of the equivalent
wrench show
(pitch p), and k is the unit vector in the direction of the axis, that the null plane at the point p(Bi + Aj) + (7k is
r(4i--Bj-k) + C = 0. Hence prove an angle
that,
with
and
to the axis
if
it, its
a plane meets the central axis in P and makes null point Q is such that PQ is perpendicular
of length
p
cot
.
two straight lines intersect in a point P, intersect and lie in the null plane at P
18. If also
19.
Show
their conjugates
that the null planes of collinear points have a
common
line of intersection.
20. Prove that any system of forces acting on a rigid body can be replaced by two equal forces equally inclined to the central axis. 21.
A
transversal intersects the lines of action of
Prove that either point of intersection
forces.
is
two conjugate
the null point
of
the plane containing the transversal and the other line of action. 22.
that
Any two
PQ
23.
A
conjugate lines intersect a plane in passes through the null point of the plane.
system of forces
is
the force at
A
A, B, C.
If
the other two in the line 24.
A
lies
P
and
Q.
Show
reduced to three, acting at fixed points is fixed in direction, prove that each of in a fixed plane. Also that these planes intersect
BC.
rigid
body
from point to point.
is
acted on by a force
If,
relative to
P
per unit mass, varying r is the position vector
an origin 0,
EXERCISES ON CHAPTER
VIII.]
of the point
P
where the density
equivalent to a force
whose torque
is
is
is
the element of volume at
P
Art. 105 the Cartesian equations of equilibrium
^
Z
+ X = 0,
ds_
is
where X, Y,
show that the whole action
IjuFcfo acting at 0, together with a couple
of a string,
etc.,
p,
l^rxFcfo, where dv
Deduce from
25.
is
159
VIII.
are the resolutes of the force per unit length
acting on the string.
A
26.
string is in equilibrium in the
tension is constant throughout the string.
any element
directly
is
A heavy string
27.
immersed in a just inside
and
is
fluid.
from the axis
of the helix.
suspended from two points, and hangs partly
Show
that the curvatures of the portions D' D, where D, D'
just outside the fluid are as
and
are the densities of the string
A
form of a helix, and the Prove that the force on
D-
:
fluid respectively.
heavy string
is suspended from two fixed points, and the such that the form of the string is an equiangular spiral. Show that the density at any point P is inversely proportional to 2 r cos t/^, where r is the distance of P from the pole and \fr the angle the tangent at P makes with the horizontal.
28.
density
is
A
heavy uniform string rests on a smooth curve in a vertical and is acted on by forces at its ends. Prove that the difference between the tensions at any two points is equal to the weight of a string whose length is the vertical distance between the points. Also find the pressure on the curve at any point. If the string is light, show that the tension is constant, and that 29.
plane,
the pressure varies as the curvature.
A
30.
where
on a rough curve in a state bordering on the ratio of the tensions at any two points is e^'. the coefficient of friction and 6 the angle between the
light string rests
Show that
motion. /j.
is
tangents at the 31. is
A
heavy
two
points.
on the point of motion.
tension 32.
and pressure at any point. rigid body is subjected to
A
able over the surface.
body
on a rough curve in a vertical plane, Write down equations for determining the
string, resting
is
Show
equivalent to a force
with a torque
fluid pressure of intensity p, varithat the total action of the fluid on the
- pn dS through the I
- \pr>
is
origin, together
the unit outward normal and
dS the area of an element of the surface.
SUMMAKY. ADDITION AND SUBTRACTION. Vectors
are
compounded according
Thus,
addition.
if
three points 0, P,
R
to
the triangle law of
are chosen so that
OP=a
PR=b, the vector OR is the sum or resultant of a and b. When several vectors are added together the commutative and
and
associative laws hold.
The individual vectors are
called the
components of the resultant.
The negative of a is the vector which Las the same length as a but the opposite direction. It is denoted by - a.
To add. If
subtract the vector b
Thus
m
is
vector in
from a reverse the direction ,
of b and
.,
,
a-b=a + (-b).
any positive real number, the same direction as a, but
ma of
A
is
defined to
mean
m times its length.
the
Thus
A
ma = m (aa) = (ma) a. A
where a
is
a unit vector and a the module of a.
- m) a is defined to be the vector obtained ( by reversing the direction of a and multiplying its length by m. The general laws of association and distribution for scalar multipliers hold as in ordinary algebra. Thus Similarly the vector
m (wa) = (mn) & = n (ma), (m + w)a=ma + wa,
m(a + b)=ma + mb.
Any
vector a can be expressed as the
any three non-coplanar
sum
of three others,
When
these three
components are mutually perpendicular they are
called the
parallel to
160
vectors.
'
ADDITION AND SUBTRACTION resolutes or resolved parts of
notation
a in those directions.
We
use the
a=a + a. +a sk, 1i
where
161
i,
2j
k are the unit vectors
j,
may be compounded by
in those directions.
adding their like components.
Vectors
Thus
Sa = (Sa 1 )i + (Sa 2 )j+(v a3 )k. The unit vector A
1
a=-a = (cosot)i + (cos /S)j +(cosy)k, where cos
<x,
cos
cos y are the direction cosines of the vector.
/S,
The rectangular with
its
coordinates x, y, z of a point are connected
by the
position vector r
relation
r=ai+yj +zk.
We
speak of this point briefly as the point
The line joining the points a and b by the point _ v J r wa+mb
r=
The centroid numbers px pi} ,
The centre r„
r 2J
.
.
.
of
is
the points
of .
.
m+n
.
,
is
aj,
is
r.
divided in the ratio
m
:
n
.
a2
,
.
.
.
,
with associated real
the point
mass (cm.)
of particles
the point
m m y,
2,
... at the points
2mf
__
Zm This point coincides with the centre of gravity of the particles.
The vector equation parallel to
b
The straight
of the straight line
is
r
line passing r
The necessary and be collinear
is
=a + *b.
through the points a and b
is
= (l-«)a + *b.
sufficient condition that three points should
that there exists a linear relation between their
position vectors, in is
through the point a
equal to zero.
which the algebraic sum
of the coefficients
+
VECTOR ANALYSIS
162
The bisectors r=
of the angles
between the straight A
A
/a
lines v
= ta. and
b\
a
and
r=«(a-b)=«( -T
The plane through the point a
and c
parallel to b
is
= a + sb + tc.
r
The plane through the three points a, b, C is r = (1 - s -t)a, sb + tc. The necessary and coplanar
sufficient condition that four points should
between their position
that, in the linear relation
is
vectors, the algebraic
sum
The necessary and
be equal to
of the coefficients
zero.
sufficient condition that a linear relation,
connecting the position vectors of any should be independent of the origin, the coefficients be zero.
The vector area
be
is
number
of fixed points,
that the algebraic
of a plane figure is specified
sum
1
of
by a vector normal
to the plane, with module equal to the measure of the area of the
The sum of the vector areas of the faces of a
figure.
closed polyhedron
is zero.
PRODUCTS OF VECTORS. The
two vectors a and
scalar product of
are inclined at an angle
d, is
written
= a&cos0=b-a.
The
a-b
the real
condition of perpendicularity of a a«b
The square
of the vector a
b,
whose
number ab
and b
directions
cos
6,
and
is
is
= 0.
is
a2
= a*a=a 2
-
Also, with the usual notation, a*b
a2
aa and where l v
a*b
mv
respectively.
n x and
l
= cos 2
,
m
2
,
= %&! + a 2 b 2 + a 3b 3
,
=a 1 2 + a 2 2 +a 3 2 6 = lj 2 + m 1 m 2 + n x n 2 ,
,
n 2 are the direction cosines
of a
and b
PBODUCTS OF VECTORS For the mutually perpendicular unit vectors
163 i,
j,
k,
i2=j2=k 2 =l, H =j'k=k-i = 0. Aiiy vector r
may
be expressed as the sum of two vectors a' r „
^a
respectively, parallel
j and
distributive
(r--2-aj\
and perpendicular to r=r«ii
The
a-r
(
a.
Also
+ r*jj +r-kk.
law holds for scalar products
+ c) = a*b + a*e.
a*(b
The vector product of two vectors a and b, whose directions are an angle 6, is the vector whose module is ab sin d, and whose direction is perpendicular to both a and b, being positive relative to a rotation from a to b. We write it inclined at
A
axb=a& Its value
may a*b
sinfl
n=
-b*a.
also be expressed
= (a 263 -a 3 6 2 )i + (a36 1 -a 1 6 3 )j + (a 1 b 2 - a 2b 1 )k 1
2
1
h
h
b3
i
i
The condition of parallelism
of a
k and b
is
axb=0. For the unit vectors
i,
j,
k we have
i*i=H =k*k = 0, while
ixj
=k=
-j*i,
jxk=i = -kxj, kxi=j = -ixk.
The
distributive
law holds for vector products also
order of the factors in each
The
;
but the
term must be maintained.
scalar triple product of three vectors a, b, c
is
the scalar
measure of the volume of the parallelepiped whose edges are determined by the three vectors. The value of the product is unaltered by interchanging the dot product of a and b^c.
It is the
VECTOR ANALYSIS
164
and the
cross, or
by changing the order Thus
of the factors, provided the
cyclic order is unaltered.
a-b>< c
and
so on.
The product
is
= a>< b*c = cb x a, generally written [abc],
a notation which indicates the three vectors and the cyclic order. If,
however, the cyclic order of the factors
of the product
is
[abc]
The value
of the
product [abc]
is
changed, the sign
Thus
changed.
is =
= - [acb].
given by the determinant «1
,
PRODUCTS OF VECTORS The a, b,
c
165
reciprocal system of vectors to the non-coplanar system v „ , „ c „ a'=-°^-- b' = c _ tate]* [abc]' [abc]'
— es-
is
•
which
satisfy the relations
= b'b' = cc' = 1 = a-c' = etc. = 0.
a-a'
and
The
a-b'
reciprocal system to
for r in
terms of
a, b, c
The above expression
a', b', c' is a, b, c.
may
be written
r=r«a'a+r'b'b + r'c'c.
The system
i, j,
k
is its
own
reciprocal.
THE PLANE AND THE STRAIGHT The standard form nis The perpendicular
LINE.
of the equation of a plane perpendicular to
i-n=q.
from the point
distance
r'
to this plane
is
<7-r''n
p=The distance measured
n
parallel to the vector b
is
n«b
The plane through the point d perpendicular
to
n
is
r*n=d*n.
The planes bisecting the angles between the two planes
rn
=q,\
.(A)
vri =q'j
n_n -+—
are
n
The equation planes (A)
is
of
n
1^' n
any plane through
n the line of intersection of
the
expressible as
r-(n-An')=g'-Ag''. This plane
may
be made to satisfy one other condition by giving
a suitable real value to the parameter X.
The plane containing the three points r«(b«c
+ c*a + a
a, b, c is
= [abc].
,
VECTOR ANALYSIS
166
parallel to b
The plane through the point a
and
c is
Hb*c = [abc] [r-a,
or
b,
c]=0.
The plane through the points a and
b,
and
parallel to c, is
r(b-a)*c = [abc].
The plane containing the c
is
straight line r
r-(a-c)*b
= a + fb and
the point
= [abc].
The perpendicular from the point
r'
to the straight line
r=a + £b p=a-r'-^2
is
The condition
b«(a-r')b.
of intersection of the straight lines
r=a + th,\ r
= a' + sb'
a-a']=0.
[b, b',
is
The length
common
of the
perpendicular to the two lines
n'(a-a')
P=— where n=b*b' and
(B)
j
1
rv =-[b,
n=mod.
n.
,
b',
is
,,
a-a'],
The common perpendicular
is
the line of intersection of the planes
[r-a, [r-a',
b,
b*b']=0,
b',
b*b']=0.
Plucker's coordinates of a line are the unit vector d parallel to the line,
m
and the moment
about the origin of
this vector
localised in the line.
The mutual moment
of the
two straight
lines d,
m and
d',
m'
is
M =m-d'+m''d. This
is
dicular
connected with the length p of their
by the equation
M
=p
if
M=
perpen-
sin 6,
6 being the angle of inclination of the two intersect
common lines.
The
lines
0.
If the position vectors of three vertices of a
to the other vertex are
a, b, c,
tetrahedron relative
the volume of the tetrahedron
F=i[abc].
is
THE SPHERE
167
THE SPHERE. The equation point c r
of the sphere of radius a with centre at the
is
r2
or
where
Jc
..
,
(r-c) 2 = a 2
= c -a 2
-2r«c+&=0,
2-
The equation
of the tangent plane at the point d
is
r*d-c(r + d)+&=0.
The condition that the plane
r«n
=q
should touch the sphere
is
(q-cn) 2 = n 2 (c 2 -k). The condition that the two spheres *Xr)=r 2 -2r-c +h =0,1 i?"( r
should cut each other orthogonally 2cc'
The polar plane
straight line
sphere
is
is
=k + k'. h with respect
of the point
r-h-C'(r
Any
Q
)= r 2_2r. c '+fc'=0j
to the first sphere
is
+ h)+&=0.
drawn through the point h
to intersect the
cut harmonically by the surface and the polar plane
of h. If
the polar plane of the point h passes through the point g
:
then the polar plane of g passes through h. The radical plane of the two spheres (C) is F(t)
that
is
= F'(i),
2v(c-c')=k-k'.
The tangents to the two spheres from any point on are equal in length.
this plane
Also
F(i)-\F'{i)=0 represents a system of spheres with a
perpendicular to the line of centres.
common
radical plane
VECTOR ANALYSIS
168
DIFFERENTIATION AND INTEGRATION. If r is
a function of a scalar variable
t,
and dv
in r corresponding to the increment dt in
value of the quotient dt/dt as dt tends to zero tive of r
with respect to
the increment
is
called the deriva-
We use the notation
t.
^I_^r
T+
«^o
The derivative and so on. The rules for
is
then the limiting
t,
dt
dt
of this function is called the
second derivative,
sums and products of vectors sums and products. Thus
differentiating
similar to those for algebraic
d
._dr ds + + '~dI Jt
,
d
di
,
,
{l ' S)
dt d_
(r«s)
=dt'
-
ds
S+I
-Jt'
ds
=
dt
are
<s+r* dt'
dt
Differentiating both sides of the equality r 2
= r 2 and ,
using the
second of these formulae, we obtain F•
dv
— V dr dt'
dt
which
is
a useful result.
In particular, 'if a
length,
is
a vector of constant
d&
*'dt~°> showing that a is perpendicular to its derivative. worth noticing that ^ d*r Ix dt
To
_ dx 1+ dy
di
and
dt~dt differentiate
.
dt
.
i+
also
dt 2
dz
dJ
each factor in turn. da..
[abc] dt
r*
is
a triple product, or one involving several
factors, differentiate d_
=
dt_
[_
It
Thus
DIFFERENTIATION AND INTEGRATION Integration
is
the
reverse
process
to
vector F, whose derivative with respect to the integral of r, and is written
F=
A
Thus
r
The
differentiation. t
is
equal to
r, is
called
[ r dt.
may
constant of integration
calculus.
169-
be introduced as in algebraic
jt
2lr-^-*=r 2 + c = r 2 + c, f
d2l
n
dv
y*w dt=v *dt The equation
dh
^= -n
multiplication of both
2
may
i
be
—
members with 2
(I)
+c
2
-
integrated
~.
We
dt
after
scalar
then obtain
212 "
A definite integral is defined as the limit of a sum,
as in ordinary
Cf. Arts. 61, 62.
calculus.
GEOMETRY OF CURVES. If r is s
is
the position vector of a current point on the curve, and
the length of the arc up to that point, dx
_dx.
dy
ds
ds
ds
dz,
.
ds
the unit vector parallel to the tangent, called briefly the unit
tangent.
Hence the equation
of the tangent
is
R = r + ut. „
d 2i
,,
Further,
di*
where n
is
=
dt
=Kn
ds
>
the unit vector parallel to the principal normal, and
k the curvature or arc-rate of turning of the tangent. briefly the unit
normal.
From
\IsV ~\dsV
The equation
of the principal
the last equation
+
WsV
normal
R = r + wn.
is
+
\di2
it
We
call
n
follows that
VECTOR ANALYSIS
170
The normal plane
is
the plane through the point r perpendicular
Its equation is
to the tangent.
(R-r)-t=0.
The
osculating plane, or plane of curvature,
the tangent and the principal normal.
[R-r, The binormal
is
t,
is
that which contains
Its equation is
n]=0.
the straight line through the point
r
The unit binormal
is
dicular to the plane of curvature.
= t*n,
b
which has a derivative
perpen-
= - Xn,
-^-
as
where X
is
the torsion, or arc-rate of turning of the binormal.
Its value is given
by "
K
The derivative
1
d 2!
_ds' ds 2
of the unit
di3J
'
normal
is
s =-*t+Ab, and the equation
of the
binormal
R_
is
^r
ds
ds 2
PAKTICLE KINEMATICS AND DYNAMICS. The
velocity of a particle,
whose position vector
is r, is
_di
v=
di'
and
its
acceleration
is
the rate of increase of
dv^d
2
its
velocity
;
or
!
~w
Velocities are is
*~dt compounded by vector
addition
;
and the same
true of accelerations.
The velocity
of a particle
moving in a curve with speed v
\=vt,
and
its
acceleration
is
a
= t- t + kv 2 ii. dt
is
PARTICLE KINEMATICS AND DYNAMICS
171
For a particle moving in a plane curve the radial and transverse resolutes of the velocity are
dr
d6T
,
and
-
3
-
r
dt
respectively
;
and those d*r r
dr>-
The
dt
of the acceleration are
fdd\ 2 (di)
d*6
,
and
r
d¥
+2n
where k
is
the unit vector perpendicular to r and
path of the linear
is
= ^k,
p being the perpendicular distance from the The
dtdf
areal velocity of a particle about the origin
Jr*v
to the
drdd
v,
and
origin to the tangent
particle.
momentum
of a
moving
particle of
mass
m is
M = »iv, and
its
rate of increase
is
dM
d\
dt
dt
Newton's second law of motion states that the force F acting on a particle has the direction of, and is proportional to, the rate of
momentum.
increase of the particle's priate unit of force,
d F = t-(mv)
Hence, with the appro-
= ma,
the mass of the particle being assumed constant.
The impulse
of a force
F
acting during the interval
Fdt=m(v1 -v
t
to
t
t is
),
'I"
and is therefore equal to the increase of momentum produced by it.
The is
activity of the force
is its
rate of working.
Its value F-v
equal to the rate of increase of the kinetic energy of the particle.
Thus it F
'
V
=
dT d n mv) -dI=dt^ „.
-
The moment or torque of a force F about the origin is r-F, The moment of momenr is any point on its line of action. tum of a particle about the origin is similarly H=r*(mv), where
VECTOR ANALYSIS
172 r
The term angular synonymous with moment of momentum.
being the position vector of the particle.
momentum (a.m.) is The moment of the force of increase of the a.m.
acting on a particle
is
equal to the rate
or
;
dR
_ =t * F
-
-dt
A central force is one acting always towards a fixed point called The
the centre of force. force
is
orbit of a particle acted
on by a
central
a plane orbit, the plane of the orbit containing the centre
The
of force.
of the
a.m.
about the centre remains
particle
constant.
In the case of a central force square of the distance, the orbit of force.
V
If
is
is
- r,
varying inversely as
a conic with focus at the centre
the speed at a point distant c from the centre
of force, the eccentricity of the orbit is given
/X
If
F <2 u/c 2
/
the
the orbit
an
is
by
CJ
\ fX
ellipse, as in
If a, b are the semi-axes of the ellipse,
the case of the planets.
the periodic time
is
Vfx
and the speed at any point
is
h2
while
given by /2
IN
\r
al
A =—
2
a
In the case of a central force - fxmx, varying distance, the orbit
0.
is
an
The periodic time
and the speed
at
OB
is
is
2
=
u(a
/
directly as the
with centre at the centre of
now
P is
any point v
where
ellipse
2
+6
2
given by
-j- 2 )=
M .OZ» 2
,
the semi-diameter conjugate to OP. h
= V/ul
.
ab.
Also
force
PARTICLE KINEMATICS AND DYNAMICS If
a particle acted on by a force
a smooth
F
is
curve, the equation of motion
173
constrained to
move on
is
ma=F + R, R
where
is
the reaction of the curve.
This
equivalent to
is
the three scalar equations
m dt =F » m,KV 2
the suffixes
1, 2,
= F2 + R2
,
= Fs + R 3
,.
3 denoting resolutes in the directions of the
tangent, principal normal and binormal respectively.
SYSTEM OF PARTICLES. The
linear
momentum
of a
system
of particles is defined as
the vector sum of the linear momenta of the separate particles that
which
is
equivalent to
M = ~Lm
where
is
the total mass of the particles, and v the
The
velocity of the centre of mass. linear
momentum
The vector sum
:
M = 2mv, M = Jtfv,
is
is
^wr
rfv
at
at
of the forces acting
rate of increase of the
on the
particles
is
2F=-f-2mv = Ma, at
an equation which determines the motion of the cm. of the system about any point The angular momentum
H
.
sum moments about the
of the a.m. of the separate particles.
the vector
Hence
is
for
origin
H = 2r> mv, and
its
rate of increase
is
^=
»
Sr><|-(mv)
at
Thus the -vector
= 2r*F.
at
rate of increase of the a.m. about
sum
of the torques
about
is
of all the forces
equal to the
on the system.
VECTOR ANALYSIS
174 This principle
may
be used for taking moments about the cm.,
regarding that point as fixed.
The vector sum is
of the impulsive forces acting
equal to the increase in the linear
that
£I=M-M
is
The vector sum fixed point
is
point produced
moments
of the
momentum
on the system system
of the
;
.
of the impulsive forces
about a
equal to the increase in the a.m. about that
by
Thus
those forces.
2r*I=H-H
.
EIGID KINEMATICS. The motion
body about a fixed point
of a rigid
is
at
any instant
one of rotation about a definite axis through that point, called
The angular
the instantaneous axis.
sented by a vector
A
velocity can then be repre-
parallel to this axis.
particle at the point r
y
is
The velocity
of the
= A*r,
the fixed point being taken as origin.
When
no point of
particle as origin
Then the
the
and
velocity of
body
let
any other
V=v
is r is
The vector
is fixed,
A
is
take the position of any
v be the velocity of that particle. particle
whose position vector
+ A*r.
independent of the origin, and
is
called the
angular velocity of the body.
Any motion
of a rigid
body
is
equivalent to a screw motion.
A and the velocity of any on the axis is along the axis, being the same for all such particles. The two invariants of the motion are A 2 and T = vA, where v is the velocity of any particle. The pitch of the screw is The
axis of the screw
is
parallel to
;
particle
p = r/A 2
.
Simultaneous angular velocities about a fixed point are compounded by vector addition. Simultaneous angular velocities about parallel axes are compounded like parallel forces. Any simultaneous motions corresponding to velocities v v v 2 of a particle chosen as origin, and angular velocities A 15 A 2 ... of the body about that point, are compounded by vector addition .
,
.
.
,
of the velocities of the origin,
velocities.
and vector addition
of the angular
RIGID DYNAMICS
175
EIGID DYNAMICS. For a body moving about a
fixed point 0, let
A = a^i + be the angular velocity.
ft)
2j
+ (03k
Then the angular momentum
body about the fixed point
of the
is
H =h
1 i+h 2 ] + A 3 k, = Au> x - Fw 2 - Ew 3 \ h 2 = Ba> 2 - Da>z ~ fa>i, h 3 = Cw 3 — Eco 1 — D002J
where
hx
,
1
A, B, C, D, E,
body
the i, j,
k.
F
being the moments and products of inertia of
with respect to coordinate axes through
parallel to
There are three mutually perpendicular axes through
which the products of inertia vanish. These are called the and the corresponding values of A, B, C An angular velocity about are the principal moments of inertia. any one of these axes makes H parallel to that axis. The kinetic energy of the body is
for
principal axes at
;
T =$/»» =i/A 2 where w = mod.
A
and I
is
,
moment
the
of
inertia
about the
Also
instantaneous axis of rotation.
T=iA'H. The moment of inertia about the axis whose direction I, m, n relative to the coordinate axes is I =Al 2 + Bm 2 +Cn 2 - 2Dmn - 2Enl-2Flm.
cosines
are
If
the body is moving with no point fixed, let
r,
v be the position
vector and velocity of the cm., and r', v' those of a particle Then the angular momentum of the body relative to the cm.
about the origin
is
H = x>M v + Sr'*mv'.
Similarly the kinetic energy of the
T=pfv
2
body
is
+P>nv' 2
.
The first term is the kinetic energy of translation body with the velocity of the cm. and the second of the motion relative to the cm. ;
is
whole
the energy
rate of increase of the kinetic energy is equal to the activity the external forces acting on the body.
The of all
of the
VECTOR ANALYSIS
176
a
Let S 1} S 2 be two frames of reference in relative motion about common fixed point, the motion of S 2 relative to S 1 being an
angular velocity
A
about that point. Then the rates of change two frames are connected by
of a vector r relative to the
= This formula
-A
reciprocal, since
is
+
\JtJ2
dt)i
is
the a.v. of
H
Applying the formula to the vector
S t relative
to
S2
.
representing the a.m.
body moving about a fixed point 0, we obtain Euler's dynamiFor the frame S 1 we take the system of surrounding cal equations. The frame S 2 we take as objects, at rest relative to each other. Let i, j, k be unit vectors, fixed fixed in the moving body. relative to S 2 and parallel to the principal axes of the body at of a
,
The angular velocity
0.
A = a>ji + of the
body
is
also the a.v. of
&) 2 j
S2
+ ^3^
relative to
Sv
If
L=i i+i +i k 2i
1
is
3
the torque of the external forces about 0, the principle of a.m.
states that
/^
L This
is
H
/^n
n
=U)rb)
+A H *
-
2
the vector equivalent of Euler's three scalar equations
A&! - (B - C)u) 2W 3 = £1,^ 5fti 2 _ (0 — A)w 3 w 1 =L i \ Cw s — (A — B)u) 1 w 2 = L 3 .) ,
For
Coriolis'
Theorem, connecting the accelerations S 1 and S 2 see Art. 95.
point relative to the frames
of a
moving
,
RIGID STATICS. The necessary and sufficient conditions of equilibrium for body acted on by forces F 1; F 2 through the points r ls r 2 ,
are
.
.
.
,
.
a .
.
2F = 0,] Sr*F = 0,|
i.e.
the vector
sum
of the forces,
origin,
must both vanish.
origin,
they are
If
satisfied for
and
their torque
about the
the conditions are satisfied for one
any
origin.
RIGID STATICS
177
A system of forces acting on a body is statically equivalent to any other system having the same vector sum, and the same torque about a
A
common
system of
points r 1; r 2
origin.
parallel forces p^a,
p z&,
.
.
acting through the
.
equivalent to a single force (p 1 acting through the point ,
.
.
.
,
is
+p 2 +
.
.
.)&
jw
'
'Ep
In particular, the centre of gravity of the body coincides with its cm. The last formula ceases to be valid when ~Zp=0, when the system
A pair
is
equivalent to a couple.
of equal
and opposite
parallel forces with different lines
The torque L
of action constitutes a couple.
the same for
It
all origins.
ing the two lines of action,
of the
system
is
perpendicular to the plane contain-
is
and equal to
L = (r1 -r 2 )xF, where
Two
tx
,
A system of couples is is
equal to the vector
That
on the
r 2 are points
lines of
couples are statically equivalent
if
F and -F
respectively.
they have equal torques.
equivalent to a single couple, whose torque
sum
of the torques of the individual couples.
to say, couples are compounded by vector addition of their
is
torques.
A system
of forces
F1 F2 ,
.
,
.
.
through the points
r l5 r 2 ,
.
.
.
may
be replaced by a single force
R = 2F through the origin, and a couple of torque
G = 2r*F. The couple varies with the
origin,
product
but
R is invariant.
The
scalar
r=R'G
on a certain straight line, The force and couple called the central axis, G is parallel to R. then constitute a wrench, equivalent to the original system of is
also
forces.
an
invariant.
The
If
pitch of the
the origin
wrench
is
p = F/R2
Any system to
two
of forces acting
forces, of
arbitrarily.
which the
is
.
on a body
is
statically equivalent
line of action of
one
may be
chosen
Such a pair are called conjugate forces of the system.
VECTOR ANALYSIS
178 If
a body, in equilibrium under any set of forces,
started off with
any
finite motion, the initial rate of
is
supposed
work
of the
the forces are such
on the system is zero. Conversely, if however the body is set in motion; the initial activity of the forces is zero, then the body is in equilibrium under the forces. forces that,
This
is
the principle of virtual work, or virtual velocities, for the
body. If
a string
is
in equilibrium under a force
F
per unit length,
T
is
found from the
varying from point to point, the tension
iw ^pt + 2Yn + F = 0,
equation
as
which
is
equivalent to the scalar equations
£'.-*} where F v F 2 are the resolutes of F along the tangent and principal normal respectively. In the case of a wire or thin rod, if S, L represent the stress and bending couple at any section, and F,
G the impressed
force and
torque per unit length, the equations of equilibrium are
+ F-0,| £ as -5=
ds as
which
may
+ txS + G=0,
'
be replaced by six Cartesian equations.
ANSWERS TO EXERCISES. CHAPTER 1.
Sum = 10i. Modules -4 3 7
I.
are n/74, 3 JlO, 2 n/46.
-5
1
.
Direction cosines
-8 _
\/74'
2.
^74'
\/74'
Module = 9
4i-5j +llk.
v'46'
Direction cosines
^/2.
-1_
and _3_ an
3VI0' 3VIo' 3\/l0'
5.
~b 1 )i + (a i -b 2 )i+(a 3 -b 3 )k, etc. Lengths are ^{(04 - 6 X 2 + (a 2 - 6 2 2 + (a 3 - 6 3 ) 2 }, a + 3b, 3b-a, 2(a + b), -(a + 3b). b - a, - a, - b, a - b.
6.
i(v'3i
q—ts>
_6_
n/4o'
\/46'
q~~7o'
3. (a 1
)
4.
7-
6V2
8. \/l7
+ j),
i(j-v/3i),
i,
()_l) '
"6 1
(l '
l2
)
-£U/3i + i), ^(i+j),
-^3))
etc.
^-i)-
-
miles an hour at tan _1 J N. of E.
— \ with the fixed diameter.
8 (—cos = -.
1
-,
+ sm a )
after lf| sec.
10.
6J ft./sec; 9
11.
Twice the vector determined by the diagonal
ft.;
of the
cube drawn from that
corner.
A force
12.
5
13.
V3P and IP lb.
lb.
14. sJ2 16.
wt.
represented by
—^ (5i + 4j + 3k).
wt.
times the former, inclined at 135° to
The point whose distances from the
W .»+V, + k)
it.
faces through
A
are }f, \, \\.
.
18.
A force represented by 6AO,
22.
The position vectors
23.
The
where
is
the centroid of the hexagon.
- 2a and 2a - b respectively.
of the points are 3b
straight line passing through
C and
the mid-point of
AB.
N
of the join of the mid-point 28. On the straight line passing through the OG = n-2 2. vacant vertices, and the centre 0, such that
NO
CHAPTER 8.
r
= (i-2j + k)+«(i-2k);
l(6i-
10J
II.
+3k).
179
:
:
VECTOK ANALYSIS
180
CHAPTER J_ ;. ,(:,j+llk-3i);
5,
6.
12.
j
;
(i
+ j -2k) + s(j -2i + k), where
cos" 1 -^.
16. r»(3i
17.
f(qa-pb)=0.
20.
r-(2i-7j-13k) = l.
27.
= a + t(* + -b + ^
8. i
+ -2k) + s(i-2j +k) equation s z + 2s - 2 = 0.
15. cos- 1 1;
26.
a/|||.
llj+9k-3i. (i
III.
s ia
a root of the
-4j +7k) + 13=0.
^(106i + 73j-23k)+«(2i-7j-13k).
19.
23.
21. -}J6.
|(3 - ^3)(i + i + k).
A plane perpendicular to the straight line joining the two given points. A straight line parallel to the vector difference of the forces, through
the
intersection of their lines of action. 28.
JZP
perpendicular to BC.
29. 40 units.
34.
A
D such that
At a point
sphere.
35.
CHAPTER 2.
(a*dce - c«da - e)b + a , b(c*de - ced).
5.
r-ax(l>xc)=0.
[abk]/mod (kxb).
13.
r-c = f[bx(a-c)]x[b'x(a'-c)].
17.
The
[r,
straight line
k,
— 'fc
and
[r
-a,
-i'n 1 ) =
—
1.
i.
iii.
v.
2r/r
+r
2(ar
+ rb)-(a r + ,b).
Zri
2. First
i
A
7.
r»bxc
plane.
= [abc].
(q 2
kxb] = 0.
b,
r-c=«ax[bx(c-a')]-
- r»n 2 ) =
11%
— ft,
(? 3
-r»n 3 ).
3
+ (dxd')xd[m'd'd]},where^=moddxd'.
CHAPTER 2
ii.
IV.
14.
l
Ja{(dxd')xd'[mdd']
32. -^-(dxd'),
sphere,
line of intersection of
kxb]=0
(q^
A
-a)*b = [aa'b].
6. r*(a'
The
11.
i.
BT> = f BC.
+ a« rb.
V. 3r 2 fr
ii.
+ r 3 r + ax'f.
vi.
g-.
*'
b - r(a )b 4-?¥ + r r st (a«r) -'
iv.
2
3
2
mr'r.
derivatives are
L
dt
x
d? J
dt
Second derivatives are
r [_
{dt*dt*)
^2 ^5J +
r |
+ t *\dt X
Wf
^ ~m\ and another^expression.
,
ANSWERS TO EXERCISES 3
2rr(r
t
+ a2
r2 5.
i.
ii.
6. r 7.
8
13.
+a
rxa
)
aV
r'arxa
1^~~{F^-
=JaJ a + ht + c, where b and c are const. &xr = b\>t 2 + ct + &, where c and d are const.
= Ja« + JW8. s
Mod
i.
r is
constant.
ii.
Direction of
rdr^r^rl_ n
4
K=
**»» + «»«.). 6(o" + e»)Bna«- » = Hence the equations of the principal normal and the plane
of curvature
be written down.
R = asec
2
a.
R=Sa(l + t 2 2 s/T+Wt2
25.
)
CHAPTER
.
VI.
+ rjxfvs-v^ + rjxfo-Vj)}.
1.
itaxfvjs-VsJ
2-
g^-{[rir > rj]-[t 1 r 8 r 4 ]
13.
The new major
25.
The speed is J2gz.
axis
+ [r 1 r 3rJ-[r2 r 8 r 1 ]K in which
is
^=
»i> etc.
double the original one.
The
and the binormal are
resolutes of the reaction along the principal
——
cos 2 a
and - mg cos a
CHAPTER 4.
constant.
r is
a="
11.
]_ds ds 2 ds 3 J
'
may
23.
+
r
j '
(r»
181
normal
respectively.
VII.
The instantaneous
centre of rotation is the point of intersection of the perpendiculars to the two straight lines at the extremities of the rod. In the second case the rotation is round an axis perpendicular to the two given straight lines.
17. If 6 is
the inclination (to the vertical) of the radius to the bead v2
= 2ga(l
- cos
0)
+ u a2 2
sin 0(2
+sin
B),
the distance of the bead from the fixed vertical tangent, the components R, R' of the reaction on the bead, along the radius and perpendicular to the plane of the wire, are given by
and
if
x
is
v — = a 2
<7 cos 6 J
— a 2x
sin 8
Id... x dt
r-re -rsm 2
20.
1
A
(r 2 0)-r
R'
m
K
2
—mR
-\
2 0 ,
sine cos
0d> 2 ,
-L-
±(r"wof8f).
r sin
at
VECTOR ANALYSIS
182
CHAPTER 31. If
B is
VIII.
the reaction per unit length,
dT = ^-
fj.E
+ pgsiaf,
KT = R-pg cos
\p,
k being the curvature at the point, p the mass per unit length, the inclination of the tangent to the horizontal.
and f
;
;;
INDEX. The numbers
refer to the articles.
Acceleration, instantaneous, 55, 66 radial and transverse, 68 ; tangential and normal, 67. Activity of a force, 74.
Distributive law, Dot product, 25.
5, 26, 28.
EUiptic motion, 78-80.
Addition of vectors, 4. Angular momentum, 76, 84, 90. Angular velocity, 41, 87, 88.
Energy, 75, 92, 93.
Equal vectors,
3.
Equilibrium of rigid body, 97 string, 105 of wire, 106. Equivalent systems of forces, 98.
Area, vector, 23. Areal velocity, 69. Associative law, 4.
;
of
;
Equivectorial quantities, 76, 83, 84. Euler's dynamical equations, 96.
Binormal, 60. Bisector of angle, 17, 30.
Fixed point, motion about,
87, 90, 92,
96.
Central axis (Poinsot), 101. Central forces, 77-80. Centre of curvature, 59. Centre of gravity, 99. Centre of mass, 11, 62, 82-4. Centroids, 9-11. Circle of curvature, 59. Circular helix, Chap. V. Exs. 10, 21. Collinear points, 16. Commutative law, 4.
Components,
Helix, Chap. V. Exs. 10, 11, 21, 23. Hodograph, Chap. VI. Ex. 5.
Impulse, 73. Impulsive forces, 73, 75, 86. velocity, 55, Instantaneous axis, 87 acceleration, 55, 66. 65 normal surface, 64 Integral, 57 ;
;
;
tangential line, 63. Integration, 57. Invariants, 88, 101. Inverse square law of force, 78.
4, 6.
Cone, tangent, 33.
Conjugate forces and Coordinates,
Coplanar points, 20 Coriolis'
lines, 103.
7.
Theorem,
;
vectors, 6, 43.
Kepler, 78, 79.
95.
Kinematics of a rigid body, 87-89.
Couples, 100. Cross product, 27. Curvature, 59. Curve, motion on, 81.
Kinetic energy, 75, 92.
Lami's theorem,
Length vector,
Definite integral, 61. Derivative of a vector, 55.
13, 54.
2.
Like vectors, 3. Limiting points, 37. Line integral, 63. Linear momentum, 71, 82, 83. Local plane, 59.
Diametral plane, 36. Differentiation, 55. Direct distance law, 80.
Direction cosines,
particle,
7.
183
65-70
;
VECTOR ANALYSIS
184
The numbers
refer to the articles.
Measure, 1. Module, 3.
;
Moment
of a force, 40 ; of a localised vector, 40. Moment of inertia, 90-92.
Moment of momentum, Momentum, 71, 82, 83. Moving Moving
76, 84.
axes, 94, 95. origin of moments, 85.
Multiplication
by a number,
Mutual moment
5.
Reflection
and
refraction
of
light,
Ex. 2 50, Ex. Relative position, 12. 41,
;
Relative velocity, 12, 65. Resolutes, resolved parts, Resultant, 4, 6.
7.
of lines, 51.
Negative vector, 3. Newton's laws of motion, 72, 83.
Normal
Radius of curvature, 59 of spherical curvature, Chap. V. Ex. 22. Rankine's theorem of four forces, 54. Reciprocal system of vectors, 47.
plane, 59.
Normal, principal,
59.
Normal
surface integral, 64. Null plane, 102.
Scalar product, 25. Scalar quantity, 1. Screw motion, 88. Simultaneous motions, 89. Sphere, geometry of, 33-38. Spherical curvature, Chap. V. Ex. 22. Spherical trigonometry, 53.
Square of a vector, Oblique coordinate axes, 26, Ex. 4. Orthogonal intersection of spheres, 34. Osculating plane, 59. Parabolic orbit, 70.
25.
Straight line, 16, 32, 49-51. String, equilibrium of, 105. Subtraction of vectors, 4. Surface integral, 64.
Parallel forces, 99. Parallelism, condition of, 27.
Tangent cone, 33. Tangent plane, 34. Tangent to curve, 58.
Parallelogram of forces, 4. Perpendicular distance from plane, between two 30 ; from line, 32
Tetrahedron, 19, 52. Torque, 40, 100.
;
straight lines, 50. Perpendicularity, condition of, 25.
Pitch of screw, 88 of wrench, 101. Plane, geometry of, 20, 29-31, 48. Plane of curvature, 59. Planetary motion, 79. ;
Torsion, 60. Triangle, 18.
Triangle law of addition, 4. Triple products, 43, 44, 56.
Unit binormal, 60 59 tangent, 58 ;
;
;
principal normal, vector, 3, 7.
Pliicker's coordinates, 51.
Poinsot's central axis, 101. Polar plane, 35.
Vector,
1, 2.
Polyhedron, closed, 23.
Vector area, 23. Vector polygon,
Position vector, 8. Principal axes of inertia, 91. Principal normal, 59.
Velocity, instantaneous, 55, 65 ; radial and transverse, 68 ; virtual, 104. Virtual work, 104.
Products of inertia, 90. Products of vectors. Scalar, 25 vector, 27 ; triple, 43, 44 quadruple, 45 ; derivative, 56. ;
Radical plane, 37.
GLASGOW
:
13.
Wire, equilibrium
Work
of, 106.
of a force, 39
;
virtual, 104.
Wrench, 101. Zero vector,
3.
PRINTED AT THE UNIVERSITY PRESS BY ROBERT MACLEHOSE AND
CO.
LTD
