Adolphe Ganot.

Elementary treatise on physics online

. (page 3 of 94)
Online LibraryAdolphe GanotElementary treatise on physics → online text (page 3 of 94)
Font size
QR-code for this ebook


that it acts from O to A. Let us suppose the force which acts on O along

OA to contain P units of force ; from O towards A measure ON con.



14 On Matter, Force, and Motion. [30-

taining P units of length, the line ON is said to represent the force. It
will be remarked that the analogy between the line and the force is very
complete ; the line ON is drawn from O in a given direction OA, and
contains a given number of units P, just as the force acts on O in the
direction OA, and contains a given number of units P. It is scarcely
necessary to add, that if an equal force were to act on O in the opposite
direction, it would be said to act in the direction OB, and would be re-
presented by OM, equal in magnitude to ON.

When we are considering several forces acting along the same line we
may indicate their directions by the positive and negative signs. Thus
the forces mentioned above would be denoted by the symbols + P and
P respectively.

31. Forces acting: along: tne same line. If forces act on the point
O in the direction OA equal to P and Q units respectively, they are
equivalent to a single force R containing as many units as P and Q
together, that is,



If the sign + in the above equation denote algebraical addition, the equation
will continue true whether one or both of the forces act along OA or OB.
It is plain that the same rule can be extended to any number of forces,
and if several forces have the same line of action they are equivalent to
one force containing the same number of units as their algebraical sum.
Thus if forces of 3 and 4 units act on O in the direction OA, and a force
of 8 in the direction OB, they are equivalent to a single force containing
R units given by the equation

R=3+4-8=-i ;

that is, R is a force containing one unit acting along OB. This force R
is called their resultant. If the forces are in equilibrium R is equal to
zero. In this case the forces have equal tendencies to move the point O
in opposite directions.

32. Resultant and components. In the last article we saw that a
single force R could be found equivalent to several others ; this is by no
means peculiar to the case in which all the forces have
the same line of action ; in fact, when a material
point, A (fig. 4), remains in equilibrium under the
action of several forces, S, P, Q, it does so because
any one of the forces, as S, is capable of neutralising
the combined effects of all the others. If the force S,
therefore, had its direction reversed, so as to act along
AR, the prolongation of AS, it would produce the
same effect as the system of forces P, Q.

Now, a force whose effect is equivalent to the com-
bined effects of several other forces is called their re-
sultant, and with respect to this resultant, the other
forces are termed components.
When the forces, P, Q, act on a point they can only have one resultant ;




-33]



Parallelogram of Forces.



but any single force can be resolved into components in an indefinite
number of ways.

If a point move from rest under the action of any number of forces it
will begin to move in the direction of their resultant.

33. Parallelogram of forces. When two forces act on a point their
resultant is found by the following theorem, known as the principle of
the parallelogram of forces: If two forces act on a point, and if lines be
drawn from that point representing the forces in magnitude and direction,
and on these lines as sides a parallelogram be constructed, their resultant
will be represented in magnitude and direction by that diagonal which
passes through the point. Thus let P and Q (fig. 5) be two forces acting
on the point A along AP and AQ respectively, and let AB and AC be
taken containing the same number of units of length that P and Q con-
tain units of force ; let the parallelogram AB DC be completed, and the
diagonal AD drawn ; then the theorem states that the resultant, R, of P
and Q is represented by AD ; that is to say, P and Q together are equal
to a single force R acting along the line AD, and containing as many
units of force as AD contains units of length.





Fig- 5-



Fie. 6.



Proofs of this theorem are given in treatises on Mechanics ; we will
here give an account of a direct experimental verification of its truth; but
before doing so we must premise an account of a very simple experiment.

Let A (fig. 6) be a small pulley, and let it turn on a smooth, hard, and
thin axle with little or no friction : let W be a weight tied to the end of a fine
thread which passes over the pulley ; let a spring CD be attached by one
end to the end C of the thread and by the end D to another piece of
thread, the other end of which is fastened to a fixed point B; a scale CE
can be fastened by one end to the point C and pass inside the spring so
that the elongation of the spring can be measured. Now it will be found
on trial that with a given weight W the elongation of the spring will be
the same whatever the angle contained between the parts of the string
WA and BA. Also it would be found that if the whole were suspended
from a fixed point, instead of passing over the pulley, the weight would
in this case stretch the spring to the same extent as before. This experi-
ment shows that when care is taken to diminish to the utmost the friction
of the axle of the pulley, and the imperfect flexibility of, the thread, the






i6



On Matter, Force, and Motion.



[33-




Fig. 7.



weight of W is transmitted without sensible diminution to B, and exerts
on that point a pull or force along the line BA virtually equal to W.

This being premised, an experimental proof, or illustration of the
parallelogram of forces, may be made as follows :

Suppose H and K (fig. 7) to be two pulleys with axles made as smooth
and fine as possible ; let P and Q be two weights suspended from fine
and flexible threads which, after passing over
H and K, are fastened at A to a third
thread AL from which hangs a weight R ;
let the three weights come to rest in the
positions shown in the figure. Now the point
A is acted on by three forces in equilibrium,
viz., P from A to H, Q from A to K, and R
from A to L, consequently any one of them
must be equal and opposite to the resultant
of the other two. Now if we suppose the
apparatus to be arranged immediately in front of a large slate, we can
draw lines upon it coinciding with AH; AK, and AL, If now we mea-
sure off along AH the part AB. containing as many inches as P contains
pounds, and along AK the part AC containing as many inches as Q con-
tains pounds, and complete the parallelogram ABCD, it will be found
that the diagonal AD is in the same line as AL, and contains as many
inches as R weighs pounds. Consequently, the resultant of P and Q is
represented by AD. Of course, any other units of length and force might
have been employed. Now it will be found that when P, Q, and R
are changed in any way whatever, consistent with equilibrium, the same
construction can be made, the point A will have different positions in
the different cases ; but when equilibrium is established, and the paral-
lelogram ABCD is constructed, it will be found that AD is vertical, and
contains as many units of length as R contains units of force, and conse-
quently it represents a force equal and opposite to R, that is, it represents
the resultant of P and Q.

34. Resultant of any number of forces acting- in one plane on
a point. Let the forces P, Q, R, S (fig. 8) act on the point A, and let
them be represented by the lines AB, AC, AD,
H AE, as shown in the figure. First, complete the
parallelogram ABFC and join AF ; this line
represents the resultant of P and Q. Secondly,
complete the parallelogram AFGD and join
AG; this line represents the resultant of P, Q, R.
Thirdly, complete the parallelogram AGHE
and join AH ; this line represents the resultant
of P, Q, R, S. It is manifest that the construc-
tion can be extended to any number of forces.
A little consideration will show that the line
AH might be determined by the following
construction : through B draw BF parallel to, equal to, and towards the
same part as AC; through F draw FG parallel to, equal to, and towards





-35] Conditions of Equilibrium of Forces. 17

the same part as AD; through G draw GH parallel to, equal to, and towards
the same part as AE; join AH, then AH represents the required resultant.
In place of the above construction, the resultant can be determined
by calculation in the following manner : Through A draw any tvvo
rectangular axes Ax and Ay (fig. 9), and let a, tf, y
be the angles made with the axis Ax by the lines
representing the pressures, then P, Q, R can be
resolved into P cos a, O cos /3, R cos y, acting
along Ax, and P sin a, Q sin /3, R sin y, acting
along Ay. Now the former set of forces can be
reduced to a single force X by addition, attention
being paid to the sign of each component ; and in
like manner the latter forces can be reduced to a
single force Y, that is, Fi s- 9-

X = P cos a + Q cos /3 + R cos y + . . .

Y = P sin a + Q sin /3 + R sin y + . . .

Since the addition denotes the algebraical sum of the quantities on the
right hand side of the equations, both sign and magnitude of X and Y are
known. Suppose U to denote the required resultant, and the angle
made by the line representing it with the axis Ax ;
then U cos = X, and U sin /> = Y.

These equations give U 2 = X 2 + Y 2 , which determines the magnitude
of the resultant, and then, since both sin y> and cos are known, is
determined without ambiguity.

Thus let P, Q, and R be forces of 100, 150, and 120 units, respectively,
and suppose xAP, xAO, and xAR to be angles of 45, 120, and 210 re-
spectively. Then their components along Ax are 707, 75, 103-9, an ^
their components along Ay are 707, 129-9, 60. The sums of these
two sets being respectively 108-2 and 140-6, we have U cos 0= 108-2
and U sin 0= 140-6.

therefore U 2 = (io8-2) 2 + (I4O-6) 2

or U = 177-4

therefore 177-4 cos = - 108-2, and 177-4 sin <j> = 140-6.

If we made use of the former of these equations only, we should obtain
equal to 232 25', or 127 35', and the result would be ambiguous : in
like manner, if we determined <f> from the second equation only, we should
have equal to 52 25', or 127 35'; but as we have both equations, we
know that equals 127 35', and consequently the force U is completely
determined as indicated by the dotted line AU.

35. Conditions of equilibrium of any force acting: in one plane
on a point. If the resultant of the forces is zero, they have no joint
tendency to move the point, and consequently are in equilibrium. This
obvious principle enables us to deduce the following constructions and
equations, which serve to ascertain whether given forces will keep a point
at rest.

Suppose that in the case represented in fig. 8, T is the force which will
balance P, Q, R, S. It is plain that T must act on A along HA produced,



i8



On Matter, Force, and Motion.



[35-



and in magnitude must be proportional to HA ; for then the resultant of
the five forces will equal zero, since the broken line ABFGHA returns to
the point A. This construction is plainly equivalent to the following :
Let P, Q, R (fig. 10) be forces acting on the point O, as indicated, their
magnitudes and directions being given. It is known that they are balanced
by a fourth force, S, and it is required to determine the magnitude and
direction of S. Take any point D, and draw any line parallel to and
towards the same part as OP, draw AB parallel to and towards the same
parts as OQ, and take AB such that P : Q : : DA : AB. Through B draw
BC parallel to and towards the same part as OR, taking BC such that
Q : R::AB : BC; join CD; through O draw OS parallel to and towards
the same part as CD, then the required force S acts along OS, and is in
magnitude proportional to CD.





Fig. 10.



It is to be observed that this construction can be extended to any
number of forces, and will apply to the case in which these directions are
not in one plane, only in this case the broken line ABCD would not lie
wholly in one plane. The above construction is frequently called the
Polygon of Forces.

The case of three forces acting on a point is, of course, included in the
above ; but its importance is such that we may give a separate statement
of it. Let P, Q, R (fig. n) be three forces in equilibrium on the point O.
From any point B draw BC parallel to and towards the same part OP,
from C draw CA parallel to and towards the same part as OQ, and take
CA such that P : Q::BC : CA; then, on joining AB, the third force R
must act along OR parallel to and towards the same part as AB, and must
be proportnal in magnitude to AB. This construction is frequently
called the Triangle of Forces. It is evident that while the sides of the
triangle are severally proportional to P, Q, R, the angles A, B, C are
supplementary to QOR, ROP, POQ respectively, consequently every
trigonometrical relation existing between the sides and angles of ABC
will equally exist between the forces P, Q, R, and the supplements of the
angles between their directions. Thus in the triangle ABC it is known that
the sides are proportional to the sines of the opposite angles ; now since
the sines of the angles are equal to the sines of their supplements, we at
once conclude that when three forces are in equilibrium, each is propor-
tional to the sine of the angle between the directions of the other two.



-36] Parallel Forces. \ 9

We can easily obtain from the equations which determine the resultant
of any number of forces (34), equations which express the conditions of
equilibrium of any number of forces acting in one plane on a point ; in
fact, if U = o we must have X = o and Y = O ; that is to say, the required
conditions of equilibrium are these :

O = P cos a + Q cos j8 + R cos y + . . .
and O = P sin a + Q sin (3 + R sin y + . . .

The first of these equations shows that no part of the motion of the point
can take place along Ax, the second that no part can take place along Ay.
In other words, the point cannot move at all.

V 36. Composition and resolution of parallel forces. The case of
the equilibrium of three parallel forces is merely a particular case of the
equilibrium of three forces acting on a point. In fact let P and Q be
two forces whose directions pass through the points A and B, and inter-
sect in O ; let them be balanced by a third force R whose direction
produced intersects the line AB in C. Now suppose
the point O to move along AO, gradually receding
from A, the magnitude and direction of R will con-
tinually change, and also the point C will continually
change its position, but will always lie between A and
B. In the limit P and Q become parallel forces,
acting towards the same part balanced by a parallel
force R acting towards the contrary part through a
point X between A and B. The question is : First,
on this limiting case what is the value of R ; secondly,

what is the position of X ? Now with regard to the A, .-

first point it is plain, that if a triangle a b c were drawn </r Q\

as in art. 35, the angles a and b in the limit will Fig. 12.

vanish, and c will become 180, consequently a b ultimately equals a c + cb;
or R = P + Q.

\Vith regard to the second point it is plain that

OC sin POR = OC sin AOC=AC sin CAO,
and OC sin ROQ = OC sin BOC = CB sin CBO;

therefore AC sin CAO : CB sin CBO :: sin FOR : sin RO<

::Q:P(35)

Now in the limit, when OA and OB become parallel, OAB and OBA
become supplementary ; that is, their sines become equal ; also AC and CB
become respectively AX and XB ; consequently

AX : XB::Q: P,

a proportion which determines the position of X. This theorem at once
leads to the rules for the composition of any two parallel forces, viz.

I. When two parallel forces P and Q act towards the same part, at
rigidly connected points A and B, their resultant is a parallel force acting
towards the same part, equal to their sum, and its direction divides the






2O On Matter, Force, and Motion. [36-

line AB into two parts AC and CB inversely proportional to the forces
P and Q.

II. When two parallel forces P and O
act towards contrary parts at rigidly con-
/ nected points A and B, of which P is the

greater, their resultant is a parallel force
acting towards the same part as P, equal
to the excess of P over Q, and its direc-
tion divides BA produced in a point C
such that CA and CB are inversely pro-
tn portional to P and Q.

In each of the above cases if we were
to apply R at the point C, in opposite
direction to those shown in the figure, it
would plainly (by the above theorem)
balance P and Q, and therefore when it
acts as shown in figs. 13 and 14 it is the
resultant of P and Q in those cases re-
spectively. It will of course follow that
the force R acting at C can be resolved
^ into P and O acting at A and B respect-

Flg< M " ively.

If the second of the above theorems be examined, it will be found that
no force R exists equivalent to P and Q when those forces are equal.
Two such forces constitute a couple, which may be defined to be two
equal parallel forces acting towards contrary parts ; they possess the
remarkable property that they are incapable of being balanced by any
single force whatsoever.

In the case of more than two parallel forces the resultant of any two
can be found, then of that and a third, and so on to any number ; it can
be shown that however great the number of forces they will either be in
v equilibrium or reduce to a single resultant or to a couple.
3< 37- Centre of parallel forces. On referring to figs. 13 and 14, it will
be remarked that if we conceive the points A and B to be fixed in the
directions AP and BQ of the forces P and Q, and if we suppose those
directions to be turned round A and B, so as to continue parallel and to
make any given angles with their original directions, then the direction
of their resultant will continue to pass through C ; that point is therefore
called the centre of the parallel forces P and Q.

It appears from investigation, that whenever a system of parallel forces
reduces to a single resultant, those forces will have a centre ; that is to
say, if we conceive each of the forces to act at a fixed point, there will be
a point through which the direction of their resultant will pass when the
directions of the forces are turned through any equal angles round their
points of application in such a manner as to retain the parallelism of
their directions.

The most familiar example of a centre of parallel forces is the case in
which the forces are the weights of the parts of a body ; in this case the



_39] Equality of Action and Reaction. 21

forces all acting towards the same part will have a resultant, viz. their
Vgum ; and their centre is called the centre of gravity of the body.

38. Moments of forces. Let P denote any force acting from B to P,
take A any point, let fall AN a perpendicular from A on BP. The
product of the number of units of force in P, and the number of units of
length in AN, is called the moment of P with respect to A. Since the
force P can be represented by a straight line, the
moment of P can be represented by an area. In fact,
if BC is the line representing P, the moment is
properly represented by twice the area of the triangle
ABC. The perpendicular AN is sometimes called
the arm of the pressure. Now if a watch were placed



with its face upward on the paper, the force P would B N"
cause the arm AN to turn round A in the contrary Fi s- x s-

direction to the hands of the watch. Under these circumstances, it is
usual to consider the moment of P with respect to the point A to be
positive. If P acted from C to B, it would turn NA in the same direction
as the hands of the watch, and now its moment is reckoned negative.

The following remarkable relation exists between any forces acting in
one plane on a body and their resultant. Take the moments of the forces
and of their resultant with respect to any one point in the plane. Then
the moment of the resultant equals the sum of the moments of the several
forces, regard being had to the signs of the moments.

If the point about which the moments are measured be taken in the
direction of the resultant, its moment with respect to that point will be
zero ; and consequently the sum of the moments with respect to such
point will be zero.

39. Equality of Action an d Reaction. We will proceed to exemplify
ome of the principles now laid down by investigating the conditions of
equilibrium of bodies in a few simple cases ; but before doing so we must
notice a law which holds good whenever a mutual action is called into
play between two bodies. Reaction is always equal and contrary to,
action : that is to say, the mutual actions of two bodies on each other are
always forces equal in amount and opposite in direction. This law is per-
fectly general, and is equally true when the bodies are in motion as well
as when they are at rest. A very instructive example of this law has
already been given (33), in which the action on the spring CD (fig. 6) is
the weight W transmitted by the spring to C, and balanced by the re-
action of the ground transmitted from B to D. Under these circum-
stances, the spring is said to be stretched by a force W. If the spring
were removed, and the thread were continuous from A to B, it is clear that
any part of it is stretched by two equal forces, viz. an action and reaction,
each equal to W, and the thread is said to sustain a tension W. When a
body is urged along a smooth surface, the mutual action can only take
place along the common perpendicular at the point of contact. If, how-
ever, the bodies are rough, this restriction is partially removed, and now
the mutual action can take place in any direction not making an angle
greater than some determinate angle with the common perpendicular.



22



On Matter, Force, and Motion.



[40-




This determinate angle has different values for different substances, and
is sometimes called the limiting angle of resistance, sometimes the angle
of repose.

40. The lever is a name given to any bar straight or curved, AB, rest-

ing on a fixed point or edge c called the
fulcrum. The forces acting on the lever
are the weight or resistance Q, fat power
P, and the reaction of the fulcrum.
Since these are in equilibrium, the re-
sultant of P and Q must act through C,
for otherwise they could not be balanced
by the reaction. Draw cb at right angles
to QB and ca to PA produced ; then ob-
serving that P x ca, and Q x cb are the
moments of P and Q with respect to c,
and that they have contrary signs, we
have by (38),

P x ca = O x cb ;

an equation commonly expressed by the
rule, that in the lever the power is to the
weight in the inverse ratio of their arms.

Levers are divided into three kinds, according to the position of the
fulcrum with respect to the points of application of the power and the
weight. In a lever of the first kind the fulcrum is between the power
and resistance, as in fig. 16, and as in a poker and in the common steel-
yard ; a pair of scissors and a carpenter's pincers are double levers of this
kind. In a lever of the second kind \hs resistance is between the power
and the fulcrum, as in a wheelbarrow, or a pair of nutcrackers, or a door;
in a lever of the third kind the power is between the fulcrum and the
resistance, as in a pair of tongs or the treadle of a lathe.

41. The single pulley. In the case of the single fixed pulley, shown
fig. 17, it follows at once from (33) that when the forces P and Q are

in equilibrium they will be equal, the axle of the pulley being supposed

perfectly smooth and the thread perfectly
! ~ V flexible. The same conclusion follows

directly from the principle of moments ;
for the resultant of P and Q must pass
through C, or otherwise they would cause
the pulley to turn ; now their moments
are respectively P x CM and Ox CN,
and since these have opposite signs we
have (38)

PxCM = QxCN.

But CM and CN being equal, this
equation shows that P and Q are

equal. In the case of the single moveable pulley, shown in fig. 18, we
have one end of the rope fastened to a point A in a beam. The pulley




Fig. 17.



Fig. 18.



-43]



The Wedge.



is consequently supported by two forces, viz. P and the reaction of the
fixed point which is equal to P ; these two forces support Q and the
weight of the pulley w. In the case represented in the figure the parts
of the rope are parallel, consequently (36)




Fig. 19.

triangle DBC which is



Online LibraryAdolphe GanotElementary treatise on physics → online text (page 3 of 94)