APPENDIX.

ON THE COMPOSITION OF TWO FORCES ACTING ON A POINT.

2

1. SINCE two forces which are in equilibrium must necessarily be equal and opposite, two forces F1 and F2 which do not act in opposite directions, must necessarily have a resultant, the position of which we shall proceed to determine.

2

(1) The resultant of two forces F1 and F2 acting on a point P, is situated in the plane F1PF2.

For if it be not in that plane, it must be either above or below. But it cannot be above; for, any reason which would assign it such a position might be used to assign it a similar position below; for these two positions are similarly situated with regard to the forces F1 and F2; there would consequently be two resultants, which is impossible. The resultant then cannot be situated above the plane of the forces; and in a similar way we may shew that it cannot be situated below, and therefore it must be in the plane.

(2) It lies within the angle FPF2.

1

2

For the tendency of F1 is to draw the particle P in the direction PF, while that of F2 is to draw it in the direction PF2, and hence it is probable the real motion, which is the result of these united tendencies, will not be in the direction of either, but intermediate to both; and therefore within the angle HPF: consequently the resultant, which is a single force that would produce the same motion, must be situated within the angle F1PF2.

What is here stated with regard to the 2nd Case, can hardly be called a proof, but is rather a strong reason for presuming that the resultant is situated within the angle included by the forces. Perhaps, after what is said in the 1st Case, it may be regarded as an axiom.

2

2. Since F1 and F2 do in part hinder each other from producing their whole effects, it appears that their resultant must be less than their sum; for their resultant can only be equal to their sum when neither interferes with the other, which is not the case unless they act in the same direction; consequently

R<F1 + F2.

3. If the forces F1 and F2 are equal, their resultant R will bisect the angle F1PF2.

For if there be a reason why PR should lie nearer to PF1 than to PF2, there must be a similar reason why it should lie nearer to PF2 than to PF1, since the forces are equal; and hence there would be two resultants, which is impossible : consequently PR bisects the angle F1PF2.

4. Having thus determined the direction of the Resultant of two equal forces, we proceed to find its magnitude.

Let F1, fi (fig. 107) be two equal forces acting on the particle P, and R their resultant bisecting the angle F1Pfi. Since R is less than the sum of the two forces F1 and fi

it is clear that

R

R

is always less than F1 + fa' or its equal F 1; and, consequently, an angle e may be found such that

The angle o is unknown at present, but from Art. 19, we learn that so long as the angle F1 Pf remains the same, θ continues unchanged; that is, if we have two sets of forces inclined at the same angle with each other respectively, we shall have R = 2F, cos 0, and R' = 2F1' cos e, and therefore

R: R':: F1: F1(Α),

1

that is, the resultants are proportional to the components.

2

Let now F2, f2 be two other equal forces acting on P whose resultant is also equal to R, the angles FPF2, f. Pfa being each equal to RPF, or RPf1. Now at P apply four forces, each equal to &, two of them respectively in the directions PF2, P.f2, and the other two in the direction PR; and let them be of such magnitude, that F1 may be the resultant of the one in the direction PF2 and one in the direction PR. Then, since these two contain the same angle as F and fi, and F1 is their resultant,

F1 = 2 x cos θ.

Also, if we substitute instead of F1 and fi, their components, we may consider R as the resultant of the forces x, x, x, and ; of which two act in the same direction as R; and, consequently, R - 2x is the resultant of the two x, x, which act in the directions PF2, Pf2; and since, by hypothesis, R is the resultant of F2 and f2, which act in the same directions as a, X,

R:R- 20 :: F2: x, from (A);

But R = 2 F1 cos 0 = 2.2x cos θ. cos 0 = 4x cos2 0;

if we double the angle at which the forces are inclined, we must also double θ.

We will now suppose, that when the angle at which the forces act is a multiple n, or any inferior multiple, of F1 Pf1, it is true that in the same formula the corresponding equimultiple of e is to be taken; so that

72

...

= 2 F1 1 cos 0.

R = 2F cos ne = 2Fn-1 cos (n − 1) 0 = Apply (fig. 108) at P, as before, four forces in the directions PFn+1, PF-19 Pfn+1, and Pfn-1 respectively, each of such a magnitude a that F may be the resultant of the two in the directions PFn+1, PF-1, and f of the other two;

But if, instead of the forces Fn, fn, we substitute their four components, we may consider R as the resultant of the forces x, x, x, and a, of which two acting in the directions PF-1, Pfn-1 will have 20 cos (n - 1) 0 for their resultant in the direction PR, and consequently R - 2x cos (n − 1) θ is the resultant of the other two which act in the same directions as Fn+1 and fn+1; consequently, from (A),

= 4 cos e cos nθ – 2 cos e cos nθ – 2 sin e sin ne

Hence the formula is true for a multiple (n + 1) if it be true for n and all inferior multiples: but it has been shewn to be true for 2 and 1, and consequently it is true for multiples 3, 4, 5, 6, and generally, by induction, for any multiple whatever.

...

It appears then, that as we increase the angle at which two equal forces (F, f) act, we must increase the angle in the same proportion, and then, that the formula

R = 2 F cos θ

still holds good. This, however, supposes the angle between the forces to be a multiple of F1 Pf, (fig. 107), which may not happen to be the case; but by taking the original angle F1 Pf1 exceedingly small, we may find a multiple of it which shall differ from FPf a proposed angle by less than any assignable quantity. It is evident then, that FPf and have an invariable ratio to each other, so that if FPf = 24,

then

θ

= constant quantity = c suppose; あ

.. R = 2F cos cф.

To determine the value of c, we observe that if F and fact at an angle w, or are opposite to each other,

Now none but angles which are odd multiples of have their cosines = 0 ;

C

an odd integer, = 1 as we shall shew.

π

2

For if c is not = 1, let the angle FPf be such that

=, which is therefore less than a right angle, and then

2c