APPENDIX. ON THE COMPOSITION OF TWO FORCES ACTING ON A POINT. 1 1. SINCE two forces which are in equilibrium must necessarily be equal and opposite, two forces F, and F, which do not act in opposite directions, must necessarily have a resultant, the position of which we shall proceed to determine. (1) The resultant of two forces F, and F, acting on a point P, is situated in the plane F ̧PF1⁄2. 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 F, and F; 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 F,PF. 1 For the tendency of F, is to draw the particle P in the direction PF1, while that of F2 is to draw it in the direction PF, 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 FPF, consequently the resultant, which is a single force that would produce the same motion, must be situated within the angle F1PF,. 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 n; consequently 1 R < F1 + F2. 2 3. If the forces F, and F2 are equal, their resultant R will bisect the angle F,PF2. For if there be a reason why PR should lie nearer to PF, than to PF2, there must be a similar reason why it should lie nearer to PF, than to PF, since the forces are equal; and hence there would be two resultants, which is impossible: consequently PR bisects the angle F、PF2. 2 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 F1Pƒ1. Since R is less than the sum of the two forces F1 and fi 1 is always less than may be found such that 2 F1 cos 0. 1 The angle is unknown at present, but from Art. 19, we learn that so long as the angle F, Pf, remains the same, continues unchanged; that is, if we have two sets of forces inclined at the same angle with each other respectively, we shall have R2F, cos 0, and R' = 2F cos 0, and therefore R: R': F, F............(A), :: that is, the resultants are proportional to the components. Let now F2, f be two other equal forces acting on P whose resultant is also equal to R, the angles F1PF2, f1Pƒ2 being each equal to RPF, or RPf. Now at P apply four forces, each equal to a, two of them respectively in the directions PF2, Pf2, and the other two in the direction PR; and let them be of such magnitude, that F, may be the resultant of the one in the direction PF, and one in the direction PR. Then, since these two contain the same angle as F1 and fi, and F, is their resultant, 2 = 2x cos 0. 1 Also, if we substitute instead of F, and f, their components, we may consider R as the resultant of the forces x, x, x, and a; 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 x, x, 2 2. from (A); if we double the angle at which the forces are inclined, we must also double 0. We will now suppose, that when the angle at which the forces act is a multiple n, or any inferior multiple, of FPf, it is true that in the same formula the corresponding equimultiple of is to be taken; so that R = 2F cos n0=2F-1 cos (n − 1)0 = ... = 2 F1 cos 0. Apply (fig. 108) at P, as before, four forces in the directions PF +1, PF-1, Pfn+1, and Pf-1 respectively, each of such a magnitude that F may be the resultant of the two in the directions PF PF and f of the other two; n+19 2-19 :. Fn = 2x cos 0.* 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 x, of which two acting in the directions PF-1, Pf-1 will have 2a cos (n - 1) 0 for their resultant in the direction PR, and consequently R - 2x cos (n − 1) 0 is the resultant of the other two which act in the same directions as Fn+1 and fn+1; consequently, from (A), R = = = R R · 2 cos (n − 1) 0 n+1 : x; 2 F2 cos n = 4x cos 0 cos n 0; = 4 cos cos n0 - 2 cos cos ne - 2 sin o sin ne Fa+1 = 2 (cos e cos n0 – sin ✪ sin n0) = 2 cos (n + 1) 0 ; .. R = 2 F2+1 cos (n + 1) 0. 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. *For the 4Fn+1 PF-1 = LF1 Pf1, (Fig. 107). 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 0 still holds good. This, however, supposes the angle between the forces to be a multiple of F, Pf, (fig. 107), which may not happen to be the case; but by taking the original angle F1Pf, 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 0 have an invariable ratio to each other, so that if FPf= 20, then To determine the value of c, we observe that if F .. c = an odd integer, = 1 as we shall shew. ET N For if c is not = 1, let the angle FPf be such that π = 2c which is therefore less than a right angle, and then |