Composition and Resolution of Forces, &c. 93 
rest, and directly opposite to it, which are evidently the conditions 
of equilibrium of a free system, when acted on by parallel forces. 
If SmF=0, but the numerators of (p) are not =0, then evidently 
any one of the forces equals the resultant of all the rest, but is not 
directly opposite to it; .’. there is properly nocenter of parallel for- 
ces in this case, although it is usual to say that the resultant =O, and 
is situated at an infinite distance. If the forces F’, F’, &c. have the 
same sign, and are equal to each other, (p) become ene y— 
Smy Smz : 
Gm? 2 Sm? (q), which are the well known formule for finding 
the center of gravity of a system of bodies; the resultant of the 
forces=F'Sm (=the weight of the system if F=geravity.) Put S 
m==M =the mass of the system, then by (q) M? X?=(Smx)?=m? 
ve? malt tm a2 + &e. +2Qmm/ xv! +m" vcr"! + Qn! mx! el! 
+&c.; but Quan’ =x? + x!* — (a! —2x)?, Qa" =a? +9? — (x —2)?, 
Qar/ao!’ =? 4-a//? — (a/’— x’)? and so on; by substituting these val- 
ues we have M? X?=MSmx*? —Smim’(x' —x)?; hence and by (q) 
soa ug set he +27) Smam’( (x’—«)? oa (y-y)? +(2/-2)*) 
(r) ; which gives the distance of the center of ne from the ori- 
gin of the coordinates by means of the distances of the bodies from 
the same point, and of their distances from each other; by finding 
in this way the distances of the center from three given points, 
which are not in the same straight line, its position in space becomes 
known; Mec. Cel. vol. 1. p. 45. 
We will now suppose that the system is to he in equilibrium about 
two fixed points which are invariably attached toit. Let the points 
be denoted by the letters m and n, then suppose the origin of the co- 
ordinates to be at m, R its reaction, ‘A, B, C the angles which the 
direction of R makes with the axes of x, y, z (as before,) R/ the re- 
action of n, A’, B’, C’ the angles which its direction makes with those 
of x, y,2; X, Y, Z the coordinates of n: then by including R and 
R/ among the forces in (h) and (i) they become SmF cos. a+-R cos. 
A+R’ cos. A’=0, SmF cos. 6+ R cos. B+ R’ cos. B/= 
SmF cos. c+R cos. C-+-R’ cos. C’=0, (s); SmF (y cos. a—2 cos. 
b) + RY cos. A’—X cos. B/)=0, SmF (z cos. a- @ cos. c)+ 
R(Z cos. A’— X cos. C’) =0, SmF (y cos. c—z cos. 6) +R/(Y cos. 
C’ —Z cos. B’)=0, (t). which are the equations of equilibrium, be- 
tween the applied forces and the reactions of the fixed points. Put 
