92. Composition and Resolution of Forces, &c. 
x, y, z we have R cos. A, Ros. B, Ros. C to be added to the first 
members of (h), hence they become X+R cos. A=0, Y+R cos. 
es. Z+Reos. C=0; ..R=vVX2+Y?+Z?, (k), cos. A= 
Y Z 
—R ©. B= —R, cos. C= — 55, (1), which give the magnitude 
and direction of the reaction of the point; but it is evident that the 
resultant of the applied forces equals the reaction of the point, and 
is directly opposite to it; .*. put A’, B’,C’ for the angles which its 
direction makes with those of x, y, z, and we have cos. A= — cos, A’, 
| x 
cos. B= —cos. B’, cos. C = —cos. C’, .’. by (1) cos. A’=R, cos. 
es cos. wes (m), which give the direction of the resultant, 
after having found its magnitude by (k). 
If F, F’; &c. act in parallel directions ; then considering those 
which act one way as positive, and those which act the contrary way 
as negative, we shall have cos. a=cos. a/==cos. a’ = &e. cos. b= 
cos. b/= &c. cos. c=cos. c= &c.; hence (i) become cos. aSm 
Fy—cos. Sm Fr=0, cos. aSmFz-cos.cSmFxr=0, cos. c Sm 
Fy=cos.6SmFz=0, (n). Supposing F,F’, &c. together with 
their points of application to be invariable, but the angles a 6 ¢ 
to be indeterminate, we have by (n) SmFx=0, SmF y=0, Sm 
F z=0, (0) which determine the center of the parallel forces: and 
by (k) their omnia: .". by (m) cos. A’=cos. a, cos. B/= 
cos. 6, cos. C’/=cos. c; .". the resultant is parallel to the compo- 
— anhits anngoitinie’’ is equal to the difference of the sums of the 
ts, and it is evidently directed the same 
way as the. greater sum. “Change x, y, 2, a’, &e. in (0) intor—X, 
y—Y, z—Z, x’—X, &c., then we have SmF (« —-X)=0, SmF 
SmF x SmFy 
i =< als Sak (z— —Z)=0, which give X= CaF? Y= Snr’ 
pa fete (p); which show (generally,) that the parallel forces 
have but one center. If in (p) we have SmF=0, SmF'x=0, Sm 
Fy=0, SmF'z=0, they will be under an indeterminate form ; but 
Sm’ Fa! S/F’ Sin’ Fz’ 
by (p) we have X/= —— , Y= aoe hi = SpF7 > for the 
center of all the forces except F ; now mF =—Sm/F’, mFr=—5 
m/F’z’, and so on, hence X’=2, Y’/=y, Z’=z; .’.in this case any 
one of the forces is equal in magnitude to the resultant of all the 
