92. Composition and Resolution of Forces, &c. 
av, y,z we have Rcos. A, Rcos.B, Ros. C to be added to the first 
members of (h), hence they become X+R cos. A=0, Y-+R cos. 
B=0, Z+Rcos.C=0; «.R=VX?+Y?+Z?, (k), cos. A= 
».¢ Me : f 
— cos. B= —p, cos. C=— , (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’=p; cos. 
¥ Z, 
B’=p, cos.C’=p, (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’/= &c. cos. b= 
cos. 0/= &c. cos. c=cos. c= &c.; hence (i) become cos. aSm 
Fy—cos. bSmFxr=0, cos. aSmFz—cos.cSmFa=0, cos. ¢Sm 
Fy—cos.bSmFz=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) SmFxv=0, SmF y=0, Sm 
F z=0, (0) which determine the center of the parallel forces: and 
by (k) their resultant=SmF, .*. by (m) cos. A’=cos. a, cos. B/= 
cos. 6, cos. C’=cos. ¢; .’. the resultant is parallel to the compo- 
nent, and its magnitude is equal to the difference of the sums of the 
positive and negative components, and it is evidently directed the same 
way as the greater sum. Change a, y, z, a’, &c. in (0) intox—X, 
y—Y,z—Z, «’—X, &c., then we have SwF (# -X)=0, SmF 
( i SmF a SmFy 
(y—Y)=0, SmF(z—Z)=0, which give X= oor’ ‘ie aie 
Smi'z P 
L=SF? (p); which show (generally,) that the parallel forces 
have but one center. If in (p) we have SnF=0, SmFx=0, Sm 
Fy=0, SmFz=0, they will be under an indeterminate form ; but 
Sil Bal Sin Blyt ) ) veiSmalzt 
by (p) we have Ee = Gain (Ler Sais for the 
center of all the forces except F; now mE =—Sm’/E’, mFx=—S 
m/F’x’, and so on, hence X’=x, Y’/=y, Z’=z; .’.in this case any 
one of the forces is equal in magnitude to the resultant of all the 
