308 On the Parallelogram of Forces. 
ing the cothponents and the resultant in the directions of x, y, 2 re- 
spectively, we shall have rcos. a + 7’ cos.a’ +, &c. = Reos. A’ 
r cos. 6b +r’ cos. 6’+, &c. = Ros. B, rcos.c+r’ cos. ce’ +, &c. = 
Ros. C, (9); whose squares, when added, give (7 cos. a+r’ cos. a, 
+, &c.)? + (rcos.b + r' cos. 6'+, &c.)?+ (rcos.¢ + r' cos.c’+, 
&c.)° =R?, (10); hence R being found, the angles A, B, C are ea- 
sily found by (9); and it may be remarked that ae known rules for 
the algebraic signs of the cosines must be observed. 
If M is free, and the conditions of equilibrium are required, then 
we must have R=O, .’. the first members of (9) being each put =0, 
will be the conditions required ; if the first members of (9) are each 
identically =0, then R=O, and M will not be affected by the forces. 
If M is not free, but is pressed by the forces against any line or sur- 
face, it will be necessary that R should be at right angles to the line 
or surface, so that it may be destroyed by the reaction. 
We will now consider the subject after the manner of La Place, at 
pp: 4; 5, Vol. I. of the Mecanique Celeste. 
Let the two forces w and y, whose directions form a right angle, 
be applied as before to M, also let z denote the resultant, 4 the angle 
which its direction makes with that of w, then .- é= the angle whick 
its direction makes with that of y. Hence we shall have (1) and (2) 
in the same manner as before; if y=0, 0=0, .”. (5 if s) becomes 
°(5) =0; if y is indefinitely small relative to z, 6 will be indefinite- 
ly small, and may be denoted by d)5 hence representing the value 
of y in this case by dy, we shall have by (2) ~ 25 (5-48) = 
(3) —kd, by neglecting quantities of the orders dj*, dj?, &c. as 
is evident by the method of indeterminate coefficients, or Taylor’s 
theorem ; but we have proved that o(5 = Pipes eee Sh ere AIU A) 
where & evidently = a constant quantity. Suppose that # and y 
are each resolved into two forces 2’, x”, and y’, y” 3 2’, y’ acting in 
the direction of z, and x”, y” perpendiculer to it; then we evident- 
ly must have a! + y'=z, (2’), e”=y", (3’), 
Since x forms the angle 4 with a’, and aoe with #”, and that ¥ 
