308 On the Parallelogram of Forces. 



ing the components and the resultant in the directions of x, y, z re- 

 spectively, we shall have r cos. a + r' cos. a' -f- , he. = R cos. A' 

 r cos. b -\- r cos. 6' + , &lc = R cos. B 7 r cos. c-fr' cos. c'+> &c. 

 Rcos. C, (9) ; whose squares, when added, give (r cos. a+r' cos. a, 

 + 3 &lc) 3 + (r cos. 6 + r' cos. 6'+, &,c.) 2 -{- (rcos. c -\- r' cos. c'+, 

 &&c.) 2 =R 2 , (10) ; hence R being found, the angles A, B, C are ea- 

 sily found by (9) ; and it may be remarked that the 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 — 0, .\ 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=0, 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 x and y, whose directions form a right angle, 

 be applied as before to M, also let z denote the resultant, 6 the angle 



"' P 



which its direction makes with that of #, then - — = the angle which 



2 b 



its direction makes with that of y. Hence we shall have (1) and (2) 



in the same manner as before; if y=0, 4=0, .\ <pj _ — 4] becomes 



<p[- ] =0 ; if y is indefinitely small relative to z, 6 will be indefinite- 



ly small, and may be denoted by rfJ ; hence representing the value 



of y in this case by dy, we shall have by (2) -¥=$( - — d& 



z \2 



cpl- ) — M), by neglecting quantities of the orders dP, dd 3 , kc. as 



is evident by the method of indeterminate coefficients, or Taylor's 



theorem ; but we have proved that p[ - j =0, .*• -1 = — kdA, (1') j 



\2 / z 



where k evidently = a constant quantity. Suppose that x and y 

 are each resolved into two forces x\ j?", and y\ y" ; #', y' acting in 

 the direction of z, and #", y" perpendicular to it ; then we evident- 

 ly must have x / +y / =z, (2 / ), x"=y", (3'). 



p 



Since x forms the angle 6 with x f , and - —6 with #", and that y 



2 



