The Rev. H. Lloyd on the mutual Action of permanent Magnets. 167 



And the forces exerted by c upon the same element, in the direction bc, and in 

 the direction perpendicular to bc, are 



2 Cm , . Cm., 

 ^^*^n«-U' - -^-sinCa-f). 



Resolving these forces, as before, in the direction of the magnetic meridian, 

 and in the direction perpendicular to it, and making the sum of the resolved 

 parts in each direction equal to nothing, the equations of equilibrium are found 

 to be 



c c . "> J. 



— 5 -< 2 cos (a — f ) cos a — sin (a — f ) sin a V -| (2 cos^ 7 — sin* 7) = 0, 



— 5 -j 2 cos (a — f ) sin a -\- sin (a — f) COS a I -j- 3 —j sin 7 COS 7 = 0. 



If we resolve the trigonometric products, and make, for abridgment, 



A B a h 



^=P, -^=Q, -=p, - = g, 



the four equations of equilibrium become 



3cos(2/3 — f) + cos^ -1-3 Qq^sm2y = 0, (10) 



3sin (2j3 — f) + sinf + ^^'(1 — 3cos27) = 0, (11) 



3cos(2a — f) + cosf +P/(14-3cos2 7) =0, (12) 



3sin(2a — f) + sin f-1- 3 Pp' sin 2 7 = 0; (13) 



of which (10) and (12) relate to the forces in the magnetic meridian, and (11) 

 and (13) to those perpendicular to it. The ratios p and q are functions of the 

 angles a, j3, 7, f, expressed by the formulae : 



sin (^-7) _ sin (g - 7) 



^~sin(a-j3)' ^~sin(a-|3)' ^^ 



The complete solution of the problem is contained in the preceding equa- 

 tions ; and it follows, in general, that they may be satisfied by means of the four 

 arbitrary angles, a, /3, 7, f, — and consequently the desired equilibrium produced 



