694 
ME. HOPKINS ON THE THEOET OF THE MOTION OF OLACIEES. 
“We have 
cos ^ = cos X cos a + cos cos /3+ cos v cos 7, 
whence we immediately obtain 
^cos^=Acos®a +B cos^|3 + C cos® y4-2Dcos/3cos 7+2E cos a cos 7+2F cos a cos |3 ; 
and since 
cos® a + cos® /3 + cos® 7=1, 
we have (L being an arbitrary multiplier), 
(A+L) COS®05 +(B+L) cos® /3+(C4-L) cos® 7+2D cos /3 cos 7+2E cos a cos 7 
H-2F cos a cos (5=maw. 
Hence 
{(A+L) cos a +E cos 7+F cos |3} sin 05=0,1 
{(B+L) cos /3+D cos 7 +F cos a} sin/3 = 0, i 
{(C+L) cos 7+D cos /3+F cos «} sin 7 = 0. J 
(L) 
( 2 .) 
(b.) 
“ To satisfy these equations together with (2.), we must equate the first brackets to 
zero. We thus have four equations from which L may be eliminated, and 05, (3, and 7 
determined. 
“ If we multiply the first factors on the left-hand sides of equations (b.) by cos «, 
cos j3, cos 7 respectively, and add them together, we have, by vh’tue of equations (a.), 
L= cos S ; 
and substituting for L in equations [b.), we have 
p cos 5 cos 05= A cos a+F cos j3+F cos 7 
cos X 
by the first of equations (a .) ; and therefore 
cosS cos£e= cos'X, 
and similarly 
cos S cos j3= cos j«/, 
cos 'b cos 7= cos V, 
whence 
cos® ^=1 
^= 0 , 
which shows that when the resultant force p is a maximum or minimum, its direction 
coincides with that of the normal to the plane s. Consequently, also, the tangential 
force, y) sin b, then becomes = zero. 
“ This value of ^ gives L= ; and substituting for L in equations (b.), we have 
(A— jp) COS05+F cos|3+E cos 7=0, 
F cos a+(B— ^) cos j3+D cos 7=0, ( (c.) 
E cos a+D cos |3+(C— ^) cos 7=0 ; 
and eliminating cos a, cos (3, and cos 7 by cross multiplication, we obtain 
(A-j?)(B-^)(C-p)-D®(A-p)~E®(B-^)-F®(C-i))+2DEF=0. . . (3.) 
