86 Composition and Resolution of Forces, &c. 
that it may be destroyed by the reaction. Let X, Y, Z, denote the 
sum of the components of all the forces which affect M, when resol- 
ved in the directions of x, y, z, severally ; Let N denote the reaction 
of the surface, p= V (x —d)?+(y—e)?+(z—f)? =the perpendic- 
ular to the surface drawn through M, d, e, f being the coordinates of 
the origin of p; we shall suppose that the origin of p is taken on 
that side of the surface, towards which N is dinaetedls 
Now by resolving N in the directions of 2 y z, 
e—y t Stas 
Nx > > NX ? > for the values of N when reduced to those di- 
rections ; hence for the equilibrium of M, we have X-++N X = =O, 
m ee 
Y+Nx 5 = Z+Nx <=, (15). By the nature of the 
= dy + (=) az = 0, 
du du du : 
(16), and by (14) Felt qyly + qg2= 0 (17), multiply (17) by 
the indeterminate L, add the product to (16), then put the coeffi- 
cients of the indeterminates, dx, dy, dz separately, =O, and we have 
du du 
ane =0, es i +L7, =0, ==L 4 LE=0, (18) ; hence L= 
a (3). +(2 a aca -'. denoting this value of L, when | 
real by N, by N’, we shall have by (18) and (15), X+ 
r—d 
perpendicular, we have (== Jae ( 
Nist=0, vine anil); ZN =0, (19); by eliminating N’ 
du du du 
from (19) we have x -VS =0, Xz, -~Z7 =0, (20), for the 
conditions of the equilibrium of M ; ay and ii are sufficient to 
find where M must be placed on the given surface, to be in equilib- 
rium, and by (19) we have N=V X2+Y2+4-Z? = the force with 
which the surface must react in order to destroy the resultant of the 
applied forces. If M is to be in equilibrium on a line, which is 
formed by the mutual intersection of two surfaces, which are deno- 
ted by u=0, u’==0, (21); then by using the same notation as be- 
ek pin emt (+(e ne 
