310 On the Parallelogram of Forces. 
Suppose then, that x is changed to x+.’, but that y is the same 
as before ; let z’ be the resultant of these forces, 6’ the angle which 
its direction makes with that of a; then it is manifest that 4’ is 24, 
and that 4 —6’= the angle formed by the directions of z’ and z. Put 
6— 4’=», (a), then by resolving z/ in the direction of z by (1), we 
shall have 2’p(v)= the resultant resolved in the direction of z; but 
x'p(4) =a resolved in the same direction, and by adding z, we have 
the components reduced to the same direction; hence 2’9(v)=z-+ 
x’p(6), (0). 
Put “=m, gy, ee Ean Aan ree (4’),m? nae 
g 
Zz z! 
m!? +-n/? =1, wee. a het ctx’ =m'2', y—=nz—= nz’ ; hence 
we shall have 2/=—, 2’ =m'z’ —o= (mm ), but by (1),<= 
n! n 
m=9(9), hence z+-x/9(d)=(n/(1— m?)-+ nmm’)= = (since 1— m* 
n! 
=n?,)=(n/n + mm!)"~ = (mm! --nn’ 2! ; hence by (0), we shall have 
n 
o(v)=mm/+nn’, (e). It is evident that @ and 4’ are independent of 
each other, and that m, n are functions of 4 without 4’, and that m’, 
n’ are functions of & only; hence by putting ao, and tak - 
v 
ing the ce eae (e) relative See 6’, we shall Be 
Hoe), me dv 
= ss =— 3 but by al, 7 
nT aoe ne a an mtn ee batby(«) di’ 
—Il,. ; avid A i= Fh (me Cee (f)- Now by (d), 
dn m_,dm dn’ oe dim! 
ca aa cvs , by substituting these values in ( 
dj n dd dd! no di’ cure f); 
/ 
we have by reduction, Onn on (¢). Since the first mem- 
ber of (g) is a function of é only, and the second of 8’, it is evident 
that each member = a const.; .°. let — /& denote the constant, and 
since by (d) n=V 1 — m?, we shall have by multiplying by ds, 
Beat =hkd), whose integral gives m =" = cos. (A3-+c), ¢ = the 
z 
V1il—m 
correction; hence (as before,) we shall have e=z cos. 6. 
