Mr. Moseley on the Equilibrium of the Funicular Curve. 325 
from the last equation 
f (Xdz4 Vdy+T++ Pec 
ao. T?4 OTE = CE —2/(Xdzx + Xdy)E 
“ T=—E+ V E74 {C—2 [(Xdz+ YdyRE 
if one extremity of the string be free so that at this extremity 
_T =0, and the integral be taken from this point as a limit, 
then C = 0, and the equilibrium becomes impossible, unless 
E>2 /(Xdx+Ydy). 
In the impossible case, a continual motion will be com- 
municated to the string by the action of the forces upon it. 
In the case in which the extremities of the string are joined: 
if M be the value of /(X dz + Ydy) taken through the arc 
(s), and N its value taken throughout the whole length plus the 
quantity s, we have, since in the latter case T becomes — T 
T? + 2TE = C — 2ME 
T? — 2TE = C — 2NE 
eT =(N—M) _ 
N-—M 
— 
- __ To determine the equation to the curve, we have generally, 
if {C —2/(Xdx+ Ydy)}{E=K. 
E E \i 
Spe ye Gar ) 
= —E+ (E?+ KE)! 
63 
+ — E+ (E+ KE) eV a ik ea gf 
/ #(E +5)? 
Whence the curve may in all cases be determined. Also 
when the force acts from a centre, we have Xy— Yr =0 
. by the equation («) 
d(Tp)=0 
—E+(E?+ XE)} 
the double value of (p) for a given value of (r), shows two 
positions of equilibrium ; in the case in which p is negative 
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