THE CATENARY 81 



Resolving horizontally, we obtain 



= 0. . (18) 



Resolving vertically, we obtain 



Tsm6-ws = Q. (19) 



To find the shape of curve in which the string will hang, we 

 must obtain a relation between 6 and s. Eliminating T, we obtain 



H tan 6 = ws, 

 or, if we replace H/w by the single constant c, 



(20) 



This is one form of the equation of the curve, s and being taken as 

 coordinates. The equation in this form is known as the intrinsic equation 

 of the curve. We require, however, to deduce the equation in its Carte- 

 sian form. 



59. If the point o in fig. 42 is taken as origin, the axes being 

 horizontal and vertical, we have at once the rela- > 



fcl ns dx : dy :ds = cos 6 : sin d : 1, (2 1) ds/ 



/ dy 



for dx and dy are the horizontal and vertical pro- 



jections of the small element ds of length of the 

 string. As a first step, let us use relations (21) to 

 change the variables of equation (20) from s and 6 to s and y. 



We have //7 \ 2 



c 2 = s 2 cot 2 = s 2 cosec 2 - s 2 = s 2 1 } - s 2 , 



\dy] 



,, ds 



so that s = 



dy. 



Thus dy = 



and integrating this, we obtain 



y = V s 2 + c 2 + a constant. (22) 



We can determine the constant of integration as soon as we 

 decide where the origin is to be taken so far we have not fixed 



