SUSPENSION BRIDGE 79 



The forces acting on the piece OP of the cable consist of 



(a) the tension at 0, of amount 77" acting horizontally; 



(b) the tension at P, of amount T acting at an angle 6 with 

 the horizontal; 



(c) the tensions of the vertical chains, all acting vertically. 



Resolving horizontally, we obtain 



H-Tcos0 = Q. (15) 



Resolving vertically, we obtain 



Tsm6- S = 0, 



where S is the sum of the tensions of all the chains which leave 

 the cable between and P. These tensions support the portion 

 op of the bridge, and if the weight of the bridge is w per unit 

 length, the weight of op will be wx. Thus S = wx y and therefore 



Tsm0 = wx. (16) 



This and equation (15), T cos = H, (17) 



will give us all the information we require. 



To find the shape which the cable must have in order that the 

 bridge may hang horizontally, we require to obtain a relation 

 between 6 and x. We accordingly eliminate Tfrom equations (16) 

 and (17), and obtain 



tan 6 = x. 

 H. 



If y is the height of the cable above the bridge, x t y may be 

 regarded as the Cartesian coordinates of a point P on the cable, 

 and we have , 



tan0 = ^. 

 ax 



Thus the coordinates x, y of P are related by 



dy _ w 

 dx = H X> 



"I f\ ij 



giving on integration y = x 2 4- O, 



U -tL 



where C is a constant of integration. 



