390 CHAIN BRIDGE, 



From the principles of statics, when three 

 forces balance each other, any one is as the sine 

 of the angle formed by the other two; hence, 

 when the angles and one force are given, we 

 may easily find the other forces, or the reverse. 

 — Example. One of the links ce in the bridge 

 inclines 6^ from the vertical*, the point c being 

 turned toward the river : what would be the 

 tensions in ce and ca, and the effect to rend the 

 pier, during the fracture of the hnk dc? the 

 inclinations of dc and ac being as in Art. 4. 



We have dcf = 90<> - 18^.32' z= 71^.28', and 

 BCF = 90*=» - 33*^.10' = 56^.50'; therefore dcb 

 zz 7r,28'+ 56'*.50'z= 128*^.18', dce = supplement 

 of 71^.28' + 6<^ = 114^.82', bce = 360° - (dce 

 4-dcb) = 117**.10'. 



Whence sin. dcb = .780, sin. dce = .910, sin. 

 BCE = .884; and since tension in CD = 351 tons, 

 by Art. 4, we have as below :— 

 .884: .910 :: 351 : 361 tons = tension in CA, 

 .884: .780 :: 351 : 309 tons = tension in ce. 



If we resolve the tension in ce just found into 

 its vertical and horizontal effects, we have 

 309 X cos. 6** = 309 x .9945 = 307.3 tons, and 

 309 X sin. 6** = .309 x .1045 =32.3 tons. Hence 



• The defect mentioned is being removed at the time of 

 printing this paper. 



