The element dS of the cable at P (or Q) is in equilibrium under the action 

 of the system of forces comprising its weight WdS, the force R sin a $ dS normal to 

 the cable, the force FdS along the cable and the tension at its ends. Denoting the 

 tension of the cable at any point by T and resolving along the cable, we obtain the 

 equation 



dT/dS = F + W sin $ (2) 



which may be combined with (1) to eliminate sin 4> and then integrated to give 



T = T Q + FS + WY (3) 



where T Q is the tension of the cable at the origin. Resolving also at right angles 

 to the cable we obtain 



Td4>/dS = W cos $ + R sin a <|> , <f><0 (at P) (4a) 



and Td(f>/dS = W cos <|> - R sin a $ , <)>>0 (at Q) (4b) 



Equations (2) and (4a) may be combined to eliminate S and then integrated 

 to give a relationship between T and <|>. For, from (2) and (4a) we obtain 



dI/ra » • 1 cLVM"s?n' t * co/»Vr"sU ■ —~ * - V* , r - U ..(5) 



Introducing the substitution Z.. = - tan -*- , (5) may be integrated (see Appendix 

 III) to give 



log t, = 2f -^fo tan-Vz, ♦ jr^g, log j * glfa - 



f pH^l°gff|^ (« 



where p a = \j4r a + 1 + 2r, and ^ = T.,/To, T, s T for <|><0 



We can now express Y in terms of T and Z.. by eliminating S between equations 

 (1) and (4a). Thus we obtain 



dY/T 1 d * = W cos ft t sin" ♦ C7) 



Making the substitution Z 1 = - tan <|>/2, (7) becomes 



Y - f 1 4Ta H!lSL (M 



Y " J TT (p a -Z 1 s )(T/p a + Z 1 a ) (8) 



o 



Equations (6) and (8) have been derived for the case $<0. Correspondingly, for 

 ♦ >0, we obtain from equations (2) and (4b) 



