Question

A uniform flexible steel cable of weight mgissuspended between two points at the same elevation (60degrees).Determine the tension in the cable at its lowest point andat thepoints of attachment.

So, I think that the weight of the cable, mg, has tobebalanced by the other forces in the y-direction, which would bethetension on both sides of the cable.

Let tension on the right side be F_T1 and tensiononthe left be F_T2.

All of the forces must sum to zero if the cable isinequilibrium,

F_T1 sin 60 + F_T2 sin 60 - mg =0

F_T1 sin 60 + F_T2 sin 60 = mg

Because the cables are in equilibrium, the tension in eachmustbe equal. (F_T1 = F_T2)

So we can write:

2F_T1 sin 60 = mg

F_T1 = mg/(2 sin 60)

F_T1 = 0.58mg

But according to the book, this is the answer to part b,notpart a. I'm not sure what I'm doing wrong?

Thanks in advance for your help.