Cable 80 m: lowest point, length, tensions

Set-up. A is the lower end and B is 10 m higher. The lowest point O sags 7.5 m below A, so B is $7.5 + 10 = 17.5$ m above O. Load $w$ = 1.5 kN/m on the horizontal span; $y = \dfrac{wx^2}{2H}$.

(i) Position of the lowest point

$$\frac{x_A}{x_B} = \sqrt{\frac{7.5}{17.5}} = 0.6547,\qquad x_A + x_B = 80$$ $$x_B = \frac{80}{1+0.6547} = 48.35\ \text{m},\qquad x_A = 31.65\ \text{m}$$

The lowest point is 31.65 m horizontally from the lower end A.

(iii) Horizontal tension and end tensions

$$H = \frac{w x_A^2}{2y_A} = \frac{1.5(31.65)^2}{2(7.5)} = 100.18\ \text{kN}$$

(Check from B: $\dfrac{1.5(48.35)^2}{2(17.5)} = 100.18$ kN.)

(ii) Length of the cable

With $s = 2y/x$ at each end and $l = \dfrac{x}{2}\left[\sqrt{1+s^2}+\dfrac{\sinh^{-1}s}{s}\right]$:

Total length = 85.09 m.

Answer: (i) lowest point 31.65 m from A; (ii) cable length = 85.09 m; (iii) $H$ = 100.18 kN, $T_A$ = 110.86 kN, $T_B$ = 123.68 kN.