A steel cable supporting a 670 kg load experiences an elastic stretch proportional to its free length. For a cable length of 12m, the calculated stretch is 0.0008m. For a longer cable length of 350m, the calculated stretch increases to 0.023m due to the cable's increased free length.

1. 44. To support a load of W = mg = (670)(9.8) = 6566 N, the steel cable must stretch an amount proportional
to its “free” length:
W
∆L = L where A = πr2
AY
and r = 0.0125 m.
(a) If L = 12 m, then
6566
∆L = (12) = 8.0 × 10−4 m .
π(0.0125)2 (2.0 × 1011 )
(b) Similarly, when L = 350 m, we find ∆L = 0.023 m.