Question

A tightrope walker who weighs 610 N walks along a steel cable. When he is halfway across, the cable makes an angle of 0.040 rad below the horizontal. (a) What is the strain in the cable? Assume the cable is horizontal with tension of 80 N before he steps onto it. Ignore the weight of the cable itself. (b) What is the tension in the cable when the tightrope walker is standing at the midpoint? (c) What is the cross-sectional area of the cable? (d) Has the cable been stretched beyond its elastic limit (2.5x108 Pa)?

Answer 1

For the change of length we need to find the difference in the length before the walker on the rope and after walker on the rope.

Initial length is L₁ and if L₂ is the length after the walker is on the rope then L₁ = L₂ cos 0.04 rad

The change in length is L₂- L₁ = L₂ - L₂ cos 0.04

Strain = change in length / initial length

   = L₂ - L₂ cos 0.04 / L₂ cos 0.04

   = (1- cos 0.04)/ cos 0.04

   = 8x10^-4

b)

Tension in the cable is

T = m g / 2 sin 0.04 = 610 / 2 sin 0.04

T = 7627 N

c) According to hooke law within elastic limit, stress is proportional to strain

F/ A = Y (strain)

A = F / Y (strain)

Y is not given in the question so googled it , 200x10^9

A = 610 / (200 x10^9)(8x10^-4)

A = 3.813x10^-6 m^2

d) Elastic limit

= T/A = 7627 /3.813x10^-6 m^2

   = 2x10^9

This is greater than 2.5x10^8 Pa so the cable has stretched beyond.