Example 5.3 from College Physics, 5.3 Elasticity: Stress and Strain

The Stretch of a Long Cable

Suspension cables are used to carry gondolas at ski resorts. (See Figure 5.14) Consider a suspension cable that includes an unsupported span of 3020 m. Calculate the amount of stretch in the steel cable. Assume that the cable has a diameter of 5.6 cm and the maximum tension it can withstand is $3 . 0 \times 10^{6} N$ .

Ski gondolas travel along suspension cables. A vast forest and snowy mountain peaks can be seen in the background.

Figure 5.14 Gondolas travel along suspension cables at the Gala Yuzawa ski resort in Japan. (credit: Rudy Herman, Flickr)

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
The force is equal to the maximum tension, or $F = 3 . 0 \times 10^{6} N$ . The cross-sectional area is ${πr}^{2} = 2 . 46 \times 10^{- 3} m^{2}$ . The equation $Δ L = \frac{1}{Y} \frac{F}{A} L_{0}$ can be used to find the change in length.
Solution
All quantities are known. Thus,

$\begin{array}{l} Δ L & = & (\frac{1}{210 \times 10^{9} {N/m}^{2}}) (\frac{3 . 0 \times 10^{6} N}{2.46 \times 10^{–3} m^{2}}) (3020 m) \\ = & 18 m . \end{array}$
5.31
Discussion
This is quite a stretch, but only about 0.6% of the unsupported length. Effects of temperature upon length might be important in these environments.

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