Question

A massless cable of diameter 2.54 cm (1 inch) is tied horizontally between two trees 18 m apart. A tightrope walker stands at the center of the cable, giving it a tension of 7300 N. The cable stretches and makes an angle of 1.50° with the horizontal. The Young's modulus of the cable is:

Answer 1

$E \approx 4.20 \times 10^{10}$ Pa

Answer 2

Solution

1. Determine the extension:

For one half of the cable, the horizontal distance is 9 m. When stretched, the half‐cable makes an angle $\theta = 1.50^\circ$ with the horizontal so that its actual length becomes

$$L' = \frac{9}{\cos\theta}.$$

The extension of each half is

$$\delta = L' - 9 = 9\left(\frac{1}{\cos\theta}-1\right).$$

With $\cos(1.50^\circ) \approx 0.999657$,

$$\delta \approx 9(1.000343-1) \approx 9(0.000343) \approx 0.003087\text{ m}.$$

2. Calculate the cross‐sectional area $A$:

The cable’s diameter is 2.54 cm, so its radius is

$$r = \frac{2.54}{2}\text{ cm} = 1.27\text{ cm} = 0.0127\text{ m}.$$

Thus,

$$A = \pi r^2 = \pi (0.0127)^2 \approx \pi (0.00016129) \approx 0.0005067\text{ m}^2.$$

3. Apply Hooke’s law to relate stress and strain:

The stress in the cable is

$$\sigma =\frac{T}{A},$$

and the strain in one half is

$$\epsilon =\frac{\delta}{9}.$$

The Young's modulus is given by

$$E=\frac{\sigma}{\epsilon} = \frac{T/A}{\delta/9} = \frac{9T}{A\delta}.$$

4. Substitute the known values:

Given $T = 7300\,\text{N}$,

$$E = \frac{9 \times 7300}{0.0005067 \times 0.003087} \approx \frac{65700}{1.565\times10^{-6}} \approx 4.20\times10^{10}\, \text{Pa}.$$

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Explanation (Minimal):

• Compute extension $\delta = 9(1/\cos1.5^\circ - 1) \approx 0.003087$ m.
• Find cross-sectional area $A = \pi (0.0127)^2 \approx 0.0005067$ m².
• Use $E = \frac{9T}{A\delta}$.
• Substitute $T = 7300$ N and compute $E \approx 4.20 \times 10^{10}$ Pa.