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Chapter 9: Problem 13

For safety in climbing, a mountaineer uses a nylon rope that is \(50 \mathrm{~m}\) long and \(1.0 \mathrm{~cm}\) in diameter. When supporting a \(90-\mathrm{kg}\) climber, the rope elongates \(1.6 \mathrm{~m}\). Find its Young's modulus.

Short Answer

Expert verified
The Young's modulus of the nylon rope is \( 1.4 \times 10^{11}\, Pa \).

Step by step solution

01

Calculate the force F

We start by calculating the force 'F' which the rope is handling. This can be found using Newton's second law of motion, \( F = ma \), where 'm' is the mass and 'a' is acceleration due to gravity. Therefore, \( F = 90\, kg \cdot 9.8\, m/s^2 = 882\, N (Newton) \).
02

Calculate the cross-sectional area A

Next, we calculate the cross-sectional area 'A' of the rope which is a cylinder, using the formula \( A = \pi r^2 \), where 'r' is the radius (diameter/2). Here, diameter is 1 cm, hence radius \( r = 0.5\, cm = 0.005\, m \), so the cross-sectional area \( A = \pi \cdot (0.005\, m)^2 =7.85 \times 10^{-5}\, m^2 \).
03

Identify change in length \( \Delta L \) and original length L

We directly have these values from the problem. The change in length \( \Delta L = 1.6\, m \) and the original length \( L = 50\, m \).
04

Calculate Young's modulus Y

Lastly, we calculate Young's modulus 'Y' using the formula \( Y = \frac{{F/A}}{{\Delta L/L}} \). Putting in the known values, \( Y = \frac{{882\, N/7.85 \times 10^{-5}\, m^2}}{{1.6\, m/50\, m}} \) which simplifies to \( Y = 1.4 \times 10^{11}\, Pa (Pascal) \).

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