Question

A nylon rope used by mountaineers elongates 1.10 m under the weight of a 65.0-kg climber. If the rope is 45.0 m in length and 7.0 mm in diameter, what is Young’s modulus for nylon?

Short answer

$6.77\times {10}^{8}Pa$

Step-by-step solution

Given information

Length: $l_0=45.0m$,

diameter: d = 7.0 mm,

Elongation: $∆l=1.10m$,

Climber mass: m = 65 kg .

Concept/Formula used

$\mathit{Y}\mathbf{=}\frac{{\mathbf{l}}_{\mathbf{0}}\mathbf{F}}{\mathbf{∆}\mathbf{l}}$

Where, Y is Young’s modulus, l₀ is length of muscle, F is muscle force, A is cross-sectional area and $\mathbf{∆}\mathbf{l}$is elongation.

Area of rope

$$\begin{gathered} Area=\left(\frac{d^2}{4}\right) \\ =\left(\frac{{(7.0\times {10}^{-3}m)}^2}{4}\right) \\ =3.85\times {10}^{-5}{m}^2 \end{gathered}$$

Weight of climber

$$\begin{gathered} W=mg \\ =(65kg)(9.81m/s^2) \\ =637N \end{gathered}$$

Young’s modulus calculation

$$\begin{gathered} Y=\frac{(45m)(637N)}{(3.85\times {10}^{-5}{m}^2)(1.10\times {10}^{-2}m)} \\ =6.77\times {10}^{8}Pa \end{gathered}$$