Q28 E - University Physics with Modern Physics

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?

Step-by-Step Solution

Answer

$6.77\times {10}^8 Pa$

Step 1: Given information

Length: $l_0=45.0 m$,

diameter: d = 7.0 mm, 

Elongation: $∆l=1.10 m$ ,

Climber mass: m = 65 kg .

Step 2: Concept/Formula used

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

Where, Y is Young’s modulus, ${\mathit{l}}_{\mathbf{0}}$ is length of muscle, F is muscle force, A is cross-sectional area and $\mathbf{∆}\mathit{l}$ is elongation.

Step 3: Area of rope

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

Step 4: Weight of climber

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

Step 5: Young’s modulus calculation

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