Q11-28E - University Physics with Modern Physics

Question

A nylon rope used by mountaineers elongates 1.10m 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

6.77x10⁸Pa

Step 1: Given information

$Length I_0=45.0m, diameter d=7.0mm , Elongation ∆I=1.10m, Climber mass m=65kg$

Step 2: Concept/Formula used

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

Where, Y is Young’s modulus, I₀ is length of muscle, F is muscle force, A is cross-sectional area and $\mathit{\Delta }\mathit{l}$ is elongation.

Step 3: Area of rope

$\begin{array}{rcl}A& =& \pi \left(\frac{d^2}{4}\right)\\ & =& \pi \left(\frac{{\left(7.0\times {10}^{-3}m\right)}^2}{4}\right)\\ & =& 3.85\times {10}^{-5}{m}^2\end{array}$

Step 4: Weight of climber

$\begin{array}{rcl}W& =& mg\\ & =& \left(65kg\right)\left(9.81m/s^2\right)\\ & =& 637N\end{array}$

Step 5: Young’s modulus calculation

$\begin{array}{rcl}Y& =& \frac{\left(45m\right)\left(637N\right)}{\left(3.85\times \times {10}^{-5}{m}^2\right)\left(1.1o\times {10}^{-2}m\right)}\\ & =& 6.77\times {10}^{8}Pa\end{array}$