Question

A 70 kg mountain climber dangling in a crevasse stretches a 50-m-long, 1.0-cm-diameter rope by 8.0 cm. What is Young’s modulus for the rope?

Short answer

Young's modulus of the rope is$5.5\times {10}^{9}N/m^2$.

Step-by-step solution

Given information

The mass of the climber is $m=70kg$

$∆L=8cm=0.08m$

The length of the rope is $L=50m$

The radius of the rope is $r=\frac{1.0cm}{2}=0.5cm=0.005m$

Tensile stress

The tensile stress is given by $\frac{F}{A}$.

role="math" localid="1650295387964" $$\begin{gathered} \frac{F}{A}=\frac{mg}{{\mathrm{\pi r}}^2} \\ =\frac{70\times 9.8}{\mathrm{\pi }{\left(0.005\right)}^2} \\ =8.73\times {10}^{6}N/m^2 \end{gathered}$$

Strain

The strain is given by $\frac{∆L}{L}$.

$$\begin{gathered} \frac{∆L}{L}=\frac{0.08}{50} \\ =0.0016 \end{gathered}$$

Young's modulus of the rope 

The Young's modulus of the rope is given by $Y=\frac{{\displaystyle \raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$A$}\right.}}{{\displaystyle \raisebox{1ex}{$∆L$}\!\left/ \!\raisebox{-1ex}{$L$}\right.}}$.

Substitute the obtained values

localid="1650296015704" $$\begin{gathered} Y=\frac{8.73\times {10}^6}{0.0016} \\ =5.5\times {10}^{9}N/m^2 \end{gathered}$$

Therefore, Young's modulus of the rope is localid="1650296024948" $5.46\times {10}^{9}N/m^2$.