Question

After a fall, a $\mathbf{95}\mathbf{\text{kg}}$rock climber finds himself dangling from the end of a rope that had been$\mathbf{15}\mathbf{\text{m}}$ long and$\mathbf{9}\mathbf{.6}\mathit{m}\mathit{m}$ in diameter but has stretched by$\mathbf{2}\mathbf{.8}\mathbf{\text{cm}}$ .For the rope, calculate (a) the strain, (b) the stress, and(c) the Young’s modulus.

Short answer

a) Strain, $ϵ=1.9\times {10}^{-3}$

b) Stress,$\sigma =1.29\times {10}^7\text{\hspace{0.17em}N}/{\text{m}}^{\text{2}}$

c) Young’s modulus, E=$6.84\times {10}^{9}N/m^2$

Step-by-step solution

Listing the given quantities

$L=15m$

$l=2.8cm\left(\frac{1m}{100cm}\right)=0.028m$

Diameter of the rope$d=9.6mm\left(\frac{1m}{100cm}\right)=9.6\times {10}^{-3}m$

Acceleration due to gravity$\text{g}=9.8m/s^2$

Understanding the concept of stress and strain

$\text{σ=}\frac{\text{F}}{\text{A}}$We have been given all the required values to calculate the stress, strain, and Young’s Modulus. We convert them into equivalent units. By plugging these values into the formula:

Strain localid="1661252112148" $\mathit{ϵ}\mathbf{\text{=}}\frac{\mathbf{l}}{\mathbf{L}}$

Stress

Young’s Modulus $E\text{=}\frac{\text{σ}}{ϵ}$.

(a) Calculation ofthe strain

By usingthe formula for strain,

$\begin{array}{c}ϵ=\frac{l}{L}\\ =\frac{0.028m}{15m}\\ =1.9\times {10}^{-3}\end{array}$

Strain,$ϵ=1.9\times {10}^{-3}$

(b) Calculation ofthe stress

We have the formula for stress,

$\sigma =\frac{\text{F}}{\text{A}}$

We calculate the area of rope,

$\begin{array}{c}A=\frac{\pi }{4}\times {\text{d}}^2\\ =\frac{\pi }{4}\times {(9.6\times {10}^{-3})}^2\\ =7.2\times {10}^{-5}{m}^2\end{array}$

The force will be the weight of the rock climber,

$\begin{array}{c}\text{F}=\text{mg}\\ =95kg\times 9.8m/s^2\\ =931N\end{array}$

Using the above-mentioned stress formula, we get

$\begin{array}{c}\sigma =\frac{931N}{7.2\times {10}^{-5}{m}^2}\\ =1.29\times {10}^{7}N/m^2\end{array}$

Stress $\sigma =1.29\times {10}^{7}N/m^2$,

(c) Calculation of Young’s Modulus

We have the formula for E,

$E=\frac{\sigma }{ϵ}$

$\begin{array}{c}=\frac{1.3\times {10}^{7}N/m^2}{1.9\times {10}^{-3}}\\ =6.84\times {10}^{9}N/m^2\end{array}$

Young’s modulus, E =$6.84\times {10}^{9}N/m^2$