Question

A mine elevator is supported by a single steel cable$\mathbf{2}\mathbf{.5}\mathbf{\text{cm}}$in diameter. The total mass of the elevator cage and occupants is$\mathbf{670}\mathbf{\text{kg}}$ .By how much does the cable stretch when the elevator hangs by (a)$\mathbf{12}\mathbf{\text{m}}$ of cableand (b) $\mathbf{362}\mathbf{\text{m}}$of cable?(Neglect the mass of the cable.)

Short answer

a) Cable stretch when elevator hung by $12\text{\hspace{0.17em}m}$is$8.0\times {10}^{-4}\text{m}$ .

b) Cable stretch when elevator hung by $362\text{\hspace{0.17em}m}$is$0.024\text{\hspace{0.17em}m}$ .

Step-by-step solution

Understanding the given information

Mass of system is$670\text{\hspace{0.17em}Kg}$.

The diameter of the wire is $2.5\text{\hspace{0.17em}cm}\left(\frac{1m}{100cm}\right)=0.025m$ .

Concept and formula used in the given question

Using the given mass, you can find the weight of the elevator. Using Hooke's law formula, you can find the distance by which the cable is stretched. The formulas used are given below.

Hooke's law,$\mathit{\Delta }\mathit{L}\mathbf{=}\mathbf{\left(}\frac{\mathbf{w}}{\mathbf{A}\mathbf{Y}}\mathbf{\right)}\mathit{L}$

Area of wire,$\mathit{A}\mathbf{=}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}$

$W=mg$

Calculation of how much the cable stretch when the elevator hangs by  12 m of cable

First, you found the weight of load from a given mass is

$\begin{array}{c}W=mg\\ =672kg\times 9.8m/s^2\\ =6566\text{\hspace{0.17em}N}\end{array}$

The area of the wire is

$\begin{array}{c}A=\pi r^2\\ =3.14\times {\left(0.0125m\right)}^2\\ =4.91\times {10}^{-4}{\text{m}}^2\end{array}$

$\begin{array}{c}\Delta L=\left(\frac{6566N}{(4.91\times {10}^{-4}{m}^2)(2.0\times {10}^{11}N/m^2)}\right)\left(12m\right)\\ =8.0\times {10}^{-4}\text{\hspace{0.17em}m}\end{array}$

(b) Calculation ofby how much the cable stretch when the elevator hangs by362 m   of cable

$\begin{array}{c}\Delta L=\left(\frac{6566N}{(4.91\times {10}^{-4}{m}^2)(2.0\times {10}^{11}N/m^2)}\right)\left(362m\right)\\ =0.024\text{\hspace{0.17em}m}\end{array}$