Question

An elevator cable is to have a maximum stress of $7\times {10}^{7}n/m^2$ to allow for appropriate safety factors. Its maximum upward acceleration is $1.5m/s^2$. If the cable has to support the total weight of $2000kg$ of a loaded elevator, the area of cross-section of the cable should be

• A. $3.22c{m}^2$
• B. $2.38c{m}^2$
• C. $0.32c{m}^2$
• D. $8.23c{m}^2$

Answer 1

Answer: (C)

$0.32c{m}^2$

$\begin{array}{cc}\text{Given:}-& \frac{F}{A}=\sigma =7\times {10}^{7}N/m^2\\ \text{acceleration}& =a=1.5m/s^2\text{going upwards}\end{array}$

$\begin{array}{c}m=2000kg\\ \text{To find:}A\text{.}\\ \text{effective acuralion}=g+a=11\cdot 3m/s^2\\ \frac{F}{A}=\frac{m(g+a)}{A}=7\times {10}^7\end{array}$

$\begin{array}{cc}\Rightarrow A& =\frac{2000\times 11.3}{7\times {10}^7}{m}^2\\ \Rightarrow A=3.22\times {10}^{-4}{m}^2& \\ \text{Answer:}\left(A\right)& \end{array}$