Question

The maximum stress that can be applied to the material of a wire employed to suspend an elevator is $\frac{3}{\pi }\times {10}^{8}N/m^2$. If the mass of the elevator is $900kg$ and it moves up with an acceleration of $2.2m/s^2$ then calculate the minimum radius of the wire.

Answer 1

$S_{max}=\frac{3}{\pi }\times {10}^{8}N/m^2$

$m=900kg$

$ma=T-mg$

$900\times 2.2=S_m\times A-900\times 10$

$S_m.A=900\times 12.2$

$A_{max}=\frac{900\times 12.2}{\frac{3}{\pi }\times {10}^8}$

$\pi r_{min}^2=\frac{900\times 12.2}{\frac{3}{\pi }\times {10}^8}$

$r_{min}^2=12.2\times 3\times {10}^{-6}$

$r_{min}=6\times {10}^{-3}$

$r_{min}=6mm$