Answer to Question #167427 in Quantum Mechanics for Kipkirui Emmanuel

Question

An 800N elevator hangs by a steel cable for which allowable stress is 1.2 ×10^8 n/m².what is the minimum diameter required if the elevator accelerates upwards at 1.5 m/s²

Answer 1

Let T be the tension in the cable.

Resolving upwards on the elevator gives: $T - 800 = 80 \times 1.5$ as (F = ma)

Solve for T: $T = 120 + 800$

$T = 920 N$

Divide this by the stress to get the cross-sectional area required:

$A = \dfrac{920}{1.2\times 10^8}$

= $766.66\times 10^{-8}$ $m^2$

Divide by pi to get $r^2$

$r^2 = 244.161\times 10^{-8}$

$r = 15.625\times 10^{-4} m$

Multiply by 2 to get the diameter

Hence, $d = 31.25\times 10^{-4}$ m