First slide

Stress and strain

Question

An elevator cable is to have a maximum stress of 7×10⁷ Nm^-2 to allow for appropriate safety factors. Its maximum upward acceleration is 15 ms^-2. If the cable has to support the total weight of 2000 kg of a loaded elevator, the area of cross-section of the cable should be

Moderate

A

3.23 cm²
B

2.38 cm²
C

0.32 cm²
D

8.23 cm²

Solution

Tension,
$\begin{array}{l} T_{max} = m (g + a) = (2000) (9.8 + 1.5) = 22600 N \\ ∵ Maximum stress = \frac{T_{max}}{Area} \\ ∴ Area = \frac{T_{max}}{Maximum stress} = \frac{22600}{7 \times 10^{7}} \\ = 3.23 \times 10^{- 4} m^{2} = 3.23 {cm}^{2} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App