COMEDK · Physics · 10. Mechanical Properties of Solids
An elevator cable is to have a maximum stress of \(7 \times 10^7 \mathrm{~N} / \mathrm{m}^2\) to allow for appropriate safety factors. Its maximum upward acceleration is \(1.5 \mathrm{~m} / \mathrm{s}^2\). 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
- A \(3.22 \mathrm{~cm}^2\)
- B \(2.38 \mathrm{~cm}^2\)
- C \(0.32 \mathrm{~cm}^2\)
- D \(8.23 \mathrm{~cm}^2\)
Answer & Solution
Correct Answer
(A) \(3.22 \mathrm{~cm}^2\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { } T_{\max }=m(g+a)=(2000)(9.8+1.5)=22600 \mathrm{~N} \\ & \text { Maximum stress }=\frac{T_{\max }}{\text { Area }} \end{aligned}\)…
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