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Question

A 900 kg elevator hangs by a steel cable for which the allowable stress is 1.15×108 N/m2. What is the minimum diameter required if the elevator accelerates upward at 1.5 m/s2. Take g=10 m/s2

A

6×10210π M
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B

6×1025π M
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C

3×1025π M
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D

6×10210π M
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Solution

The correct option is A
6×10210π M

Calculate tension for elevator

Given, mass of the lift, M=900 kg
Acceleration of the lift, a=1.5 m/s2
T=M(g+a)
T=9000(10+1.2)
=900(11.2) N

Calculate diameter required by elevator

Formula used: stress =TA
Given, stress in the cable =1.15×108 N/M2
Let the diameter of the cable be d. Therefore,
stress =TA
1.15×108=900(11.2)(πd24)
d=6×10210π m

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