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Question

The maximum stress that can be applied to the material of a wire used to suspend an elevator is 108 Nm2. If the mass of the elevator is 1000 kg and it moves up with an acceleration of 0.2 ms2, what is the minimum diameter of the wire required? (Take g=9.8 m/s2)

A
2×102π m
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B
102π m
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C
2×1022π m
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D
4×102 m
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Solution

The correct option is A 2×102π m
As the elevator moves up, the tension in the wire is
F=mg+ma=m(g+a)
[Here, ma= pseudo force acting on the elevator]
F=1000×(9.8+0.2)=10,000 N
Stress in the wire =FA=Fπr2
Clearly, when the stress is maximum, r is minimum.
Maximum stress =Fπr2min
or r2min=Fπ×Maximum stress
r2min=10000π×108=104π
rmin=102π m
Hence, minimum diameter
dmin=2rmin=2×102π
dmin=2×102π m

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