The elastic limit of an elevator cable is 2 × 109 N/m2. The maximum upward acceleration that an elevator of mass 2 × 103 kg can have when supported by a cable whose cross-sectional area is 10-4 m2, provided the stress in cable would not exceed half of the elastic limit would be
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a
10 ms−2
b
50 ms−2
c
40 ms−2
d
Not possible to move up
answer is C.
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Detailed Solution
From the free-body diagram of the elevator, T - mg = ma T = m(g + a) Stress in cable is α=TA=m(g+a)aFrom the given condition, α≤αmax2m(g+a)A≤σmax2,g+a≤2×1092×10−42×103=50
The elastic limit of an elevator cable is 2 × 109 N/m2. The maximum upward acceleration that an elevator of mass 2 × 103 kg can have when supported by a cable whose cross-sectional area is 10-4 m2, provided the stress in cable would not exceed half of the elastic limit would be