Question

The elastic limit of an elevator cable is $2\times {10}^{9}N/m^2$. The maximum upward acceleration that an elevator of mass $2\times {10}^{3}kg$ can have when supported by a cable whose cross-sectional area is ${10}^{-4}{m}^2$, provided the stress in cable would not exceed half of the elastic limit would be

• A. $10m{s}^{-2}$
• B. $50m{s}^{-2}$
• C. $40m{s}^{-2}$
• D. Not possible to move up

Answer 1

Answer: (C)

$40m{s}^{-2}$

If the acceleration is upwards, pseudo force acts downwards. Thus using FBD, T=mg+ma, or $T=2000\times 10+2000\times a=2000(10+a)$
Now, $Stress=\frac{Force}{Area}$
Thus. given max. stress= half of elastic limit or ${10}^{9}N/m^2$
Thus, ${10}^9\times {10}^{-4}=Force=2000(a+10)$
Thus, $a+10=50$ or $a=40m/s^2$