An elevator cable is to have a maximum stress of `7xx10^(7) Nm^(-2)` to allow for appropriate safety factors. Its maximum upward acceleration is `1.5 ms^(-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

Options

A. `3.22 cm^(2)`
B. `2.38 cm^(2)`
C. `0.32 cm^(2)`
D. `8.23 cm^(2)`

Answer

A

Explanation

(a) Tension, `T_("max")=m(g+a)=(2000)(9.8+1.5)=22600 N`
Maximum stress `=(T_("max"))/("Area")`
`therefore` Area`=(T_("max"))/("Maximum stress")=(22600)/(7xx10^(7))`
`=3.22xx10^(-4)m^(2)=3.22 cm^(2)`