An elevator cable can have a maximum stress of $7 \times 10^7\,N/m^2$ for appropriate safety factors. Its maximum upward acceleration is $1.5\,m/s^2$ . If the cable has to support the total weight of $2000\,kg$ of a loaded elevator, the minimum area of crosssection of the cable should be ....... $cm^2$ $(g = 10\,m/s^2)$
Diffcult
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$\frac{\mathrm{F}}{\mathrm{A}}=7 \times 10^{7}$
$\mathrm{F}=\mathrm{A} \times 7 \times 10^{7}$
$\mathrm{F}-2000 \mathrm{g}=2000 \times \mathrm{a}$
$\left(7 \times 10^{7}\right) \mathrm{A}=2000(\mathrm{a}+\mathrm{g})$
$\mathrm{A}=\frac{2000}{7 \times 10^{7}}(10+1.5)$
$\mathrm{A}=3.28 \times 10^{-4} \mathrm{m}^{2}$
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