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An elevator cable is to have a maximum stress of  7\times 10^{7} N/m^{2} to allow 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 area of cross- section of the cable should be -

Option: 1

3.22 \, cm^{2}


Option: 2

2.38 \, cm^{2}


Option: 3

0.32 \, cm^{2}


Option: 4

8.23 \, cm^{2}


Answers (1)

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\begin{aligned}T_{\text {max }}=m(g+a) \\ T_{\text {max }}=(2000)(9.8+1.5) \\ T_{\text {max }}=22600 \mathrm{~N} \end{aligned}\\ \begin{aligned} \text { Maximum stress } & =\frac{T_{\text {max }}}{\text { Area }} \\ \therefore \text { Area } & =\frac{{T_{\text {max }}}}{\text { Maximum Stress }} \\ \text { Area } & =\frac{22600}{7 \times 10^7} \\ & =3.22 \times 10^{-4} \mathrm{~m}^2 \\ & =3.22 \mathrm{~cm}^2 \end{aligned}

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Nehul

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