• # question_answer 20) 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 $1.5\text{ }m/{{s}^{2}}.$ 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: A) $~3.22\,c{{m}^{2}}$                    B) $~2.38\,c{{m}^{2}}$C) $~0.32\,c{{m}^{2}}$                    D) $~8.23\,c{{m}^{2}}$

Invoking Newton's second law, we get $T-mg=ma$ Let, the area of cross seion of the cable be A $\therefore$    $7\times {{10}^{7}}A-mg=ma$ or,        $A=\frac{mg+ma}{7\times {{10}^{7}}}=\frac{m(g+a)}{7\times {{10}^{7}}}$ $=\frac{2000(9.8+1.5)}{7\times {{10}^{7}}}=3.22\times {{10}^{-4}}{{m}^{2}}$ $=3.22\,c{{m}^{2}}$ Hence, the correction option is [a].