An elevator cable can have a maximum stress o | vedclass

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Question

(A) The maximum stress allowed in the cable is $\sigma_{max} = 7 	imes 10^7\,N/m^2$.The tension $T$ in the cable when the elevator is accelerating up

Vedclass · Accepted Answer

$3.28$

Vedclass · Answer

(A) The maximum stress allowed in the cable is $\sigma_{max} = 7 	imes 10^7\,N/m^2$.The tension $T$ in the cable when the elevator is accelerating upwards with acceleration $a = 1.5\,m/s^2$ is given by Newton's second law:$T - mg = ma$$T = m(g + a)$Given $m = 2000\,kg$,$g = 10\,m/s^2$,and $a = 1.5\,m/s^2$:$T = 2000(10 + 1.5) = 2000 	imes 11.5 = 23000\,N$.The stress $\sigma$ is defined as $\sigma = \frac{T}{A}$,where $A$ is the cross-sectional area.To ensure safety,$\sigma \leq \sigma_{max}$,so $A \geq \frac{T}{\sigma_{max}}$.$A = \frac{23000}{7 	imes 10^7} = \frac{2.3 	imes 10^4}{7 	imes 10^7} \approx 0.32857 	imes 10^{-3}\,m^2$.$A = 3.2857 	imes 10^{-4}\,m^2$.Since $1\,m^2 = 10^4\,cm^2$,we have:$A = 3.2857 	imes 10^{-4} 	imes 10^4\,cm^2 = 3.2857\,cm^2$.Rounding to the nearest option,the minimum area is $3.28\,cm^2$.

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