Answer
A) We need the area to be at least $3.7*10^{-5}m^2$
B) The cable will stretch by about $2.7*10^{-3}m$
Work Step by Step
A) Using the maximum tensile strength of steel we can solve for the minimum area required to hold the 270kg load with a factor of safety of 7.
$7*F=500*10^6N/m^2*A$
$A=\frac{7*F}{500*10^6N/m^2}$
$A=\frac{7*270kg*9.8m/s^2}{500*10^6N/m^2}$
$A\approx3.7*10^{-5}m^2$
We need the area to be at least $3.7*10^{-5}m^2$.
B) First, we will solve for the change in length required using the following formula:
$\frac{F}{A}=\frac{\Delta l}{l_0}E$
$\Delta l=\frac{F}{A}l_0\frac{1}{E_{steel}}$
$\Delta l=\frac{270kg*9.8m/s^2}{3.7*10^{-5}m^2}*7.5m*\frac{1}{200*10^9N/m^2}$
$\Delta l\approx2.7*10^{-3}m$
The cable will stretch by about $2.7*10^{-3}m$.
