Answer
$$\delta = 0.0303 \space in$$
$$\sigma_{AVG} = 15.28 \space ksi$$
Work Step by Step
1. Average Tensile Stress:
$\sigma_{AVG} = \frac{P}{A}$
For a constant, circular cross section: $A = \frac{\pi D^4}{4}$
Substituting all known values into the tensile stress equation, obtain average tensile stress:
$\sigma_{AVG} = \frac{P}{A} = \frac{4 * 750 \space lb}{\pi * (1/4)^2 \space in^2} = 15278\space psi = \boxed{15.28\space ksi}$ $\leftarrow ANS (b)$
2. Young's Law (strain calculation)
$\sigma = E\epsilon$
Substituting all known values and solving for the strain:
$\epsilon = \frac{\sigma}{E} = \frac{15.28 \space ksi}{29000 \space ksi} = 5.269*10^-4
\space in/in$
Finally the actual elongation of the rope is calculated as follows:
$\delta = \epsilon * L = 5.269*10^-4 \space ft/ft * 4.8 ft = 2.529*10^-3 \space ft = \boxed{0.0303 \space in}$ $\leftarrow ANS (a)$
