Chapter 11 - Equilibrium and Elasticity - Problems - Exercises - Page 359: 11.28

$Y=6.77*10^8Pa$

$\Delta l=1.1m\qquad l_\circ=45m\qquad F=(65)(9.8)=637N\qquad d=7*10^-3m$ $A=\pi r^2=\pi(3.5*10^{-3})^2=3.85*10^{-5}m^2$ $Y=\frac{Tansile Stress}{Tansile Strain}=\frac{\frac{F_\perp}{A}}{\frac{\Delta l_\circ}{l}}=\frac{(637)(3.85*10^{-5})}{(1.1)(45)}=6.77*10^8Pa$