Chapter 11 - Equilibrium and Elasticity - Problems - Exercises - Page 360: 11.37

$$S=7.36\times10^{6}Pa$$

$F=1375 N\qquad\qquad\Delta\theta=1.24^{\circ}\frac{2\pi}{360}=0.0216 rad$ $Fcos\theta=(1375)cos8.5=1360 N$ $S=\frac{\frac{F}{A}}{\Delta\theta}=\frac{(1360)}{(0.0216)(9.25\times10^{-2})^{2}}=7.36\times10^{6}Pa$