Answer
a). $9.4\times10^{4}N/m^{2}$
b). $1.25\times10^{5}N/m^{2}$
Work Step by Step
a). The compressive stress is given by the ratio of the force perpendicular to surface of square face to area of cross-section.
Thus, compressive stress = $\frac{F sin37^{\circ}}{A}=\frac{250\times sin37^{\circ}}{0.04\times 0.04}=9.4\times10^{4}N/m^{2}$
b). The shear stress is the ratio of force parallel to the surface to the cross-section area.
Shear stress $=\frac{Fcos37^{\circ}}{A}=\frac{250\times cos37^{\circ}}{0.04\times 0.04}=1.25\times 10^{5}N/m^{2}$
