Two equal and opposite forces are exerted on a bar (rod) as shown in the figure. The plane $PQ$ makes an angle $\theta$ with the cross-section $PR$ (having area $a$). Find the tensile stress on the plane $PQ$.

(N/A) In $\Delta PRQ$,the angle between the normal to the plane $PQ$ and the direction of the force $F$ is $\theta$.
The area of the cross-section $PR$ is $a$.
The area of the inclined plane $PQ$ is $A_{PQ} = \frac{a}{\cos \theta}$.
The force $F$ acts perpendicular to the cross-section $PR$. The component of this force $F$ acting normal to the plane $PQ$ is $F' = F \cos \theta$.
The tensile stress on the plane $PQ$ is defined as the normal force divided by the area of the plane:
$\text{Tensile stress} = \frac{F'}{A_{PQ}} = \frac{F \cos \theta}{a / \cos \theta} = \frac{F \cos^2 \theta}{a}$.