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(A) Consider the equilibrium of the plane $BB'$. $A$ force $F$ acts on this plane. Let $N$ be the normal to the plane $BB'$. The angle between the for

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$	an\,	heta$

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(A) Consider the equilibrium of the plane $BB'$. $A$ force $F$ acts on this plane. Let $N$ be the normal to the plane $BB'$. The angle between the force $F$ and the normal $N$ is $(90^{\circ} - 	heta)$.Resolving $F$ into two components:$1$. Component of force $F$ normal to the plane $BB'$ (responsible for tensile stress): $F_N = F \cos(90^{\circ} - 	heta) = F \sin 	heta$.$2$. Component of force $F$ along the plane $BB'$ (responsible for shearing stress): $F_P = F \sin(90^{\circ} - 	heta) = F \cos 	heta$.Let the area of the cross-section $BB'$ be $A'$. From the geometry,$A = A' \sin 	heta$,so $A' = A / \sin 	heta$.Tensile stress (normal stress) $\sigma = F_N / A' = (F \sin 	heta) / (A / \sin 	heta) = (F/A) \sin^2 	heta$.Shearing stress $	au = F_P / A' = (F \cos 	heta) / (A / \sin 	heta) = (F/A) \sin 	heta \cos 	heta$.The ratio of tensile stress to shearing stress is:$\frac{\sigma}{	au} = \frac{(F/A) \sin^2 	heta}{(F/A) \sin 	heta \cos 	heta} = \frac{\sin 	heta}{\cos 	heta} = 	an 	heta$.

$A$ bar of cross-sectional area $A$ is subjected to two equal and opposite tensile forces at its ends as shown in the figure. Consider a plane $BB'$ making an angle $\theta$ with the length. The ratio of tensile stress to the shearing stress on the plane $BB'$ is

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