Two equal and opposite forces `F` and `-F` act on a rod of unifrom cross-sectional ara `A`, as shown in the figure. Find the (i) shearing (ii) longitual stress on the section `AB`.

Explanation

(i) As the net force acting on the rod is zero, it is in equilibrium. Let the tesnion in the segment `AB` be `F^(-1)`. Applying Newton's 2nd law for the segment
`F_("net") = F^(1) - F = ma`

where, `a = 0`, this given, `F^(1) = F`
Resoliving the force `F^(1)` parallel and perpendicular to the given area `AB(= A^(1), "say")`,
We have `F_(tan) = F^(1) cos theta, F_("long") = F^(1) sin theta`
Then, `P_("shearing") = (F_(tan))/(A^(1)) = (F^(1) cos theta)/(A^(1))`
where, `A^(1) = (A)/(sin theta)` and `F^(1) = F`
This gives, `P_("shearing") = (F sin theta cos theta)/(A)`
(ii) Similarly, `P_("long") = (F_("long"))/(A^(1))`
`= (F^(1) sin theta)/(A^(1))` where `A^(1) = (A)/(sin theta)` and `F^(1) = F`
This gives, `P_(long) = (F)/(A) sin^(2) theta`