A bar of cross- sectional area A is is subjected two equal and opposite tensile forces at its ends as shown in figure. Consider a plane BB' making an angle `theta` with length

Options

A. `tan theta`
B. `sec theta`
C. `cot theta`
D. ` cos theta`

Answer

A

Explanation

Consider the equilibrium of the palne BB.

A force F must be acting on this on this plane making an angle `(90^(@) - theta)` with the normal ON. Resolving F into two components, along the palne and normal to the plane.
Component of force F along the palne, `F_(p) = F cos theta` Component of force F normal to the palne ,
`F_(N) = F cos (90^(@) - theta) = F sin theta`
Let the are of the face BB be A . Then `(A)/(A)= sin theta`
`therefore A' =(A)/(sin theta)`
`therefore` Tensile strss `= (F sin theta theta)/(A') =(F)/(A) sin^(2) theta`
Shearing stress `= (F cos theta)/(A') =(F)/(A) cos theta sin theta =(F sin theta)/(2A)`
Their corresponding ratio is
`("Tensile stress")/("Shearing stress")=(F)/(A) sin^(2) theta xx (A)/(F sin theta cos theta) = tan theta`