In the figure shown, `A` and `B` are two short steel rods each of cross-sectional area `5 cm^(2)`. The lower ends of `A` and `B` are welded to a fixed plate `CD`. The upper end of `A` is welded to the `L`-shaped piece `EFG`, which can slide without friction on upper end of `B`. A horizontal pull of `1200 N` is exerted at `G` as shown. Neglect the weight of `EFG`.

Longitudinal stress in `A` is

Options

A. Tensile in nature and having magnitude `180N//an^(2)`
B. Tensile in nature and having magnitude `240N//an^(2)`
C. Compressive in nature and having magnitude `180N//an^(2)`
D. Compressive in nature and having magnitude `240N//an^(2)`

Answer

A

Explanation

A will be applying pulling force to make EFG in equilibrium.

For vertical equilibrium of EFG, `T=N_(2)`
As A is in equilibrium `f_(1)=f_(2)=f` (suppose)
For horizontal equilibrium of EFG, `f_(1)=1200N`
For rotational equilibrium , `tau_(A)=0rArrN_(2)xx4-1200xx3=0rArrN_(2)=900N`
Shearing stress in `A=(F)/(A)=(1200)/(5)=240N//cm^(2)`
Shearing stress in B = 0
Longitudinal stress in `A=(T)/(A)=180N//cm^(2)`
Longitudinal stress in `B=(N_(2))/(A)=180N//cm^(2)` & it will be compressive in nature