A uniform rod of length L and density `rho` is being pulled along a smooth floor with a horizontal acceleration `alpha` (see Fig.) The magnitude of the stress at the transverse cross-section through the mid-point of the rod is……..

Answer

`((rhoLalpha)/(2))`

Explanation

`((rhoLalpha)/(2))`
Let A be area of cross - section of the rod.

Mass of half rod , m= volume`xx` density `=((L)/(2)A)rho`
FBD of rod at mid - point

`thereforeT=malpha=((L)/(2)Arhoalpha)therefore"Stress"=(T)/(A)=(1)/(2)rhoalphaL`