A uniform square block of side 'a' and mass 'm' is lying on a horizontal surface. A small block of mass `m` is placed on the top block. Find the distance by which normal reaction is shifted.

Explanation

For the force equilibirum
`N_(1)=2mg`
for torque equilibrium
`tau_(0)=(N_(1)x-mg a//2)=0`
`x=(mga//2)/(2 mg)`
`x=(a)/(4)`
Here the normal reaction is still distributied but now its distribution is not uniform