Why doesn't the Normal reaction force on surface and the weight $mg$ add up? I was learning about normal reaction force today and I saw a diagram like this:

My questions about this are:

*

*Are $\mathbf {N_1}$ and $\mathbf {N_2}$ same in magnitude?


*Why don't $\mathbf{N_2}$ and weight($m\mathbf g$) add up? And if they do so, will they be bigger than $\mathbf {N_1}$?
Here $m\mathbf g$= weight of mass $m$.
$\mathbf {N_1}$ is the normal reaction force on mass $m$ due to surface.
$\mathbf {N_2}$ is the normal reaction force on surface due to ground).
 A: Yes, $|\vec{N_1}|=|\vec {N_2}|$, from Newton's third law of motion. But note that they are acting on two different bodies: block and earth respectively.
Your confusion seems to be around role of these forces in F.B.D of our block. Our block is in contact with earth's surface, and because of weight of block, contact forces come in play between the two. Our block exerts force of weight on earth, due to which, the earth, exerts an opposite and equal force on our block.
Now, when drawing FBD solely of block, we wouldn't include $\vec{N_2}$ force, since it is exerted on earth not on block. If we draw FBD of earth, we would include our $\vec{N_2}$ force, in same direction as you shown in drawing.
Note that when we draw FBD, we depict forces acting on the body, not by the body on some other body. Here $\vec{N_2}$ is exerted by the block, not on the block, so we will not include it in our block's FBD!
Hope it helped you out!
A: The net force downward on the mass is $mg - N_1 = 0$. $N_1$ is upward. $N_2$ is the force on the ground downward, not a force on the mass.  By Newton's third law, the forces $N_1$ and $N_2$ are equal in magnitude, but these two forces are opposite in direction.
