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I was learning about normal reaction force today and I saw a diagram like this:

The diagram

My questions about this are:

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

  2. 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).

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    $\begingroup$ N2 in your sketch does nothing on the body. It is just the weight of the crust on the (under)ground, which in turns applies force upwards. N1 is a kind of "excess" force as for also the body and not only the surface don't sink. I don't see why they put N2 there. $\endgroup$
    – Alchimista
    Commented Oct 23, 2021 at 8:52
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    $\begingroup$ Normal reaction on surface due to ground --Think about it. $\endgroup$
    – Laxmi Narayan Bhandari
    Commented Oct 23, 2021 at 9:02

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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!

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  • $\begingroup$ Thanks for the explanation :) , But I have one doubt as the magnitude of N1 is equal to magnitude of N2 why don't they both cancel out (as they are in opposite directions).I guess if they do so then only mg would act on the mass, but we don't consider N1 and N2 to be cancelling each other out? $\endgroup$
    – Oitanny
    Commented Oct 23, 2021 at 12:33
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    $\begingroup$ @According to Newton's Third Law, the pair of reaction forces act on two different bodies. We cancel equal forces, (example) when they are acting on a single body (by drawing FBD and resolving forces). Here reaction forces occur on two different bodies, so you can't cancel them! Think of it by drawing FBDs and you will get it. Coincidently when I was new to physics, I had same doubt, but it got resolved as I analyzed FBDs :) $\endgroup$
    – Kshitij Kumar
    Commented Oct 23, 2021 at 13:54
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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.

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