I am aware that the normal reaction and gravitational force on a body are not action-reaction pairs. However, Newton's Third Law implies that to every action there is an equal and opposite reaction.

My question is:

If we imagine two bodies A and B such that A rests on B, then the weight of B is definitely an action force on B's body. The electromagnetic interaction between A and B results in a normal force on A, through which A remains at rest.

However, A's weight (the action force I mentioned above) doesn't seem to be having a reaction to it. If the normal force isn't a reaction force to the weight (which I am aware is not), then what is the reaction force to A's weight? Why is this force not included in a Free Body Diagram?

I have attached a pictorial representation of my thought for those who may find my language confusing (for which I apologise profusely).

Please let me know of the drawbacks of my arguments. I shall be deeply grateful for any help at all, since I have been struggling with this idea for quite some time. I am acquainted with Newtonian mechanics.

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1 Answer 1


Reaction forces do not act on the same object. That might be the key in your question.

A in your example rests on B and exert a downwards force on B. As a reaction, B exerts a force upwards on A that is numerically identical.

Now, in general, gravity and normal force are not action/reaction force pairs, as you rightly say, and they do not in general have to have anything to do with each other. But in this special case that you have defined, they do happen to be numerically equal to each other. The downwards force on B by A is equal to the weight of A (the gravitational force). The upwards force on A by B is what we call the normal force on A.

A scenario where gravity and normal force is not equivalent to an action/reaction pair would be if you imagine pressing a block horizontally onto a wall. You can try setting that up and rethinking the forces. Here, the normal force is a reaction to your pressing force, and a vertical friction will be a reaction to gravity.

In general, you can usually try to look for reaction forces by looking for normal forces and friction forces.

  • $\begingroup$ Thanks for this answer, it did help me clear my thoughts. It is true that in this particular case the normal reaction is equal to the weight. Is it okay to call them an action-reaction pair in this case? $\endgroup$ Apr 3, 2023 at 8:12
  • $\begingroup$ @SmarikaSingh I never use the term "normal reaction", but I understand what you mean. And yes, in this particular case the normal force is equal to the weight (a normal force is a "holding back" force that in this case "holds back" against the entire weight but nothing more). $\endgroup$
    – Steeven
    Apr 3, 2023 at 8:33
  • $\begingroup$ @SmarikaSingh Saying "action/reaction" is not perfectly well-defined terminology in physics but is very often used. Strictly speaking, the normal force by B on A equals a normal force by A on B, and these two constitute the action/reaction force pair. That normal force turns out to be equal to the weight of A since the entire gravitational force is propagating to the contact surface. Thus, the normal force by B on A is equal to the weight of A. Since they are identical, I might not mind if you call the normal force and the weight an action/reaction pair - some rigid teachers might, though. $\endgroup$
    – Steeven
    Apr 3, 2023 at 8:36
  • $\begingroup$ I see. So, I may state that if two bodies lie upon each other and both of them are at rest from an inertial Frame Of Reference, then the weight and normal force are equal. Also, didn't Newton himself use the term 'action/reaction forces'? Why must it be undefined? It's okay, though, thank you for all your help! $\endgroup$ Apr 3, 2023 at 11:01
  • $\begingroup$ @SmarikaSingh If "lie upon" means one on top of the other within a vertical gravitational field, then yes. The terms are not undefined, just not well-defined. Colloquially, we have the idea of one force acting and the other reacting - but from a physics perspective, one does not act before the other reacts. There is no actual distinction between an action and the corresponding reaction force, and you cannot in generel tell which is which. $\endgroup$
    – Steeven
    Apr 3, 2023 at 11:21

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