This is a question on a worksheet that I'm giving to a student:
A man of weight $W$ steps into the lift. The lift then moves downwards with a constant speed. If $R$ represents the force acting on the man by the lift floor, which of the following statements is/are correct?
(3) $R$ and $W$ are an action-reaction pair according to Newton's third law
Why isn't this statement (3) correct?
Action/reaction pairs are describing momentum flow. Momentum is a vector, so it is more difficult to explain intuitively, so you should start with money, which is a scalar. Let me call a "payment" money that enters your posession. A payment can be negative, in which case you lose money, like when you buy a hat.
Newton's third law of finance says: for any payment, there is a negative equal payment associated to it on somone else (if you aren't a central bank!). So if you have a payment of -100 dollars, someone else got 100 dollars. This should be completely intuitive, because, outside of banking, on the personal level, money is a conserved quantity.
Newton's law is the same: the conserved quantity is momentum, and the momentum is flowing between objects. The flow is called the force, and the force is the "payment", it tells you how many units of momentum are incoming per unit time. The third law says that every payment is associated with a reverse payment going the other way (just like money, except the quantity is a vector).
So when the Earth pulls on you, it is paying you downward momentum, which means that you are paying the Earth upward momentum. That's the action reaction pair. If you are on a scale, the scale pays you up momentum (it pushes you up), and you pay the scale down-momentum (you push the scale down). The end result is that the force from the Earth and the scale cancel out, and the gravitational force on the Earth from you plus the downward force you exert on the Earth through the scale cancel out, and nothing ends up moving.
This is like a closed circuit of momentum, and elucidating the way in which momentum is flowing, even though the objects don't move, is the subject of statics. Newton's laws add to this the interpretation of momentum as a dynamical quantity, mass times velocity, so when an object accumulates momentum, you know how fast it is going.
This point of view is very useful, but it is not often explicitly taught.
Here is my try. The weight W is due is to the earth pulling on him. It is part of a pair action-reaction with the force the man exerts on the Earth which is not R. The floor is pushing the man with force R and the man exerts a force with the same intensity as R but in the opposite direction on the floor because it belongs to the same pair action-reaction. The forces exerted on the man (W and R) cancel each other out: they are opposites with same intensities (W - R = 0). The man doesn't move with respect to the lift.
An important point about Third Law pairs that I've noticed seems to escape many students of intro physics is that our action-reaction forces are acting on different objects. For example, the electrostatic force of $q_1$ acting on $q_2$ is paired with the equal and opposite force of $q_2$ acting on $q_1$. The two force vectors we are pairing are acting on two different charges. With that in mind, it should be trivial to see that $R$ and $W$ from your example cannot be a Third Law pair.
Do not use the terms "action" or "reaction" in this context. It is best to understand that Newton's third law always involve two, and only two, agents. Let's call them A and B. Newton's third law should be stated as:
A exerts a force on B. B exerts an equally strong, but oppositely directed, force on A. These two forces constitute an "interaction pair."
In your example, you introduce three agents: the man, the floor, and Earth. The man's weight has nothing to do with the floor because "weight" is the nickname for the force on the man exerted by Earth. The corresponding force that forms an interaction pair would be the force on Earth exerted by the man. So if you involve the floor and the man, then you must not mention Earth at all.
Finally, note that two force vectors cannot be "equal and opposite" because by definition, two vectors are equal if and only if they have the same magnitude AND the same direction. Too many instructors let students get away with this sloppy terminology and it promotes much confusion.
