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I have read so many forums on this and I still do not understand and it's affecting my ability to move forward with learning physics right now.

Imagine the following scenario: a person on a frictionless surface pushes on a stationary object. Both the person and the object have the same mass. I understand that if the person pushes with 100N both the person and the object will feel that force (though in opposite directions) due to Newton's third law (though I'm not sure WHY the reaction occurs, but that is a question for another day).

Now imagine the same scenario but instead of an object with the same mass it is another person with the same mass. Both people push off each other with 100N. I have been told that this will look the same; both people will feel 100N. However in my mind I am imagining that each person will feel the 100N push from the other person as well as the 100N reaction force from their own push. Should this not total to 200N? And if not, WHY NOT? I do not understand this and I can't stop ignoring it anymore.

Thank you for your help, I'm sorry that this is probably a really obvious concept that I'm just not grasping but I cannot wrap my mind around it at all. Thank you.

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  • $\begingroup$ More on forces and factors of two: physics.stackexchange.com/q/41291/2451 and links therein. $\endgroup$ – Qmechanic Nov 16 '13 at 23:50
  • $\begingroup$ I deleted a long comment discussion that was distracting from the question. It's fine to have such discussions but in Physics Chat or another chat room, not in comment threads on the main site. $\endgroup$ – David Z Nov 19 '13 at 17:25
  • $\begingroup$ I would justify this the same way I justify colliding cars: Pushing a perfect immovable mirror not only looks the same as pushing your perfect mirror image, it actually is the same. $\endgroup$ – Fax Apr 6 at 11:10
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The force that a block exerts on a person is more similar to the force a person exerts on another person than you might initially realize.

Most of the time when a solid surface like a block exerts a normal force, it deforms a little bit, like a spring. The stress that results from the deformation is providing the force, similar (although not totally identical) to the way a person's muscles change their shape to provide a force. The harder the person pushes, the more the solid surface deforms and the higher the force that each of them exerts (at least up to a point where the solid is pushed so hard that it deforms irreversibly, which you're not considering here).

When viewed this way, there's no reason to expect a different answer for the block + person system than the case of the person + person system.

EDIT: Here's another way to say the same thing. The reaction force mentioned in statements of Newton's 3rd law is not a mysterious force that exists in addition to other forces that you might learn about. The reaction force will always come about in a way that relates to forces that can be studied in other contexts: deformation stresses, gravity, air pressure, and so on, depending on the specific system under consideration.

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As has been said in the long chain of comments, when you push your friend with a force of 100N, it automatically entails your friend pushing you back with 100N. This is what is meant by action and reaction.

What I mean to say is that the 100N you are counting as your friend's push has already been accounted for as the reaction to your push.

To make this simpler, consider you're pushing a block, but the block is held in place by your friend from the other side. For the block to remain in place, your friend must equal your force. Now suppose you and your friend have choreographed the entire stunt and arranged that 30 seconds into the stunt, he will apply 100N more than what he's applying. For the block to remain stationary, you're going to have to apply 100N too. In this case, the force applied by either one of you is 200N totally, but you're applying the action and he's applying the reaction (or the other way around), so it isn't right to add your action to his reaction.

Hope this clears your confusion.

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You seem to understand the first case pretty well. A person pushes on a block with a force $\vec{F}$ and by newton's third law, the block pushes back with the force $-\vec{F}$.

Now let's consider the second case. Instead of a block, it is now a person.They are both pushing on each other with the same force $F$ as before (by assumption). Therefore each person feels a force of strength $F$, exactly as before.

I think maybe the confusion arises because it is counter-intuitve that you don't need to work hard to exert a force on something else: even though the guy is pushing on the block and the block is just doing nothing, the force the guy exerts on the block is the same as the force the block exerts on the guy.

It might be good to know that the role of work is more intuitive. In the case with the block, the energy stored in the guys body and transferred to his own kinetic energy and the energy of the block. Therefore, the guy does work on the block. Now in the case where there are two people pushing against each other, they do no work on each other by symmetry. Instead they both convert their own stored energy to kinetic energy. When it comes to work, we see that the amount of work done does reflect the apparent "effort" put forth.

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I see what you are saying as striking at the heart of Physics.

Though answering with a simple "no" or "that facetious" would seem suitable and sufficient to most, I think you are trying to bring out a point everyone should be aware of.

Forces are not simple, they are the combining of the paths of two groups of matter for all of existence.

Quantum Mechanics more significantly recognizes this with an amazingly concise discovery. That is, the basic act of observing the existence of an object will alter that object's path for the rest of its existence.

If you don't understand this, think of it as an infinite potential type problem first. That is, the person is pushing off a wall. The wall does not move, or really do any work, yet it does. The wall supplies a force equal and opposite to the force put on it. Now think of the center of the two people pushing off each other, if they both supply the same force, this center will not move. Like the wall the center of the two people remains in place.

If a double force did exist on either person, after the push they would be required to move away at double speed. So in the case of the wall we could say the momentum was converted into energy in the wall. While in the case of the two people of equal mass the absolute value of the remaining momentum is double that of the case with the wall. This is where your factor of two got away, not in the initial force provided.

So as for the answer, why not...

Because we cannot create more momentum(energy) than what we used. Both people could provide twice the force, then move away at twice the speed, yes. But what does that get us, they still both feel an equal and opposite force.

Hope this helps...

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It is very trivial experimentally, to measure BOTH the force YOU apply to another object, whether block or another person, and the Newton's third law reaction force, back on you.

Apparatus required: one small portable bathroom scale, and a small mirror.

Stand the bathroom scale on edge, with the back up against a chair leg, and place the mirror behind the chair leg, so you can see the back of the scale, and read the scale at the same time.

Press on the scale , and observe the scale reading of the applied force. (you might want your laboratory assistant to sit on the chair).

Note the reading as the force you are applying against the scale AND the chair leg.

Reverse the scale, so you can read it in the mirror.

Repeat the pushing action, this time on the back of the scale.

Note the scale reading, which is the reaction force that the chair leg is applying to the scale in response to your push on the back.

If you use two identical scales back to back, you can read both of them simultaneously, and you will find they both read the same.

You can use this technique to weigh the entire planet earth.

Place the scale upside down overhanging a step, with the mirror on the ground, under the readout.

Stand on the back of the scale, so you can look down in the mirror, and read the weight of the planet. When I do the experiment, I get 180 pounds for the weight of the earth. The result of course depends on the local gravity, and I don't create a very big gravitational force, since my mass is quite low. You may get a different answer if you have a different gravity.

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