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So I'm just wondering about three scenarios:

You've got a book on the table. It's got a gravitational force acting downwards. This is the force on the book by the earth. Apparently there is an equal and opposite force on the earth by the book.

But then there is also a normal force involved. The normal force is the force of the table pushing back against the book. Does this have a reaction force of the book pushing back on the table? And isn't this already accounted for with the gravitational force? What?

And then another massive question: If everything has equal and opposite forces on each other then how can there be net forces? Seriously that is pretty weird. For example when you push a cube with a non-constant velocity along a horizontal plane then there are net forces which is confusing because according to the third law there are equal and opposite forces on every interaction.

Lastly if you walk along a floor it pushes out against you with an equal and opposite reaction normal force. But surely if you are really heavy such that the surface breaks and you fall through then there is no equal and opposite reaction force? So what's up with that?

Update: Thanks for all the good answers guys but I am still low key really confused so I hope you don't mind if I don't accept anyone's answer as of yet. The quality of your answers are great but I just want to make sure I understand these concepts 100% before I resolve my issue

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The normal force is the force of the table pushing back against the book. Does this have a reaction force of the book pushing back on the table? And isn't this already accounted for with the gravitational force? What?

  • The book feels weight downwards. Reaction: the Earth feels a gravitational pull upwards.
  • The book feels a normal force upwards. Reaction: the Earth feels a push downwards.

Another example:

  • An billard ball feels a "push" backwards at impact (equivalent to the normal force before). Reaction: The other billard ball feels the same "push" but oppositely (equivalent to the push before).

If everything has equal and opposite forces on each other then how can there be net forces? Seriously that is pretty weird. For example when you push a cube with a non-constant velocity along a horizontal plane then there are net forces which is confusing because according to the third law there are equal and opposite forces on every interaction.

There is a reaction force to every force exerted, yes. Just remember that you are talking about another object then.

  • The book feels the normal force, but it's reaction is not felt by the book; the reaction is felt by the Earth.
  • The billard ball feels the impact "push" and thus has a net force and flies back (accelerates). The reaction is not felt by the same billard ball but by the other ball, which then also has a net force and flies off (accelerates).

When setting up Newton's 2nd law, always look at only one object/system at a time.

Lastly if you walk along a floor it pushes out against you with an equal and opposite reaction normal force. But surely if you are really heavy such that the surface breaks and you fall through then there is no equal and opposite reaction force? So what's up with that?

  • If a ninja karate-chops a plank without breaking it by applying force $F$, then the plank is able to hold back with the entire same force $-F$.
  • If the ninja karate-chops a plank that breaks, then he did not need all the force $F$. He only applied force up until the plank stopped reacting with the same force. After that point he did not increase his force. So the force he exerted equals the force the plank was able to do (plus the acceleration term it also causes), which is smaller than the $F$ that he could have done.

Think of the difference between throwing your karate-hand through the air vs. throwing it into a falling piece of paper vs. throwing it into a wall: When hitting the wall, you exert a force on that wall. When hitting the paper, you exert a force but much smaller (and you move it as well). When not hitting anything, you don't exert any force on anything - you just move.

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    $\begingroup$ Oh Steeven you beautiful boy I will accept your answer shortly I just need some time to read over and make sure I don't have any additional questions or misunderstandings with you answer but thank you sooooo much $\endgroup$ – Scott Simmons Jul 17 '17 at 8:18
  • $\begingroup$ I don't understand the second part to your answer. Could you explain what is happening when a man pushes a box along a horizontal plane with a net force. $\endgroup$ – Scott Simmons Jul 18 '17 at 1:15
  • $\begingroup$ @ScottSimmons The box feels the man's push. Let's say that is $10\;\mathrm N$. The man feels the same push backwards, but that doesn't change the fact that the box feels $10\;\mathrm N$ of push. The box also feels friction from the ground. Let's say that is $7\;\mathrm N$. The ground also feels friction oppositely, but that doesn't change the fact that the box feels $7\;\mathrm N$ of friction. The net-force is $\sum F=10\;\mathrm N-7\;\mathrm N=3$ and so the box accelerates. The fact that there are equal and opposite ... $\endgroup$ – Steeven Jul 18 '17 at 7:36
  • $\begingroup$ ... forces appearing doesn't matter, because they do not act on the box but on the man and the ground. Those reaction-forces do not have any influence on the net-force on the box itself, they only have an influence on the surroundings of the box. @ScottSimmons I hope this was enough explanation? Otherwise please let me know $\endgroup$ – Steeven Jul 18 '17 at 7:36
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    $\begingroup$ It is not a surprise that you can charge $40usd per hour when you are able to communicate key concepts as effectively as you did with this question. Stack exchange is how I got into the university course I wanted to do. Awesome help Steeven $\endgroup$ – Scott Simmons Feb 8 '18 at 2:21
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Consider the following diagram,

enter image description here

The upper block pushes the black box down with a normal reaction and the black box provides the normal force (reaction) to the red box so that the net force on the red box becomes $0$, but these two forces (normal force due to red box and the normal force due to black box ) act on two different bodies, so they cannot cancel each other out.

The red box pushes the black box by the normal reaction not by its weight.

If the system is at rest, then it would mean that the normal force due to the black box on the red box cancels out the weight of the later. This is turn means that the normal force between the boxes in equal to the weight of the red box.

Its not at all necessary that the normal force is equal to the weight of the red box. There can be many other cases where such a system may be accelerating (like in a lift), where the normal force won't be equal to the weight of the red box.

When the system is at rest, the normal force due to the red box (it's weight) adds to the weight of the black box. So now the ground has to provide the black box the normal force equal to the weight of both the boxes if the system has to remain at rest.

For you last question, if your weight is more than what the floor can handle, it will indeed break. Since you are in equilibrium with the floor, the normal reaction provided by the floor to you cancels out your weight. But you will break the floor if the normal reaction provided by you to the floor exceeds its limits.

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The main concept you have to keep in mind here are the following:

  • Action/Reaction pairs never act on the same body.

To think that normal and weight form an action/reaction pair is one of the most common misconception among fresh students in physics. By the statement above you conclude that the normal and gravitational forces acting on the book or in the walking person cannot form an action/reaction pair since they act on the same body. By the same reason, we see that it is perfectly possible to have net forces on a body.

Just for completeness, it is not trues that all forces have a reaction. This holds only for interaction forces (such as gravity, normal, electrical,...). Pseudo forces (such as centrifugal force, Coriolis force,...) do not have a reaction since they are not originated from a body.

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  • $\begingroup$ Well, strictly speaking, pseudo forces are not really forces. So saying any force causes an equal but opposite reaction force is still valid in every situation. $\endgroup$ – Steeven Jul 17 '17 at 12:22
  • $\begingroup$ @Steeven Well, they are not interaction forces. But I see your point and that goes in the direction of how to define force. $\endgroup$ – Diracology Jul 17 '17 at 12:24
  • $\begingroup$ Wait centrifugal forces are another way of saying centripetal forces right. $\endgroup$ – Scott Simmons Jul 17 '17 at 23:05
  • $\begingroup$ Technically, there are NO reaction forces. There are Newton's 3rd law pairs. If you say a force is a reaction, then it must be caused by an action. How do you know which force acted first? Answer: You can't because they happen simultaneously. They are PAIRED. Not action/reaction. Reaction is just poor language that has gotten scooped along. It connotes and perpetuates a very poor concept. Let's drop that term. By the way, Newton never said action/reaction. He wrote about pairs of forces. $\endgroup$ – Bill N Jul 18 '17 at 2:50
  • $\begingroup$ @BillN Yes, that's a good remark. The nomenclature of action/reaction can be very misleading. $\endgroup$ – Diracology Jul 18 '17 at 13:21

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