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My textbook explains Newton's Third Law like this:

If an object A exterts a force on object B, then object B exerts an equal but opposite force on object A

It then says:

Newton's 3rd law applies in all situations and to all types of force. But the pair of forces are always the same type, eg both gravitational or both electrical.

And: If you have a book on a table the book is exerted a force on the table (weight due to gravity), and the table reacts with an equal and opposite force. But the force acting on the table is due to gravity (is this the same as a gravitational force?), and the forcing acting from the table to the book is a reaction force. So one is a gravitational, and the other is not. Therefore this is not Newton's Third Law as the forces must be of the same type.

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    $\begingroup$ You've been given a rather confusing and imprecise explanation. The answer to this question is wrapped up in the same issues as the answer to your question about the ball. The Newtonian pair are the force of the book on the table and the force of the table on the book. They are both equal in magnitude to the weight of the book, but that is because the problem is static (nothing undergoing acceleration). I recommend that you try to understand the other question first, and then come back to this one. $\endgroup$
    – dmckee --- ex-moderator kitten
    May 24, 2012 at 20:12
  • $\begingroup$ Sorry I got the question slightly wrong, gravity is acting on the book, and the table pushing upwards is acting on the book. So they are both acting on the book. $\endgroup$
    – Jonathan.
    May 24, 2012 at 20:15
  • $\begingroup$ @dmckee, I have edited my question and I think it is different? $\endgroup$
    – Jonathan.
    May 24, 2012 at 20:18
  • $\begingroup$ Yes. And because the book is not accelerating you know the $F_g = -F_N$. You also know that the table feels a force from the book equal to $-F_N = F_g$. Got it? $\endgroup$
    – dmckee --- ex-moderator kitten
    May 24, 2012 at 20:19
  • $\begingroup$ @dmckee, I've ended up getting confused so I rewrote the question from scratch. $\endgroup$
    – Jonathan.
    May 24, 2012 at 20:26

6 Answers 6

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And: If you have a book on a table the book is exerted a force on the table (weight due to gravity),

That's where you went wrong. The force that the book exerts on the table is not a gravitational force, it's a normal force.

and the table reacts with an equal and opposite force.

That's also a normal force. So the book exerts a (normal) force on the table, and the table exerts a (normal) force on the book.

But the force acting on the table is due to gravity (is this the same as a gravitational force?),

No, it's not, and in fact this force (the normal force) is only indirectly due to gravity. The only relevant gravitational force is the force exerted by the Earth on the book. And the book also exerts a gravitational force back on the Earth, but because the Earth is so heavy, that force has no noticeable effect. (The Earth also exerts a gravitational force on the table, and the table on the Earth, but those don't matter so much in this particular scenario.)

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This is common misconception with my students too, and the only way to understand it you must draw all forces that act on both objects (in total five forces)!

In order to make things clearer, I will label the force with which table acts on book as $F_{12}$ and not $F_\text{N}$! Also suppose that $z$ axis is vertically up, so positive forces push upward and negative forces push downward.

There are two forces acting on book, its gravitational force $-F_\text{g,book}$ (downward) and the force of table on the book $F_{12}$ (upward). According to first Newton law for the book they are equal by magnitude

$$F_{12} - F_\text{g,book} = 0.$$

According to the third Newton law book must be acting on table with the force $-F_{12}$ (downward). So there are three forces acting on table: its gravitational force $-F_\text{g,table}$, force of the book $-F_{12}$ (both downward) and the force of the ground $F_\text{N}$ (upward)!

Now let's write the first Newton's law for the table

$$F_\text{N} - F_{12} - F_\text{g,table} = 0.$$

Consequently

$$F_\text{N} = F_{12} + F_\text{g,table} = F_\text{g,book} + F_\text{g,table}$$

The ground force must support both book and table! Isn't that obvious?

Conclusion: So third Newton's law is perfectly valid for this case as well!

If you still do not understand, write on the paper book, table, and all five forces (two acting on the book and three acting on the table).

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  • $\begingroup$ Why isn't $F_g$ and $F_N$ the same force, as gravity cuases the book to push down on the table. $\endgroup$
    – Jonathan.
    May 24, 2012 at 20:28
  • $\begingroup$ $-F_\text{g,book}$ is gravitational (downward) force of the book and $F_\text{N}$ is (upward) force of the table. According to first Newton's law, they are equal by the magnitude and opposite in direction. These are two separate forces. $\endgroup$
    – Pygmalion
    May 24, 2012 at 20:32
  • $\begingroup$ @Jonathan. I edited the answer to distinguish between inter-force $F_{12}$ between book and table and ground force to table. $\endgroup$
    – Pygmalion
    May 24, 2012 at 20:36
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One way to make it obvious is think about how the down-momentum is flowing. The book is getting down-momentum from the Earth (through action-at-a-distance gravity), and this down-momentum then flows downwards to the table, and across the table to the legs, then through the legs of the table back down to the Earth, making a closed circuit of down-momentum, like a closed electrical circuit.

Each time momentum leaves an object A and enters another object B, we say a force is acting from A to B, and simultaneously that a reaction force is acting from B to A (since the momentum gained by B is the momentum lost by A). This is Newton's third law.

In this circuit, the down-momentum goes

Earth $\rightarrow$ book $\rightarrow$ table $\rightarrow$ Earth

So there is an action/reaction pair from the Earth to the book (the Earth is pulling the book and transferring down-momentum to it, and the book is pulling the Earth, transferring an equal amount of negative down-momentum--- or up momentum--- to the Earth). There is an action reaction pair from the book to the table ( the book is transferring down-momentum to the table through a contact normal force, and the table is transferring negative down-momentum to the book by the same contact normal force), then the table has an action/reaction pair with the Earth (the table sends the down-momentum into the Earth, and the Earth sends negative down-momentum into the table)

Each of these flows is describing how a conserved quantity, namely down-momentum is going from place to place. It is easiest to sort this out with flows of charge, because unlike charge, momentum is a vector.

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Newton's third law is about pairs of objects interacting. The force that acts on one object is equal and opposite to the force acting on the other object. So you can never have a third law pair acting on the same object.

The equality of the reaction force and the weight force is nothing to do with the third law, and is just as a result of the first law applied to the forces acting on the book.

Let's look at some third law pairs in this scenario:

  1. The weight of the book and the weight of the earth. Yup, the earth is pulled up by the book, but because $F=ma$ and the earth is more than a little heavier, it doesn't result in a great deal of movement on the earth's part when the book is released!
  2. The normal force of the table on the book and the book on the table. The force that the book exerts on the table is a normal force, not a weight force. (The book's weight doesn't act on the table, it acts on the book.) It's equal in magnitude to the weight of the book, again, because of the first law. The book and the table press on each other. It's probably better to think of the normal force as being generated by the electromagnetic forces between molecules in the table and book. You get a normal pair like this in the man-leaning-on-wall example.
  3. The normal forces between the desk and the earth
  4. The weight forces between the desk and the earth
  5. (The gravitational forces between the book and the table are negligable.)

Force 1=Force 2 in magnitude by law 1, not by law 3. (Same for forces 3 and 4.)

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  • $\begingroup$ In the book-table normal force can we think that both book and table exert normal/contact force? Or is only 1 of them? In all books it is stated that the normal force is exerted by the table. Why the book doesn't exert a contact force in the table so the table "feels" the force from book and the force from the reaction force which the book exerts on the table? $\endgroup$
    – Antonios Sarikas
    May 28, 2020 at 9:42
  • $\begingroup$ @AntoniosSarikas Read the answer. "The force that acts on one object is equal and opposite to the force acting on the other object." Key words: "OTHER OBJECTS". $\endgroup$
    – CountDOOKU
    Aug 1, 2020 at 12:48
  • $\begingroup$ @AntoniosSarikas Please read the answer. The book exerts a normal force on to the table and the table exerts a normal force back on the book. The normal is the support force. $\endgroup$
    – CountDOOKU
    Aug 2, 2020 at 3:03
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A lot of questions here talk about "normal force", but I get the feeling that you're still confused about what that is.

First consider the book - Whether it is resting on the table or not, it has a weight. Here weight is different from mass. The weight is the mass $m$ times the acceleration due to Earth's gravity $g$, or more familiarly $$F = mg$$

The same goes for the table. Now this is the important part - The weight isn't gravitational force. The gravitational force that you are thinking of is expressed as $$F_g = \frac{Gm_1 m_2}{r^2}$$ and that is the force due to the gravitational attraction between two bodies.

In the case of the table and the book, the gravitational attraction is absolutely negligible, since they are both so tiny. The force that the table experiences because of the book is what is being called normal force.

The table then exerts an equal and opposite force. This is also clearly seen, because if the table didn't exert an equal and opposite force, the book would be accelerating downward. But the whole system is at rest, therefore the total force on the book-table system must be zero.

EDIT: @AndrewC has mentioned in the comments below why my earlier reasoning was wrong. Basically normal force is only indirectly due to gravity. Khan Academy has a brilliant explanation of these concepts.

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  • $\begingroup$ Nonono, the "if the table didn't exert an equal and opposite force" argument is Newton's first law. If that's what Newton's third law said (every action has an equal and opposite reaction), it would mean nothing ever moved! My trailer exerts an equal and opposite tension force on my car, even when I'm accelerating. $\endgroup$
    – AndrewC
    Nov 28, 2012 at 21:38
  • $\begingroup$ Would you like to explain your interesting statement about Weight force not being gravitational force? $\endgroup$
    – AndrewC
    Nov 28, 2012 at 21:39
  • $\begingroup$ Newton's first law says that anything that's moving keeps moving, and anything that's at rest stays at rest, unless you have an external force. In this case, the external force is gravity, which is trying to pull the book down. That force is nicely cancelled with the force the table exerts on the book. $\endgroup$
    – Kitchi
    Nov 29, 2012 at 5:57
  • $\begingroup$ My point is that your last paragraph sounds like it's talking about Newton's third law by using the phrase equal and opposite, but you're actually using Newton's first law. That's exactly the confusion the textbook was trying to avoid and the question is trying to unpick, so it's unhelpful in this context. $\endgroup$
    – AndrewC
    Nov 29, 2012 at 22:44
  • $\begingroup$ I thought you were making an interesting point in distinguishing weight force from gravitational force (perhaps about the discrepency between $g=9.81m/s^2$ and $Gm_E/r_E^2$ in practice) but actually I think you were just making a mistake. Weight is the force due to gravity in the sense you're using it in your answer, calling the distinction important is misleading in this context. $\endgroup$
    – AndrewC
    Nov 29, 2012 at 22:45
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You need to sort out these ideas.

1 Free body diagrams: Book table Book and earth Table and earth

2 sort the force pairs by 'kind' of force:

Interaction is contact (due to electric forces) Gravity is force due to each of the bodies

So book-table has force pairs due to interaction forces, balanced and oppsite, call them normal due to book, normal due to table. Both same kind. Sorted.

Book-earth has force pair due to gravity of each acting on other. Both same kind of forces, equal and opposite, and on different bodies

Table-earth, there is contact, which is electric interaction at electronic charge level. Equal, opposite yet same kind of force.

Finally, each mass has gravity and the mass exerts force on other mass - NOTE: "on other mass!!!!" Same kind of force again.

Conditions for N3: Equal magnitude Opposite direction Same kind of force

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