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Assuming you place an object so heavy on a table that it breaks it, then according to newtons third law the forces must cancel out (equal magnitude and opposite direction), but if this is true, then how can the object break the table in the first place?

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  • $\begingroup$ "for every Action, there is an opposite and equal re-Action." What do you think is the "Action" in this case? Be specific. $\endgroup$ – RBarryYoung Oct 19 '13 at 20:57
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according to newtons third law the forces must cancel out (equal magnitude and > opposite direction)

That is not what the third law says. It says that the force of A on B has the same magnitude but opposite orientation to the force of B on A.

The two forces act on different bodies, so they do not "cancel out". In common situations like the one you mentioned, the 3rd law is valid irrespective of whether the object or table move.

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  • $\begingroup$ but if 'the force of A on B has the same magnitude but opposite orientation to the force of B on A' then doesn't that mean they would cancel out since the magnitude on A is equal to the magnitude on B and opposite orientation? Logically speaking...? $\endgroup$ – fYre Oct 19 '13 at 22:38
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    $\begingroup$ @HaniSayegh: They would cancel out if it made sense to add them. Similarly, the money you have in the bank might cancel out my gambling debts, but I don't think you intend to add those two numbers. $\endgroup$ – Ben Crowell Oct 19 '13 at 23:11
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If you put an object on the table the forces cancel out, if you take the WHOLE system into account.

You forget to look at the entire system: table-object-earth. The breaking of your table happens because your object is to heavy for your table (or that is at least what I assume). The net force of your object on the table can be nonzero, but then you need to take the earth (which causes the gravitational pull on your object) into account.

In the case of the non-breaking table you also need to take the earth into account (in principle), but then te contributions of the earth cancel out since the push of your object on the table is neutralised by the push of the earth on the table.

In adittion to the comments, consider n different particles with momentum $\mathbf{p}_i$, which feel a force $\mathbf{F}_i$, then we can apply Newton's third law:

$\mathbf{F}_i = \dot{\mathbf{p}_i}$, and sum for all of the particles:

$\sum\limits_{i=1}^n\mathbf{F}_i = \sum\limits_{i=1}^n\dot{\mathbf{p}_i}$

Now we define the total momentum as $\mathbf{P} = \sum\limits_{i=1}^n$ and the total force $\mathbf{F} = \sum\limits_{i=1}^n\mathbf{F}_i$, this yields:

$\mathbf{F}=\dot{\mathbf{P}}$.

Now since gravity (I will consider gravity since this was the force of your origional question) is a central force, the Lagrangian has translation-invariance and we have conservation of momentum, so $\dot{\mathbf{P}} = 0$, which tells us that:

$\mathbf{F} = 0 \Leftrightarrow \sum\limits_{i=1}^n\mathbf{F}_i = 0$.

So the sum of all forces should vanish! To come back to your origional question, your sum of the forces didn't equal zero because you didn't include the contributions $\mathbf{F}_i$ which came from the earth.

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  • $\begingroup$ Not quite. Two asteroids can collide and one break the other with no direct contact with other celestial bodies. And the third law still applies. $\endgroup$ – legrojan Oct 20 '13 at 13:49
  • $\begingroup$ @legrojan correct, then your entire system are the two asteroids (and after the collission the different pieces). If you'd sum all forces this should give zero due to conservation of momentum (see adjusted comment) $\endgroup$ – Nick Oct 20 '13 at 13:59
  • $\begingroup$ Precisely. It is therefore not necessary to include the Earth in the original system since Earth and table are in equilibrium (at least as long as the table is still in one piece). The only drawback of that is having a floating table :) $\endgroup$ – legrojan Oct 20 '13 at 14:08
  • $\begingroup$ @legrojan I don't see why Newtons laws would only apply in equilibrium ? Newtons laws are used do discribe dynamics (certainly not an equilibrium situation). In equilibrium you would have that $\dot{\mathbf{p}}_i = 0$ for the individual particles and the result would be evident! The whole point of my answer is the fact that you need to take into account your ENTIRE system, of a closed system as it's called nowadays. Simply go to the wikipedia-page (en.wikipedia.org/wiki/Newton's_laws_of_motion) and search the wordt equilibrium. You won't find it (unless you ofcourse edit it ...) $\endgroup$ – Nick Oct 20 '13 at 14:15
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    $\begingroup$ No, that's not my point, I expressed myself wrong. Newton's laws apply out of equilibrium as well (see my answer). I mean that the force the earth exerts on the table and the force the table exerts back on the earth are equal and opposite, thus their sum is zero. That's what I meant with equilibrium. As I understand it now, you mean that you cannot include gravity in the OP's picture unless you include earth as well, and that is true. But if you remove both the gravity and the earth, you have equilibrium as well. $\endgroup$ – legrojan Oct 20 '13 at 14:22
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In your example of the heavy object O on the table T, the 3rd-law pair is O on T and T on O. One of these acts on T. The other acts on O. If you want to determine the motion of T, you only include forces that act on T. If you want to determine the motion of O, you only include forces that act on O. In either case, you don't add the two forces, because they act on different objects.

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The main issue here is that you are mixing the statics problem of two rigid bodies with the real life problem of the table breaking.

In an idealised world, which is what the elementary physics books deal with, the table is infinitely sturdy, as is the weight you place on it; the system reaches immediate equilibrium and nothing is broken. No need to consider the Earth here as well, but you can do that if you want to.

In the real world, however, the table is made of a real material, say wood. If the impulse it receives from the object on top of it is too strong, it will not be able to dissipate it and it will break. The third law still applies, but the table cannot be considered a rigid body. So instead you would have to look at all the "table fragments" as the whole system and see their individual forces if you would like to apply the third law here.

It is therefore much simpler to consider the problem in simple terms of energy dissipation as the table is breaking. It cannot absorb the impact so it vibrates, splits, etc. until a new equilibrium is reached. Then we find ourselves in the static case once more, only with many more objects than the original two.

So it is convenient to apply the third law only for static cases, even though if you analysed the system with all its components, you would still find that the system as a whole, which is the weight plus the table fragments, fulfills the third law to the letter.

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