# Newtons 3 Law

1. Sep 9, 2004

### omin

Newton's 3 Law: Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first.

Since the magnitude of force applied is said to be exactly the magnitude returned and the direction is said to be exactly opposite, shouldn't no truly equal and opposite collision change the velocity of the second object?

It seems to me that the slightest change of velocity of the second object implies that equal and opposite forces didn't occur. For a velocity change of the second object to be sensed after the collision, it seems the objects must disconnect before equal and opposite actions may finish (during the time the objects touch, the action/reaction acceleration periods).

2. Sep 9, 2004

### Tom Mattson

Staff Emeritus
That's definitely wrong. If a force acts on the second object, then according to Newton's second law, there is a change in momentum. This must either manifest itself as a change in mass, or as a change in velocity.

3. Sep 9, 2004

### Staff: Mentor

What do you mean by that? That's not how collisions are typically described, so I'm thinking thats where the misunderstanding lies. Maybe you could be more specific about this collision...

4. Sep 9, 2004

### robphy

Let's clarify the Third Law.

$\vec F_\text{on A due to B}=-\vec F_\text{on B due to A}$
Assuming no other forces act,
$\vec F_\text{net on A}=\vec F_\text{on A due to B}$ and $\vec F_\text{net on B}=\vec F_\text{on B due to A}$.

So, (assuming constant masses)
\begin{align*} m_A\vec a_A\stackrel{N2}{=}\vec F_\text{net on A}=\vec F_\text{on A due to B} \stackrel{N3}{=}-\vec F_\text{on B due to A}=-\vec F_\text{net on B}\stackrel{N2}{=}-m_B\vec a_B \end{align*}
That is, assuming nonzero forces, each object in this collision experiences its own acceleration, and, thus, a change in its own velocity.

5. Sep 10, 2004

### joyful55

Well actually, a change in momentum could also manifest itself as a change in both mass and velocity as in the case of rocket fuel using up and thrsuting itself into space.

6. Sep 10, 2004

### Tom Mattson

Staff Emeritus

Right, take my "or" statement in the inclusive sense.

7. Sep 10, 2004

### Cyrus

your problem is the frame of refrence. You can push on an object but the object pushes back on YOU too. So If were on an ice rink, and I give YOU as shove, you feel a force I gave to you, and you go drifting off. But at the same time, I feel you pushing me equally and oppositely, so I also go flying off in the OPPOsite direction as you do. So we both go flying, even though I pushed on you. The force will act equally on each of us, just in different directions. The equal and opposite forces act on each body, not the same body. WoW tom is a SUPER mentor, I think he has a cape, all the other mentors are so jealous :rofl:

Last edited: Sep 10, 2004
8. Sep 10, 2004

### omin

Here's how I'm seeing this in steps:

1. Two objects not touching but on a collision course.

2. Objects impact, but no force exists yet.

3. Object one exerts a force upon object two.

4. Object two exerts a force on object one.

5. Objects separate.

I've read action and reaction may be simultaneous, but that seems to ignore spacial dimension, so I'm assuming there are spatial coordinates which implies an order in action and reaction.

In step 3 the first object impress a force upon the second object (action). If this force is returned equal and opposite in step 4 (reaction), why should the second object have any change in momentum, except during the impulse period (where force is exerted upon it but returned in exact magnitude and opposite direction, resulting in no velocity change?

9. Sep 10, 2004

### Tom Mattson

Staff Emeritus
You just answered your own question. The second object suffers a velocity change because there is a force acting on it.

10. Sep 10, 2004

### omin

Do you mean the force that occurs in the action period in step three? What the first object exerts is what it loses, so what the first object loses could be termed to be the opposite reaction of the second object, or not?

I can see that as simultaneous, but not the entire process I've tried to delineate here. What did Newton mean?

11. Sep 10, 2004

### Cyrus

Hey omin, is my post talking about the same topic as your question, or are you talking about something different? Im not sure if your asking what I responded too or not.

12. Sep 11, 2004

### joyful55

Try picturing your step 3 & 4 as a single step which occurs at the same time.

13. Sep 11, 2004

### Staff: Mentor

Newton's 3rd applies to each object individually. It seems you are thinking that the action happens to one and the reaction happens to another. Each object experiences a force, so each object experiences an acceleration. The forces are equal, the masses are equal, so the accelerations are equal.

14. Sep 11, 2004

### omin

I apologize, I didn't mean to seem to ignore your comment. I get an hour a day at the library and I don't have the time to respond to all comments and my posts get sloppy if I don't take my time with them.

You said it like the book, I admit. I'm just not sure where the equal and opposite simultaneousness assertion is in Law 3. I don't know if it's meant to be in step three or meant to occur over the course of the whole the collision phenomenon.

I tried and I have a bit of a problem fusing them. In step three, the acceleration object one loses is the acceleration object two gains. This I see as simultaneously equal and opposite. But, I don't know if that is what law 3 is supposed to represent, because of step 4 where the second object seems to me not to be able to react until acted upon which implies time duration rather than an instant of simultaneousness.

Last edited: Sep 11, 2004
15. Sep 11, 2004

### Cyrus

Well, whenever any two bodies touch they have to have equal and opposite forces. the moment they no longer act on eachother, there is no longer an equal and opposite force. I just like to think about my hand. If i push on somthing i give it a force. But I FEEL myself pushing on that object, so it HAS to be pushing back on me too, otherwise, how would I feel it? The moment I stop touching it, the force goes down to zero on both me and the thing I was pushing.

16. Sep 11, 2004

### Tom Mattson

Staff Emeritus
It doesn't matter which force you call "action" or "reaction". The simple fact of the matter is that each object exerts a force on the other. Since each object has a nonzero force acting on it, each object is subjected to a change in velocity. It's just that simple.

This doesn't make any sense to me. What exactly does either object "lose"?

He meant that forces come in pairs. And yes, they do appear simultaneously.

17. Sep 11, 2004

### omin

I'll have to think about this more, because my original question just isn't being pinpointed and it's my fault because the slight digression on other fine points sourrounding it is moving away from the original point.

18. Sep 11, 2004

### Tide

The "digression" was necessary and not slight because from your original post it was clear you had a fundamental misunderstanding of the principles involved. From my perspective the "other fine points" are right on the mark and necessary to help you understand your original point - or the incorrectness of it.

19. Sep 12, 2004

### joyful55

First of all, the equal and opposite assertion is meant to occur over the course of the whole collision. Try reading up and relationship b/w impulse, momentum and net force.

Secondly, forget what I said about steps. The whole 3rd law, has only one single mechanism is that the force when acted by a body on another could also be viewed in relative of the other body as in it was acted on the first body. Just infer from Einstein's Relativity.

20. Sep 12, 2004

### ballooza

let me tell u something omin, try to punch someone so hard on his shoulder-for example-, ofcourse u exert force there...u were asking like: where is the feedback force? well, don't u feel ur hands get hurt??? that's the equal opposite power... sometimes that's unimaginable (if that's even a word) but there's no other explannation....i just read the wuestion of yours and not the answers so maybe u know that now.... uneed anything for this answer ask away.