# Questions about Newton's third law

1. Jun 21, 2006

### Byrgg

I'm not questioning his law or anything, I just need to understand a few things about it. I understand the basic concept of the law, for every force, there is an equal force in the oppostite direction.

But what provides the reaction force in the following situations?

A car driving, has an applied force forward, but what's the reaction force backwards? Also, in the same situation, what about friction? It applies a force backward, so what's the reaction force forward provided here?

Second, a person lifting weights, they apply a force upwards to lift the weight, what is the reaction force here provided by?

Also, how would the motion work in these cases? Assuming there's a car applying force forward, and the reaction force going backwards, the car accelerates forward, so the car must have a lesser mass than the object providing the reaction force, right? Or is it the other way around, and also, same with friction, it accelerates the car backwards, so then the mass of the object providing friction must be smaller than the mass of the object providing the reaction force forward, or is this backwards too?

Last edited: Jun 21, 2006
2. Jun 21, 2006

### scott1

No it's:
For every action there is an equal and oppositite reaction.

3. Jun 21, 2006

### Byrgg

I'm not wording it exactly, just summarizing it. Actually I think I figured it out, I was reading my physics textbook and it mentioned the forces being on the two objects. Therefore would I be right to believe that the reaction force of the applied force from the car is provided by the ground, and the reaction force of the friction is provided by the car?

If that's true, then how does it all work out exactly, the car does not accelerate earth much, so the reaction force on the car from the applied force is really what accelerates it forward right? And then the same for ground friction, the car has less mass, so the car is accelerated more than the earth for friction? What about air friction, are you comparing the mass of the atmosphere to the mass of the car or something?

And then for my second question, the reaction force to the applied ofrce is provided by the weights pushing back on the weightlifter, while the reaction force of air friction is provided by the weights pushing back on the air, right?

And what about the 'normal force', and object on a surface, say a chair on a floor, gravity pulls the chair down, the chair pulls earth up, then what does that make the normal force?

4. Jun 21, 2006

### Byrgg

Ok, I think the normal force is the reaction force to gravity, but then can someone explain something? When you are in free fall, you only experience the force of gravity, right? There still needs to be a reaction force here, you are acclerating the earth very slightly during free fall. You pull Earth up, while it pulls you down, but on a surface, gravity still pulls you down, and therefore there's an equal reaction force opposite in direction. This means that you are still pulling Earth up, right? If this is so, how is it any different than free fall? I'm sure I'm missing something really obvious here, but someone please help.

5. Jun 21, 2006

### Staff: Mentor

Forget about the terms "action" and "reaction"; they are old-fashioned and invite confusion. A more accurate term is "3rd law pairs". Think of the 3rd law like this: Forces always occur in pairs. If object A exerts a force on object B, then object B exerts an equal and opposite force on object A.

So, to identify 3rd law pairs, start by identifying the objects that are interacting. For example: The normal force is the contact force that the ground (for example) exerts on you; so the 3rd law pair is the contact force that you exert on the ground.

The normal force and your weight are not 3rd law pairs. Your weight is the gravitational force of the earth on you; the correct 3rd law pair to your weight is the gravitational force that you exert on the earth.

Whether in free fall or not, you are always pulling up on the earth with a force equal to your weight.

6. Jun 21, 2006

### rcgldr

Assuming you mean a car driving forwards, at the contact patch of the tires, the tires exert a backwards force on the pavement. The pavement exerts an equal and opposite forwards force. If the car is cornering, at the contact patch, the tires exert an outwards force, and the pavement an inwards force. Since the earth's mass is so much larger than the car, the change in angular velocity of the earth is insignificant.

Getting back to the car moving forwards, if the sum of the friction and drag forces are less than the forwards force from the pavement, then the car accelerates. If the friction and drag forces are the same then the car maintains a constant speed. Lastly if friction, and drag forces are higher, the car decelerates.

7. Jun 21, 2006

### rcgldr

This seems to imply that the gravitational field created by a person would be significant compared to that of the earth.

The force that draw a person towards the earth, and the earth towards the person is almost all due to the gravitational field created by the mass of the earth.

8. Jun 21, 2006

### Staff: Mentor

Not sure what you mean. What is important is that person and earth exert equal and opposite forces on each other. Of course, the same force produces different effects on each.

Not sure what you mean by this.

9. Jun 21, 2006

### Staff: Mentor

The earth and you may have vastly different gravitational field strengths, but there can be only one force between you and both masses are equally important: decrease either one by half and the force is decreased by half.

10. Jun 22, 2006

### rcgldr

My point was that the person's mass contributes little to the strength of the gravitational field. I agree that the mass of the person corresponds to the amount of force generated, but just wanted to note that the gravational field component due to the mass of the person is very small compared to the field component due to the earth's mass.

11. Jun 22, 2006

### Byrgg

Yeah, I thought of this last night, the reaction force of gravity is the force you exert on Earth, not the force exerted upwards on you. The upwards force on you is the reaction force of gravity pulling you into the ground, right? Not the reaction force of gravity simply pulling you down.

I was kind of confused because another site mentioned that the normal force occured because of newton's third law, and this is true, but the normal force and force of gravity(weight) are not pairs.

Hmm, but what about air friction, it exerts a backwards force on the car, and the car exerts a forward force on the air, since you analyze it as the air accelerating the car backwards(but only slightly), this means the car must have a smaller mass than the air right? Is this because you are considering the entire atmosphere or what?

Last edited: Jun 22, 2006
12. Jun 22, 2006

### Staff: Mentor

Right: The upwards force on you (the normal force) is the reaction force to you pressing down onto the ground.

This example might clarify why normal force and weight are not 3rd law pairs. You stand in an elevator car which is not accelerating. Let's describe the forces involved in terms of 3rd law pairs. The earth pulls you down (your weight); in turn, you pull the earth up. You also press down onto the floor of the elevator; in turn, the elevator pushes you up (the normal force).

If you cut the cable so that the elevator is now in free fall (forget about brakes and friction): As far as your weight goes, nothing changes: The earth still pulls you down (your weight); in turn, you still pull the earth up. However, the amount by which you press onto the floor of the elevator drops to zero; in turn, the normal force of the elevator pushing you up drops to zero.

If the air pushes on the car (air resistance), then the car pushes on the air. That's what Newton's 3rd law tells us. This has nothing to do with the relative masses of the car verses the air, or whatever. To determine the acceleration of an object (such as the car), you must consider all the forces acting on it and its mass. This is Newton's 2nd law.

13. Jun 22, 2006

### Byrgg

But say you are standing on the ground, you accelerate earth, earth accelerates you, same with the ground. Can't the same thing be said here? For air friction, it accelerates backwards, the car accelerates the air forwards. Similar to the previously mentioned situation. The only thing is, earth has a much larger mass, so it doesn't acclerate anywhere near as much.

14. Jun 22, 2006

### Hootenanny

Staff Emeritus
Whats your point? It would be better if we spoke in terms of forces rather than accelerations. The air doesn't accelerate the car backwards, it produces a force which opposes motion. As Doc Al says, it is the sum of the forces which is important.

15. Jun 22, 2006

### Byrgg

The air doesn't accelerate the car bakcwards? But if it's applying a force, doesn't it mean there's an accleration? Like I read in my textbook, standing on the earth, you accelerate earth, just as it accelerates you, because you are both applying forces on one another, can't I apply the same idea here? I'm really sorry, I'm not completely understanding all of this, but I ask that you please continue to help until I've sorted this out.

16. Jun 22, 2006

### Hootenanny

Staff Emeritus
If the air always accelerates the car backwards, as you say, then how does a car move forwards? How does a car accelerate?

17. Jun 22, 2006

### nrqed

It is a very bad idea to mix forces and accelerations like this. The key point is that Newton's second law states that the acceleration is proportional to the NET (vector) sum of all the forces.

There can be 2,3, 20 or even 500 forces acting on a an object but the object can only have one acceleration! Or there may be 500 forces and the acceleration may be zero! Or there may be two forces on an object (which don't cancel) but the acceleration is not even in the direction of either of the two forces!
So it is really bad to think of an acceleration on the object for each force!

One should not even think about "9.8 m/s^2" when thinking about gravity. This is the acceleration only if the force of gravity is the only one acting (and one is near the surface of the Earth). One should rather think in terms of the gravitational force, which is mg. The acceleration will simply be the vector sum of the forces divided by the mass. *If* only gravity is acting, this will come out to be 9.8 m/s^2 downward.

Patrick

Last edited: Jun 22, 2006
18. Jun 22, 2006

### Byrgg

So basically, is it incorrect to consider that you and the earth are acclerating one another if you are standing on a surface? The earth is not accelerating you because the normal force exactly opposes the gravitational force? And you are not accelerating the earth because the force you exert on earth(up) is cancelled out by the the force that is exerted by you pushing the ground(down)?

So then you only analyze acceleration if there's a net force on the object? Did I get all that right?

19. Jun 22, 2006

### Hootenanny

Staff Emeritus
That is correct. Only unbalanced forces create an acceleration, therefore the expression for Newton's second law is usually presented as;

$$\sum\vec{F} = m\vec{a}$$

Notice the vector symbols above the force and acceleration, direction matters.

20. Jun 22, 2006

### Byrgg

Ugh, thanks for all your time everyone, I guess it was a little simply then I thought, so then the only acceleration that occurs is a result of the net force than? So then if air friction was the only force applied horizontally, than you would consider that as the acceleration? But of course that isn't really possible is it? Because air friction could only occur if some force was a applied forward... Then again if the force applied forward was smaller than air friction, then the acceleration would be backwards, right? In that case, the air would be accelerating the car, and the car would be accelerating the air, right? Of course I know that the car is moving forwards, but the acceleration is backwards(therefore the car's velocity is decreasing), but , what role do the relative masses play here?

Last edited: Jun 22, 2006