Understanding Newton's Third Law: Exploring Confusion and Misconceptions

[EddiePhys]

I am not the best at drawing, I know.
If A exerts a force of 200N on B, then by Newton's third law, 200N would be exerted on it. Similarly, if B exerts 100N on A, by Newton's third law, 100N would be exerted on it. Then why wouldn't both accelerate in opposite directions with accelerations equal to 300/m_a and 300/m_b respectively?

 

[vanhees71]

Why should they? Newton's 2nd Law tells you that
$$m_1 \vec{a}_1=\vec{F}_{1}, \quad m_2 \vec{a}_2=\vec{F}_2.$$
Now, if the forces are interaction forces between the two bodies, i.e., if the system is closed, you have (3rd Law)
$$\vec{F}_1=-\vec{F}_2$$
and thus
$$m_1 \vec{a}_1+m_2 \vec{a}_2=0 \; \Rightarrow \; m_1 \vec{v}_1+m_2 \vec{v}_2=\vec{P}=\text{const},$$
i.e., that the total momentum is a conserved quantity. So Newton's 3rd Law is, from a modern perspective, just the conservation of total momentum of a closed system, which is a necessary consequence of the homogeneity of space in the Newtonian space-time model (via Noether's theorem).

 

[sidt36]

Remember F = MA
F1 = -F2 is a perfect interpretation of the 3rd law

How ever

your forgetting

The masses aren't equal

So something which might help you over her
m1a1 = m2a2

Hope that helped

 

[jbriggs444]

View attachment 100306
I am not the best at drawing, I know.
If A exerts a force of 200N on B, then by Newton's third law, 200N would be exerted on it. Similarly, if B exerts 100N on A, by Newton's third law, 100N would be exerted on it. Then why wouldn't both accelerate in opposite directions with accelerations equal to 300/m_a and 300/m_b respectively?

You exert either 100N or 200N and you then expect an acceleration as if you applied 100N and 200N together? But let that slide.

Edit: Perhaps you have a violation of Newton's third law in mind so that A pushes on B with 100N at the same time that B pushes on A with 200N. How is that supposed to work?

What about the rest of the picture? One imagines that hand A and hand B are both connected to the same person and that the person is pressing his hands together (with either 100N or 200N of force). That means that his arms are pushing his hands together. The F in Newton's second law is the net force on the object from all the individual forces combined. It is not the single force that happens to appear on a drawing.

When you do a "free body drawing", you are expected to include all relevant forces.

 

[EddiePhys]

You exert either 100N or 200N and you then expect an acceleration as if you applied 100N and 200N together? But let that slide.

Person A is exerting a 200N force on B and B is exerting 100N on A

 

[jbriggs444]

Person A is exerting a 200N force on B and B is exerting 100N on A

Try to do that in real life. You can't.

 

[EddiePhys]

Try to do that in real life. You can't.

Why not? Okay, maybe two pistons or some machine instead of human? One exerting 100N and the other exerting 200N acting on each other. Surely that's possible

 

[jbriggs444]

Why not? Okay, maybe two pistons or some machine instead of human? One exerting 100N and the other exerting 200N acting on each other. Surely that's possible

Nope. Cannot be done. The force the one exerts on the other will always be exactly equal to the force the other exerts on the one.

 

[lychette]

I think that you are misunderstanding Newton's 3rd Law...read it carefully!
If A is exerting a force of 200N on B then B is exerting a force of 200N on A
The force on A cannot be different to the force on B

 

[nasu]

Your problem is not even third law so much as the notion of force as a measure of interaction.
In order to have a force you need to have two objects interacting in some way. It's not that one does something on the other but rather a mutual process.
For one interaction there is one pair of forces. If you have another pair of forces, you have some other interaction, either between another pair of objects or a different kind of interaction between the same objects (like gravitational and electric interaction, for example).
It would be less confusing if you stop thinking in terms of "A did this to B" and see it as interaction between A and B, measured by a pair of equal and opposite forces.

 

[EddiePhys]

Nope. Cannot be done. The force the one exerts on the other will always be exactly equal to the force the other exerts on the one.

So if I take a piston that can push with 200N and one that can with 100N and place their pushing ends together and activate them at the same time, what would happen?

 

[jbriggs444]

So if I take a piston that can push with 200N and one that can with 100N and place their pushing ends together and activate them at the same time, what would happen?

It depends on the pistons. A piston that can push with 200N against a static load may not be able to achieve that against a load that is retreating. A piston that can push with 100 N against a static load may do better or worse against a load that is encroaching. Whatever occurs, the force of the pistons on one another will at all times be exactly equal.

Consider a 1 gram piece of paper placed between the two pistons and suppose, for the sake of argument, that the one piston succeeds in exerting 200N and the other piston succeeds in exerting 100N. Please calculate the resulting acceleration of the piece of paper.

 

[EddiePhys]

Consider a 1 gram piece of paper placed between the two pistons and suppose, for the sake of argument, that the one piston succeeds in exerting 200N and the other piston succeeds in exerting 100N. Please calculate the resulting acceleration of the piece of paper.

200-100/mass => 100/0.001 wait what is going on. The acceleration can't be this much.

 

[jbriggs444]

wait what is going on. The acceleration can't be this much

BINGO!. That's exactly right. The acceleration cannot be that much. And the only way to make it not that much is for the two pistons to be exerting [nearly] equal and opposite forces.

 

[EddiePhys]

BINGO!. That's exactly right. The acceleration cannot be that much. And the only way to make it not that much is for the two pistons to be exerting [nearly] equal and opposite forces.

This is, for some reason very unintuitive for me. If two people are pushing on each other, wouldn't the stronger one cause the weaker one to accelerate? If the forces are equal and opposite how is this possible?

 

[jbriggs444]

This is, for some reason very unintuitive for me. If two people are pushing on each other, wouldn't the stronger one cause the weaker one to accelerate? If the forces are equal and opposite how is this possible?

The two forces do not act on the same object. https://www.lhup.edu/~dsimanek/physics/horsecart.htm

 

[nasu]

This is, for some reason very unintuitive for me. If two people are pushing on each other, wouldn't the stronger one cause the weaker one to accelerate? If the forces are equal and opposite how is this possible?

If the forces due to interaction between the two people are the only forces, they will both accelerate, in opposite directions. The one with lower mass will have higher acceleration. If there are other forces present, the accelerations will depend on the net force, for each person.
If the "strong" person pushes the "weak" one against a wall, there will be no acceleration, for example.

 

[EddiePhys]

If the forces due to interaction between the two people are the only forces, they will both accelerate, in opposite directions. The one with lower mass will have higher acceleration. If there are other forces present, the accelerations will depend on the net force, for each person.
If the "strong" person pushes the "weak" one against a wall, there will be no acceleration, for example.

What will be the force with which both of them will accelerate in opposite directions? Say the strong person is pushing with 200N and the the weak one with 100N

 

[jbriggs444]

What will be the force with which both of them will accelerate in opposite directions? Say the strong person is pushing with 200N and the the weak one with 100N

Newton's third law. They can't.

 

[nasu]

You still persist in this nonsense?
You were told many times that this is not possible.
The interaction between the two is characterized by a pair of forces equal in magnitude.
Each person is pushed with the same force due to the interaction between them.

What each does (how accelerate or maybe not) depends on all forces acting on each one, forces that have as a source interactions with other objects (ground, walls, chairs, etc)

 

[EddiePhys]

Newton's third law. They can't.

You still persist in this nonsense?
You were told many times that this is not possible.
The interaction between the two is characterized by a pair of forces equal in magnitude.
Each person is pushed with the same force due to the interaction between them.

What each does (how accelerate or maybe not) depends on all forces acting on each one, forces that have as a source interactions with other objects (ground, walls, chairs, etc)

Is this correct? Even though the force applied by the muscles is different, F_x is the same(action-reaction). In this case what would be the magnitude of F_x?

 

[jbriggs444]

View attachment 100335
Is this correct? Even though the force applied by the muscles is different, F_x is the same(action-reaction). In this case what would be the magnitude of F_x?

The drawing is not clear enough to allow details to be extracted. One possibility is that the force by the left arm on the left hand is 200 N and the force by the right arm on the right hand is 100 N. In that case, the hands will jointly accelerate rightward as the left arm wins. The acceleration could be calculated if the mass of the two hands is known.

Edit: Another possibility is that the left arm muscles are tensed so that they would produce a 200 N force if applied to a motionless target and the right arm muscles are tensed so that they would produce a 100 N force if applied to a motionless target. But, In fact, the hands move rightward. The left arm is able to develop less than 200 N of force on the left hand and the right muscle is able to develop more than 100 N of force on the right hand, so the acceleration and movement is lower than a simple calculation would suggest.

 

[EddiePhys]

The drawing is not clear enough to allow details to be extracted. One possibility is that the force by the left arm on the left hand is 200 N and the force by the right arm on the right hand is 100 N. In that case, the hands will jointly accelerate rightward as the left arm wins. The acceleration could be calculated if the mass of the two hands is known.

Edit: Another possibility is that the left arm muscles are tensed so that they would produce a 200 N force if applied to a motionless target and the right arm muscles are tensed so that they would produce a 100 N force if applied to a motionless target. But, In fact, the hands move rightward. The left arm is able to develop less than 200 N of force on the left hand and the right muscle is able to develop more than 100 N of force on the right hand, so the acceleration and movement is lower than a simple calculation would suggest.

Alright. But is it correct that F_x force would act on both at the point of contact?
Thus the acceleration of the right and left hand would be 200-F_x/mass and F_x-100/mass respectively?

 

[jbriggs444]

Alright. But is it correct that F_x force would act on both at the point of contact?
Thus the acceleration of the right and left hand would be 200-F_x/mass and F_x-100/mass respectively?

Yes. Assuming that "mass" is taken to mean the mass of each hand.

Since the two accelerations are equal and if the masses of the two hands are equal then it is easy to solve for F_x.

 

[EddiePhys]

Yes. Assuming that "mass" is taken to mean the mass of each hand.

Since the two accelerations are equal and if the masses of the two hands are equal then it is easy to solve for F_x.

Alright that clears things up. Thanks a lot man, I really appreciate your help.