B Newton's third law confusion

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).

 

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

 

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.

 

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

 

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.

 

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

 

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.

 

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.

 

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.

 

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

 

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.

 

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)

 

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.

 

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.