Newton's third law in ropes and tension

[davidbenari]

For some reason I couldn't answer this, and I hope you can help me:

a) Two people pull on a rope, one with 200 N , the other 100 N. Why is the tension on the rope the shared magnitude of both forces?

b) If two people pull on a rope with equal force (say, 100 N), each person pulling feels their own force (Newton's 3rd law) plus the force the other guy exerts. Why is the tension not 200 N then?

c) If you could also explain what really happens in (b) with another combination of forces (non-equal), addressing how Newton's third law works in that case, I'd appreciate it very much. (say, 200 N and 100 N).

Thank you very much.

 

[Doc Al]

a) Two people pull on a rope, one with 200 N , the other 100 N. Why is the tension on the rope the shared magnitude of both forces?

What do you mean by "shared magnitude of both forces". If the rope is presumed massless (as typically done in physics exercises), then the net force on the rope must be zero. So it would be impossible for the two people to pull with different forces.

b) If two people pull on a rope with equal force (say, 100 N), each person pulling feels their own force (Newton's 3rd law) plus the force the other guy exerts.

Says who?

Why is the tension not 200 N then?

If you pull the rope with a force X, then the tension in the rope is X.

c) If you could also explain what really happens in (b) with another combination of forces (non-equal), addressing how Newton's third law works in that case, I'd appreciate it very much. (say, 200 N and 100 N).

Again, for a massless rope they cannot pull with unequal forces. For a real rope with mass, if they were to pull with those forces then the tension would vary from 200 N on one end to 100 N on the other. Realize that there would be a net force on the rope, which will cause it to accelerate.

 

[olivermsun]

For some reason I couldn't answer this, and I hope you can help me:

a) Two people pull on a rope, one with 200 N , the other 100 N. Why is the tension on the rope the shared magnitude of both forces?

This is like the "man walks into a bar" joke.

If two people pull on a rope, the first pulls with 100 N and the second pulls back with 200 N, then the first person will probably fall down. They need to both be pulling with the same force (e.g., 200 N) but in opposite directions to keep the rope stationary between them.

b) If two people pull on a rope with equal force (say, 100 N), each person pulling feels their own force (Newton's 3rd law) plus the force the other guy exerts. Why is the tension not 200 N then?

You pull "back" on the rope, it pulls "forward" on you (indirectly because of the other person). That's Newton's 3rd law.

c) If you could also explain what really happens in (b) with another combination of forces (non-equal), addressing how Newton's third law works in that case, I'd appreciate it very much. (say, 200 N and 100 N).

The forces are always equal. That's why the 3rd law is as stated.

 

[davidbenari]

Why can't a massless rope have a net force? I know it's a little bit weird because of F=m*a but without caring much about how it's accelerating why can't we say it has a net force? I think this is what happens in the atwood problems from typical physics exercises where unequal weights are hanging on each side.

Also, you're saying the felt tension is only equal to the reactive force?

What I mean by shared magnitude is that if one person exerts 200 N and the other 100 N the tension is only 100 N because that's the magnitude they both share. That's what I read somewhere else (in another forum).

Is this problem hard because it's suggesting something which isn't really possible?

 

[olivermsun]

Why can't a massless rope have a net force? I know it's a little bit weird because of F=m*a but without caring much about how it's accelerating why can't we say it has a net force? I think this is what happens in the atwood problems from typical physics exercises where unequal weights are hanging on each side.

It's kind of a bad idealization when what you care about is the acceleration of the rope.
If you care about the acceleration of some dangling weight, then you can at least say the mass of the rope is "negligibly small" compared to the weight.

Also, you're saying the felt tension is only equal to the reactive force?

It should be.

 

[davidbenari]

Ok so assuming a rope with mass held between two astronauts in space, and one astronaut pulls with 200 N while the other pulls 100 N. What happens to the system? To each astronaut? To the rope?


Also, I'm interested in what made you think the scenario was impossible for the massless rope. I know it was the fact that it's massless, but I'd like to know what you think would've happened that made you say this was unreal.

Thanks.

 

[olivermsun]

I don't think it's possible for the astronauts to pull on the rope with different tensions. I dunno, unless there's something I'm overlooking.

My earlier example might have been a bit misleading: suppose you're holding on to the rope and the other astronaut yanks on it with 200 N. You will have to pull back with 200 N if you plan to keep holding on to the rope.

If you both pull on the rope with 200 N then I guess you'll both accelerate toward each other at whatever rate 200 N produces on your respective bodies.

 

[Doc Al]

Why can't a massless rope have a net force? I know it's a little bit weird because of F=m*a but without caring much about how it's accelerating why can't we say it has a net force?

You can't "not care" about its acceleration.

I think this is what happens in the atwood problems from typical physics exercises where unequal weights are hanging on each side.

As long as the pulley is massless and frictionless, the tension is the same throughout the rope.

Also, you're saying the felt tension is only equal to the reactive force?

Newton's 3rd law applies. By definition, the tension is the force that the rope exerts on whatever is pulling it. So if the tension is 100 N, that means you are pulling with 100 N.

What I mean by shared magnitude is that if one person exerts 200 N and the other 100 N the tension is only 100 N because that's the magnitude they both share. That's what I read somewhere else (in another forum).

As I said before, if one person pulls with 100 N and the other with 200 N, then the tension will have to vary across the rope. (If the rope has no mass, this will be impossible.)

Is this problem hard because it's suggesting something which isn't really possible?

Not clear what the problem is that you need to solve.

 

[olivermsun]

As I said before, if one person pulls with 100 N and the other with 200 N, then the tension will have to vary across the rope. (If the rope has no mass, this will be impossible.)

Ah yes—this is another possibility. The rope can be stretchy/springy.

 

[Doc Al]

Ok so assuming a rope with mass held between two astronauts in space, and one astronaut pulls with 200 N while the other pulls 100 N. What happens to the system? To each astronaut? To the rope?

You cannot assume that because they may want to pull with those forces that it is actually possible. But if they did pull with such forces, the rope would accelerate. (I don't see how that would be possible, except for a short time, since the astronauts will be pulling themselves as well. Regardless of how they pull on the rope, the center of mass of the system--rope plus astronauts--cannot accelerate.)

 

[davidbenari]

Maybe this question is stupid and I should work out the answer but:

If the tension is just the reactive force, then how can it be the same throughout a rope that holds two unequal weights in an atwood machine, and how could these masses even accelerate? I.e. I'm imagining that since one weight pulls down with Mg and the other mg, tension will just pull back on each side with an equal force that's respective to each mass. Why isn't this the case?

 

[olivermsun]

Yeah so now I think I understand what you're asking.

Think of this:
Two weights, each of mass M, connected through pulleys. If the situation is totally stationary then would you agree or disagree that each half of the system is identical to a system with no pulley and the mass M just dangling from a fixed rope?

 

[davidbenari]

I would agree to that.

 

[Doc Al]

Maybe this question is stupid and I should work out the answer but:

If the tension is just the reactive force, then how can it be the same throughout a rope that holds two unequal weights in an atwood machine, and how could these masses even accelerate? I.e. I'm imagining that since one weight pulls down with Mg and the other mg, tension will just pull back on each side with an equal force that's respective to each mass. Why isn't this the case?

Your mistake is assuming that the the mass pulls down on the rope with a force equal to mg (its weight). But that is incorrect. Realize that mg is the force that the Earth pulls on the mass, not the force that the mass pulls on the rope. The actual force that the mass pulls on the rope depends on the mass on the other end of the rope. (For example, remove that other mass and the tension in the rope goes to zero.) Regardless, in the massless rope approximation, the tension will be the same on both ends of the rope.

Also: If the mass pulled the rope with a force equal to its weight, then by Newton's 3rd law the rope would have to pull up on the mass with an equal force. That would make the net force on the mass always equal to zero! (Of course, if the two masses are equal then the tension would equal their weight. But that's a special case.)

 

[olivermsun]

Okay, so Doc Al has already explained what I was going to say.

 

[Dale]

Why can't a massless rope have a net force?

If it did the acceleration would be infinite. Although it is an approximation, it is usually a good approximation to think about. In real life, if a rope is not massless but is very light compared to the masses of the other bodies, then the difference in the forces will be close to zero.

For example, in your 200 N astronaut and 100 N astronaut example, by Newton's 3rd law the net force on the rope would be 100 N. If the rope weighed 10 grams then it would accelerate at 10000 m/s^2.

Presumably it would be difficult for an astronaut to accelerate a rope at that rate, so using the massless approximation and ensuring that the forces are equal helps catch little mistakes.

 

[davidbenari]

And would you say the rope in an atwood machine is accelerating? (the rope is massless and unequal masses hang on both sides). Also: Why is the tension in a massless string assumed to be constant all throughout even if its moving like in the case of the atwood pulley?

 

[jbriggs444]

And would you say the rope in an atwood machine is accelerating? (the rope is massless and unequal masses hang on both sides). Also: Why is the tension in a massless string assumed to be constant all throughout even if its moving like in the case of the atwood pulley?

What is the rate of change of the momentum of a section of massless rope?

How does the net force on a section relate to the rate of change of momentum in that section?

 

[Dale]

And would you say the rope in an atwood machine is accelerating? (the rope is massless and unequal masses hang on both sides). Also: Why is the tension in a massless string assumed to be constant all throughout even if its moving like in the case of the atwood pulley?

Yes, the rope in an Atwood machine is accelerating. Since the mass is zero it doesn't require any net force to accelerate.

In reality, of course, it will have some weight, but if it is small relative to the other masses then it is negligible and using the massless approximation (equal tension) won't be too big of an error.

 

[davidbenari]

What is the rate of change of the momentum of a section of massless rope?

How does the net force on a section relate to the rate of change of momentum in that section?

I guess the rate of change of momentum = 0 given that the rope is massless. If the net force on a section is 0, then the rate of change in momentum is 0 too. But the rope is accelerating in atwood pulley problems!

 

[davidbenari]

DaleSpam: and why is the tension assumed to be equal throughout, what's the rationale here? Thanks.

 

[Doc Al]

DaleSpam: and why is the tension assumed to be equal throughout, what's the rationale here? Thanks.

The net force on any segment of the massless rope must be zero. That implies that the tension is the same throughout.

 

[davidbenari]

The net force on any segment of the massless rope must be zero. That implies that the tension is the same throughout.

Then how is it accelerating in an atwood pulley problem (if there's no net force)?

 

[Doc Al]

Then how is it accelerating in an atwood pulley problem (if there's no net force)?

It's just an approximation. If the rope is treated as massless, then no net force is required to accelerate it. It just goes along for the ride.

 

[davidbenari]

:uhh:

 

[Dale]

DaleSpam: and why is the tension assumed to be equal throughout, what's the rationale here? Thanks.

Draw a free-body diagram for any segment of the rope. Since ∑f=ma and since m=0 we have that ∑f=0 regardless of a. Therefore the force is the same on each end of any segment, therefore the tension is constant.

 

[olivermsun]

Then how is it accelerating in an atwood pulley problem (if there's no net force)?

You have two bodies, M and m. Gravity is pulling (down) with force -Mg on the one and -mg on the other. Between them is a rope that pulls (up) on each with tension T.

If M = m then the entire system is not moving, so it must be true that the net force T - Mg = 0 and therefore T = Mg.

If M > m, then you expect that the larger weight to fall and the smaller weight to rise. This happens if mg < T < Mg. Then there's a net force on each body: T < Mg (hence the M weight falls) and similarly T > mg (hence the m weight rises).

You can work out the acceleration since it has to be the same throughout the system: the two weights and the rope all move together.