# An Atwoods Machine hanging from the ceiling

1. Jun 13, 2014

### x86

1. The problem statement, all variables and given/known data
This isn't a homework problem, but I thought it would fit best in this section because it is an actual problem. Therefore, I don't know what the solution is, but it interests me.

First let me start off by saying I've solved Atwoods machines before but now that we've introduced a rope A that the Atwoods machine is hanging from, it gets very confusing for me.

The question that I'd like to figure out is:
What is the tension in rope A if the Atwoods machine is at rest on the ceiling, but the masses are not? m2 is heavier than m1. The rope is mass less and has no friction.

2. Relevant equations
Fnet = ma
w = mg
g = |9.8 m/s^2|

3. The attempt at a solution

T1 = T2 = T for m1 and m2

The force acting on m1
Fnet = T - m1a1

The force acting on the mass m2
Fnet = T - m2a2

Up is the positive direction, and m2 is going down, so our new equations are

Fnet = T - m1a
Fnet = m2a - T

But this is sort of where I get confused, because the Atwoods machine is hanging from the ceiling.

I know that m2 has a tension force acting on it in the up direction, and the rope has a resultant force from this tension which is pulling the string down- this causes mass m1 to have a Force in the up direction and its resultant force is what causes m2's force in the up direction.

So the rope has no net force acting on it, because both of the tensions from m1 and m2 on the rope cancel out.

Now if we have a rope A that is applying a tension in the up direction to our rope.

So our rope has a force pulling it upwards.

Since the rope is mass less we have to "skip" it and pretend its not there and apply this tension of force A onto both of the masses.

But I have no idea how this tension applies.

Do I have to divide it in half because I have two masses?

And if I did do that, it would change the acceleration of the two masses, but that doesn't make sense to me- because if we just pretended the Atwoods machine was hanging from a ceiling and ignored the tension A, then the acceleration would be different from what I would get if I didn't ignore it. And I know this is wrong.

2. Jun 13, 2014

### Nathanael

True that those forces cancel out. False that it has no net force acting on it. What about the weight of the objects, is that not a force acting on the rope?

3. Jun 13, 2014

### Nathanael

The tension from A acts on to the 2 objects (and rope) together, as if they were 1 object

4. Jun 13, 2014

### x86

Doesn't the weight not act on the rope? The reactant force of the weight is the force of the mass on the earth.

If it is (m1+m2)g then doesn't that basically cancel the force of gravity acting on the masses, thus making the whole system not move at all?

5. Jun 13, 2014

### Nathanael

Yes that is the reactant force, but the weight is still attatched to the rope, and therefore exerts a force on the rope.

If I pull down on one of the objects, (with my hand,) the reactant force will be that the object pulls up on my hand. But there will still be a force on the rope, right?

Basically, since the rope is attatched to the object, it's essentially a part of the object, so the force of gravity on the object is also a force on the rope.

It does cancel the force of gravity of the system, thus keeping the whole system from falling down. But it only acts on the entire system as a whole.

But rope A is not directly attached to the other rope/weights. If it were, then the objects would not move at all like you said. Rope A is holding up the pulley-thing, so it keeps it (and the rope/weights on it) from falling down, but the rope/weights still have freedom to slide over it.

(The reason I kept saying "rope/weights" is because it's essentially a single object)

6. Jun 13, 2014

### x86

Yes, that is true. But for the first part of your post, if the force of gravity for a mass, say m1, acts on the rope and pulls it down, doesn't that essentially cause m2 to receive an upward force of m1g plus an upward force of T?

Therefore the force on say m2 would now be

Fnet = T + m1g - m2g

That means that we would receive a different answer for the accelerations.

Last edited: Jun 13, 2014
7. Jun 13, 2014

### Nathanael

But the mass pulling it down is what causes the tension, so it's already accounted for in the tension term

8. Jun 13, 2014

### x86

True. But aren't both of those forces canceled out? There are two tensions, each in opposite directions. Therefore the net force on the rope is 0. How does that effect the rope A?

You said in your prior post it cancels out, but then what causes the pulley to have a Force downwards?

Also, A = 2T basically?

9. Jun 13, 2014

### Nathanael

The tension forces do cancel out, so if you're looking at the net force on the entire system, tension is irrelevant (because it cancels out) so you just have $F_{system.net}=g(m_2-m_1)$

But that doesn't necessarily mean that the net force on the rope is zero, because tension is not the only force involved in the situation.

(If you're looking at a single block, though, instead of the entire system, then tension becomes relevant again, because it no longer cancels out.)
No, that wouldn't be true. Tension is just with how much force the rope is being pulled apart (or, another way of looking at it is, with how much force the rope is trying to pull itself back)

The tension in the lower rope doesn't have any special relationship with the tension in rope A

I'm sorry, I don't quite understand what you are saying. Can you elaborate?

Last edited: Jun 13, 2014
10. Jun 13, 2014

### x86

Okay, thanks. I think I understand it now. But isnt the net force g(m1 + m2) and not g(m2 - m1)?

11. Jun 13, 2014

### Nathanael

Well they're in opposite directions, that's why I subtracted.

(In a sense, they're in the same direction. But the reason why it's actually in opposite directions is because one is trying to move the rope "to the right" and the other is trying to move the rope "to the left")

12. Jun 13, 2014

### Tanya Sharma

Are you talking about the net force on the system comprising the two blocks and the pulley ? If yes , net force is neither g(m1 + m2) nor g(m2 - m1) .

13. Jun 13, 2014

### Nathanael

What would the net force on the rope and weights be? We're talking about a simplified situation where the pulley only serves the purpose of changing the direction.

14. Jun 13, 2014

### Tanya Sharma

(m1+m2)g - 2T

15. Jun 13, 2014

### BiGyElLoWhAt

Since the objects are connected by a rope, which travels over a pulley, both WEIGHT forces act down. The weight force from one block exerts a force on the rope which exerts a force on the other block, therefore the system is comparable to something like this (see attached) ignoring the mass of the pulley of course. What's the net force on this system?

#### Attached Files:

• ###### atwoods analogy.png
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16. Jun 13, 2014

### BiGyElLoWhAt

The system comprising of the 2 blocks and rope of course.

17. Jun 13, 2014

### Nathanael

Ok yes I suppose that would be the net force on the center of mass. So we weren't technically talking about the entire system

But in a sense we were talking about the whole system, just over a different path (the pulley changes the direction of one of the mg's and one T leaving you (m1-m2)g)

I don't know what this "net force" would be technically called

Last edited: Jun 13, 2014
18. Jun 13, 2014

### Tanya Sharma

Even if you consider only the blocks ,then too (m1-m2)g doesn't make any sense .

Whichever system you want to consider , only one block ,both the blocks ,both the blocks + pulley ; (m1-m2)g is not the net force on any of them .

19. Jun 13, 2014

### Nathanael

The system is 2 blocks and a rope.

Do you really not understand where I'm coming from at all? (m2-m1)g doesn't make ANY sense?

The path it moves is an upside down U. Imagine you could somehow "bend space" to make the path straight. Then you would have the picture that BiGyElLoWhAt attatched a few posts ago, with both of the tensions cancelling out and the net force being (m2-m1)g

Sure, the net force on the Center-of-Mass would be (m2+m1)g-2T but what I am talking about is the Net force "along the path of motion" (so to speak).

20. Jun 13, 2014

### Tanya Sharma

So you believe that net force (m2-m1)g acts on the system of 2 blocks and rope . Then what is the acceleration of the COM of the system (2 blocks and rope) ?