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Tension on Atwoods machine

by x86
Tags: atwoods, machine, tension
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x86
#1
Jun12-14, 01:02 AM
P: 94
I'm having trouble with the concept. I have two questions. According to various internet posts, the force of gravity of the masses attached to the Atwood machine does not act on the rope whatsoever.

Is this correct? It is kind of confusing to me, because the rope is attached to these masses. So shouldn't the force of gravity on the masses effect the rope too?

This means the only force acting on the rope is reactant force of Tension from the mass. But these two forces cancel each other out.

If one of the masses goes up and the other goes down, this means the rope HAS TO MOVE.

So, then how does the rope move? There are no forces acting on the rope, so how does it move?
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UltrafastPED
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Jun12-14, 06:39 AM
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It's an idealized massless, unstretchable rope with negligible size. If you were using a real line ... imagine heaving too on a sailing ship: lines are heavy, they stretch, they have friction going through the blocks, and you get blisters on your hands until the skin toughens up!
UltrafastPED
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Jun12-14, 06:44 AM
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Oh, and the tension on the rope doesn't cancel out - the opposing forces of Newton's third law are acting on two different bodies. In the case of tension the rope is the intermediate body, carrying the force from the first two the second, and from the second to the first. So no cancelation.

You can solve these problems with real ropes and chains - but they are more complicated.

x86
#4
Jun12-14, 12:36 PM
P: 94
Tension on Atwoods machine

Quote Quote by UltrafastPED View Post
Oh, and the tension on the rope doesn't cancel out - the opposing forces of Newton's third law are acting on two different bodies. In the case of tension the rope is the intermediate body, carrying the force from the first two the second, and from the second to the first. So no cancelation.

You can solve these problems with real ropes and chains - but they are more complicated.
I drew a picture



You are saying T1 does not cancel T2, because both of these Tension forces are transferred to the masses connected to the ropes?

Also, what causes the rope to move with the masses? If tension is equal on the rope, then the rope would not move- only the masses would move.

I've thought that the heavier mass is what makes the rope move, that it "pulls" on the rope somehow, and that this pull has something to do with the force of gravity acting on the mass.
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Jun12-14, 01:26 PM
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Quote Quote by x86 View Post
You are saying T1 does not cancel T2, because both of these Tension forces are transferred to the masses connected to the ropes?

Also, what causes the rope to move with the masses? If tension is equal on the rope, then the rope would not move- only the masses would move.

I've thought that the heavier mass is what makes the rope move, that it "pulls" on the rope somehow, and that this pull has something to do with the force of gravity acting on the mass.
Start by assuming that the rope has no mass. That's not true, of course, but it's a good approximation (and we can always make it better by using larger masses so that the mass of the rope becomes relatively less significant), and it makes the solution much much easier:

Let the tension in the rope be ##T##, the force of gravity on mass 1 be ##w_1##, and the force of gravity on mass 2 be ##w_2##.

What forces are acting on mass 1? Gravity pulling down and the tension of the rope pulling up. The net force on mass 1 is ##F_1=w_1-T## after I've made the (arbitrary, but i have to be consistent about it) choice to make the positive direction downwards.

What forces are acting on mass 2? Again, gravity pulling down and the tension of the rope pulling up, so we have ##F_2=w_2-T##.

Now we can use ##F=ma## to calculate the accelerations ##a_1## and ##a_2## in terms of ##w_1##, ##w_2##, and ##T##. Then we use the requirement that ##a_1=-a_2## (if one side is going down the other side is going up) to solve for ##T## and we have the accelerations of both objects in terms of their weights.

This is the simple solution, in which we assume that the rope is massless, or at least so close to massless that we can ignore its weight and mass. Now when you ask what force makes the rope move, I answer that the rope doesn't need any force to move. It's massless, so ##F=ma## reduces to ##F=0## no matter what the value of ##a## is; I can accelerate the rope at any rate that I please even with a zero force.

But of course in the real world, the rope isn't massless. So consider how you'd solve the problem if you weren't allowed to ignore the mass of the rope... The first thing you would discover is that the tension in the rope is no longer constant along the length of the rope; at every point on the rope, the tension must be sufficient to accelerate the object at the end of the rope AND the mass of the rope below that point. Furthermore, the tension in the rope will vary in a complicated way with time, as the amount of rope and hence the mass on each side of the pulley changes with time.

You can still solve this more complicated problem, but we don't need to here. Instead, we just need to remember that the tension in the rope is no longer constant along its length. In particular, the tension in the rope at mass 1, which we'll call ##T_1##, is different from the tension at mass 2, which we'll call ##T_2##.

Ok, what's the force that mass 1 exerts on the rope? Well, the rope is pulling on mass 1 with a force of ##T_1##, so by Newton's second law, mass 1 is exerting an equal and opposite force on the rope. Likewise, mass 2 is exerting a force that is equal and opposite to ##T_2## on the other end of the rope. ##T_1## and ##T_2## are not the same so their equal and opposite forces are also not the same; they don't cancel and there's a net force on the rope. It's just enough to accelerate the rope as needed to keep everything moving together.


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