Assume the weights are resting on a surface, the pulleys and rope are massless, the pulleys are frictionless, and the system is maintained in quasi-equilibrium as the rope is pulled (steady state & small accelerations). In this case, the tension T in the rope is constant everywhere. Now, take a horizontal section through the ropes. ("Cut" them and replace the missing portions of the rope with the tension.) Each weight is experiencing an upward force of 2T.
When 2T >= 20, or T = 10, weight A begins to rise. Once a hits a stop, T must be increased to just above 20 to get Weight B to rise. Similarly, T just above 30 causes C to rise after B stops.
The "trick" with these pulley problems is to section the problem through the cables and show the tension, T. Then you've just got free body problems, in this case subject to the floor constraint.
Oh, also, while the first weight is being lifted, the floor beneath weight B experiences 40 - 20 = 20 units of force, and the floor under C experiences 60 - 20 = 40 units of force. Once B is lifted, the floor under C experiences 60 - 40 = 20 units of force. (Presuming that the labels are weights, and not masses.)
No; the Tension the guy is supporting at rest is bigger than 10 (there are 2 other weights) so at T = 10 he would be moving backwards. He needs to apply a bit more than the tension at rest.
(edit: this is for when all 3 weights are in the air, I see now some people see them in a floor that is not drawn)
The bottom of each weight is at about the same level as the bottom of the man's feet. A floor is the natural thing to assume. Otherwise it would be difficult to get them into this configuration.
A good way to think about this is to replace the man with a fourth weight, fixed to the rope (as opposed to a pulley). Now it should be clear that the weights would not rest in that configuration unless they were supported by a floor.
It's slightly foolish to think about such an idealized scenario, but I believe that the answer is no, until mass A rises fully, the rope can still slide around masses B and C without needing to apply its tension to moving them upward.
If the system is frictionless, the rope is weightless, and the weights are not supported, then the lighter weight will rise and the heavier weight(s) will fall. Therefore if the rope is pulled very slowly, the lighter weight will rise first.
If the rope really is weightless and the pulleys really are fictionless (and inertialess) then it doesn't matter how hard or fast you pull, the lighter weight will rise first. This is at odds with your intuition simply because you have no (or insufficient) experience with weightless and frictionless environments. This is one reason why space is so bloody dangerous, in addition to the dangers posed by, say, diving, where similarly to space, your equipment has to work perfectly or you die.
In the real world, pulling fast enough will make the closer weight rise first.
>If the rope really is weightless and the pulleys really are fictionless (and inertialess) then it doesn't matter how hard or fast you pull, the lighter weight will rise first.
I'm considering a thought experiments that make me believe that this is not the whole story.
Imagine a two weight system set up similar to the original diagram in which the weights are the same weight, and the gravity is very little. Yanking on the rope I imagine them rising at the same speed. Now, we take a very small flake off of one weight and repeat the experiment. It seems clear that both weights will still rise from the start, just the lighter weight will rise at a faster speed.
I think this should extend to three weights of any positive mass -- if you yank the rope fast enough (and it might be very fast) all three should rise from the start.
If you take the rope from 0T to >30T instantly, then you will have all the weights lift up. As long as you keep the rope tense, they will all rise with different accelerations until they hit the top.
If you apply tension over time, even very quickly, then the lighter weight will rise first.
In your thought experiment you make the weight difference so small that otherwise insignificant factors (e.g. friction) take over. Go the opposite direction and make the weight difference enormous.
Assuming no friction and given a fast tug of the rope I still don't think the near equal weight would hold still or sink while the other near equal weight flew up quickly.
With the large weight difference, I think the required speed of pulling the rope to make them both rise just becomes impractically fast.
Imagine the system in space - all the weights would move up, so they all have an upward force applied to them from the rope tugging. It's just a matter of pulling so fast that that upward force overcomes gravity.
Your conclusion is incorrect given your assumptions. You are assuming no floor, in which case all the weights will rise. Only the man will fall. I think the problem was intended to include a floor that simply isn't drawn.
Do you really want me to write the full six page analysis? No, I didn't think so. Unfortunately, concise replies such as the one I gave and which require the reader to think a little often result in replies such as yours where it is difficult to discern whether you are being deliberately clever, deliberately trollish, or genuinely confused.
Let me expand.
I only assumed the weights were unsupported, not the man. I did that to assist the reader in understanding the analysis. I did not assume the man was unsupported - I had hoped my initial description of what happens implied that. Possibly it didn't.
I would expect that the problem is intended to include the floor - that's not my point. Having made the analysis for the unsupported weights, the evolution of the situation when there is a floor becomes obvious.
I'm not trolling, and only half joking. Your first paragraph where you talked about the heavier weight falling seemed like a joke about the drawing not including a floor. Which would be fine, except someone making that silly assumption should go all the way with it and assume the man is unsupported too.
Now I understand that you were helping people understand the analysis, and I'm sorry my reply sounded trollish.
Well, the diagram is impossible unless the weights are all resting on the floor, so let's assume they are. This system will find a steady state only at a local minimum of potential energy, so the lightest weight will be lifted before the heavier two get off the ground.
As tension is applied to the rope, Weight A will be lifted first, until it is lifted to the ceiling. Then Weight B, and finally Weight A.
It helps to visualize Weight A as being massless. In that case, there would just be extra slack in the rope, and B&C would not move until the slack was taken in.
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This tricks people because it doesn't start out in static equilibrium. For it to look like this, the weights would have to be resting on a table to relieve the tension in the rope.
If the man does nothing, the heaviest weight will fall and the lightest will rise. If it's frictionless and he starts pulling, the same thing will happen, only the lengths will lessen.
Could it not be in a meta stable equilibrium like this if the weights were of equal mass? The numbers of the weights are unit-less. We assume they represent the mass, but for all I know the number is the serial number of the weight.
For all we know, it could be in space, they could be heavily magnetized, 100m/s wind, .5 CoF in the pulleys, the rope could stretch, and the system could be traveling near the speed of light.
I'm with Rider, if you view it as a "classic" physics class problem, ie weightless ropes and pulleys (and hence no angular inertia for the pulleys, and no inertia due to the ropes), and no friction, weight A would rise even without pulling the rope, weight B would remain static, and weight C would drop at the same rate that weight A would go up. If you pulled on the rope you could never make weight C go up faster than weight A until it hit the top and stopped moving. So the pulling bit is a red herring, as weight A would always hit the top before the other two, either pulling or holding it still. Assuming the weights are in kilograms you would have to let the rope go at a rate of 3.13 m/s (11.27 kph, 7 mph) to prevent weight A from rising.
1. There is a floor that the weights (and the man) are standing on.
2. Weight C is dropping whether he pulls the rope or not.
I think that the diagram is meant to show all the weights resting on a floor, and all the confusion is due to misunderstanding of that. In the other case, the diagram neglects to mention something to the effect of "supports have just been removed".
Is the person currently pulling, are the weights on the ground, are the weigths currently moving to get into a stable position? Friction, weight of cord, acceleration, etc. etc. Any combination of the previous?
Depending on your level with maths/physics you'll probably give different answers and make different assumptions.
I didn't know physics problems are assumed to be idealized... When you build a bridge for example, will the physics try to be simplified to the max? (Honest question here).
Velocity would matter -- it seems to me -- if there is friction and probably a few other factors included like elasticity. No?
Personally, I always considered physicists to be applied mathematicians (not the other way around although I've seen physics problems thrown in university level math classes). That's why I put that there, so assuming a high level of math skills, you'd probably change your way of thinking quite a bit.
When you build a bridge, that's engineering, not physics. It's a physics convention that unspecified factors are assumed to be unimportant. If they were important, they would be specified.
Anyway, it's implausible that the elasticity of the rope or the friction of the pulleys are going to matter. Unless you have really rusty pulleys, or something.
This reminds me of the airplane on the conveyor belt, in that the only confusion comes from the question being insufficiently specified.
- Friction of the pulleys
- Mass of the pulleys
- Moment of inertia of the pulleys
- Mass of the rope
- Unit of mass of the weights
- Is there a surface that the weights are resting on?
- What's the local gravity like?
- Others
If we assume the things we're likely supposed to (rope mass, pulley friction, pulley mass and moment of inertia all insignificant, gravity tending down, resting on a surface), it's clear that the lightest weight will rise first. If, on the other hand, we make ridiculous assumptions (weights mass in AMU, in a no-gravity environment, high moment of inertia pulleys), then the "heavy" weight will lift first (because it's easier to lift the weight than to spin the pulleys).
The airplane on the conveyor belt is more interesting in this respect, in that different people have different ideas about how the question should be interpreted.
For this question, I think everyone will agree on the likely expected assumptions. Incidentally, the mass of the rope doesn't affect the answer so long as it's uniform, and the mass of the pulleys doesn't matter so long as the pulleys directly attached to the weights all have the same mass. And once you assume the pulleys are frictionless, their moment of inertia doesn't affect the answer either.
The length of the rope is constant. For every weigh with a pulley the weigh moves 1/2 length unit for every length unit it's being pulled. Weights don't matter; C will go up first, then B then A (at a relative 1/2 length ratio).
Assuming frictionless rope, pulleys, non-stretchable rope, etc. etc., and assuming the system is at rest, with weights on the ground.
The tension (call it T) in the rope is the force that is acting on each pulley; since the rope is wrapped around all pulleys exactly once, the force applied to every weight is 2T (assuming the fixed pulleys attached to the ceiling aren't going anywhere). If 2T isn't larger than the force of gravity for any one of the weights, then nothing is going to happen. If the man applies enough force (T), the first weight to lift up is the weight whose force of gravity (mass times g) is exceeded by 2T. Since the force of gravity is proportional to the mass, the weights do indeed matter; the lightest weight rises first.
You go against the consensus without a solid explanation. For instance, why don't the weights matter? Would they matter if they were respectively 1mg, 1kg, and 1megaton?
Fair enough. I gave the summarized version. I'll try to explain better.
I'll only take basic classic mechanics assumptions: that the rope is of constant length (ie is like a cable that doesn't compress or expand). The framework is quasi-static classical mechanics; the results of the guy pulling slowly a little bit can solve the problem or be generalized (I won't consider the situation of the guy jerking quickly the rope etc).
The guy's hand is under a calculable tension Tw that will be 60 units or whatever, it doesn't matter. The problem asks what happens when the guy pulls, so we suppose that he's not the one being pulled but he moves forward to the right. The tension Tg that he applies doesn't matter; as long as it's bigger than the one from the weights (Tg > Tw) he'll move the rope (we discard friction since he moves slowly or if you take into account friction he just needs more force, it doesn't matter). So the distribution of weights or their actual measure don't matter so far. (of course if you have a million tons and you blow the guy away weighs matter).
Now since the cable/rope has constant length (there are no slacks etc since it's moving) when the guy pulls 1mm then than length needs to be taken from somewhere in the pulley systems.
The effect of a pulley is to divide the length of rope you take in two (one has to go to the left vertical part of rope and the other one to the right one); this is why the tension in each side of a pulley is 1/2 of the total tension and you can pull with 1/2T a weight of T with a pulley. So with this we can straightforwardly calculate the tensions everywhere but we don't need that.
So the that 1mm is taken from the system and the more pulleys the rope has to go through the less is taken (because of this 1/2 I explained above), so the weight closer to the guy (with less pulleys) will raise first, then the next one in the middle, then the next one etc; weighs don't matter.
clarification update: poor conclusion wording: if weights are in the air all 3 weights go up at the same time but C will move more than B and B more than A.
When a weight is at rest, the tension in the rope must be less then or equal to half the weight. The problem asks which weight will be first to not be at rest. The answer should now be obvious.
When a weight is at rest the tension in the rope is exactly half its weight.
Problem asks which weight would be first to raise (not "not be at rest" that could be going down) so that's why I'm supposing the guy can pull the whole thing. I'm also saying the first one is C (closer to guy).
Basically if the weights are at rest on the floor then mkn's answer is the correct one. If they are at rest on the air, then my answer is the correct one.
I supposed the system was at equilibrium without making the numbers. If it's not (as it seems) then sure, my reasoning is invalid and he's right, no problem.
You had the right answer for one set of assumptions. But, at the extreme weight is important EX: if the numbers where solar masses then gravity would suck them into each other.
If they pulley’s had sufficient friction C may move first. Or if you pull a real rope fast enough C will also move first (think KM/s speeds). Etc.
Assume the weights are resting on a surface, the pulleys and rope are massless, the pulleys are frictionless, and the system is maintained in quasi-equilibrium as the rope is pulled (steady state & small accelerations). In this case, the tension T in the rope is constant everywhere. Now, take a horizontal section through the ropes. ("Cut" them and replace the missing portions of the rope with the tension.) Each weight is experiencing an upward force of 2T.
When 2T >= 20, or T = 10, weight A begins to rise. Once a hits a stop, T must be increased to just above 20 to get Weight B to rise. Similarly, T just above 30 causes C to rise after B stops.
The "trick" with these pulley problems is to section the problem through the cables and show the tension, T. Then you've just got free body problems, in this case subject to the floor constraint.
Oh, also, while the first weight is being lifted, the floor beneath weight B experiences 40 - 20 = 20 units of force, and the floor under C experiences 60 - 20 = 40 units of force. Once B is lifted, the floor under C experiences 60 - 40 = 20 units of force. (Presuming that the labels are weights, and not masses.)