Feynman Turnscrew Example Question

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Feynman's screwjack example illustrates the concept of equilibrium in lifting a one-ton mass, requiring 1.6 pounds of force to maintain that state. However, this force does not initiate movement; an infinitesimal additional force is needed to overcome inertia and start lifting the mass. The discussion highlights the ambiguity in interpreting "lift," as the term can imply different scenarios, such as maintaining equilibrium versus initiating upward acceleration. The participants acknowledge that Feynman's problems often contain underlying complexities that challenge straightforward interpretations. Ultimately, understanding the physics principles behind the scenario is crucial, even if the definitions of terms like "lift" vary.
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Very early on in Feynman's Lectures on Physics, he offers the example of a screwjack with a mass on it to demonstrate the conservation of energy (picture attached).

He asks how much force would be needed to lift the one ton sitting on top of the screwjack. He then comes up with 1.6 pounds applied to the handle.

Just to make clear I understand, he is saying that when a force of 1.6 pounds is applied to the handle, the system is in equilibrium, right? The mass will technically not lift then, it needs some infinitesimal additional force for the ton to start moving up, right?
 

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DocZaius said:
Very early on in Feynman's Lectures on Physics, he offers the example of a screwjack with a mass on it to demonstrate the conservation of energy (picture attached).

He asks how much force would be needed to lift the one ton sitting on top of the screwjack. He then comes up with 1.6 pounds applied to the handle.

Just to make clear I understand, he is saying that when a force of 1.6 pounds is applied to the handle, the system is in equilibrium, right? The mass will technically not lift then, it needs some infinitesimal additional force for the ton to start moving up, right?

Hmmm, someone confused by a Feynman problem. How unusual. :smile: (kidding)

Well, you are both right and wrong.

If you think about it, the force needed to maintain equilibrium when the weight is stationary is the same force needed to maintain equilibrium when the system is moving. I'm guessing that friction is ignored in this problem, so if you use more than the equilibrium force, the weight would actually accellerate. However, there is no need to accellerate since lifting can be done with constant speed, which only requires equilibrium once the weight starts moving.

However, you are correct in the sense that you need to apply more than the equilibrium force just to get the weight moving at an acceptable speed.

Keep in mind that this example is very unrealistic because friction is ignored. Unrealistic problems tend to confuse us all, and require deeper thinking about the principles sometimes.
 
stevenb said:
Hmmm, someone confused by a Feynman problem. How unusual. :smile: (kidding)

Well, you are both right and wrong.

If you think about it, the force needed to maintain equilibrium when the weight is stationary is the same force needed to maintain equilibrium when the system is moving. I'm guessing that friction is ignored in this problem, so if you use more than the equilibrium force, the weight would actually accellerate. However, there is no need to accellerate since lifting can be done with constant speed, which only requires equilibrium once the weight starts moving.

However, you are correct in the sense that you need to apply more than the equilibrium force just to get the weight moving at an acceptable speed.

Keep in mind that this example is very unrealistic because friction is ignored. Unrealistic problems tend to confuse us all, and require deeper thinking about the principles sometimes.

Thanks for your reply. I am not sure how I was wrong, though. The infinitesimal additional force I mentioned was never meant to last for more than an infinitesimal period of time.

edit: Hmmm I think I need to stop using the word "infinitesimal" since there are issues regarding how distinguishible it is from 0. Just assume I mean "very small."

All I meant is that at some point a very small additional force (in addition to the 1.6 pounds) would need to be applied, for a very short time, for the ton to move up. I think you inferred that I meant that this additional force would have to be applied for a considerable length of time, which I did not mean to imply.

Speaking of unrealistic problems, let me pose one which is sort of a follow up to this. Say you have a mass being pushed up by some force which is exactly equal to g. Ignore air friction. The mass is hovering. Now barely push that mass up with your finger and watch it move upwards. As the mass moves up, the force of gravity reduces and the mass would thus be accelerating, despite the force pushing it up remaining constant right? The net force would increase towards "up". So not only would the mass leave the Earth's gravitational pull, but it would do so while accelerating?
 
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DocZaius said:
I am not sure how I was wrong, though. The infinitesimal additional force I mentioned was never meant to last for more than an infinitesimal period of time.

All I meant is that at some point a very small additional force (in addition to the 1.6 pounds) would need to be applied, for a very short time, for the ton to move up. I think you inferred that I meant that this additional force would have to be applied for a considerable length of time, which I did not mean to imply.

Yes, this is why I said you were both right and wrong (don't you hate it when people claim two mutually exclusive things are true :smile:). You seem to understand the physics here, so you are correct. But, as typical of Feynman, the question has an underlying trick to it. You stated his question as, "How much force would be needed to lift the one ton sitting on top of the screwjack?". Based on his answer, we should interpret "would be needed to lift" as a statement of the minimum value that can lift the weight, not how much is needed to start lifting the weight from a dead stop, or to stop it from falling and turn it around into a lift mode.

DocZaius said:
Speaking of unrealistic problems, let me pose one which is sort of a follow up to this. Say you have a mass being pushed up by some force which is exactly equal to g. Ignore air friction. The mass is hovering. Now barely push that mass up with your finger and watch it move upwards. As the mass moves up, the force of gravity reduces and the mass would thus be accelerating, despite the force pushing it up remaining constant right? The net force would increase towards "up". So not only would the mass leave the Earth's gravitational pull, but it would do so while accelerating?
Interesting.
 
stevenb said:
Yes, this is why I said you were both right and wrong (don't you hate it when people claim two mutually exclusive things are true :smile:). You seem to understand the physics here, so you are correct. But, as typical of Feynman, the question has an underlying trick to it. You stated his question as, "How much force would be needed to lift the one ton sitting on top of the screwjack?". Based on his answer, we should interpret "would be needed to lift" as a statement of the minimum value that can lift the weight, not how much is needed to start lifting the weight from a dead stop, or to stop it from falling and turn it around into a lift mode.

Sorry to be pedantic here, but I need to know how I will interpret the word "lift" in future physics problems.

As the problem is stated, the weight seems to be immobile, and we are trying to figure out a force (it is implied that we search for the minimum force) which will move the weight up (thus "lifting" it).

1.6 pounds of force is not that number. Rather 1.6 pounds of force will enter the system into equilibrium. It will not accelerate the weight upwards.

The answer I find to be more correct (but which obviously isn't since Feynman doesn't agree) is that the minimum amount of force which would lift the weight is:

"The smallest number which is > 1.6 pounds."

So either:

1. I am actually wrong in my thinking somewhere (please tell me where!)
or
2. I am technically right, but it assumed in physics problems that when we say "lift" we mean "enter into vertical equilibrium"
 
DocZaius said:
Sorry to be pedantic here, but I need to know how I will interpret the word "lift" in future physics problems.

Personally, I wouldn't worry too much about it. Feynman has a history of causing some ambiguity despite his success at provided great insight, and as good as his lectures are, there are points of confusion created, if you take some things he says too literally.

I would answer your question if I thought I could, but I can't be sure any definition of "lift" would hold up in all circumstances. In contrived problems like this, it's acceptable to answer the question with your own definition of ambiguous terms, and to even provide more than one answer based on the various possible interpretations. I remember doing this on more than one test when I was back in school. It saved me more than a few points in total, and once got me some extra credit. As an example, you could answer, "a force of x will maintain the lifting at constant velocity once the object is moving upward, and if one wants to lift a falling or stationary object, a force >x is needed to achieve upward accelleration for long enough time to provide an upward velocity, after which a force of x will maintain that lifting velocity". This is clearly your understanding, and this answer would prove that you understand.

One thing is clear to me. You seem to understand the physics of this problem, and I'm sure Feynman would be happy about that. Although, he might say you are wrong if you answer with a simple answer "The smallest number which is > 1.6 pounds.". -But who cares what dead physicists think anyway? Some mathematicians might complain too, but I worry about them even less. (:smile: kidding)
 
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