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Equilibrium vs. Work Done? |
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| Mar11-13, 01:41 PM | #1 |
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Equilibrium vs. Work Done?
Hello physics whizzes.
How much force is required to raise an item weighing 5,000kg to a height of 50 meters in a frictionless environment? It seems that all you would do is F = MA = (5000) (9.80665) = 49,033.25 Newtons, and that the height is irrelevant assuming that the gravity stays constant for the height raised. Questions: Wouldn't that simply put it into equilibrium? Wouldn't one need an infinitesimally larger amount of force on the driving end? Yet wouldn't that infinitesimally larger amount of force on the driving end be for all practical purposes, the same numerically as the force required to lift the load in the first place? If so, then what would be the force required for equilibrium versus the force required to lift the load? Surely they cannot be exactly the same in a frictionless environment? Very much thanks, and this is my first post (or very close to it)! |
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| Mar11-13, 02:03 PM | #2 |
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On the other hand the acceleration of 9.8xxxx is not constant either....so there are various approximations here. And air pressure [a minor buoyancy] also declines with height.... |
| Mar11-13, 02:05 PM | #3 |
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The force that you calculated would keep the object moving at constant velocity (upwards or downwards) or stationary.
In order to start the object moving, you have to apply at least a bit more force at the beginning. How much more force, and for how long, depends on how fast you want the object to move. In order to stop the object when it reaches its target height, you have to "ease off" on the force at least a bit. How much you have to ease off, and for how long, depends on how fast the object was moving. |
| Mar12-13, 09:59 AM | #4 |
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Equilibrium vs. Work Done? |
| Mar12-13, 10:09 AM | #5 |
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I am not concerned about acceleration of the load, only that the load must (I am pretty sure) rise at some inevitably increasing rate (ignoring relativistic calculations). What would be the force required for equilibrium versus the minimum force required to lift the object? What precisely is this infinitesimally larger amount of force on the driving end actually overcoming? It surely cannot be friction or buoyancy or other similar variables as I have eliminated those as mentioned. How would you precisely calculate the minimum amount of additional force to "apply at least a bit more force at the beginning" in order to start the object moving? Also as discussed, given that an infinitesimally larger amount of force on the driving end would be "a bit more force at the beginning" and given that an infinitesimally larger amount of force on the driving end would be virtually the same numerically as the force required to put the system into equilibrium then how can you need "a bit more force at the beginning"? |
| Mar12-13, 10:30 AM | #6 |
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(I'm basically reiterating what jtbell already said.) |
| Mar12-13, 11:27 AM | #7 |
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By definition, infinitesimal is arbitrarily small. In other words, there is no minimum.
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| Mar12-13, 12:52 PM | #8 |
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Yes OK I'm cool with that, however: a) given that an infinitesimally larger amount of force on the driving end would be "a bit more force at the beginning", b) given that an infinitesimally larger amount of force on the driving end would be virtually the same numerically as the force required to put the system into equilibrium, then how can you need "a bit more force at the beginning" if numerically the force required to put the system into equilibrium is the same as the minimum force required to lift the object? Also if you can't precisely calculate the minimum amount of additional force to "apply at least a bit more force at the beginning" in order to start the object moving because it's infinitely small, then how do you prove it exists mathematically? Also what precisely is this infinitely small additional force called, and why do we need it? Very much thanks! |
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