Equilibrium and Minimum Force Requirements in a Frictionless Environment

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Discussion Overview

The discussion revolves around the force required to raise a 5,000 kg object to a height of 50 meters in a frictionless environment. Participants explore concepts of equilibrium, minimum force requirements, and the implications of applying force to initiate movement.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the force required to lift the object using F = MA, suggesting that the height does not affect the force needed as gravity remains constant.
  • Another participant questions whether an infinitesimally larger force is needed to initiate movement, implying that this force might be numerically similar to the force required for equilibrium.
  • Some participants argue that to start moving the object, a bit more force is necessary, and the amount of additional force depends on the desired speed of movement.
  • Concerns are raised about the implications of acceleration not being constant and the effects of air pressure, although one participant insists these factors have been eliminated from consideration.
  • There is a discussion about the definition of "infinitesimal" and whether it can be applied to force in this context, with some participants questioning how to mathematically prove the existence of an additional force if it is infinitely small.

Areas of Agreement / Disagreement

Participants express differing views on the necessity and calculation of additional force to initiate movement, with no consensus reached on whether the force required for equilibrium and the minimum force to lift the object are the same.

Contextual Notes

Participants mention various approximations and assumptions regarding acceleration and external factors, indicating that the discussion is limited by these unresolved elements.

Chumly
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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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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?

yes,yes.
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...
 
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.
 
Naty1 said:
yes,yes.
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...
Helllo Naty1, it cannot be friction or buoyancy and other similar variables as I have eliminated those as mentioned.
 
jtbell said:
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.
Hello jtbell,

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"?
 
Chumly said:
Hello jtbell,

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"?
If you just applied a force equal to the weight, then the net force would be zero and it would just sit there. You obviously have to get it moving! That requires a bit more force. How much? Up to you! It depends on how fast you want to lift this thing.

(I'm basically reiterating what jtbell already said.)
 
By definition, infinitesimal is arbitrarily small. In other words, there is no minimum.
 
russ_watters said:
By definition, infinitesimal is arbitrarily small. In other words, there is no minimum.
Hello russ_watters (and all others kind enough to take the time to chat!)

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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