Equilibrium and Minimum Force Requirements in a Frictionless Environment

In summary, the question being discussed is how much force is required to raise a 5,000kg item to a height of 50 meters in a frictionless environment. The equation F = MA can be used to calculate the required force, which in this case would be 49,033.25 Newtons. However, this force would only maintain the object's velocity or keep it stationary, not lift it. To start the object moving, a bit more force would need to be applied at the beginning. While there is no minimum amount for this additional force, it would need to be at least slightly greater than the weight of the object in order to overcome the net force of zero and get the object moving. Additionally, in order to
  • #1
Chumly
5
0
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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  • #2
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...
 
  • #3
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.
 
  • #4
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.
 
  • #5
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"?
 
  • #6
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.)
 
  • #7
By definition, infinitesimal is arbitrarily small. In other words, there is no minimum.
 
  • #8
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!
 

1. What is the difference between equilibrium and work done?

Equilibrium refers to a state where there is no net change in the system, while work done refers to the energy transferred to or from a system. In other words, equilibrium is a static state, while work done involves a change in the system's energy.

2. How is equilibrium achieved?

Equilibrium can be achieved in a system when there is a balance between the opposing forces or processes. For example, in a chemical reaction, equilibrium is reached when the rate of the forward reaction is equal to the rate of the reverse reaction.

3. What factors affect equilibrium?

The factors that affect equilibrium include temperature, pressure, and concentration of reactants and products. These factors can shift the equilibrium position by changing the rate of the forward and reverse reactions.

4. Can work be done in an equilibrium system?

Yes, work can still be done in an equilibrium system. For example, in a chemical reaction at equilibrium, work can be done to increase the concentration of one of the reactants, which would shift the equilibrium position.

5. How is work done related to the laws of thermodynamics?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred from one form to another. Work done is a form of energy transfer, so it follows the first law of thermodynamics. The second law of thermodynamics states that the total entropy of a closed system will never decrease over time. Work done can increase the entropy of a system, but the overall entropy of the closed system will always increase.

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