7
   

Equilibrium vs. Work Done?

 
 
Chumly
 
Reply Wed 13 Mar, 2013 11:46 am
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? Assume gravity stays constant for the height raised, there is no friction, buoyancy or other similar variables.

It seems that all you would do is F = MA = (5000) (9.80665) = 49,033.25 Newtons.

Questions:
Wouldn't 49,033.25 Newtons simply put it into equilibrium?

Wouldn't you 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?

If you can't precisely calculate the minimum amount of additional force 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!
 
Ice Demon
 
  3  
Reply Wed 13 Mar, 2013 12:06 pm
@Chumly,
What you've demonstrated is that one must exert more than 49,033.25N to lift the weight, but not necessarily how much energy needed to lift fifty meters.
Although related, it will be easier to find the energy required to lift the weight and translate that to the needed force that must be applied.
In your case WF = ΔEk + ΔEp
But since initial and final velocity equals zero, there is no change in kinetic energy. So, work become WF=F *x=mgh.
You'll have the answer in newton meter, or joules.
You can use E=mc^2 to convert joules to kilogram if you wish.
dalehileman
 
  0  
Reply Wed 13 Mar, 2013 02:07 pm
@Ice Demon,
But Ice; Chum asks, "How much force is required to raise an item weighing 5,000kg," whereupon my immediate response is "5,000kg"

Forgive
engineer
 
  1  
Reply Wed 13 Mar, 2013 02:10 pm
@dalehileman,
The real question is how much work or energy is required. There are many ways to apply different forces and get the same result as long as you get the same amount of work done. As posted above, the energy change is mgh where m is the mass, g is the gravitational constant and h is the change in height.

dE = (5000 kg)(9.8 m/sec^2)(50m) = 2,450,000 N.m

If you really want a force, you could apply a force of 2,450,000 N until the mass moves 1m. The mass will accelerate for the first meter then slow until it peaks at 50m above the ground.

(Edit: I think you have to add in force to counter gravity.)

0 Replies
 
Ice Demon
 
  1  
Reply Wed 13 Mar, 2013 02:18 pm
@dalehileman,
Force is just the method and as engineer said, there are many ways to apply it.
To capture the capacity needed to lift it to the given height, one must look into the energy because energy is a property which you have when a force is applied over a certain distance, and basically it is the counting up of all the force which has acted on the object. For example in order to lift an object with weight 1 Newton a height of 1 meter, I must do 1Newton* 1 meter= 1 Joule of work.
dalehileman
 
  0  
Reply Wed 13 Mar, 2013 02:25 pm
@Ice Demon,
Quote:
counting up of all the force from the initial height to the final height.
Ice we might be embroiled in another mere semantic issue. To me the force is 5000kg all the way to the top

Depending of course on the rate at which we lift the thing
Ice Demon
 
  1  
Reply Wed 13 Mar, 2013 02:36 pm
@dalehileman,
I have made a change if it makes you understand it better, and if it is a semantic issue for you. The object may 49050 newton all the way to the top, but per meter, one must exert the force of 49050 newton, or 49050 joules, and as engineer did the calculation, refer to his post to the energy spent to lift this item 5o meters.
Here is the definition of work done by a variable force, if you can understand it better.
http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/imgmec/wovf2c.gif
maxdancona
 
  1  
Reply Wed 13 Mar, 2013 03:03 pm
@Chumly,
Hi Chumly,

This is a good question, with a fairly straightforward answer. There is nothing infintessimal about it. You will likely use a measurable additional force to get the thing moving (unless you want to spend an infinitely long time to raise it).

The answer is exactly 2,450,000 Joules as long as the starting velocity and the ending velocity are the same (so that Kinetic Energy doesn't require more work). That means in this case that you must end up stopped.

Then the problem becomes simple. You must use more force in the beginning to accelerate the object. You must use less force in the end to get it to decelerate. This means you are doing a more work per centimeter in the beginning and a little less work at the end to decelerate (remember that gravity is involved).

In the end, per the energy calculations done by Engineer and Ice Demon, the total work will be Force * distance.

In the middle the object has two types of energy... potential (based on height) and kinetic (based on speed). At the end the object only has potential energy.
0 Replies
 
Chumly
 
  1  
Reply Wed 13 Mar, 2013 07:20 pm
Thanks so much guys, truly appreciated! I'll digest this for a few days and get back to y'all.
0 Replies
 
DrewDad
 
  1  
Reply Thu 14 Mar, 2013 07:31 am
@Chumly,
The amount of force necessary is pretty arbitrary. You can strong-arm it against gravity, as you're demonstrating here, or you could use an extremely shallow ramp, which would get your* force down to almost nothing.

For your force to be almost nothing, you have to apply that force over a really long distance, though.

I suspect the question is designed to highlight the difference between force and work.




*The total vertical vector has to be equal or greater than what you've calculated, but if the floor is already applying most of that vertical component, then you only have to add a little force to push it up the inclined plane. How you're going to do this in a frictionless environment is left as an exercise for the student.
maxdancona
 
  1  
Reply Thu 14 Mar, 2013 07:41 am
@DrewDad,
No Drew, the distance can't be more than 50m. It is the time it takes to raise the object that is most affected by the force you apply.

Consider the following trip

He has to put more than the weight (49,033.25 N) to overcome gravity. If he puts .000000001 N more than the weight, then the object will accelerate (very slowly). You could put that very small amount of extra force for a second (during which time the weight has moved a very small distance) but now the object has a very small velocity.

You continue putting the exact force equal to the weight (no more and no less) to maintain this very small velocity until the last second.

During the last second of the trip, you put .000000001 N less then the weight to decelerate the object until the object is at rest at the end of the trip.

In this case the distance you are applying the force over will be very short. The time it takes to raise the object will be very, very long.
DrewDad
 
  1  
Reply Thu 14 Mar, 2013 07:45 am
@maxdancona,
I can gently push that mass for miles (horizontal miles) up an inclined plane. The total vertical distance is 50 m. The problem says nothing about the horizontal distance traveled.

And if you're going to be persnickety, then I can design a frictionless screw that does the same thing. I walk in circles pushing that screw, which lifts the mass, and the only thing that affects how far I walk in circles, or how much force I apply, is the pitch of the threads.
0 Replies
 
DrewDad
 
  1  
Reply Thu 14 Mar, 2013 07:47 am
@maxdancona,
The time is immaterial to how much work is done. Work is force applied over a distance. Work applied over a time is power.

I can use very little power, and do the work over a long time, or I can apply a lot of power and do the work over a short time.

But the question doesn't say anything about how long the task takes.
maxdancona
 
  1  
Reply Thu 14 Mar, 2013 07:53 am
@DrewDad,
Quote:
The time is immaterial to how much work is done.


Agreed. That is the whole point. No matter how we do this, the work is the same.

It is clear that in any case, even your inclined plane obfuscation, the more excess force (above the weight) initially the faster the object will go and the shorter the trip will be.

We all agree that this doesn't impact the amount of work done in any case.
DrewDad
 
  1  
Reply Thu 14 Mar, 2013 07:55 am
@Chumly,
In a frictionless environment, the force is literally arbitrarily low.

On a flat plane, I can use any force I choose, applied in a horizontal direction, to accelerate the mass. I accelerate the mass until it has the kinetic energy required, hits a ramp, and then halts at 50 m.

It has to be a perfectly designed ramp, though, at the top.
0 Replies
 
DrewDad
 
  1  
Reply Thu 14 Mar, 2013 07:56 am
@maxdancona,
"Obfuscation?"

Clearly it's been some time since you had a classical physics class. Profs love to discuss this stuff.
maxdancona
 
  1  
Reply Thu 14 Mar, 2013 08:00 am
@DrewDad,
It has been a little while since I last taught a classical physics class. Perhaps we profs love obfuscation.

DrewDad
 
  2  
Reply Thu 14 Mar, 2013 08:07 am
@maxdancona,
Sure. Everyone's a Ph.D. on the Internet.
dalehileman
 
  1  
Reply Thu 14 Mar, 2013 11:04 am
@Ice Demon,
Well Ice, thanks most kindly for that clarification. In my own defense tho,

http://www.google.ca/#hl=en&sclient=psy-ab&q=force+isn%27t+measured+in+pounds&oq=force+isn%27t+measured+in+pounds&gs_l=hp.12...89735.108778.4.110888.31.23.8.0.0.0.136.2599.2j21.23.0.les%3B..0.0...1c.1.5.hp.fiiocAsV9aI&psj=1&bav=on.2,or.r_qf.&bvm=bv.43287494,d.aWc&fp=28590474882e63b5&biw=1499&bih=575
0 Replies
 
maxdancona
 
  1  
Reply Thu 14 Mar, 2013 04:25 pm
@DrewDad,
It is an obfuscation because the math is the same whether it is on an inclined plane or not. All you do multiply the equilibrium force by a factor (the sine of the angle to the horizontal) and divide the length the the same factor.

It is still true pushing on the object at the eqilbrium force will keep the object at constant speed.

It is still true that pushing with a little more than equilibrium force will accelerate the object.

It is still true that if you push with a very tiny bit more than equilbrium force for a very tiny time before going back to pushing with exactly equilibrium force will cause the object to move very slowly, meaning that you will have to push a very long time before you reach the goal. And if you start by pushing a fair amount more than the equilibrium force to get the object moving quickly early (of course compensating to make sure that it is stopped at the end), the object will reach the goal much more quickly.

You haven't changed the problem at all by adding this needless complication. And you haven't made it any more informative, at least as far as answering the original question. That is why I called it an obfuscation.
0 Replies
 
 

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