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Alright guys, it's *summer break* and I promise this isn't for homework. I just wanted to calculate something theoretical and suddenly got stuck. Now I am incredibly disappointed and somewhat confused. Please do not laugh  Well here goes:
How does one calculate the minimum theoretical power required to keep an object stationary against constant force?
I realize that the solution to this problem should probably be potential energy. The conceptual problem I am having is this: If you are standing on the ground and holding a rock, it takes no more power than if you were standing on a tree and holding a rock (over the ground). The potential energies as measured from the ground are not the same, but the forces and powers are the same. I cannot however seem to derive the power directly from force. As I write this I am suddenly inspired to study springs. I think I will find the answer I seek there. But nevertheless this will be posted!
*Winter break
EDIT (Post 12): Let me provide a concrete example. Let's say that a rocket is stationary in mid-air under its own power. If we model the rocket's mass as being approximately constant, is all the power it is producing being lost to heat, sound, and other non-mechanical inefficiency?
EDIT 2: Alright people stop posing the same thing over and over. Thanks.
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On December 25 2010 16:41 Coagulation wrote:
its summer break?
Perhaps he lives down under?
It requires zero power (energy per second) to hold something stationary (with some minor nuances/caveats).
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No my bad it is winter break. He is right.
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Doesn't Newton's third law tell you u just need the equal amount of force in the other direction to make everything = 0?
And changing the frame of reference would help ur example make more sense
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Yeah I thought about that, but then what is the extra energy you are expending while holding up a rock vs holding up your hand?
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Guhh, this is unfortunately how power/work/force works in physics.
So work only happens when there is a force over some distance. Power only happens when there is a change in work with time. Work is energy, so power is the rate of energy change. But no energy is changed since the object does not move (it is stationary), so the power of the whole system is zero.
Therefore, it takes zero power to keep an object held against a constant force.
Yeah I know, it's a really unsatisfying answer, since we humans feel like we expend energy (we do) in order to hold something. But that is not as easily answer with just work/power - you probably need to know some biology for that.
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On December 25 2010 16:47 Ecrilon wrote:
Yeah I thought about that, but then what is the extra energy you are expending while holding up a rock vs holding up your hand?
Yeah as I said, that gets us into a bit of biology.
Here is some explanation on the physics forum: http://www.physicsforums.com/showthread.php?t=119026
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You're not expending extra energy holding up the rock if ur up in a tree, assuming you're keeping gravity constant (so you don't like go into outer space or something) and theres a person holding a rock and exerting energy to keep that rock there.
Both the person and object gain more potential energy going up, but if your system is the person and the rock, then there is no change in potential energy, so the amount of energy is the same.
If your system is the rock to the ground, then yea potential energy increases.
You can't have the system of a person vs ground to the system of a rock vs person...they're separate.
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On December 25 2010 16:47 Ecrilon wrote:
Yeah I thought about that, but then what is the extra energy you are expending while holding up a rock vs holding up your hand?
You aren't performing work on the rock, but your body has to keep spending energy to generate the force that keeps your arm stationary. Due to biology/chemistry greater force from your muscles (to keep a greater mass aloft) costs more energy.
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What then is the thermodynamic description of a free-floating object in a gravitational field? If left to its own devices, it has potential energy proportional to its height above a reference point, but it technically takes no energy to keep it aloft?
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On December 25 2010 16:55 Ecrilon wrote:
What then is the thermodynamic description of a free-floating object in a gravitational field? If left to its own devices, it has potential energy proportional to its height above a reference point, but it technically takes no energy to keep it aloft?
I'm not quite sure why you're bringing thermodynamics into this discussion.
If you assume a vacuum, say a planet orbiting a star, then the object's total energy does not change as it falls. If left to its own devices, the object would orbit the 'source of gravity' and eventually return to its original 'height'.
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Let me provide a concrete example. Let's say that a rocket is stationary in mid-air under its own power. If we model the rocket's mass as being approximately constant, is all the power it is producing being lost to heat, sound, and other non-mechanical inefficiency?
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On December 25 2010 16:55 Ecrilon wrote:
What then is the thermodynamic description of a free-floating object in a gravitational field? If left to its own devices, it has potential energy proportional to its height above a reference point, but it technically takes no energy to keep it aloft?
Don't misunderstand the concept of potential energy. Potential energy exists as a purely relative measurement. To say I have 50 J of potential energy means absolutely nothing on its own.
I could have X potential energy relative to the ground. Maybe I have Y potential energy relative to something else. Just having potential energy doesn't DO anything.
So yeah, that object can have X potential energy relative to the ground, but if some other holds it up, then that energy is useless. That is, the potential energy is never released so it does nothing.
So if the object is held stationary by something else, don't even worry about gravitational potential energy as it doesn't DO anything.
On December 25 2010 17:01 Ecrilon wrote:
Let me provide a concrete example. Let's say that a rocket is stationary in mid-air under its own power. If we model the rocket's mass as being approximately constant, is all the power it is producing being lost to heat, sound, and other non-mechanical inefficiency?
I don't like this example, because it won't quite answer your question, but here it goes anyways. The rocket is expending power by converting it's fuel into thrust. The exhaust that pushes the rocket up gets the energy. The fuels chemical energy is converted into kinetic energy of the exhaust. There is no change in potential energy of the rocket.
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On December 25 2010 17:01 Ecrilon wrote:
Let me provide a concrete example. Let's say that a rocket is stationary in mid-air under its own power. If we model the rocket's mass as being approximately constant, is all the power it is producing being lost to heat, sound, and other non-mechanical inefficiency?
I would say the simple answer is that the power goes into accelerating part of its mass downward (to achieve an equal and opposite reaction upward).
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Alright, yeah, After that I am able to reason the situation out. Many thanks for your time!
I hereby dedicate my zealot post to you two.
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On December 25 2010 17:10 oxidized wrote:Show nested quote +On December 25 2010 17:01 Ecrilon wrote:
Let me provide a concrete example. Let's say that a rocket is stationary in mid-air under its own power. If we model the rocket's mass as being approximately constant, is all the power it is producing being lost to heat, sound, and other non-mechanical inefficiency? I don't like this example, because it won't quite answer your question, but here it goes anyways. The rocket is expending power by converting it's fuel into thrust. The exhaust that pushes the rocket up gets the energy. The fuels chemical energy is converted into kinetic energy of the exhaust. There is no change in potential energy of the rocket.
I need to quickly correct this as I forgot something. The mass of the rocket does change and therefore its potential energy does change. Therefore the rocket does have a power expenditure.
This is different from the example of something stationary holding something else up. When a table is holding up a book, there is no power expenditure.
And congrats on the Zealot! 
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On December 25 2010 16:50 oxidized wrote:
Guhh, this is unfortunately how power/work/force works in physics.
So work only happens when there is a force over some distance. Power only happens when there is a change in work with time. Work is energy, so power is the rate of energy change. But no energy is changed since the object does not move (it is stationary), so the power of the whole system is zero.
Therefore, it takes zero power to keep an object held against a constant force.
Yeah I know, it's a really unsatisfying answer, since we humans feel like we expend energy (we do) in order to hold something. But that is not as easily answer with just work/power - you probably need to know some biology for that.
I believe this is the right answer, although it's been a few years since I've taken a real science course (stupid nursing). I remember coming across the same question in my physics class. It has to do with the formulas (one of the triangle ones, if you will). Searches google.....Here we go,
W= F.d
Since the d (displacement), then the work must be 0. Next formula.
P=delta W/delta t
Since the net work is 0, the power should equal 0.
Take this with a grain of salt because I haven't done this in a long time, hope it helps.
EDIT: Modifying my teacher's analogy, say you're pushing a book against a wall. The book's not falling through the wall (normal opposing force), but you are applying force to keep it there. However, since the book is not moving (not enough force), no displacement is occurring, and therefore, the work does has to be 0. And since power is defined in terms of work over time, if no work is done, there is no power present, as defined in physics terms.
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AHA FUCK YES I FEEL SO SMART. LOOK HERE!!!!!.
You need to find the µ sub s (coefficient of static friction) of the two surfaces. You need to know the weight of the object you want to push so you can calulate its natural force. Then you simply do the calculation of maximum exertion of static friction (aka how much force the object can take without moving) = Static coefficent of friction * natural force. Any force greater than the result will push the object.
If you need help calculating natural force or w/e just lemme know.
.....errrr Just read the whole post... nvm =(
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Well, I'm no physics expert, but I read the forum link posted by oxidized, and I think I can explain this fairly well.
1. Imagine you put a rock on the ground. The force of gravity is counteracted by an opposite force, which is provided by whatever forces make the ground solid. Put the rock on water and you don't get this force.
2. No imagine that you put the rock at the end of a metal rod which is perpendicular to the force of gravity. Perhaps the tin-man is holding the rock instead of you. What holds the rock up? The answer is the same as #1: whatever forces hold the metal rod together and make the metal what it is.
3. Now imagine that you are holding a rock. What holds the rock up? Whatever forces hold your arm solid and upright. Some of these forces are the same as those involved in #1 and #2. Lie your arm flat on the ground and you can hold a rock with no work at all. But if you want to hold your arm upright, then you need to use your muscles (this involves many tiny contractions). That's why it feels like you are doing work, even though nothing is moving.
Summary: the simple model of mechanics doesn't take into account internal forces in materials.
Now I'm curious what kinesiologists' models look like. I also want to mention that Laerties post is brilliantly funny (whether intentional or not).
Edit: also, forgot to mention that the rocket example is misleading. A ton of work is being done there, as the propellant shooting out the bottom of the rocket is definitely going places. The only way this example is helpful is in pointing out that there are no energy sinks. Rocket fuel being consumed is balanced by the fact that propellant is accelerated. The energy your body is using to hold a rock also has to end up somewhere. At least some of it becomes heat; the energy may also end up in other places, but I don't know enough biology to say exactly what happens.
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An unintuitive 0 because you are doing no work on the rock. The energy you are exerting is internal keeping your muscle contracted rather than keeping the rock up. This question is a paradox that arises due to using an overly simplistic physical model and confusing the scientific and everyday meanings of words like power/work/weight etc. To answer the question properly you probably have to dive into the realm of squishy science.
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for yout question, ZERO. just dun make things so complicated.
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i didnt read all posts so i dont know if it was answered, but im gonna try anyway.
A force (F1) is being exerted on an object and you want to counter that force with your own force (F2). Naturally those 2 forces would have to be equal because you want your object to be stationary hence:
F = force
F1 = F2
But you want your answer to be in power .. kk
In order to answer this we must see how far the object travelled because
![[image loading]](http://upload.wikimedia.org/math/7/6/8/7680d79cfc1c61f21fe00e1089a9493b.png)
where C is the path the object travelled.
So we see that d is 0 because it was stationary, that is, the power for the complete system is 0 - but that wasnt what you asked about. You wanted to know how much power you had to exert to keep it in balance.
Well in order to answer this we must set F1 = 0 so there is no force. Now that the forces are no longer in equilibrium the object begins to move, and it begins to accelerate. Now we see that d is no longer 0. Actually d is ∞. That is because you have not specified for how long time you want to exert your power.
Should you exert power for a period of time then you can find d and d will be a valid number somewhere between 0 and ∞..
Then you can calculate how much power YOU and not the system exerted.
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356 Posts
erhm... dx is infinitesimal, not a value.
The path length is zero- so your integral has the same lower bound as upper bound, so it subtracts to zero.
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If the object travelled on a path that is straight then it doesnt matter - and the path is straight because we didnt adjusted on the direction of the force F2.
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Yup, the answer's zero. Think of some object sitting on a table: gravity is being constantly exerted on the object, but the table is keeping it still with no power expended.
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On December 25 2010 23:35 matjlav wrote:
Yup, the answer's zero. Think of some object sitting on a table: gravity is being constantly exerted on the object, but the table is keeping it still with no power expended.
The energy that bonds and keeps the atoms solidly together is what's keeping the object stationary.
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On December 26 2010 00:36 Zilver wrote:Show nested quote +On December 25 2010 23:35 matjlav wrote:
Yup, the answer's zero. Think of some object sitting on a table: gravity is being constantly exerted on the object, but the table is keeping it still with no power expended. The energy that bonds and keeps the atoms solidly together is what's keeping the object stationary.
The energy isn't expended to holding up the object. It just exists, so no power is required. That potential energy isn't holding up the object; the forces of intermolecular attractions and repulsions is what's holding the object up, and force is a distinctly separate concept from energy.
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its 0 assuming all force are balanced
your confusing power work and energy
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EPT
