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Forces and work .

  1. Apr 12, 2009 #1
    Forces and "work".

    I've read that no matter what force is acting upon an object, no actual "work" is being done unless the object moves.

    This seems easy to accept if one is thinking about a bookshelf keeping a book from falling (the shelf is propping the book up and keeping it from falling, and the falling that would occur if the shelf disappeared is called "potential energy".

    What I have a problem with is this concept as it applies to forces "at work". Say we have a steel bar tied to one end of a string, with the other end of the string fastened to the floor. We then attach an electromagnet to the ceiling, and feed it with just enough current that the steel bar is suspended in the air, completely immobile, several inches below the electromagnet, with the string (pulled tight) preventing it from reaching said electromagnet. There is an air-gap between the steel bar and the electromagnet (this gap isn't actually necessary for the purposes of my question, but just makes it easier to visualize the issue, at least for me).

    Ok, so we can measure the electrical current being constantly fed into the electromagnet, and if we lessen the current by any amount, the steel bar falls to the ground. This implies to me that the current flowing through the coil of the electromagnet must be "working" to keep the steel bar suspended in mid-air below the electromagnet. If we lower (or completely eliminate) the current flowing through the coil, it no longer "works" to keep the steel bar where it is and the bar falls.

    So am I confused over simple semantics where "work" is concerned? When a force is the only thing keeping an object from moving (ie. electromagnet prevents bar from succumbing to another force (gravity) and falling due to the electricity being constantly fed into the electromagnet), and the removal of said force instantly allows the object to move, how can the force (or the electricity which creates the force) be thought of as not doing any work? Does my hydro bill not prove that work has been done?
  2. jcsd
  3. Apr 12, 2009 #2
    Re: Forces and "work".

    A few things, it depends on how to you define your system. It takes work to lift the steel bar up into the air.

    Your circuit must also be doing work because you have moving charge carriers creating a current, which means that there is an electric potential which means that is work being done.

    However no work is being done on the steel bar by the magnet once it is suspended in air. Work is defined as the path integral of the dot product of a force and your path. If you're not moving anywhere then there can be no work in that sense.
  4. Apr 12, 2009 #3
    Re: Forces and "work".

    Replace the electromagnet with a permanent magnet (or at least use superconducting coils so that the current stops being converted to heat).
  5. Apr 12, 2009 #4


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    Re: Forces and "work".

    This is an efficiency problem. Efficiency is work done on something divided by work done by something. For your electromagnet obviously, electrical work is being lost, but no mechanical work is being generated. That's just zero efficiency.
  6. Apr 12, 2009 #5
    Re: Forces and "work".

    This is what I mean when I wonder if I'm not just confused over semantics -If my objective is to continuously counter the force of gravity and keep the bar in the air, then it cannot be a case of "zero efficiency"...can it? The magnet would seem to be continuously working (or continuously fighting the force of gravity) to keep the bar in the air...successfully so. Furthermore, is the work being done on the bar not evidenced by the electrons in the atoms of the steel bar being forced into an unnatural state (ordered as in magnetic domains)?
  7. Apr 12, 2009 #6


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    Re: Forces and "work".

    No, there is still no work being done, because it isn't moving. To visualize this, you could replace the electromagnet with a permanent magnet, or even just stick a shelf under the weight. All of these solutions will achieve the same result with zero input power at all.
  8. Apr 13, 2009 #7
    Re: Forces and "work".

    The electromagnet exerts a force on the steel bar that is equal and opposite to the gravitational force. The forces cancel each other out, so there is no net force. If there is no net force (and no displacement), no work is being done by the definition of work.
  9. Apr 13, 2009 #8
    Re: Forces and "work".

    Hey Robert I'm no authority or anything close to that in physics but I guess what they are trying to say is that there is still energy being expended but in the form of heat rather than work. Not sure if that is correct...

    That leads me to another question though: if you have a stationary book on a table obviously the forces of gravity on the book cancels out with the reaction force of the table so does that mean that there is some "tiny" amount of heat being generated?
  10. Apr 13, 2009 #9

    Doc Al

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    Re: Forces and "work".

    Since the steel bar doesn't move, no mechanical work is being done on it. Period.

    Depending upon the nature of whatever mechanism you are using to create the force that holds up the steel bar against gravity, you may be required to expend energy to create and maintain that force. But that energy is not expended by doing work on the steel bar.

    Another way to create a force that requires energy to sustain is to use muscle power. Just support the steel bar with your hand. Your muscles will be doing work internally (as the fibers contract and relax) and you'll be expending energy (and you'll fatigue), but no work will be done on the steel bar. And the expended energy will end up as internal energy ("heat").

    No, why would you think that?
  11. Apr 13, 2009 #10
    Re: Forces and "work".

    By the way, while the steel bar is being lifted work is being done. The work done is equal to the increase in potential energy.
  12. Apr 13, 2009 #11
    Re: Forces and "work".

    Here are some guidelines:

    F*dx = work (joules)
    F*dx/dt = F v = power (joules/sec or watts)
    F*dt = momentum change

    You have to move a distance dx to do F*dx amount of work. When using vectors (force, dx,dx/dt or v), the equations are vector dot products.
  13. Apr 13, 2009 #12
    Re: Forces and "work".

    I was thinking of it as being the same thing as your example of using muscle power to support a steel bar with your hand in which you concluded that the expended energy will end up as internal energy ("heat"). Wouldn't there be some sort of compression of the book and table which causing molecules to collide quicker which would insignificantly raise the temperature or am I completely off? If so I guess I'm having problems accepting that nothing comes out of 2 forces that are counteracting eachother.
  14. Apr 13, 2009 #13

    Doc Al

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    Staff: Mentor

    Re: Forces and "work".

    When you first put the book on the table there will be a bit of compression and thus work done, which will ultimately lead to a minuscule increase in temperature. But unlike the example of muscle power, which requires a continuous input of energy to maintain the force, once the book compresses the table the force is maintained without further energy usage.
  15. Apr 15, 2009 #14
    Re: Forces and "work".

    Woops! I hope I didn't seem rude or ungrateful for not acknowledging all of these replies; I figured that since I wasn't getting any notifications of replies via email, the thread had died. Evidently the auto-notification stops working after a few replies(?).

    Doc Al, I had, of course, considered the muscle-power angle -If I'm holding a bucket of sand perfectly still with my arm extended, the burning muscles in my arm tell me I'm doing a lot of work (albeit for nothing), but I guess I'm just not doing any work on the bucket per se. I guess it just doesn't seem natural for me to think of it that way, that is, wether I'm spending energy causing something to move or spending the same amount of energy preventing it from moving (in this case, by countering another force), it just feels natural to assume work is being done on the object in question. Sometimes I feel like a complete idiot for getting confused over such simple concepts as this...it really does just sound like semantics to me. Ok, maybe someone will have fun explaining more to this imbecile (me) - I'll try another example and try to bring relativity (or more accurately, frames of reference) into it:

    Let's say I'm suspended by a rope a mile about ground-level (yes, I know that sounds silly). Let's also say that right next to me, there is a tiny rocket suspended from another rope. Both ropes are suddenly cut, but at that precise moment, the rocket's engine ignites with precisely enough thrust to keep the rocket perfectly stationary above the ground. From the frame of reference of someone standing on the ground, there is an impressive fire display doing absolutely no work on the rocket (since it remains perfectly stationary above the ground) and there is also an idiot plummeting to his death (since he's accelerating toward the ground). But from my frame of reference, the rocket is not only moving, it's also accelerating. Would I, in my frame of reference, be able to say that the thrust from that impressive fire display is indeed doing work on the rocket since it is zooming away from me?

    Anyways, thanks to everyone for all your replies!

  16. Apr 15, 2009 #15
    Re: Forces and "work".

    The potential energy of the rocket is mgh, where h is the height (distance from the surface to the rocket). Does this potential energy differ in different frames of reference? No, because if we were to take the frame of reference of the poor falling fellow, the earth would be accelerating to you as much as the rocket accelerates away from you. The distance between the rocket and the earth remain the same. So there is no change in potential energy. Why is that important to know?

    Well, work is a kind of energy that can be converted into kinetic or potential energy. This is not happening, so there is no work being done.

    You just have to accept that work in a physical sense is different from the kind of work as you would 'normally' define it. You're making it harder than it is :) .
  17. Apr 15, 2009 #16

    Doc Al

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    Re: Forces and "work".

    I'm sure you realize that while the engine is doing no work on the rocket, it's sure doing work (and expending energy) in expelling the fuel to maintain the thrust.
    Work and energy are frame-dependent quantities. (Realize also that this particular choice of frame is a non-inertial one.)
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