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I've read this question and answers, but neither of the posts does seem to address my concerns. The problem is as follows:

Assuming a device that keeps something in a gravitational field on a constant height, like an electromagnet, how does one measure the required power?

I'm stumbling on that because the equation $P = Fv$ would mean that, assuming the object stays in place, the power is zero. That might be true for, say, a rigid construction holding something up. What if I am actively trying to keep something up, like holding it in my hand, or powering an electromagnet to create an electromagnetic field to keep something in the air? I am applying a constant force, but the object doesn't move!

I've tried to imagine it from the point of theoretical loss of potential energy I'm preventing, but it only got me to something like $P = g^2 m t$, which depends on time. In fact, every attempt to tackle the energy was in vain, because the velocity of the object in a gravitational field, if left without the support, wouldn't be constant (thus making the power depend on time), and is apparently not fit to describe the problem formally.

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Bartek Banachewicz
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  • possible duplicate of Why does holding something up cost energy while no work is being done? – DumpsterDoofus Mar 04 '14 at 21:13
  • You're right that holding an object does no "work". However, that doesn't necessarily mean no energy is expended, since real systems often are inefficient. – DumpsterDoofus Mar 04 '14 at 21:15
  • @DumpsterDoofus Thanks for the link! Skimming trough the answers, I still think this question touches something different. It is obvious that by powering an electromagnet, we are spending some energy. Does it mean, though, that the only way of calculating power is studying, for example, how much power does an electromagnet need to output a constant force? That would make more sense, I guess. – Bartek Banachewicz Mar 04 '14 at 21:17
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    Yes. It all depends on how inefficient the supporting device is - and nothing else - because an efficient device (such as "the floor") expends exactly zero power. – Sideshow Bob Mar 04 '14 at 21:20
  • @BartekBanachewicz: Yes. For example, both a permanent magnet or an electromagnet could be used to give the exact same effect. The permanent magnet uses no power, the electromagnet uses power. Therefore, it's impossible to determine the power usage just by observing the levitating object alone. – DumpsterDoofus Mar 04 '14 at 21:20
  • I was missing just that observation! Thanks a lot. I think I'll keep that question here, with a duplicate link it should be OK. – Bartek Banachewicz Mar 04 '14 at 21:21

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