Why Gravity wheels don't violate the laws of Physics

[rmd3]

Having trouble with basic physics, so please try this scenario on for size...

Let's say I have a 10 kg ball on a table 1 meter off the ground. I go under it and exert a force just to the point where it still isn't
being lifted and I've kicked away the table to get it out of my way.

What's the force I am exerting to keep it at the same height?
(F = ma = 10kg * 9.8 m/s^2 = 98 N)... it's still not moving yet - conventionally, no energy has been added.

Now I apply just .05 m/s^2 more acceleration and I raise the ball 10 meters higher and hold it there (again, applying just the right force to counter gravity).

If energy is assumed to be Fd, then the energy I expended to raise the ball would only be 10kg * 0.05 m/s^2 * 10m = 5 Nm because the rest of the force countering gravity was just to maintain no change in height and thus no change in energy.

Now at the 10m position (I have long arms, btw) when I remove my hand and it comes down it does so at 10kg * 9.8 m/s^2 * 10m = 980 Nm.

That's a considerable difference... That seems like strong evidence that a PMM is possible to me.

I'd like an explanation from anyone who thinks my argument is wrong. I'd like to know why I must add in the force opposing gravity's pull when it doesn't add/subtract anything to an object's PE.

-Randall

[ovyyus]

Randall, the 10kg ball is held suspended at 1m by a force (the table) of 10kg. As soon as the ball is raised to a position higher than 1M (the surface of the table) it is obviously no longer supported by the table and an applied force of 10kg+ is required to move it higher . Force x distance = work.

[rmd3]

Bill,

Actually, I say I apply a counter force to gravity which is the 98 N (not 10kg as you say) to keep it at the same height. So no argument there.

The starting height is irrelevant because I am looking at the raising from one height to the next and dropping it the height it was raised. If it was 1000 m as a starting height and raised 10 m the difference in PE is the same and 1m from ground rasied 10 m, right?

It also doesn't matter what height I am holding the ball at, I still only apply a 98N to keep it at any fixed height.

My argument still stands. Please re-read it and clarify again. Thanks.

(my edits on the original post were adding the word acceleration and fixing energy units - I put it as Nm, if that's okay).

[ovyyus]

Sorry, you added the above while I made a slight edit. The force required to maintain the ball at any height is obviously going to be equivalent to the weight of the ball - 10kg.

Therefore, in order to lift the ball from any one height to any other height will require a force of 10kg (minimum suspension force) + any additional force applied to accelerate the ball upwards. The minimum applied suspension force (10kg) isn't free when it's applied over distance against an equal and opposite counteracting force (in this case gravity).

[rmd3]

Bill,
I know that's the convention, but that's exactly what I'm calling that into question. I trying to show using conventional physics understanding of energy that there is a discrepency unaccounted for.

Conventionally, if there is no addition of energy when holding a ball at a fixed height (i.e. applying a force countering gravity), how can that force be part of the force that changes the height? If you stop anywhere along the way and hold the ball at the stopped height, the force countering gravity has to stay in order to maintain the height stopped at, right? That means the force that is doing the work is the force above and beyond the counter force to gravity. No? If not, I need a reasoned answer.

Try it yourself. Pick up a weight and hold it at a height. When you lift it higher you don't add much force (edit: to the counter force you are already supplying) to make the change and then when you hold it at a higher height you do so with the same counter force to gravity as you did at the lower height. You can feel that the amount of work you have to do to raise it is much less than that of maintaining a height.

Does anybody understand what I'm saying?

[ovyyus]

Conventionally, if there is no addition of energy when holding a ball at a fixed height (i.e. applying a force countering gravity), how can that force be part of the force that changes the height?

Because the force isn't just holding the ball at a fixed height when it is being used to lift the ball higher. You're talking about two totally different things here - holding the ball at a fixed height and lifting the ball to a greater height. One is not the other.

If I clamp my car the the back wall of the garage with a force of several tons, that force still won't help my car travel down the road and 100 MPH into a strong head wind. I'm not suggesting wind here as an analogy for gravity, am I? :D

[jim_mich]

You're talking about two different situations. While the table is in place then a long line of molecules hold the weight up. As soon as the table is kicked away then a constant force is required to hold the weight in place or to move it upward. Even though people sometimes say that the table is excerting a force to hold the weight up, this is not exactly true. The table just conducts the force.

[rlortie]

what the heck s going on with this board

dp

Michael, You are not the only one experiencing problems. I have them too!

Submissions freezing up then duplicating out of sequence. Half downloading and then freezing. Hitting the refresh button and having the page you were seeking show up. loosing posts after a copy backup only to have them double after resubmission ETC.

[rmd3]

Bill,
If I understood your analogy correctly, you're saying that applying a forward force to maintain a position in your car won't help your car travel against the wind that's pushing with an equal force against you. Isn't that exactly what I am saying - i.e. the force for maintaining a height against gravity "wind" (your's and John's analogy - not mine!) won't help you climb any higher, that's why it is the additional force that does the work.

If I didn't understand you're analogy, could you rephrase it?

Jim_Mich,
I don't think it matters much what is conducting the counter force to gravity. The point is that there is a counter force to maintain a non-changing PE and that is equal and opposite to gravity, and the additional force added does the work beyond going against gravity.

Forces are summed via superposition, right? If so, then the counter force to gravity contributes nothing to the movement - how could it? The movement must come from the additional force. This is clear logic, no?

I don't think I am being blind here, am I? I think the logic is sound. Isn't it? Aaargh!

-Randall

[Bessler007]

hello rmd3,

I'm wondering if the situation you're describing is similar to the one you posted yesterday?

I have a mass resting on a carton of eggs. I raise it to say 100 ft. and drop it. Intuitively, did it return to the top of the carton of eggs recouping all the "energy" I put into it, and my eggs are safe? Please explain why they broke and where that extra "energy" came from? (perhaps someones answer to this will make me change my mind)

One thing that I noticed about your second example is

Now I apply just .05 m/s^2 more acceleration and I raise the ball 10 meters higher and hold it there (again, applying just the right force to counter gravity).

and I think what you're noticing is the rate of doing work or power.

You are slowly lifting the mass to a higher potential energy level then pulling the stops out and letting gravity do its thing. Power is the ratio of work to time. When gravity accelerates the mass it doesn't take as much time for the mass to do it's work. There's more power in it falling because the energy develops at a faster rate. The faster work is accomplished the more power is necessary.

[Bessler007]

Hello Mr. Collins,

I thought that analogy might be the way to go but perhaps my analogies aren't making things clear enough.

The analogy of gravity and a stream works to a point. Both the stream and gravity have the potential to move mass.

No matter which medium moves the mass to a lower level of potential the problem of causing that mass to swim back up stream doesn't go away. As ovyyus put it:

If gravity acts like a constant stream or a river carrying everything with it then how does a cork carried from the head of the river all the way down to the ocean ever get back up to the head again? Isn't that the full cycle?

The analogy doesn't add any more insight into that dilemma than we originally had.

[rmd3]

Bessler007,

Hello there. I think you're on to something... I want to think about it some more. I am definitely looking for something that matches the intuitive observation. I think that might just fit the falling/raising object example.... I have to give it deeper thought.

Is power always positive? I'm thinking of the pendulum oscillating. Is the average power 0? That would seem bizarre because it keeps going up one side then the other... seeming to always reuse the energy (so the rate of work would be sinusoidal) but how can there be negative power? I'm not sure I grasp that (if it is even the case).

Thanks Bessler007 - you've oiled the gears,
Randall

[Tinhead]

Having trouble with basic physics, so please try this scenario on for size...

Let's say I have a 10 kg ball on a table 1 meter off the ground. I go under it and exert a force just to the point where it still isn't
being lifted and I've kicked away the table to get it out of my way.

What's the force I am exerting to keep it at the same height?
(F = ma = 10kg * 9.8 m/s^2 = 98 N)... it's still not moving yet - conventionally, no energy has been added.

Now I apply just .05 m/s^2 more acceleration and I raise the ball 10 meters higher and hold it there (again, applying just the right force to counter gravity).

That's the catch, you apply .05 m/s^2 MORE ... that ends up as 9.85m/s^2 ...

If energy is assumed to be Fd, then the energy I expended to raise the ball would only be 10kg * 0.05 m/s^2 * 10m = 5 Nm because the rest of the force countering gravity was just to maintain no change in height and thus no change in energy.

So you expend 10kg * 9.85 m/s^2 *10m = 985Nm

Just my point of view :)

Cheers,
Rainer

[rmd3]

Thanks everyone who responded regarding the raising of a weight. Surprisingly, nobody convinced me that my observation was inaccurate or illogical. For sure, it doesn't match the math of conventional physics (at least the way I word the problems). I'll be watching the board to see if there are any more comments, but I make no promise to respond because...

I discovered two things. (1) this board is addictive, and (2) this board is really addictive. And to think I didn't even read all the active threads.... wow! What are you guys doing that are posting sooooo many post a day?!!! I've got other stuff that's a higher priority, but, man, it's nice to throw unconventional ideas around and not be totally dismissed (partial dismissal expected).

Thanks for indulging my questions.
-Randall

[rlortie]

Randall,

I would be the first to readily admit that yes! this forum is addictive!

Most of the time I find it that is also therapeutic. I spend a lot of time here, I find it motivates my innovative thinking and retards tunnel vision regarding a design that I am either scrutinizing fo a member or working on my own.

A good debate is just that.. good but occasionally a fly finds its way into the therapeutic ointment and good things go astray. Its the yin and Yang of life. We cannot enjoy one without experiencing the other.

Ralph