Energy, Distance and Force

I'm drawing a blank trying to remember how to use (IIRC) Hamilton's Principle.

Isn't there some simple-minded way to take the energy at point A and energy at point B, and calculate the force required to get from point A to point B?

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
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Jim Thompson
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Not the force, but possibly the work.

Energy at point B = energy at point A + work to get to point B.

In a perfect physics world that means you could either go one inch while exerting 10 tons, or you could go 20 000 inches while exerting one pound.

------------------------------------------- Tim Wescott Wescott Design Services

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Tim Wescott

Defining the problem more simplistically... what is the force between the two plates of a capacitor, spaced "d", area "A", voltage "V"?

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

Oh, electrostatic force?

Don't know why you'd need to know this...this is first year textbook stuff, Jim... ;-)

Lesse, one definition for electric field is the force divided by the charge, E = F/q where E and F are vectors.

q = VC and C = e0*A / d (e0 = epsilon naught, permittivity of space), so: E = F*d / e0*V*A > F = E*e0*V*A / d E is also defined in volts per meter, so E = V/d. Thus: F = e0 * V^2 * A / d^2

Force goes down inverse square with distance, and linear with area, I'd think voltage would be linear as well though. Eh, what do you expect for a usenet reply mmmh?

Tim

-- "California is the breakfast state: fruits, nuts and flakes." Website:

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Reply to
Tim Williams

Of course it is. It's just that I'm coming up on 50 years since high school physics and, you know, use it or lose it ;-)

[snip]

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

[snip]

Wow! Old times. I should buy that book as a keepsake if nothing else.

I have White & Woodson, the predecessor.

I worked as a technician in Professor Woodson's lab in MIT Building 20 while I was a student at MIT.

Melcher was a graduate student studying MHD under Woodson.

Melcher also taught some EE courses that I attended.

Melcher went on to become head of the EE Department, but died of colon cancer, IIRC, mid 80's.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

Woodson & Melcher (part 1) gives:

F = dW/dx (where the d's have melted and leaned to the left a bit)

W = energy = .5CV^2

C = Ae/x (3.1.54)

F = AeV^2 (3.1.55) ----- 2x

which is about what I guessed.

Cheers Terry

Reply to
Terry Given

and agrees with Tim, so must be right :)

Cheers Terry

Reply to
Terry Given

Hi Jim,

I bought vol. 2 & 3, brand new, for $1 each while I lived in Taxachusetts - from a book clearing place, Hamiltons IIRC. Vol. 1 arrived about a month ago, is older than I am, and cost about $20. They are an excellent treatment of electromechanical systems. I knew you would recognise them, which is why I used them ;)

My copy of Zverev turned up this morning, and I eagerly await Motchenbacher (its in the mail....).

Cheers Terry

Reply to
Terry Given

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