Energy, Distance and Force

J

Jim Thompson

Guest
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
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
 
Jim Thompson wrote:

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?
Click to expand...
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
http://www.wescottdesign.com
 
On Tue, 17 May 2005 21:33:51 -0700, Tim Wescott <tim@seemywebsite.com>
wrote:

Jim Thompson wrote:

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?


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
http://www.wescottdesign.com
Click to expand...
Defining the problem more simplistically... what is the force between
the two plates of a capacitor, spaced "d", area "A", voltage "V"?

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
 
"Jim Thompson" <thegreatone@example.com> wrote in message
news:mlhl811sfj9cip8anspsc4t6q8pimqi9mn@4ax.com...
Defining the problem more simplistically... what is the force between
the two plates of a capacitor, spaced "d", area "A", voltage "V"?
Click to expand...
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: http://webpages.charter.net/dawill/tmoranwms
 
Jim Thompson wrote:
On Tue, 17 May 2005 21:33:51 -0700, Tim Wescott <tim@seemywebsite.com
wrote:


Jim Thompson wrote:


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?


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
http://www.wescottdesign.com


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

...Jim Thompson
Click to expand...

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
 
Terry Given wrote:
Jim Thompson wrote:

On Tue, 17 May 2005 21:33:51 -0700, Tim Wescott <tim@seemywebsite.com
wrote:


Jim Thompson wrote:


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?


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
http://www.wescottdesign.com



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

...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
Click to expand...
and agrees with Tim, so must be right :)

Cheers
Terry
 
On Wed, 18 May 2005 00:13:37 -0500, "Tim Williams"
<tmoranwms@charter.net> wrote:

"Jim Thompson" <thegreatone@example.com> wrote in message
news:mlhl811sfj9cip8anspsc4t6q8pimqi9mn@4ax.com...
Defining the problem more simplistically... what is the force between
the two plates of a capacitor, spaced "d", area "A", voltage "V"?

Oh, electrostatic force?

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

Tim
Click to expand...

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
 
On Wed, 18 May 2005 17:15:06 +1200, Terry Given <my_name@ieee.org>
wrote:

Jim Thompson wrote:
On Tue, 17 May 2005 21:33:51 -0700, Tim Wescott <tim@seemywebsite.com
wrote:


Jim Thompson wrote:


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?


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
http://www.wescottdesign.com


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

...Jim Thompson


Woodson & Melcher (part 1) gives:


[snip]
Click to expand...

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 |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
 
Jim Thompson wrote:
On Wed, 18 May 2005 17:15:06 +1200, Terry Given <my_name@ieee.org
wrote:


Jim Thompson wrote:

On Tue, 17 May 2005 21:33:51 -0700, Tim Wescott <tim@seemywebsite.com
wrote:



Jim Thompson wrote:



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?


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
http://www.wescottdesign.com


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

...Jim Thompson


Woodson & Melcher (part 1) gives:



[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
Click to expand...
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
 
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