Stop this charade, Tony.
You can not POSSIBLY be this incompetent. Therefore, you have clearly decided to simply obfuscate in order to pander to a very few clueless truthers.
The first 3 or 4 times you dodged and mislead, it was simply hand-waving & obfuscation. Now is has elevated to simpl lying, Tony. Why?
You did give your real name, Tony. And then you besmirch your own name, Tony, with a High School FAIL like this. In Public...!
How humiliating, Tony, to voluntarily reduce your credibility to zero, because you choose to not simply say "OK, you're right."
Weight is NOT acceleration, Tony.
Weight is a force.
A force is NOT an acceleration.
You do NOT need an acceleration in order to generate a force, Tony. Walk up to the side of any building. Push as hard as you can. What is your acceleration, Tony? It is zero. What is the building's acceleration, Tony? It is zero. Zero acceleration. And yet you still are generating a force between you & the building.
BY DEFINITION, a rigid body can have any number of forces (and any number of moments).
BY DEFINITION, a rigid body can have only ONE linear acceleration (and one angular acceleration), Tony. That is the linear (& angular) acceleration of the center of gravity.
Say that to yourself over & over & over again.
The ONE linear acceleration of any rigid body is the result of the sum of all forces.
The ONE angular acceleration of any rigid body is the result of the sum of all moments.
BY DEFINITION, a rigid body in static equilibrium has all of its forces (& moments) sum to zero.
BY DEFINITION, a rigid body in static equilibrium has one, AND ONLY ONE, linear acceleration, which equals ZERO. (So does the angular acceleration.)
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Here is your claimed case, Tony. An object with mass & weight.
[qimg]http://www.internationalskeptics.com/forums/picture.php?albumid=176&pictureid=2255[/qimg]
How many forces can this rigid body have? Any number.
How many does this one have? 1.
How many linear accelerations can a rigid body have? 1.
∑forces = m a
∑forces = W
a = W / m = mg / m = g
Result #1: This object is falling with an acceleration = g
Result #2: This rigid body is not in static equilibrium.
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Here is the real situation we are discussing. A mass in static equilibrium.
[qimg]http://www.internationalskeptics.com/forums/picture.php?albumid=176&pictureid=2256[/qimg]
How many forces can this rigid body have? Any number.
How many does this one have? 3.
How many linear accelerations can a rigid body have? 1.
∑forces = m a
∑forces = W + F1 +F2
Static equilibrium ➔ ∑forces = 0
W + F1 + F2 = 0
Static equilibrium ➔ ∑Moments = 0
∑Moments = 0 ➔ F1 = F2
F1 = F2 = -W/2
-W + F1/2 + F2/2 = 0
∑forces = 0 = m a
a = 0
Result #1:
This object is in static equilibrium.
Its one and only linear acceleration: a = 0
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You have already shown everyone what a coward you are, Tony, by refusing to answer these very simple questions the last 3 times I've asked them. You've also shown everyone what a weasel you are by your repeated claims that I'm wrong, while you KNOW that this isn't true. And yet you choose to say it anyway.
Will you continue to show everyone what a coward & weasel are you are, by choosing, ONCE AGAIN, to refuse to answer them? These simple questions that answer this issue with simple clarity & certainty.
My bet is "yes, you will". Prove me wrong, Tony.
1. What was the vertical velocity of the upper block in 1975?
2. What was the vertical velocity of the upper block in 2000?
3. What was the CHANGE IN VELOCITY between 1975 & 2000?
4. What is the "change in velocity with respect to time called"?
5. What was the linear acceleration of the upper block between 1975 & 2000?
Why are you smearing your own name, Tony?
I am providing you with my credentials by producing simple & accurate explanations of rudimentary terms like force & acceleration & static equilibrium.
I don't give a flying fig about your name, Tony. That is yours to drag thru the mud of incompetence.
I care about accurate, competent engineering.
WHY DON'T YOU TRY GIVING US SOME OF THAT, Tony? Instead of the crap that you've been shoveling.