How does this Domino Tower Collapse relate to 9/11 Collapses

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Mick West

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So we have ma - the upwards force

No we don't. There's no "a". Acceleration comes from net force, so if the net force is zero there's no acceleration.

You need to describe things in terms of forces, not accelerations.

And there's no homogenous "stiffness of the structure". Most of the stiffness is provided by the columns, which were essentially bypassed in the collapse (at the top tllted), they were not "in the way", and hence largely irrelevant to the speed of the collapse front.

This domino model on the other hand has the main support structure in the way all the way down, and is not really falling in any macro sense (dominos are largely just toppling, and then falling off).
 

aka

Member
There's no "a".
If there is no a, there is only g, and it falls.
Acceleration comes from net force, so if the net force is zero there's no acceleration.
No net acceleration. The towers are being accelerated all the time. So they must be stable. But they will and must displace, even if only the tiniest amount. So of course there is a, lots of a even. At least as long as it stands.


And there's no homogenous "stiffness of the structure".
I know that and you know I know that. That it is not a constant spring is not the point. The point is that the load-displacement curve F(u) more or less already dictates the downwards acceleration (and hence, by necessity the "stiffness") and there is Oysteins Dirac impulse function that gives a clue as to how F(u) must look like. In any case, on average, it must be smaller than mg for it to fall.
This domino model on the other hand has the main support structure in the way all the way down, and is not really falling in any macro sense (dominos are largely just toppling, and then falling off).
On that I agree, the lateral ejections of the real thing were much more impressive.
 

econ41

Senior Member
I am not aware of when, where and through what mechanism Bazant arrived at g/3, but that's what I got: By modelling pure momentum transfer of floors pancaking inelastically. No strength, force or energy of anything considered (connections can be thought of as holding the floors just barely, but requiring practically zero force/energy to disconnect once tapped on ever so lightly).
And that is a valid representation of the real WTC events at least to first order significant quantification.
That seems as far away from anything Bazant has modeled in his WTC papers as I can think of.
It is - diametrically opposite. Bazant's modelling - all of it AFAIK - is based on the abstract "Limit Case" model from B&Z 2002 which simply didn't happen . The same "take the abstract literally" error which led T Szamboti astray followed by or in company with far too many debunkers.

Or I missed that part (quite likely, given that I never truly studied any of them after B&Z :D)
What Bazant modelled in his later papers - principally B&V 2007 with "crush down crush up" {"cd/cu"] - was definitively NOT applicable to WTC real event. Cd/cu has at least THREE fatal errors in that context.

I've been making and explaining that assertion for about 5 years in the face of entrenched "Bazantophile" denialsim.

I'll return you to the normal program. ;)
 
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econ41

Senior Member
Excellent point. Clearly the mechanism is entirely different than the tower collapses, which might lead to premature dismissal of the instructive value it offers.
Agreed, agreed strongly, agreed with my usual proviso - respectively.

"instructive value" for what? Understanding/explaining WTC collapses OR generic abstract principles? No problem with either but...

...you know the rest of the advice. ;)
 

Mick West

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Staff member
If there is no a, there is only g, and it falls.

Gravity is a force. The use of "g" as an acceleration is just a convenience, as things normally fall with that acceleration. But the force of gravity is G*m1*m2/(r^2) - i.e. it's a force.

That force, absent opposing forces, will produce an acceleration downwards.

When something rests on the ground then it is being opposed by an upwards force. It's not accelerating upwards.

There is no a, there isn't really a g either. If nothing is moving there is no acceleration. It's nonsensical to say something is accelerating in two directions at the same time.
 

OneWhiteEye

Senior Member
Yes,,,, IF the tower is tall enough.
Analogue would be skydiver in free fall. Eventually air friction force equals gravitational force and net acceleration reduces to zero. It is of little consequence if the jumper is just leaping off of a six foot step ladder. A shorter structure than the towers may well not see acceleration drop to zero.
Good point. One which got me thinking... I have to think more before speaking.
 

OneWhiteEye

Senior Member
Here is how I understand "Mechanics of Progressive Collapse" and I hope OWE will be so kind to correct me where I'm wrong.
This was an excellent rundown. There is a problem in there, but very nice all the same.

...the effective acceleration only depends on how much mass is shed...
At this point, you have accounted for the external force due to structural resistance but not momentum change from inelastic accretion. Mass shedding is one side of the coin, mass accumulation is another. It turns out that it is this component which largely determines the eventual acceleration limit. It is of course possible to specify a form of F(u) which overrides that behavior but it wouldn't be realistic. This acceleration limit we discuss also applies to a uniform mass distribution without shedding, so in some respects you're ahead of the discussion.

And since the area under F(u), let us call it E[elasticplasticpot], is smaller than E[gravpot], this is in essence a mathematical way of saying "for every meter height, there is on average less Newtons upwards force in the way than necessary to decelerate the kilograms of the tower against the gravity of the planet."
Yes.

So far, Oysteins model (if I understand correctly) derives U, an optimum of the energy going into deformation/pulverization/etc/friction (in short: heat) to achieve optimum v, from perfectly inelastic collisions.
Yes. Slab model. Bam-bam-bam down the line.

So far, a Dirac impulse is assumed here, which is of course unrealistic (which does not matter because the Twin Tower floors did not hover mid-air) but extremely useful for analysis ...
Also because it allows a neat trick. Do you remember the crazy back-and-forth about momentum conservation we had a few years back? Is it or is it not conserved?

Yes, on the global level, it is.
No, for the structure as a whole, yet isolated system.
Yes, for imaginary Dirac-like impulse.

It's the last which permits a simple analysis to do so much. It essentially allows decoupling the structural and momentum-change components of retarding force.
 

aka

Member
so in some respects you're ahead of the discussion.
I agree. I guess I just wanted to have one "free" parameter in there for fine-tuning of motion history, because the others are dictated somewhat by the conditions given.
Do you remember the crazy back-and-forth about momentum conservation we had a few years back? Is it or is it not conserved?
How could I ever forget it! Yes, it is conserved in a closed system. Always.
It's the last which permits a simple analysis to do so much. It essentially allows decoupling the structural and momentum-change components of retarding force.
It is the approach of finding the ideal, bestest possible collapse time! If the collisions are not perfectly inelastic, the mass does not accrete optimally, and a sole slab will fly ahead of the loosely accreted mass and impact the next one with less momentum, which will therefor then fall slower and so forth. If, on the other hand, more energy than needed is dissipated during the momentum exchange, less kinetic energy than possible is available and the rate of fall also decreases. Perfect inelasticity and therefor ideal mass accretion is ONE condition utterly favorable towards optimal rate of fall.

Assuming a Dirac function is only the cherry on top to keep things simple here - of course we know that it is impossible to have infinitely high force for infinitely short displacement and infinite transition steepness, so that the area under the curve just happens to be equal to the energy that needs to be converted into heat. The Dirac function is not even a function mathematically, it is just a convenient tool for physicists.

In the real world, the force would be finite and stretched out over a displacement - be it micrometers or centimeters. This, in turn, would take time, and this in turn would result in less-than-optimal drop height, which in turn would mean the accreted mass cannot impact the next slab with optimum velocity - resulting in a smaller effective downwards acceleration overall.

Gravity is a force. The use of "g" as an acceleration is just a convenience, as things normally fall with that acceleration. But the force of gravity is G*m1*m2/(r^2) - i.e. it's a force.
Yes, gravity - the acceleration of a mass - is a force. g is the acceleration of that force.
That force, absent opposing forces, will produce an acceleration downwards.
Correct (for all things near the surface of the planet).
When something rests on the ground then it is being opposed by an upwards force. It's not accelerating upwards
This is also correct.
There is no a, there isn't really a g either. If nothing is moving there is no acceleration. It's nonsensical to say something is accelerating in two directions at the same time.
This is not correct. There was a mathematician once who calculated that when two elephants push against each other with equal force, the resulting net force equals zero. He figured it must be safe to stand between them.

There is both an a and a g, and if ma and mg are equal, and just pointing into opposite directions, there is no displacement - the body is at rest in static equilibrium, that is all. It is still bathing in the vast force field resulting from the planet's mass.
 
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Mick West

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This is not correct. There was a mathematician once who calculated that when two elephants push against each other with equal force, the resulting net force equals zero. He figured it must be safe to stand between them.

There is both an a and a g, and if ma and mg are equal, and just pointing into opposite directions, there is no displacement - the body is at rest in static equilibrium, that is all. It is still bathing in the vast force field resulting from the planet's mass.

The a you use here is just a number you get when you divide the force opposing gravity by the mass. As the building is not actually accelerating then it's a meaningless number.

Acceleration is the result of net forces. It does not have components. Force has components.

Where is a? Somehow "stored" in the subatomic bond of the structure? If you remove the top nine bricks from a stack of ten then will the bottom brick accelerate upwards at 9g?
 

jaydeehess

Senior Member.
Deceleration is generally understood as a reduction in speed (where "speed" is the scalar value of velocity in the direction of movement). I.e. slowing down. So:
Respectfully, that is simply another way of saying exactly what I said. Acceleration in the opposite direction to positive convention. Thus a reduction instead of an increase, in velocity.

However many times I see people using 'decelleration' to imply a lessening, but still with the same vector direction, of acceleration.

I do not in fact recall my physics prof in university ever using this term/ It was acceleration, a number with attendant direction. In the case of an object that only moves along a single dimension a +/- will suffice, in other situations such as a pendulum swinging, it moves in two dimensions, an electron in a CRT moves in all three directions. At that level the term 'decelleration' moves from ambiguous to downright meaningless.
IMHO
 
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jaydeehess

Senior Member.
The strawman is simply a result of misapplication of physics by others. I guess you also missed it when Oystein ( I now note my new tablet'so annoying autocorrect) said the same thing.

I am not "explaining" the WTC collapses with the TNB collapse. I am pointing out that a design peculiarity that allows for collapse need not be a deliberate feature the engineer built in.
I understand. And I refute it by pointing out, by means of the domino tower model, that there is a measurable, energetic difference between "shit happening" and "a plan coming together", between failure due to stupidity and neglect and a perfect chain reaction, between dumb accident and blatant purpose, between ordo ab chao and intelligent design. .
I'm sorry what exactly are you "refuting"?

The domino collapse illustrates the way a specific design collapses, a design that has lower 'floors' supporting the mass of upper structure. That is a feature diametrically opposite of how the towers, and indeed most high rise structures, are built.

Mick's model demonstrates something much more akin to the design of the towers and indeed the columns do not even enter into the mode of collapse. Its the destruction of floor to column connections, thus the loss of lateral support offered by the floors to the columns, that causes collapse. Progression is solely due to loss of floors.

Your beef seems to be that the columns are two flimsy. Like I believe I said above, make it 3d and use 4X4 columns and the structure will still fall apart if the floors progressively are no longer attached to the columns.

The only way this could not result would be if the individual columns are, as put above, "meta-stable". That is to say that individual columns could stand upright on their own. In the TTs they fairly obviously, could not.
 
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Mick West

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Staff member
Respectfully, that is simply another way of saying exactly what I said. Acceleration in the opposite direction to positive convention.

I wasn't disagreeing with you, I was putting what you said in another way to explain it to @aka.

However now I do actually disagree slightly with "Deceleration is ... a change in the sign of acceleration." The sign comes from the velocity, not acceleration. For example if a train is traveling at constant speed and then brakes, then it's decelerating, but the prior acceleration was zero. It's really just slowing down, it's a convenience word, not some kind of anti-acceleration.

Niggling things, but I think ultimately a lot of trutherism, especially the appeals to Newton, come from a crippled epistemology in the realm of physics.
 

aka

Member
As the building is not actually accelerating then it's a meaningless number.
It is accelerating the mass opposite the acceleration of gravity, resulting in mechanical equilibrium. If it would not do that, then only gravity would act on the mass and displace it downwards.

I know the debate about whether "slowing down the acceleration" is "deceleration" has been had already in the debates with Tony Szamboti. While it is purely of semantical nature, I have learned it the way it has been presented by those who disagreed with him: if g = 9.81m/s², and a thing falls only at 5m/s², then it is decelerated, or accelerated in the opposite direction, for all intents and purposes, by 4.81m/s². I think technically, it has something to do with velocity and acceleration being vector quantities or something in that spirit, but don't quote me on that.
Where is a? Somehow "stored" in the subatomic bond of the structure?
Allow me to repost the diagram I made for the same discussion we had before in another thread:



So yes, a is kx/m.
If you remove the top nine bricks from a stack of ten then will the bottom brick accelerate upwards at 9g?
Of course not. But it will release the elastic energy that has previously been stored in it to return to mechanical equilibrium. Let me think about that. In a sense, yes.

And if you try the same with a linear spring, yes, it will accelerate upwards if you suddenly unload it.
 
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Mick West

Administrator
Staff member
While it is purely of semantical nature, I have learned it the way it has been presented by those who disagreed with him: if g = 9.81m/s², and a thing falls only at 5m/s², then it is decelerated, or accelerated in the opposite direction, for all intents and purposes, by 4.81m/s². I think technically, it has something to do with velocity and acceleration being vector quantities or something in that spirit, but don't quote me on that.

Beating a dead horse here. You can't sum acceleration vectors, you sum force vectors. Acceleration is the result of the net force, divided by the mass.

Saying a falling object is being accelerated upwards is just nonsense. If something is falling with an acceleration less than g than that means there's a retarding force perpendicular to the force of gravity.

Force is the important thing here. Acceleration is a measure of the rate of change in velocity. I think your point is being lost because of your strange insistence on this upwards acceleration of an object whose downward speed is increasing. Just make your point using force, and net acceleration.
 

aka

Member
You can't sum acceleration vectors, you sum force vectors.
https://en.wikipedia.org/wiki/Acceleration
Acceleration is the result of the net force, divided by the mass.
Force is the product of mass and acceleration.
Saying a falling object is being accelerated upwards is just nonsense.
This is what I said:

"I have learned it the way it has been presented by those who disagreed with [Tony Sz]: if g = 9.81m/s², and a thing falls only at 5m/s², then it is decelerated, or accelerated in the opposite direction, for all intents and purposes, by 4.81m/s²."

If something is falling with an acceleration less than g than that means there's a retarding force perpendicular to the force of gravity.
A "retarding force"! A force, that would be the product of mass and acceleration! So if something is falling with an acceleration less than g that means there is an acceleration in the opposite direction of g!
I am quoting Wikipedia's article on acceleration because the disagreement with the most fundamental concepts is starting to make me doubt my sanity and google things I thought are elementary.
I think your point is being lost because of your strange insistence on this upwards acceleration of an object whose downward speed is increasing.
I think my point is quite clear. Gravity does not magically get switched off once the building stands. It is being accelerated all the time, which is why its mass has a weight, which is why there needs to be a normal force to keep it up all the time, decelerating it, so that the effective acceleration remains zero and the displacement within given parameters.
Just make your point using force
I will make my point using force, and Wikipedias diagram in the mechanical equilibrium article:

mg is the downwards force, resulting from the mass and the acceleration due to gravity. N is the normal force, acting upwards. It results from there being something in the way of the green object that has a "retarding force". The green object does not move. It is in mechanical equilbrium. It does not fall, it does not move, it is not displaced, although there are still 6371 km to go to the center of gravity of the planet and thus, it has huge gravitational potential energy. Because the yellow stuff is too stiff and is in the way, the green object does not move, because mg = N (technically, mg = -N, but we can't do vector signs here).

ETA

Just make your point using force, and net acceleration.
"Net" only applies to forces.

:confused:
 
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Mick West

Administrator
Staff member
I will make my point using force, and Wikipedias diagram in the mechanical equilibrium article:

mg is the downwards force, resulting from the mass and the acceleration due to gravity. N is the normal force, acting upwards. It results from there being something in the way of the green object that has a "retarding force". The green object does not move. It is in mechanical equilbrium. It does not fall, it does not move, it is not displaced, although there are still 6371 km to go to the center of gravity of the planet and thus, it has huge gravitational potential energy E=mgh. Because the yellow stuff is too stiff and is in the way, the green object does not move.


Think of it another way. Does acceleration cause force, or does force cause acceleration?

I think it's clear that force causes things to accelerate. Acceleration is just a measure of the rate of change of velocity. It's not a thing in itself. it does not do anything.

Perhaps this is confusing because F = ma suggests that Force is the result of mass multiplied by acceleration. This suggests the mass and the acceleration are creating a force.

But really one or more forces acts on a mass, giving a net force, and this creates an acceleration.

Look at the convolution in the sentence I bolded from your post

"mg is the downwards force, resulting from the mass and the acceleration due to gravity"

That's essentially circular. You are saying the downwards force is caused by the acceleration caused by the downwards force. And yet nothing is moving.

g is not gravity. It's the acceleration resulting from the force of gravity. It's a handy approximate constant for the resultant acceleration in the absence of any other force. And since the force of gravity is also a product of the mass, we can handily use g to calculate the force Fgravity​ = mg

If we've got another force acting on a body, like Fspring​ = kx (stiffness*displacement) then you can subtract the forces, assuming they act in opposing directions (although it's more correct to use -kx as the force, and add them)

Fnet​ = Fgravity​ - Fspring​

Then you can calculate the acceleration as (Fgravity​ - Fspring​)/m

Now you might argue that you could expand this equation into two accelerations Fgravity​/m and Fspring​)/m (i.e. your g and a) and add them together, and get the same result. But you are introducing a meaningless term a, which would be equal to the acceleration from the spring if there was no gravity. But there isn't.

It makes no sense to say there's an upwards acceleration. However there IS an upwards force.
 

aka

Member
Think of it another way. Does acceleration cause force, or does force cause acceleration?
I will not think of it any other way, because the way I think it is correct, as I have shown. A force accelerates a mass and a mass exerts a force when it is accelerated. actio=reactio. This is fundamental and elementary. There can be not much use in an ontological debate about the prima causa of the universe's inner workings except to distract from the point being made: If a mass does not accelerate, it does not mean no force is acting upon it, it only means that the vector sum of all forces acting upon it is zero.

Perhaps this is confusing because F = ma suggests that Force is the result of mass multiplied by acceleration. This suggests the mass and the acceleration are creating a force.

But really one or more forces acts on a mass, giving a net force, and this creates an acceleration.
It is not confusing at all. Yes, the mass and the acceleration are creating a force. If the mass were not accelerated, it would exert no force. And when the net force is zero, the acceleration of the mass equals zero. Thus, a = 0 means not that no force is acting at all.
Look at the convolution in the sentence I bolded from your post

"mg is the downwards force, resulting from the mass and the acceleration due to gravity"

That's essentially circular. You are saying the downwards force is caused by the acceleration caused by the downwards force.
This reminds me of the time they asked Feynman to explain how magnets work.

Objects attract each other. That phenomenon is called gravity. It is a force. It depends on their mass and their distance.

So, I am saying the downwards force results from the mass being accelerated due to gravity. You said yourself gravity is a force. You gave the formula.

F=G*m1*m2/(r^2)

m1 is the mass of the planet, m2 is the mass of the green object, r1 is the distance of the surface to the middle of the planet. If you make a unit check, you arrive at Newtons. This is a force. Hence g is G*m1/r² and it for most m1 >> m2 and most r ~ 6000km relatively constant. The unit will be meters per second squared. It is the rate at which objects fall. And it says what a given mass weighs. mg is the downwards force, resulting from the weight of the object, it is its gravity. It is the force with which the planet and the object attract each other, and the force that must at least be applied in the opposite direction to lift it - to increase their distance - the mass must be accelerated!
And yet nothing is moving.
That is what I am saying. Because there is an equal and opposite force acting against gravity.
g is not gravity. It's the acceleration resulting from the force of gravity.
So we agree on that at least.
Then you can calculate the acceleration as (Fgravity - Fspring)/m
Correct. And since we know roughly how fast, at which rates and velocities, the object we are analyzing falls and when, all we are asking for now is the relation between its F[gravity] and its F[spring]. We know a, we know g, we have x and h so we can approximate m and we get k.

This refutes your next statement:
Now you might argue that you could expand this equation into two accelerations Fgravity/m and Fspring)/m (i.e. your g and a) and add them together, and get the same result. But you are introducing a meaningless term a, which would be equal to the acceleration from the spring if there was no gravity. But there isn't.
a is not meaningless, it is our given, our measurement, our observation. We already have it, roughly, every step of the way, even if there is a little margin of error.
It makes no sense to say there's an upwards acceleration. However there IS an upwards force.
Shall we simply call it "retardation" then? There must be a name for when a thing accelerates less than it would in freefall, even if not at all, when it is in mechanical equilibrium.
 
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Mick West

Administrator
Staff member
Shall we simply call it "retardation" then? There must be a name for when a thing accelerates less than it would in freefall, even if not at all, when it is in mechanical equilibrium.

Yes, it's when there's some upwards force.
 

jaydeehess

Senior Member.
I wasn't disagreeing with you, I was putting what you said in another way to explain it to @aka.

However now I do actually disagree slightly with "Deceleration is ... a change in the sign of acceleration." The sign comes from the velocity, not acceleration. For example if a train is traveling at constant speed and then brakes, then it's decelerating, but the prior acceleration was zero. It's really just slowing down, it's a convenience word, not some kind of anti-acceleration.

Niggling things, but I think ultimately a lot of trutherism, especially the appeals to Newton, come from a crippled epistemology in the realm of physics.
I see your point.
The convention + is the same for velocity and acceleration. Deceleration is acceleration with - sign.

My main point is that the term is ambiguous given it is used interchangeably for decreased acceleration in the positive direction, and acceleration with - sign in the acceleration vector.
 

aka

Member
Shall we simply call it "retardation" then? There must be a name for when a thing accelerates less than it would in freefall, even if not at all, when it is in mechanical equilibrium.
Yes, it's when there's some upwards force.
I am glad we could finally, not despite, but through what might at first seem like a needlessly pedantic and nonsensical debate, to a mutual understanding of each other's... "epistemologies", however... "crippled". If you insist that what is an "acceleration" be named "retardation" just because it is the rate of change in velocity - or the force per mass - in a different direction, I will gladly comply, if it allows us to get back on topic.

We know that the "retardation" of the structure must equal the gravitational acceleration so it stands up. If additional forces act on the structure - a Tae Bo class, a subtropical hurricane, a library full of heavy books - the structure must still be able to "retard" the accelerations resulting from those forces so the structure remains in mechanical equilibrium.

Expressed in terms of forces, the forces keeping the structure up must equal the gravitation resulting from its mass. If additional forces act on the structure, it must still be able to exert forces in the opposite direction - "push back" - so the structure remains in mechanical equilibrium.

Expressed in terms of energy, the elastic potential energy must do the virtual work of keeping the displacements due to additional inputs of mechanical energy within a given margin so that the structure does not convert its gravitational potential energy into kinetic energy.

We also know, by observation, that when the structure falls, the "retardation" is smaller than half the gravitational acceleration on average. In terms of forces, the forces acting on the structure during the fall - the friction force - are smaller than half the weight of the structure on average. In terms of energy, all that keeps the gravitational potential energy from being completely converted into kinetic energy is the energy of friction.

This leads us to a fool-proof way of describing the system objectively, mathematically and physically.

We have the Bazantian computational model, we have Oysteins computational model, and we have the domino tower and the Twin Towers. I am convinced that we can mold these approaches into a grand unified theory of tower self-disassembly, simply by taking Oysteins computational model and, instead of letting the masses hover mid-air, rest them on "springs" with known load-displacement curves (à la Bazant) so the structure stands up. Instead of a Dirac function, we only have to "smear" the function a little so its area equals the energy of friction, with still high enough a peak so that small displacements can be balanced to remain in mechanical equilibrium.

If we now allow the "mass shedding" parameter to follow an arbitrary function, this computational model will be able to describe both the domino tower and the Twin Towers, even the "NMSR does the Heiwa Challenge" "weights on toothpicks on a broom stick" model and psikeyhackrs "Momentum Interference Test" model, and additionally describe the possibility of arrest as is the case in the crushing experiments "Collapse onto cumulative supports" and Coles' models with the concrete slabs and paper loops and pizza box columns - and the real-world "experiments" (botched demolitions), and even vérinages - simply by adjusting the load-displacement curve relative to mg.

Any objections?
 
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Mick West

Administrator
Staff member
We know that the "retardation" of the structure must equal the gravitational acceleration so it stands up. If additional forces act on the structure - a Tae Bo class, a subtropical hurricane, a library full of heavy books - the structure must still be able to "retard" the accelerations resulting from those forces so the structure remains in mechanical equilibrium.

What accelerations? If it's not moving then there is no acceleration.

We can't have a discussion if you call things what they are not.
 

aka

Member
If it's not moving then there is no acceleration.
I already showed that that is not true. If it is not moving (EDIT: or more accurately, moving at constant velocity), the sum of all accelerations (or "retardations", or forces per mass) is zero, that is all, it does mean there is no acceleration at all. Accelerations are vector quantities and add up according to the parallelogram law. "Crippled epistemology" or not, this is how classical mechanics work.

To throw a famous experiment back at "debunkerism": hold a bowling ball straight away from your body by an outstretched arm. Keep it still. It is not moving. According to your epistemology, there is no acceleration, because it is not moving. So its weight is zero, because F=m*0=0, and you can hold it up for all eternity.

That is clearly wrong. The bowling ball is still being accelerated by gravitation, giving it weight F=mg, which you will have to acknowledge when your muscles begin to hurt.

We can't have a discussion when you insist on denying a fundamental concept of classical mechanics.
 
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Mick West

Administrator
Staff member
Sorry it seems you are immune to understanding this (you can vector sum forces, and the acceleration is the resultant net force, divided by the mass). So I'm done trying to explain it.
 

jonnyH

Senior Member.
To throw a famous experiment back at "debunkerism": hold a bowling ball straight away from your body by an outstretched arm. Keep it still. It is not moving. According to your epistemology, there is no acceleration, because it is not moving. So its weight is zero, because F=m*0=0, and you can hold it up for all eternity.

That is clearly wrong. The bowling ball is still being accelerated by gravitation, giving it weight F=mg, which you will have to acknowledge when your muscles begin to hurt.
No, in that scenario F is the sum of the force of gravity and the equal and opposite force applied by your arm, so;

F=mg+FArm=0

acceleration is a measure of the rate of change of velocity, so an object that is motionless like the bowling ball has no acceleration. When your arm gives up, i.e. when FArm=0, the bowling ball will accelerate towards the floor at g.
 

aka

Member
Sorry it seems you are immune to understanding this
Still no need for insults.
you can vector sum forces
...because velocities, and hence accelerations, and hence momenta and forces are vector quantities that add according to the parellelogram law.
and the acceleration is the resultant net force, divided by the mass
This statement and that the net force is the resultant acceleration multiplied with the mass are equivalent. F=ma, hence a=F/m. a=F/m, hence F=ma.
So I'm done trying to explain it.
Then please address the point I've been trying to make instead of trying to prove I am unable to understand although I backed my assertions.
No, in that scenario F is the sum of the force of gravity and the equal and opposite force applied by your arm, so;

F=mg+FArm=0
F is the net force here. It is only zero because mg and F[Arm], the two forces acting on the mass, have equal magnitude and point into opposite directions. You still have g=9.81m/s², and you have F[Arm], so a = F[Arm]/m[Bowlingball] = -9.81m/s², so that F[net]=m(g+a)=m(9.81m/s²-9.81m/s²)=m*0=0.

I am glad you agree that F[Arm] is necessary to keep the bowling ball up. A huge step forward.
acceleration is a measure of the rate of change of velocity
Correct.
so an object that is motionless like the bowling ball has no acceleration
...or, to be more precise, the vector sum of all accelerations (=forces per mass) is zero.
When your arm gives up, i.e. when FArm=0, the bowling ball will accelerate towards the floor at g.
Precisely. Keeping the bowling ball where it is, keeping its gravitational potential energy, requires work to be done.
 

Keith Beachy

Senior Member
... famous experiment back at "debunkerism": hold a bowling ball straight away from your body by an outstretched arm. Keep it still. It is not moving. According to your epistemology, there is no acceleration, because it is not moving. So its weight is zero, because F=m*0=0, and you can hold it up for all eternity.
... We can't have a discussion when you insist on denying a fundamental concept of classical mechanics.
If the bowling ball is 1kg, it weighs 9.81 newtons (on earth), more on Staturn, less on Mars. There is no acceleration of the ball standing still, but the ball has mass, and the ball in on earth, thus to say it has no weight is not understanding physics. F=ma, the ball is not moving, a=0, but it has weight. The problem is not a "debunkerism" one.

Truth is not an insult

On 9/11 the arm failed in fire..

""requires work to be done."" nope, not in physics... the work would be moving the ball from the floor to standing up... you don't get credit for work in physics for standing still holding a bowling ball... it could be torture
 
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Trailspotter

Senior Member.
hold a bowling ball straight away from your body by an outstretched arm. Keep it still. It is not moving. According to your epistemology, there is no acceleration, because it is not moving. So its weight is zero, because F=m*0=0, and you can hold it up for all eternity.

That is clearly wrong. The bowling ball is still being accelerated by gravitation, giving it weight F=mg, which you will have to acknowledge when your muscles begin to hurt.

That is clearly wrong. Weight of the ball is NOT the force that acts on the ball, it is the force which the ball exerts on the rest, in this case a human arm. According to the Newton's third law, the arm exerts equal and opposite force on the ball. The sum of this force and the force of gravity acting on the ball is zero.

Keeping the bowling ball where it is, keeping its gravitational potential energy, requires work to be done.
That is also wrong. The work has already been done to put the ball where it is. There is no additional mechanical work required to keep it where. Your muscles work to keep the skeleton in the fixed position, which is far from the human body resting state. A body in rigour mortis state or a wired skeleton will serve as the ball's rest perfectly; no muscle work required.
 

Mick West

Administrator
Staff member
Sorry, this is really wasting everyone's time. I'm going to ban @aka for one month, or until he demonstrates an understanding of the fact that acceleration is the result of the net force.

I realize that this might seem a bit harsh, but there's really no point just endless rehashing it here. We can all pick up again in a month if @aka so wishes.
 

Mick West

Administrator
Staff member
How very unexpected.

Yes, it's unfortunate because it creates the misconception to some that that I'm cracking down on what he's saying because don't like the implications.

But really @aka is simply laboring under a very simple misconception, that there is an "upwards acceleration" keeping a building from falling down. Really there's an upwards force, matching the force due to gravity, which results in zero net force, and hence zero acceleration.

But I've already said the same thing several times in this thread, and he keeps insisting that something can "be accelerated" by gravity and by normal forces, etc, when its velocity is not changing.

Simply a semantic point? Sure, you could say there's a "imaginary acceleration", for the purposes of doing the math. The problem here is one of being able to communicate clearly. "Acceleration" means that there's a change in velocity. So if we are talking about the "acceleration" of a structure, then the casual reader who is familiar with physics will assume you mean that the structure is moving.

@aka can make any point he has to make here by discussing the sum of forces and then net acceleration, rather than a sum of "accelerations". The math works out the same. It's simply an issue of communications.

@aka also keeps mentioning how velocity and acceleration are vector quantities and so can be added to create a net vector. Now I do actually understand vector arithmetic in this context. In fact I had a job (video game programming)for 20 years, of which a significant percentage (video game physics) involved vector arithmetic with position, velocity, force, and acceleration vector. Sometimes I'd spend weeks doing essentially nothing but vector arithmetic. It's foundational to video game physics:

upload_2016-4-24_6-30-50.png

So I understand the point he is trying to make.

But adding together velocity (or acceleration) vectors only makes sense if the vectors are in different frames of reference. For example, you are on a train moving at 50mph, you walk backwards on the train at 3 mph, you can add the velocity (one dimensional here) and get a net velocity (relative to the ground) of 47 mph. You can do this because you are measuring the velocities in different frames of reference. One is relative to the ground, and the other relative to the train.

But in a building, not only is nothing moving, but if things start moving then the velocity and acceleration we are interested in are all in the same frame of reference (i.e. relative to the ground).

In a single frame of reference, an object can have multiple forces acting on it. But it only has one velocity, and one acceleration.
 

Mick West

Administrator
Staff member
...because velocities, and hence accelerations, and hence momenta and forces are vector quantities that add according to the parellelogram law.

And a very minor point for clarity: the parallelogram law is irrelevant here. You add vectors visually head to tail, or in practice just add their individual x,y,z components.

upload_2016-4-24_6-52-49.png

The parallelogram law of vector addition is really just an illustration of the commutative nature of vector addition (a+b = b+a). It's a trivial bit of geometry that's helpful in understanding vector addition, but it's not how you add vectors.

upload_2016-4-24_6-50-24.png
 
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