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On the curved track, the radial forces nearly always had a component in the - direction. In this device the outward forces are always straight outward. They have no component in the - direction.



It is free force, but not free energy (energy isn't force!). There's nothing wrong with free force, why should there be? Without it a lever couldn't work.


Based on what physical theory? You're assuming that the total magnitude of the force can't change. Show me a reference which supports that idea.



Noway!!!! Any hint of a valid force diagram will instantly destroy your concept. You can do that. Use a free body diagram.


There's my challenge to you. Draw a free-body diagram of it. Be sure to look up Wikipedia to find out the rules for free-body diagrams.

I had to use a force field to redirect the force in the (-)
direction and when I did look who showed up !!!
it was pac man all along.

but hes just imaginary isnt he?

it never is a mechanical advantage there is no
pivot or fulcrum.

it gives more force to the springs as it is flattened down but never more than the 4000N (-) force divided by 4 = 1000N each spring.

it starts off at 45 degrees , and the (-) force is greatest there , then at 0 degrees the (-) force is weakest.

and at 45 degrees the lateral force is weakest and strongest at 0 degrees.

but you cant treat the forces the same as you did with your excell calculations.
because the extra force you got in the excell calculations
was due to the the pipe supplying a force to turn the mass around 180 degrees.

the springs do not turn the mass.

the mass only presents a 4000N force to the center.

not 8000N.

You didn't really look up free-body diagrams did you?

Where are the forces applied to the ramp? Where are the reaction forces?



Here (above) you've shown a single force on the right-hand ball. Under that force it'll accelerate towards the bottom-left. But the ramp's in the way. Where does it go without any other forces to redirect it?

You can look it up in the dictionary



I agree with that part at least.





If it's stopping in 1s then it's an average of 4000N.
If it's changing direction in 1s then it's an average of 8000N.
That's applied to both the 'center' and the pipe.
You keep changing the design so I forget if it's still reversing or maybe it's stopping now?

I cant seem to find any use for a free body diagram
its more like filler for physics books , and lecture times.

were you going to use one to show that there really are no
forces involved?

just say 1-1=0 and 1-1=0 then your done.

all they do is show the exact reverse of a force.

you apply a force
at -0
and the same force is applied back
at +0

whats the point?




no its not , use the force kallog.

F = ma

unless f=ma is wrong , are you saying that?


you must be trying to average something out again , trying to get an angle on it.

I suppose that since the 100kg mass was accelerated for
8 seconds then its average would be 500N right?

so I suppose you could say that the average could be only
2000N if you had an extra second , but you only have 1 to work with.

but if you cut the second in half you could get 2 1/2 seconds , this way you could get an average of 2000N
over the 2 1/2 seconds.

nope that wouldnt even be right you would still have
4000N because a Newton is measured in seconds not 1/2 seconds.

I guess I should elaborate on that 1N = 1kg meter/s^2
to avoid the comments from the peanut gallery.

a force that will accelerate a mass of 1 kg
at a rate of 1 meter per second per second.

if you cut the second in half it is only half the newtons.



it changes direction in 2 seconds not 1 second.
so its not 8000N its 4000N

I will re-post that part.

You answered your own question. So are you going to do it or are you afraid it'll reveal what you're trying to sweep under the rug?



Instead of supposing, why don't you actually work things out?

I did you just forgot.

I was refering to you wanting to average out the 4000N
to fit into another scheme.


the free body diagram is crap.

it doesnt calculate anything.

you cant determine if anything can or cannot be accelerated
with it , so whats the point.

and your sudgestion that I use a nothing to prove something
just falls in line with what you are trying to accomplish.

heres a free body diagram , can you use it to
calculate anything?



according to the above free body diagram the block should fly upwards off of the incline , because the brunt of the
forces are in that direction.

and that is exactly why I use the word crap when describing such crap.

http://en.wikipedia.org/wiki/Free_body_diagram
All external contacts and constraints are left out and replaced with force arrows as described above.

Forces which the free body applies to other objects are not included. For example, if a ball rests on a table, the
ball applies a force to the table, and the table applies
an equal and opposite force to the ball. The FBD of
the ball only includes the force that the table causes on the ball.


therefore a FBD is useless when trying to
calculate forces in nature as the ball would
apply a force to the table , like the block
above would provide a force downward opposite of F3,
and the block would provide a force downward opposite
of force F2 and the incline would provide a force
upward opposite force F1 , your sudgestion of using
a FBD is senceless and only shows your intent to
cloud the subject matter with crap.

of course using crap like that would have worked
out just fine you wouldnt it.

Yes, if you make up values for the forces. Work it out correctly and it'll tell you what actually happens. If F3 is zero, the diagram tells you it acceleartes down the slope, and how fast. If you pick another value F3 it tells you which direction it'll go and with what acceleration. If you want to hold it stationary the diagram tells you what value of F3 to use for that too. F1 and F2 are determined by the block's weight.



It's useless when you're trying to fool people. You can't easily hide force components with a free-body diagram. More relevent here, you can't easily hide them _from yourself_. If you do the diagram you'll discover your device is very different from what you've described.

I know it's a simple idea. That's why it's so powerful. It's often easy to misjudge how forces work, to mix up reactions with actions, to reverse directions, to omit force components, to guess wrong values for magnitudes, etc. But it's hard to make a mistake with a free body diagram.

By the way, it's not always trivially easy to do. The block picture above requires some trig to calculate F2. You do actually have to do maths sometimes!

You've found a mistake in Wikipedia, the part you underlined is wrong. The rest of the paragraph is correct.

"The FBD of the (anything) only includes forces applied to the (thing)"

Kind of make sense really. The ball doesn't care what's happening to some bat on the other side of the world. It's only influenced by forces applied to itself.

I dont think so , I have watched several lectures on the
free body diagrams and I never considered this type of diagram valuable as all forces are not included.

in the diagram I posted with the block and incline the incline
is showing as a force , but not the force that the block would
exert in the opposite direction.

I think its best to include all forces in our
discussion as you might try to slip something around
a corner to lean the discussion in your favor.

That diagram would be used to analyse the behaviour of the block, not the incline. So it includes all the forces on the block, and none of the forces on the incline.

If you wanted to analyze the incline, you'd do a separate FBD for that.

If you wanted to analyze the complete system, you'd still have 2 seperate FBDs, but you'd find that some forces have the same magnitude in both diagrams, so you can connect them together with equations to describe the complete system thoroughly.

Come on, do FBDs for the components of your mechanism. Do each moving part with a seperate diagram. The diagrams you've shown are incomplete and ambiguous. FBDs are crystal clear because they're done with such simple rules.

this thing is really pretty easy to figure out
the 4000 N force is applied to the center.

it is then divided by 4 because there are 4 arms extending
downwards from the center.

so each arm gets 1000N

at the end of each arm there is force at a 45 degree angle
this force continues straight but cannot apply itself in that
direction but the force can apply itself in a 45 degree angle
to the spring.

so the 4 1000N forces must pass through 2 45 degree angles before
they reach the springs.

since the original force was at 180 degrees the 180 degree force
is directed to the 90 degree angle , so the 4000N force is all
directed outward.

so the 1st 45 degree angle results in a force of 1000N at 45 degrees and
the 2nd 45 degree angle results in a force of 1000N at a 90 degree angle

these 4 1000N forces are lateral or at 90 degrees from the center.

the pipe wouldnt even feel any (-) force as the mass applies the 4000N force.

it seems to easy but this is the way forces work.

theres no need to draw a bunch of diagrams.

its a done deal , even it you do try and complicate it
its just too simple a machine.

its hard to complicate simplicity.

What physical theory are you using to find that?

Electric currents behave like that. Put 4000A into a conductor, split it 4 ways, and you'll find the total current in those 4 conductors is also 4000A. If it's symmetric (as yours is), then there'll be 1000A in each.

There's no such law for magnitudes of forces.



Give me a reference.


A FBD would show what actually happens. I know you're afraid to draw one because you'll find you have to include forces that you pretended didn't exist.

that doesnt need a theory its basic math.

4000N/4=1000N

no theory needed.

Hey why don't we all put one of these on the front of our car? Then when we crash into a tree, we don't get hurt because hardly any force is transmitted through the device to the cabin.

Skydivers wouldn't need parachutes. Just tie these to their feet and when they land, they only get a fraction of the force from the ground.

heres a pretty good one , if someone sits on it
there weight is divided equally between the 4 legs.

you could put 4 weight scales under the 4 table legs
and each weight scale would read 1/4 a persons weight plus
1/4 the weight of the table.

but put the 4 scales and table in zero g
and someone sitting on the table will not
register any weight on the 4 scales

same same

This equation works for static forces, right?

Put in on a scales and measure the weight of something sitting on the center.

Here all the forces are in the same direction.

It does indeed work when they're all in the same direction.

In fact you can see the legs are sloping a bit. Would you say the total vertical force on the floor is _less_ than the weight of somebody sitting on it? Because some of the force must be transmitted outward, right?

those arent bad ideas , only you would need them to be very large but they would work on the same principle.

you would need a much larger one for a car because we are
only talking about a 100kg mass impact and
thats only 220 lbs.