View Full Version : I'm questioning the validity of gravity and Coulomb's law?
"The ratio of the lifting weight and the weight lifted is inverse the ratio of their distances to the center" - Aristotle's followers.
"Magnitudes are in equilibrium on distances reciprocally proportional to their weights" - Archimedes.
See any exceptions in these statements?
Where is the rigid bar in these two statements?
www.geocities.com/dedanoe
The law of gravity according to Newton is:
F=G\frac{M_1M_2}{R^2}
The Coulomb's law is formally same:
F=-k\frac{Q_1Q_2}{R^2}
The gravity law that respects the law of lever must look like this:
\frac{F_1}{D_2}=\frac{F_2}{D_1}=G\sqrt{\frac{M_1M_ 2}{D_1D_2}}
The Coulomb's law that respects the law of lever must look like this:
\frac{F_1}{D_2}=\frac{F_2}{D_1}=k\sqrt{\frac{Q_1Q_ 2}{D_1D_2}}
Until we resolve this we cannot recall these laws in their previous forms!
www.geocities.com/dedanoe
by the way the law of lever has this form:
\frac{F_1}{F_2}=\frac{M_1}{M_2}=\frac{Q_1}{Q_2}=\f rac{D_2}{D_1}
Once I dated this girl I thought was a bit heavy. I tested my hypothesis on a teeter-totter. I measured the lever distance difference using a Stanley tape measure [which is NBS traceable]. By lever formula calculations, she was at least 3 standard deviations heavier than I. I tried to lift my own weight off the ground, and could, I tried to lift her off the ground and could not. I therefore conclude the lever formula is correct [on the light side] within two standard deviations.
the Earth / Moon is not a "lever system" so i'm not sure where you're getting that idea.
"where is the rigid bar?"
the Earth / Moon is not a "lever system" so i'm not sure where you're getting that idea.
"where is the rigid bar?"
My bad, did not read link. Wrong formula. The lever formula only works in 2 dimensional space.
Is the law of lever something we should have learned in school ( I missed it if so ) or is it a contribution of your own? :smile:
Keep on chuggin !!
Vern
Until we resolve this we cannot recall these laws in their previous forms!
The way to resolve it is for you to realize that the "law of the lever" is very limited and only applies to real levers with forces perpendicular to the rigid bar. It has nothing to say about Newton's law of gravity or Coulomb's law.
more importantly if the lever only applies force directly to "the bottom" of the object then inertia dictates that they spiral outwards away from each other.
2. it doesn't explain why they would be rotating around each other in the first place.
3. if you assume the two objects ARE connected rigid to a hypothetical "bar" then there is measurable centripetal force which would cause all the water on the planet to fly off it
i don't think it holds together, but maybe you have a few arguments in context that i have overlooked...
the Earth / Moon is not a "lever system" so i'm not sure where you're getting that idea.
"where is the rigid bar?"
The law of lever needs no rigid bar.
Where in the Archimedes' statement do you see rigid bar?
The force always points in direction of displacement!
Assume the earth has:
F:=(10,0,0);D=(0,1,0)
The moon would have:
F:=(-5,0,0);D=(0,-2,0)
Now, for each apply:
NewF=Fcos(alpha)-aDsin(alpha)
NewD=Fsin(alpha)+aDcos(alpha)
It's only another form of the law of lever where NewF*NewD=F*D is equivalent with NewF= aD and F= a*NewD
www.geocities.com/dedaNoe
The way to resolve it is for you to realize that the "law of the lever" is very limited and only applies to real levers with forces perpendicular to the rigid bar. It has nothing to say about Newton's law of gravity or Coulomb's law.
Archimedes says:
"Magnitudes are in equilibrium on distances reciprocally proportional to their weights".
Where in this law do you see a rigid bar or necessity of it?
Vern:
The law of lever is not mentioned in school because if so Newton would've fall down earlier.
This nonsense from your own website says it all:
"The energy and the lever
According to the widely accepted physics, the vector product of the force and the distance of one weight is rotational momentum. I disagree with that because I am sure that this kind of product is already defined in physics as energy. This way Newton times meters give nothing but Jules - counter parts for energy."
This nonsense from your own website says it all:
"The energy and the lever
According to the widely accepted physics, the vector product of the force and the distance of one weight is rotational momentum. I disagree with that because I am sure that this kind of product is already defined in physics as energy. This way Newton times meters give nothing but Jules - counter parts for energy."
Yes it says extreme force - extreme energy; extreme energy potential - extreme distance but...
Whatever that part of my page says it's not in the context of this topic, right?
Newton's gravity and Coulomb's law don't respect the law of lever!
Prove that wrong or right if you want to contribute in this discussion?
Magnitudes are in equilibrium on distances reciprocally proportional to their weights - The law of lever according to Archimedes.
Where in this statement do you see any rigid bar?
Archimedes says:
"Magnitudes are in equilibrium on distances reciprocally proportional to their weights".
Where in this law do you see a rigid bar or necessity of it?
Perhaps Archimedes just assumed that by calling this "the law of the lever" that you would realize that it applied to levers.
You do know what a lever is, don't you?
Magnitudes are in equilibrium on distances reciprocally proportional to their weights - The law of lever according to Archimedes.
Where in this statement do you see any rigid bar?
Do you know what a lever is?
dedaNoe
Aug10-04, 04:02 AM
All is nothing but highly complex lever in equilibrium.
dedaNoe
Aug10-04, 04:09 AM
I do know what lever is and my idea of "lever" happens to be the widest and most general one. I think that all is nothing but highly complex lever in equilibrium. Even the earth and the moon make a lever although there is no rigid bar there to support them. The center of that invisible lever is the center of mass in the system. With respect to the center every one in the system has its own central distance. Their forces are same by direction as their displacements. The law of lever covers the interaction between the earth and the moon.
Nonsense. No rigid bar (or equivalent) = no lever. It's as simple as that. The "law of the lever" applies to real levers only. Real ones, not invisible, imaginary ones.
JoeWade
Aug10-04, 10:51 AM
Magnitudes are in equilibrium on distances reciprocally proportional to their weights - The law of lever according to Archimedes
this is law of the LEVER, not law of things in the universe not connected by a lever...
the statement assumes the reader knows what a lever is, that is, two downwards forces on a rigid beam supported by a fulcrum.
so the question remains. where is the rigid beam in the Earth Moon system?
Tom Mattson
Aug10-04, 11:02 AM
One silly thread on levers is enough. I'm merging these two.
TOKAMAK
Aug10-04, 12:31 PM
Since the law of the lever can be derived from Newtonian Classical physics, how can "The Law of the Lever" be more fundamental?
dedaNoe
Aug10-04, 03:09 PM
so the question remains. where is the rigid beam in the Earth Moon system?
The earth has position R_e;
The earth has force F_e in direction of its displacement;
The moon has position R_m;
The moon has force F_m in direction of its displacement;
We assume that F_e and F_m are parallel and opposite;
The center CF can be found like this:
-Switch the forces;
-Draw the line thru the top of F_e and F_m;
-Find the intersection for that line and the line thru R_m and R_e;
-The intersection is the center CF;
The earth has distance D_e = R_e - CF from the center CF;
The moon has distance D_m = R_m - CF from the center CF;
The law of lever is F_e * D_e = F_m * D_m;
You just need to open up your eyes to new things!
Every two different forces make a lever!
dedaNoe
Aug10-04, 03:14 PM
Since the law of the lever can be derived from Newtonian Classical physics, how can "The Law of the Lever" be more fundamental?
No member. All Newton's laws are only special cases of the law of lever.
Take Newton III for instance:
Newton III says: F_a = -F_r
The law of lever says: F_a * D_a = F_r * D_r
So Newton III is special case of the law of lever when D_a = -D_r.
Weights don't compare only on same distances from the fixed point.
In Macedonia we call this instrument (widely used on markets) as KANTAR.
russ_watters
Aug10-04, 03:32 PM
No member. All Newton's laws are only special cases of the law of lever. In that case, would you be so kind as to derive one of Newton's laws from 'the law of the lever'? How 'bout f=ma?
dedaNoe
Aug10-04, 03:40 PM
In that case, would you be so kind as to derive one of Newton's laws from 'the law of the lever'? How 'bout f=ma?
No problem.
As I said earlier the law of lever has this form:
\frac{F_1}{F_2}=\frac{D_2}{D_1}=\frac{M_1}{M_2}
So F_1 = a * M_1 and F_2 = a * M_2.
Done.
Newton only don't pay attention that this 'a' has to be same for the weight and anti-weight and that it's no acceleration at all. You see, although the weight and anti-weight have same 'a' the weight if heavier will displace less.
TOKAMAK
Aug10-04, 10:36 PM
http://en.wikipedia.org/wiki/Torque
Well that explains torque anyway, which is really the basis for the operation of most simple machines, including the lever.
dedaNoe
Aug14-04, 08:25 AM
Torque has nothing to do with it. Archimedes' law of lever governs magnitudes only.
F and D can also be colinear and then torq1 = torq2 is trivial case of Ort = Ort yet magnitudes are still related as the law of lever suggests.
Also distance times force equals energy potential times energy: D * F = P * E. Otherwise the harmonic oscillations would not represent closed physical system.
dedaNoe
Aug15-04, 05:57 AM
I want you to try this dynamical law:
F' = F cos(k) - aD sin(k)
aD' = F sin(k) + aD cos(k)
It is another form of the law of lever, where F and D are the old force and distance, F' and D' are the new force and distance and a is force / distance ratio.
http://www.geocities.com/dedanoe/_files/fizika18.gif
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.