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Newton's gravity and Coulomb's law are not quite accurate->

  1. Jan 17, 2004 #1
    If you combine Archimedes' laws of lever with Newton's gravity and Coulomb's law you'll see that the last are not quite accurate.

    is the Archimedes' law of lever.
    The "wrong laws" will say that each mass or charge is subjected to absolutely same force but it's not always how it is:

  2. jcsd
  3. Jan 17, 2004 #2
    Let me give you an example of how this works.

    Example: Consider two weights on a bar which is supported by a pivot in a uniform gravitational field which has a gravitational acceleration of g. Let the weights be at opposite ends of the bar. Let the weight on the left end be #1 and have mass M1 and the one on the right end be #2 and have mass M2. Assume the masses are, in general, different and the distance each is from the pivot are, in general, different. Assume the system is in static equalibrium. Then the total torque on the system will equal zero.

    Let the force on end #1 be F1 and the force on end #2 be F2. Let the magnitude of the distance between pivot and end #1 be D1 and the magnitude of the distance between pivot and end #2 be D1. Use the x,y,z coordinate system using the unit vectors i,j,k in those directions. The g-field is parallel to the z-axis and points in the -z direction.

    Calculate the torque using the pivot point as the origin of the cordinate system

    [tex]\overrightarrow {T} = \overrightarrow {T}_{1} + \overrightarrow {T}_{2} = \overrightarrow {F}_{1}\times \overrightarrow {r}_{1} + \overrightarrow {F}_{2}\times \overrightarrow {r}_{2}[/tex]

    This reduces to

    [tex]\overrightarrow {T} = [F_{1}D_{1} - F_{2}D_{2}]\hat{k} = 0[/tex]

    [tex]F_{1}D_{1} = F_{2}D_{2}[/tex]

    Since F1 = M1g and = F2 = M2g then

    [tex](M_{1}g)D_{1} = (M_{2}g)D_{2}[/tex]


    [tex]M_{1}D_{1} = M_{2}D_{2}[/tex]

    So that is how and why and when that relationship holds true.

    In your equations above you incorrectly applied the gravitational force law since you seem to think that one of the either M1 or M2 is the mass of the Earth where in fact neither is. We can go back before I put the masses in and let

    [tex]g = \frac{GM}{R^{2}}[/tex]

    where R = radius of Earth and M = mass of Earth and use this as an approximation to the uniform g-field if you'd like. But the answer will be the same.

    One needs to be very careful when claiming that something is wrong. One should first ask oneself Am I missing something that I can't see?. I do that when I come to a conclusion in a derivation. I e-mail a friend when I have high confidence in my result and I get feedback.
  4. Jan 17, 2004 #3
    Re: Re: Newton's gravity and Coulomb's law are not quite accurate->

    I went far more further than that:
    I 1st removed the influence of the earth.
    I 2nd removed the bar.
    I 3rd noticed that the equi. point remains the same.
    I 4th noticed that Archimedes' law applies the same.
    And then I applied the gravity law between the two masses (M1 and M2 (neither is the mass of the earth)).
    Finally, I came to my point and it holds.
  5. Jan 17, 2004 #4
    Re: Re: Re: Newton's gravity and Coulomb's law are not quite accurate->

    It does hold. You're using incorreclty. The law of which you speak refers only to the relationship between the forces applied perpendicular to the line between the pivot point and the point of application of the force,l i.e. perpendicular to the lever arm. That is why a cross product is used. What you've tried to do is apply the parallel components of the force and then claim the same relationship holds true. That is probably how you came to such an erroneous conclusion, i.e. you started with an erroneous assumption.

    See the diagram in these notes so that you understand what the law means and the terms in that law

    In actuallity it is no longer a law in the true sense of the word since it can be derived from more basic principles, namely consevation of angular momentum. So its more like a theorem.

    The relations that you've provided are incorrect as they stand, i.e. the relationship


    is invalid. The ratio of forces is not equal to the ratio of masses.
  6. Jan 18, 2004 #5
    Over mine and Archimedes' dead bodies->

    In the time of Archimedes (287-211BC) there was no angular momentum.
    It comes from Newton's 2nd law some 18 centures later.
    Here is my little math manuver to support my claims.
    I'll use the conventional equations.

    pick origin and put x-axes on it.

    for i=1 to 2 put Mi in xi0.

    for i=1 to 2 put the equations of motion with no initial speeds

    eliminate time from both and get

    now put the equations for the work
    [tex]\Delta E_i=F_iD_i[/tex]

    now put the conservation of energy
    [tex]\Delta E_1=F_1D_1=-F_2D_2=-\Delta E_2[/tex]

    now put Newton the 3rd and get

    now get back to

    and implement D1=D2 to get
    Newton's laws are valid for trivial cases.
    Now I'd like you to provide prove for any of:
    -Newton's 2nd
    -Newton's 3rd
    -Newton's gravity law.
    They might be obviuosly right for you.
    But for me they are obviously wrong.
    Convince me.
  7. Jan 18, 2004 #6
    The proof lies in experiment and observation, not mathematical manipulation. And besides, since you think they're obviously wrong, what would be the point in illustrating such a proof?
  8. Jan 18, 2004 #7
    Re: Over mine and Archimedes' dead bodies->

    Nobody suggested there was. I simply explained a minor point on semantics as far as whether this is, today, considered a law or a theorem. Then it was a law. Today it is a theorem
    At that time (and the time that followed) there was angular momentum, torque etc. At which point these things took on a more precise form and it was proved derived from Newton's laws and it is thus a theorem. The term theorem is defined as follows
    You go on to explain
    That equation holds only when the acceleration is constant. You've chosen a problem in which the acceleration is not constant and therefore that expression is invalid.

    You continue ..
    These equations are invalid as well. You wrote

    [tex]\Delta E_i=F_iD_i[/tex]

    This is the definition of work. As such E in such an equatoin represents kinetic energy. Kinetic energy is not a conserved quantity in physics and the sum of the kinetic energy of the particles is a system is also not a conserved quantity either. Only the total energy of a particle or system is coserved. And then the total energy of a particle is conserved only when the force is conservative.

    You continue...
    This is invalid was well. As I've explained above, the forces in the Archimedes lever law refer to the component of force which is perpendicular to the line connecting the two points of application of the force. You're using the parallel components.

    The rest of your equations are wrong due to the errors I've described above. Did you look at the diagram that I suggsted??

    You write
    Those are Newton's laws of physics. As such they are not provable by definition. They are postulates, i.e. that which is assumed to be true. Even then Newton's Third Law fails in many cases such as action at a distance of two charges which are in relative motion.

    Newton's Second Law is not really considered a law in today. It's considered a definition of force.

    Please examine the diagram that I provided a link to so that you can obtain a better understanding of the meaning of the terms. Please note that the forces in that equation are not forces which are equal and opposite forces but they are forces which are perpendicular to the lever arm. Otherwise you're going to keep making this mistake.
  9. Jan 18, 2004 #8
  10. Jan 18, 2004 #9
    Re: Re: Re: Over mine and Archimedes' dead bodies->

    How do you find then the equi.point between two masses in free space?
    Why there cannot be a lever even if it has no actual physical or visible appearence?
  11. Jan 18, 2004 #10
    Re: Re: Re: Re: Over mine and Archimedes' dead bodies->

    Because a lever has a fulcrum - a pivot point. That pivot point is applying a force. The forces at each end are external forces. The total force on everything is zero since otherwise something would be accelerating and the lever equation that you've posted is an equation of static equilibrium.
  12. Jan 18, 2004 #11
    my term equilibrium point is same as yours pivot point.
    when interacting two masses there is also one pivot point between them. it's the point in which if you put any 3rd mass the influences of the first two on the 3rd cancel. that point remains static despite the displacements of the other two. how do you find that point between two masses in free space?
  13. Jan 18, 2004 #12
    I don't know what you mean by "pivot point" as you've described things. It seems to me that you have two objects at the end of a stick and you're letting the mutual forces of gravity attract them. As such the objects don't move since the force of gravity is balanced by the force of the stick. But you can't use that relation you want to since it does not apply.

    For two free particles which are accelerating towards each other in a straight line see

    By the way - you are most definitely not using the term "pivot point" in the same way I am. You're speaking of something radically different.
  14. Jan 18, 2004 #13
    For instance it can be:
    a point where the forces cancel.
    a point where the energies cancel.
    a point where the gravity potentials cancel.
    I leave it up to you to decide.
    not on stick but in free space-in vacuum.
    I'm speaking as if even though in the case of interaction of two masses in vacuum there is no physical presence of bar-lever the lever law yet applies.

    You can understand me better with this example:
    Choose one equilibrium point.
    Pull one mass away from that point.
    You've lost some energy to do that.
    Pull another different mass in oposite direction of the 1st one so that you regain the same energy you lost while pulling the 1st mass.
    Ain't there a lever there?
  15. Jan 18, 2004 #14
    let's be concrete,OK?

    put the mass M1=6kg some where in empty space.
    on distance R=7m put the mass M2=15kg.
    mark the current positions as starting.
    let the initial speeds be zeros.
    let go the masses towards each other, driven only by their internal gravity pull.
    freez the action when the masses colide.
    make Di (for i=1 to 2) be the distance from the starting position of i-th mass to the colision point.
    finally, find D1 and D2.
    use any method that you think is the most accurate.

    here is my simple math:
    with concrete values the same equation is
    [tex]\frac{21 kg}{7 m}=\frac{6 kg}{D_2}=\frac{15 kg}{D_1}[/tex]
    the results are D1=5m and D2=2m.
  16. Jan 18, 2004 #15
    Center of Mass Vector

    Ah! I see where you're comming from now. Then Archimedes' lever relation does not apply. However this is now a well defined problem. Unfortunately I'm exhausted right now. I'll get to it tommorow.

    Off hand I'd say that the center of mass relation plays a crucial role here. The center-of-mass theorem states that the center of mass of a closed system does not change if there are no external forces acting on the system. What you've just described is the center of mass relation and not the Archimedes' lever relation. They may look the same but the meaning is different.

    In the case of that diagram that I showed you the center of mass is supported by the fulcrum and that is the reason it remains in equilibrium. The center of mass vector is defined here


    Note: That is a special relativity relation. However it holds in Newtonian mechanics as it, of course, must.

    Simply place the particles at two arbitrary locations on the x = axis and then use the formular in the link above to find the center of mass.

    More tommorow when I'm more awake
  17. Jan 19, 2004 #16

    i saw the page...
    and i'm waiting for your result.
  18. Jan 19, 2004 #17
    I've decided that I've been putting this off for some time so I made a web page on the center-of-mass theorem. See

    I think it would be worthwhile to consider the general case of two particles. The total momentum of two partilces is given by

    [tex]p = p_{1} + p_{2}[/tex]

    where pk is the momentum vector of particle k.
    One can always choose a frame of reference such that the total momentum is zero in that frame. Choose this frame to observe the motion in. Then

    [tex]p_{1} + p_{2} = 0[/tex]

    Consider only the situation for which the motion of each particle is along the x-axis. As such this expression may be written as

    [tex]\frac {d}{dt}(m_{1} x_{1}) + \frac {d}{dt}(m_{2} x_{2}) = 0[/tex]

    This can be simplified to

    [tex]\frac {d}{dt}(m_{1} x_{1} + m_{2} x_{2})= 0[/tex]


    [tex]m_{1}x_{1} + m_{2}x_{2} = constant[/tex]

    Since this is always constant then we can choose our coordinate system such that this constant is zero. I.e.

    [tex]m_{1}x_{1} + m_{2}x_{2} = 0[/tex]


    [tex]m_{1}x_{1} = -m_{2}x_{2}[/tex]

    If we take x1 as being located on the + side of the origin then x2 < 0. Let x2 = - d2 where d2 is the magnitude of x2 and is a positive quantity. Let the magnitude of x1 = d1. Then

    [tex]m_{1}d_{1} = m_{2}d_{2}[/tex]

    Notice that this assumes only that the total mechanical momentum is constant and nothing else. This will hold for any force you choose so long as the total mechanical momentum is conserved at all times. This will hold true in all cases where Newton's third law holds in the weak form.
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