Speed of Force or 'Transfer of Momentum'

Discussion in 'Physics & Math' started by hansda, Feb 14, 2013.

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  1. hansda Valued Senior Member

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    I also said that, all Newtonian equations will remain same.
     
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  3. Tach Banned Banned

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    You say a lot of things, most of them nonsense.
     
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  5. hansda Valued Senior Member

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    Give example.

    Basically my "energy point of view" follows from "conservation of energy".
     
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  7. Tach Banned Banned

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    I already did, look no further than your post #215.
     
  8. Russ_Watters Not a Trump supporter... Valued Senior Member

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    Physics is mathematical. You need to prove that mathematically. As stated, it appears to contradict Newton's laws, but it is too imprecisely stated to be sure. So you need to show what the words mean in the language of math.

    If you use the standard mathematical/scientific definitions of words it would help, but I have no confidence that you can do that (ie, "space derivative of energy" is just gibberish). So math would be clearer.

    If you are trying to say that a continuous application of force requires a continuous expenditure of energy (power), that would be an explicit violation of Newton's laws.
     
  9. Russ_Watters Not a Trump supporter... Valued Senior Member

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    So what happens when you have a force but no distance...?
     
  10. Robittybob1 Banned Banned

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    I think when you look at it closely there is a little change in distance (microscopic deformation)
     
  11. eram Sciengineer Valued Senior Member

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    Not really. You just have force, no distance.
     
  12. Russ_Watters Not a Trump supporter... Valued Senior Member

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    Not continuously, there isn't.
     
  13. Robittybob1 Banned Banned

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    "Look at it closely there is a little change in distance (microscopic deformation)", as might happen when continually pushing against a compression spring, continuous force to keep it continuously compressed. Why not continuously?
     
  14. Russ_Watters Not a Trump supporter... Valued Senior Member

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    It can't be stationary and moving at the same time. What you've said is self-contradictory.

    This problem was explicitly defined to be one showing a continuous application of force with no power absorption/dissipation. If you compress and hold a spring, it is completely stationary after the initial compression, there is no further motion and therefore no further energy absorption/dissipation.

    An object in a circular orbit is also an example of a force being applied to an object without power being applied to the object. That's the answer to the first question I asked a few posts ago that Max declined to answer: zero.
     
  15. Robittybob1 Banned Banned

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    As I understand it the gravitational force is balanced by the centripetal forces, so while they balance no further energy changes occur. But the system may still be radiating gravitational waves so the power/energy velocities has to readjust back into an orbit that fits only to lose more energy and gradually the system goes into an orbital decay. So it is losing power all while being in a circular orbit.

    The spring will stay compressed provided the forces balance too. If it was a biological system energy will still have to be used to keep the spring under tension. You try and hold a compression spring compressed for a minute.
     
  16. eram Sciengineer Valued Senior Member

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    In this case we're only discussing classical mechanics.


    hansda has used Newton's cradle to come up with a messed up definition of action and reaction and has created an association-based argument involving force and energy.

    He then came up with a wave-reflection argument for energy, and combined this with his association-based argument to support his original messed up definition.
     
  17. Russ_Watters Not a Trump supporter... Valued Senior Member

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    Yes, in real life there are a bunch of different ways in which energy is dissipated for an object in orbit. But that has no bearing on the fact that the earth does not expend energy to maintain its gravitational field/ apply the force that keeps the object in orbit.

    Yes, a person pushing on a wall is an example of a zero efficiency machine: some input power, but no displacement so no output work. That fits Max's logic for the person but contradicts it for the wall.
     
  18. hansda Valued Senior Member

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    Which statement do you think is wrong in my post #215?

    Here "force" F can also be measured in terms of energy as, F= dE/dx; where dE is small eneregy change in the direction of small distance dx.
     
  19. Russ_Watters Not a Trump supporter... Valued Senior Member

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    Please clarify: are you claiming there is a one-time expenditure of energy or a continuous expenditure in the application of a force?
     
  20. hansda Valued Senior Member

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    In action-reaction as per Newton's Third Law, it is one-time expenditure of energy.
     
  21. hansda Valued Senior Member

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    Which statement you observed as contradicting with Newton's Laws?

    Here you can consider F=dE/dx; where F is force in the direction of dx(small distance) and dE is small energy change in this direction.

    How is this a violation? Can you get a force without energy?

    Here you will get "compression" with atleast "infinitesimal deformation".

    Correct.


    Wrong.


    This is a case of conservation of energy.

    Can you define a "Reaction Force" with an example? Also identify "Action Force" in this example.
     
  22. Tach Banned Banned

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    Err, there are many things wrong as already explained. You simply tried to cover up your errors by using the particular case when \(\vec{F}\) and \(d \vec{x}\) have the same direction and sense. In general, this is not the case. Besides, as explained, \(\vec{F}\) is a VECTOR, what you wrote down is a SCALAR.
     
  23. hansda Valued Senior Member

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    I said, this way also force can be measured as it also can be measured from momentum change or acceleration.

    I wrote this way, because i don't have much expertise on TEX.
     
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