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Random task error

  1. Apr 1, 2004 #1
    I have very sharp sense. I can smell fishy logic miles away. Ok. I’m dropping 1 ton from 1 km above the surface of the earth. I presume all along the way until the ton hits the ground the ton and the earth are subjected to equal and opposite forces. Ignore the resistance from the air. I choose to view the process from the impact point. Since the earth is much heavier than the ton it will accelerate less and therefore pass smaller distance than the ton in the same time between the start and the impact. The system is isolated so the energy must be preserved meaning the work done by both must cancel. The work is defined like this: [tex] W = \int_{x_1}^{x_2} Fdx [/tex] and since the ton is not scared of falling it is not wasting additional energy on screaming. So we have equal forces and different distances in the equation for the work done by the two.

    What is wrong here?
    Why the work of the two doesn’t cancel?
     
  2. jcsd
  3. Apr 1, 2004 #2

    Doc Al

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    Yes. From Newton's 3rd law the forces that Earth and object exert on each other are equal and opposite.
    The work done on each doesn't cancel, it adds up. Energy is conserved. You start with gravitational potential energy and end up with kinetic energy. Since the earth is massive, the object gets most (essentially all) of the kinetic energy.
     
  4. Apr 1, 2004 #3
    I start and end with one type of work i.e. the one done by their forces on the distances they pass. Are they doing some other work.

    Can you show me your math?
     
  5. Apr 1, 2004 #4

    Doc Al

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    The work done is just as you wrote it:
    [tex] W = \int_{x_1}^{x_2} Fdx [/tex]
    In this case the force is that of gravity: F = GMm/r2
    For what? What do you wish to calculate?
    The work done is equal to the change in gravitational potential energy as the object falls from position R1 to position R2:
    ΔKE = Work = GMm(1/R2 - 1/R1)

    Edit: Just to be clear, R is the distance from object to center of earth.
     
    Last edited: Apr 1, 2004
  6. Apr 1, 2004 #5
    So your saying that they function as independent isolated systems doing no work at all. Instead, the change of each potential energy cancels with the change of its own kinetic energy.
     
  7. Apr 1, 2004 #6
    I know better Doc Al

    Building my faith is based on dealing well with my worst doubts.
    You sure gave me doubts saying:
    [tex]\Delta (KE + PE) = 0 = W = \int_{x_1}^{x_2} Fdx <> 0[/tex]
    Now it’s my time to give you some doubts. Lets see how you’ll handle them!
    W is the UNIQUE work done by one force on particular distance. Each object is subjected to one force only and that force is moving it. There is no way that the two can represent separate isolated systems.
    It’s not as if:
    [tex] \Delta KE = \int_{x_1}^{x_2} Fdx <> 0[/tex]
    [tex] \Delta PE = Minus \int_{x_1}^{x_2} Fdx <> 0[/tex]
    So in the end:
    [tex] \Delta ME = \Delta (KE + PE) = 0[/tex]
    There is no sense in it cause the object traveling the same distance is actually subjected on two equal and opposite forces. In that case that object will not move at all. Instead it IS:
    [tex] \Delta ME = \Delta (KE + PE) = \int_{x_1}^{x_2} Fdx <> 0[/tex]
    This Delta ME must cancel with another Delta ME from the rest of the system.
     
  8. Apr 1, 2004 #7

    Doc Al

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    The two bodies (object and earth) exert forces on each other. So I would hardly call them independent isolated systems.
    It's better to think of the PE as a property of the entire two-body system due to the relative position of the parts.
     
  9. Apr 1, 2004 #8
    Then you agree that Delta ME = Delta (KE + PE) <> 0
    The two objects are subjected to force and therefore they both must displace. Changing the viewpoint will disturb the balance of the system. We must view the process from the collision point. It is the center of the lever they make. The center of oscillation for each. This center must remain immovable to keep the system in balance.
     
  10. Apr 1, 2004 #9

    Doc Al

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    That's not something I said. I would agree that mechanical energy is conserved:
    [tex]\Delta (KE + PE) = 0[/tex]
    I would also say that the change in PE can be calculated by finding the work done against the force of gravity (F):
    [tex]\Delta PE = -W = -\int_{x_1}^{x_2} Fdx[/tex]
    The equal and opposite forces act on different objects--they don't cancel!
    Nope. See my comments about.
     
  11. Apr 1, 2004 #10

    Doc Al

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    As I said in my first post, I agree that mechanical energy is conserved.
    Yes, both objects displace. Just enough so that the center of mass of the Earth-object system remains unchanged. Of course, the Earth's displacement is infinitesimal compared to that of the object, so it can be ignored.

    I have no idea what you mean by "Changing the viewpoint will disturb the balance of the system". Or what you think is so special about the collision point.

    And, of course, this has nothing to do with levers! :smile:
     
  12. Apr 2, 2004 #11
    You are now talking as if this Delta ME belongs to the entire system consisted of the two.
    In my post Delta ME was of one object alone.
    It is sum of KE & PE and W was the total work done by the force of one object on the distance it makes.
    Because it is nonzero I said it must cancel with an opposite W from the other object but that other object's W doesn't match the first and it cannot cancel it.

    Because both objects are subjected to their own force they both must displace. If you fix your viewpoint on one of them then you're making it's force zero. It is because in such your system that object will perform no motion. No motion means no force. The force of the other object will be the only force in the system and it could not have with what to possibly balance. That's is why our only viewpoint available is the collision point. Otherwise the forces in the system will be different or there will be no balance at all.
     
  13. Apr 2, 2004 #12

    Doc Al

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    If you wish to treat the earth and object separately, no problem. One error you are making is including both a PE term and a Work term. Since the only force is gravity, the work done against gravity is already included in the PE term. Use PE or Work, but not both at the same time.
    The reason it's nonzero is because you are including a bogus Work term!
    You are correct that by taking the earth as a reference frame one is ignoring the small displacement that the earth undergoes. So what?
    I'm not sure what you mean by "balance": momentum of the Earth-object system is conserved, since there are no external forces acting. And mechanical energy (KE + PE) is conserved. Certainly the forces acting between the Earth and object do not balance!
     
  14. Apr 3, 2004 #13
    I wanna check once more if I understand you correctly.
    You say:
    Delta PE = - Delta KE = W
    It is Delta ME = Delta KE + Delta PE
    So Delta ME = W_1 + W_2 = -W + W = 0.
    This mean that in that time there are summarily two works done for the ton:
    W_1 is work the earth does with F on dx.
    W_2 is the resistance of the ton with -F on the same dx.
    This also means that the ton is isolated cause its total energy ME doesn't change.
    If isolated then it doesn't interact with earth.
    Subjected to F and -F the ton is immovable.

    What will be my work for draggin' the ton acrros the space with the same force the earth does? You are probably confused because you break appart ME on PE & KE. The earth must do work proportional to its pull on the ton and the displacement it makes. This work must show up as difference in the total mechanical energy of the ton.

    You know the way the physics you defend describes is exactly how our intellects interact. You entirely oppose my intention to bring your intellect close to mine. That's why we are on the same damn initial distance after all those replies.
     
  15. Apr 3, 2004 #14
    About the lever I see between the ton and the earth:

    This lever has invisible rigid bar that you can witness only if you imagine it.
    The center of this lever is exactly the collision point.
    With respect to that point each has distance D_i:
    D_1 the dist of the earth
    D_2 the dist of the ton
    If the ton carries force F_2 then the earth must carry force F_1 such that
    F_1 * D_1 = F_2 * D_2 - the Archimedes's condition for lever in balance.
    So it cannot be F_1 = -F_2 cause then it has to be D_1 = -D_2. It's not how it is.
    The ton and the earth must oscillate around the center of the lever with equal phase difference.

    Please try to understand that. It's the greatest thing I've ever come up with.
    Don't worry that we might hurt Newton's feelings if we find his mechanics expired.
    He's dead thus has no feelings at all.
     
  16. Apr 3, 2004 #15

    Doc Al

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    What two works? There is only one force acting on the ton.
    Huh? What is this "resistance"? Stop making things up! :smile:
    What do you mean by "isolated"? To me, the ton is obviously not isolated: there is obviously something interacting with it exerting a force on it--the earth.
    Well, obviously wrong!
    So... are you seriously saying that the ton will not fall?
    It does!
    The work done equals the change in KE. This is the Work-Energy theorem. If you wish to treat gravity as a force doing work on the object, then you can say Work done = ΔKE. But, if you wish to include a gravitational PE term, then you must realize that that term already accounts for the work done by gravity. If you insist that W = ΔKE + ΔPE, then you will continue to get nonsense.
    You got me there! :eek:
    Since you continue making the same basic errors, we will continue getting nowhere until you learn a little physics.
     
  17. Apr 3, 2004 #16

    Doc Al

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    You do realize that with levers--real levers, which obey ordinary Newtonian physics, not imaginary levers--the forces are exerted perpendicular to the lever bar?

    Perhaps you are confusing this with the center of mass of the Earth-ton system? It is true that M_1*D_1 = - M_2*D_2, and that the center of mass does not change. By the way, this center of mass is not the collision point but is much closer to the center of the Earth.
    I'm sorry to hear that since it is based on a fundamental misunderstanding of how levers work.
     
    Last edited: Apr 3, 2004
  18. Apr 3, 2004 #17
    The total energy the ton has remains zero all the time. That's the condition necessary to say it's isolated. It gains KE on account on loosing PE. At least that's how you put it.
    That is the only conclusion that I can come up with, with the physics you defend
    You agree definitly that work is difference in energy.
    Delta ME, Delta KE and Delta PE are all works on their onw.
    The whole work done is Delta KE and let it be so.
    Delta KE is done by the earth's pull alone.
    If Delta ME = 0 => Delta PE = - Delta KE and then
    what kind of difference in energy i.e. work is Delta PE?
    Which force is doing Delta PE?
    What is the difference of i.e. the work done with respect to ME?
    Which force could do Delta ME = 0?
     
  19. Apr 3, 2004 #18
    What does Archimedes means by:
    "The magnitudes are in equilibrium on reciprocally proportional distances from the center"?
    What does Newton means by:
    "For every action there is an equal and opposite reaction"?
    If Newton's law covers Archimedes lever then:
    Why these two laws are NOT simultaneously valid for different distances?
     
  20. Apr 3, 2004 #19

    Doc Al

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    This is just an old-fashioned way of saying that for a class 1 lever (fulcrum in the middle) to be in equilibrium, the torques about the fulcrum must balance. It is a trivial consequence of applying Newton's 2nd law for rotational motion.
    This is Newton's 3rd law. Applied to your Earth-ton example, it means that the force that the Earth exerts on the ton is equal and opposite to the force that the ton exerts on the Earth. Note that these action/reaction forces act on different bodies and never in themselves produce equilibrium.
    I have no idea what you are talking about. Newton's laws are perfectly valid for "different distances". Learn them and you will be able to solve any kind of lever problem, if that interests you, and many other kinds of problems as well. On the other hand, Archimedes's statement about levers only applies to certain kinds of levers.
     
  21. Apr 4, 2004 #20
    Archimedes's law is in this form:
    [tex]\frac {F_1}{F_2} = \frac {D_2}{D_1} = \frac {M_1}{M_2} [/tex]
    Newton 3 is in this form:
    [tex]F_1 = -F_2[/tex]
    The two all valid at the same time only if:
    [tex]D_1 = - D_2[/tex]
    [tex]M_1 = - M_2[/tex]
    No other time. Newton's one is special case of Archimedes's one.
     
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