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The center of Mass perfectly match the center of Force->

  1. Yes

    0 vote(s)
  2. No

    0 vote(s)
  3. Depends (show us on what)

    4 vote(s)
  1. Jan 23, 2004 #1
    In a system of several masses the center of force has same position with the center of mass in the geometrical space;

    Yes or No?

    Please explain why!!
  2. jcsd
  3. Jan 23, 2004 #2
    I think the answer is no.

    In my physics notes, it says, "If a body is placed in a uniform field of gravity then its centre of gracity will concide with the centre of mass. However, this may not be true for a very large object."

    Well, I don't know what is one of the examples of "a very large object".
  4. Jan 23, 2004 #3
    Does your notes contain equations to support that?
  5. Jan 23, 2004 #4
    I think some logic can be useful.
    Think an object which you can hold. Keep it steady and release it. You'll see that it does't turn around(neglecting air friction).
    But think a very large spacecraft(its length is two times the diameter of the earth) falling(!) to the Earth. A much greater force is exerted to the side which is nearer the earth, so it starts to turn. Obviously CM doesn't match FM.
    Am I wrong?

    But did you mean where there is no gravitational force between the masses by "center of force"?
  6. Jan 23, 2004 #5
    I think you are right, kishtik. The question is just too vague. Are we talking about a situation in which there is a uniform gravitational field and there are no other forces acting on an object?
  7. Jan 23, 2004 #6
    Please defined center of force
  8. Jan 24, 2004 #7
    Actually, that was something that I wanned to request from some of you. Now that I'm being ask for it I'll say:

    We have a system of several masses in empty space.
    There is force field in every point of space.
    The force center will then be exactly that point which contains the sum of all forces each mass is subjected to.
    But don't count one force twice as Newton's III sugests while summing the forces.This requires that you relate geometry with force.
    My only answer will then be based on summing energies.

    [tex]\sum E_i = E_{cf}[/tex]

    Since for me [tex]E_i=F_iD_i[/tex] where D is the distance and F is force in D from CF then

    [tex]\sum F_iD_i=F_{cf}D_{cf}[/tex]


    [tex]\sum F_i=F_{cf}[/tex] and [tex]\sum D_i=D_{cf}[/tex]
    => [tex]\sum_{i<>j} F_iD_j=0[/tex]
    is the condition for balance in the system.

    Now, be soft with your critics. I'm breaking the ice here.
  9. Jan 25, 2004 #8
    I'm not a native speaker of English. I only tried to say that I did't know deeply what KLscilevothma's notes were about and I would do some guessing and mental experiments etc. I am shocked as I realised what "Some logic can be useful." meant. They do not teach things like this at school or course. I'm sorry.
  10. Jan 25, 2004 #9
    Okay, here goes a more "mathematical-style" proof of a physics problem I can't honestly claim I understand.

    Suppose center of force = center of mass. Then take a system of such objects where this is true, and for which there is a single (nonzero) center of force.

    Add an object which has no force applied to it that is not centered on the center of mass of the system. Then the collective new system has a different center of mass, but the same center of force. So the center of force and the center of mass are not the same.

    Quel Est Des-probleme?
  11. Jan 25, 2004 #10


    User Avatar

    Staff: Mentor

    Not enough information given: you can apply a force whereever you want it.

    If you're talking about gravity though, the answer is no.
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