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When two objects Attract/Repel(Charges/magnets/electromagnets/etc )

  1. Apr 16, 2013 #1
    Hi,


    In any case where there is a force of attraction/repulsion between two objects, those forces are due to those TWO objects?
    Two charges/two magnets/two electromagnets

    When they attract/repel the force is due to the both acting on the other?


    Much obliged
    Phz.
     
  2. jcsd
  3. Apr 17, 2013 #2

    Simon Bridge

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    The force between two objects is mutual, yes.
     
  4. Apr 17, 2013 #3
    The force is mutual, the force is due from both of them. The net force is dependent on both objects not one only.
    Attraction/Repulsion is to me the net force of both forces between the objects. To calculate them, I need to know the force of each first! Then figure out the net force(attraction/repulsion).
     
  5. Apr 17, 2013 #4

    Simon Bridge

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    By "mutual" I mean that if object A is attracted to object B, then object B is also attracted to object A.
    You are attracted to the Earth by gravity - and the earth is attracted to you too. The Moon does not, properly, orbit the Earth - the Earth and Moon orbit each other about their common center of mass.
    That better?

    Two masses m and M are attracted to each other - show me how you would go about calculating the force experienced by m due to M?
     
  6. Apr 18, 2013 #5
    Much better.

    I have no idea.
    I understand the concept that m Is attracted M, and M is attracted to m is based of Newton's law(Not sure which one), but is the force equally the same though? Always? Aren't there any factors that determine that force?

    For example, lets say X is attracted to Y.
    X can attract with 5N
    and Y can attract with 50N.

    Is X attracted to Y with 5N, and Y is attracted to X with 50N?
    I'm positively sure 100% that's wrong but are there any examples of the above?

    Because I still think that the attraction/repulsion forces are dependent on each object's applied force on the other object. I treat both of those forces like the net force between those two object I guess.

    Due tell if I'm mistaken(Which I'm sure I am, the idea of force confuses me till this day).
    Phz.
     
  7. Apr 18, 2013 #6

    Nugatory

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    Newton's third law is the answer that you're looking for. Google for it, or check the wikipedia article on "Newton's Laws".
     
  8. Apr 18, 2013 #7

    Simon Bridge

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    Taking gravity as an example:
    You have this idea that a small mass has less gravity, so it should have a weaker attraction.
    So in the example, if X has the small mass, and Y the bigger mass, then you'd think that Y attracts X more strongly than X attracts Y.

    This intuition comes from seeing that the bigger mass clearly moves less.
    If I drop a spanner (small mass), I experience the spanner falling to the Earth (big mass), not the Earth rising to meet the spanner... so it is tempting to think that the Earth attracts the spanner more than the spanner attracts the Earth.

    However - the fact is that if X is attracted to Y, then Y is also attracted to X with equal strength.
    X and Y experience the same strength force, in opposite directions. As Nugatory says, it is law #3.

    Since F=ma, and the forces are the same, the smaller mass will experience a bigger acceleration, and, so, will have the biggest movement ... giving rise to the intuitive feeling.

    There is still a valid idea that big masses have more gravity somehow ... this is taken up by the concept of the gravitational field. To find the force on Y due to X, you take the gravitational field of X, and multiply it by the mass of Y. To find the force on X due to Y, you take the gravitational field of Y and multiply by the mass of X.

    So, if X has mass m and Y has mass M, M > m, the gravitational field due to X is ##g_X=Gm/r^2##, and the gravitational field due to Y is ##g_Y = GM/r^2## (notice that ##g_Y > g_X##)... and you can work out the forces yourself to see how they end up the same even though the strength of gravity differs.
     
  9. Apr 20, 2013 #8
    I have a question though.
    When two magnet's attract. Is the net force cancelled when they are fully attracted? If so, how is it cancelled and they are still attracted? I only notice the motion cancelled.
     
  10. Apr 20, 2013 #9

    Nugatory

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    You mean when they're touching so cannot move any closer? The magnetic force between them is still at work: the first magnet is pulling on the second, and the second magnet is pulling on the first with an equal and opposite force, just as when the two magnets were moving closer.

    However, there is another force at work when the magnets are touching: the surface of each magnet is resisting being compressed by the other magnet being pulled into it. This force exactly cancels the force that's pulling on the first magnet, so the first magnet no longer moves.
     
  11. Apr 20, 2013 #10

    Simon Bridge

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    Thanks Nugatory.
    @Phztastic: he is exactly correct - the magnets still have magnetic attraction for each other when they are touching. The magnets stop moving when they are touching because they are touching ... solid objects do not normally pass through each other.

    Similarly, when you hit the ground you stop falling... but you still have weight right? Gravity is still acting on you. If you hit water, you'd continue falling through the water at a different rate - not because you have less weight inside the water but because the water buoys you up.

    When you are stationary, or moving at a constant speed, we say that the forces acting on you are in balance so there is no net force.
     
  12. Apr 21, 2013 #11
    Finally I get the idea!!
    I was struggling with it for a while!
     
  13. Apr 21, 2013 #12
    Electromagnet(1) and electromagnet(2) are equally the same in everything but their pole strengths.
    Because(1) has more power, it would generated a stronger force than (2).
    To make it easier, (1) has a force of 10N while (2) has a force of 1 N.
    What is he force between them when attracted/repelled? How does Newton's third apply here?
     
  14. Apr 21, 2013 #13

    Simon Bridge

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    That is not a correct description of the electromagnets.
    Please read post #7 again.

    There is no way to assign a particular force to a particular magnet in the way you describe.
    See: http://en.wikipedia.org/wiki/Electromagnet#Force_between_electromagnets for how to compute the force between two electromagnets.

    The upshot is: if (2) experiences a 10N force from (1), then (1) also experiences a 10N force from (2).
     
    Last edited: Apr 21, 2013
  15. Apr 21, 2013 #14
    I know this might be stated wrongly, but I wanted to bring an example using some numbers(e.g 10N/1N) and its related to magnetsim so I can understand.

    But you're right.

    (2) would experience 10N and (1) will to.
    But in this case where the forces are not equally the same initially (1) has a stronger force than (2), shouldn't I add the forces? So that it would be 11N for both of them? What happens to the force that (2) applies that is 1N?
    Or I should consider the greater force only that is from (2) and know that it's equally applied to both of them?

    I thought that I should add/multiply the force from both of them then use the total as te "equal" force between them.

    What happens to that (2)'s 1N of force?
     
    Last edited: Apr 21, 2013
  16. Apr 21, 2013 #15

    Simon Bridge

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    That will never happen. The forces will always be the same. There is no "initial" inequality.

    You are confusing the strength of the magnet with the force of the magnet.

    The force between two magnets of different strength is proportional to the product of their strengths and inversely to the square of their separation (its a bit more complicated because magnets have two poles and electromagnets have a magnetic field inside them - see that link I gave you above).

    So the force between a magnet of strength 1 unit and another of strength 10units will be 10x the force between two 1unit magnets with the same separation.
     
    Last edited: Apr 21, 2013
  17. Apr 22, 2013 #16

    I assume your using F = m(1)m(2)/r^2 (I think I'm missing a constant)?


    My misconception about forces in magnetism is, if you had two magnets/electromagnets with each one rated different forces. I should add them up... I think I was wrong here.

    Like the example above, I thought I should add 1N + 10N = 11N of force that is equally applied on both of um.
    What I ment with "Initial inequality" is just that you would have different strengths what should I due with the values that are rated in N/Kg/Pounds etc... Should I multiply them? ex 1NX10N or should I add them up?
    Or just take the largest force in this case that's 10N and assume thats the force.

    I didn't really know how to calculate this properly...
     
  18. Apr 22, 2013 #17

    Simon Bridge

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    Yes, there should be a constant there.
    Where have you seen magnets with rated forces?

    A rating in Newtons for a magnet is useless without knowing how the force was measured and under what circumstances.
     
  19. Apr 22, 2013 #18
    You're right.
    Because I looked at most sites that sell magnet's they have a rated "pull force" or "lift force".
    That force is not a constant, it changes are their are circumstances that are different that what they the manufacture have tested.

    But I'd have to figure out the force(N) of each magnet/electromagnet, how do I deal with the forces in attract/repulsion? If I have a set a values for example:

    Attraction:
    Magnet = 10N
    Electromagnet = 100N

    Repulsion
    Same as above.
     
  20. Apr 22, 2013 #19

    Simon Bridge

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    You keep telling me I'm right and you've got what Im saying and then proceed to totally ignore what I've told you!

    You will never have magnets which have a meaningful rating in Newtons or any other unit of force.
    Your question has no meaning.

    Magnet strength is rated in units of Gauss at the poles - or by some method related to the Gauss rating by a formula.

    Please provide a link to a site which gives a rated force in Newtons for it's magnets.
     
  21. Apr 22, 2013 #20

    russ_watters

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