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Moment of force

  1. May 19, 2009 #1
    Moment of force= F*d*sin(alpha).
    Now, Must d be the perpedicular distance from the force to the axis , or any distance?
    Thanks in advance,
    Aladdin
     
  2. jcsd
  3. May 19, 2009 #2

    ZapperZ

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    You have given very little information here. For example, what is "alpha"? Is this the angle between F and d?

    I think there's quite a bit of confusion here. You may want to look at this and see if you've misunderstood something important.

    http://hyperphysics.phy-astr.gsu.edu/hbase/woang.html#waa

    Please consider formulating as clear and complete of a question next time.

    Zz.
     
  4. May 19, 2009 #3
    No, Moment of force with respect to an axis,
    Alpha is the angle betwen the force and d.
     
  5. May 19, 2009 #4

    ZapperZ

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  6. May 19, 2009 #5
    You mean that it doesn't matter if d is the perpendicular distance, because you included the angle?
     
  7. May 19, 2009 #6

    ZapperZ

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    If they both have to be perpendicular, then the angle will always be 90 degrees, and it is a constant equal to one! So then why even bother writing "sin(alpha)"?

    Zz.
     
  8. May 19, 2009 #7
    Yeah I know, but thats not my point, my point is if d is any distance or it must me the perpendicular distance? You got me Mr.
     
  9. May 19, 2009 #8

    ZapperZ

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    Did you look at the figure in the link that I showed? I thought that is self-explanatory?

    Zz.
     
  10. May 19, 2009 #9
    So , it must be perpendicular.
     
  11. May 19, 2009 #10

    cjl

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    Actually, it doesn't need to be perpendicular.

    To be slightly more accurate, your equation already takes that into account. The moment is the force multiplied by the perpendicular distance. In your equation, d*sin(alpha) is the perpendicular distance.
     
  12. May 19, 2009 #11

    ZapperZ

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    I'm going to correct this and say no, it doesn't, which is what I said already. But obviously, it is not getting through to you, but I'm going to make sure others reading this do NOT get the same wrong information.

    I have no idea why you are fixated with this "perpendicular".

    Zz.
     
  13. May 19, 2009 #12

    ZapperZ

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    We need to be careful here because that is the perpendicular component of the distance vector. If you read what the OP wrote, he/she is simply not considering that, and somehow, refuses to accept that it can be ANY direction.

    Zz.
     
  14. May 19, 2009 #13

    rock.freak667

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    isn't the moment of a force about an axis given by

    [tex]M=\hat{\lambda} \cdot (\vec{r} \times \vec{F})[/tex]


    where [itex]\hat{\lambda}[/itex] the unit vector in the direction of the axis?
     
  15. May 19, 2009 #14
    Nope, we didn' took it this way.
    M=F*d

    Ohhhhh, I got it , thanks ZappperZ, and for all :d.
    Sorry for confusion.
     
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