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Damn balancing forces, can someone help me get this straight.

  1. May 17, 2008 #1
    http://img.photobucket.com/albums/v345/crimsonflames/ssd.jpg

    theres the diagram and question taken from OCR past paper.

    I know it's simple but I'm unsure where it all comes from ATM. not 100%

    I already know the answer.. it's in the question just not how to get it. And and im not sure how the whole thing would change if the positions of the holder things changed, could someone explain it please.
     
  2. jcsd
  3. May 17, 2008 #2
    halp!
     
  4. May 17, 2008 #3

    tiny-tim

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    Hi DeanBH! :smile:

    Oh come on … you know you have to show us something! :wink:

    What have you got so far? :smile:
     
  5. May 18, 2008 #4
    OK, well i thought about it for like 20 minuets and came to the conclusion that.


    Y is 1m away from center of gravity so 1x3600.
    X is 2.5m away from Y so it would take less force to hold to 1x3600 up.
    so 2.5 X force = 3600 x1
    3600/2.5 = force
    = 1440

    But that's just a guess. Is that right?
     
  6. May 18, 2008 #5

    tiny-tim

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    Yes :smile: … why so surprised? :confused:

    What you've done is take moments about a point on the line of B (it doesn't matter which point, since all the forces are parallel).

    Though, in an exam, you should start by saying that you're doing that!
    Are you unclear as to whether to multiply or divide by the distance?

    For moments, you always multiply! :smile:

    (You do know what the "principle of moments" is, don't you?)
     
  7. May 18, 2008 #6
    I missed the lessons to do with all of this.

    But recall the basics on a see-saw from GCSE physics.

    other than that I just thought about it for 20 minutes then gave it a good guess. and when is a guess ever 100% sure.
     
  8. May 18, 2008 #7

    tiny-tim

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    ah … I thought so … you were just balancing everything.

    (10/10 for good guessing, though … your "guess" was really the applicatoin of a physical principle … and you got it right! :smile:)

    ok … quick lesson … principle of moments

    First … do you know what a cross-product of two forces is (or of two vectors)? :smile:
     
  9. May 18, 2008 #8
    I might know the science, but i don't know what you mean by cross-product.
     
  10. May 20, 2008 #9

    tiny-tim

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    Hi DeanBH! :smile:

    I'm sorry I've taken so long to reply. :redface:

    ok … every force F has a moment about any point P.

    The moment is a vector, just like the force.

    To find the moment, draw L, the line of application of the force, and draw the perpendicular line PQ from P to L.

    Then the moment of F about P is the vector written "F x P" (pronounced "F cross P"). Its direction is perpendicular to both L and the line PQ. And its magnitude is F times PQ.

    Note that if P is on the line L, then P = Q, so PQ = 0, so the moment of the force is 0.

    In nearly all exam problems, everything is in the same plane (the plane of the examination paper!), so all the moment vectors are vertically out of the page.

    In other words, they're all parallel to each other, so we can forget that they're vectors, and treat them simply as numbers, F times PQ. :smile:

    You can take moments about any point, so you always choose whatever point makes the calculations easiest.

    Usually, it's the point of application of an unknown force, so that the moment of that force is 0, making the equation shorter!

    In your example, the point P was on the line of one of the two ropes, so the moment of the force from that rope was zero, and the moment of the other force and of the weight was the amount of the force times the resepective horizontal distance … exactly as in your seesaw principle!

    Any questions? :smile:
     
  11. May 20, 2008 #10

    tiny-tim

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    oops!

    oops! that's wrong! :redface:

    it should read …
    Then the Moment of F about P is the vector written "F x r" (pronounced "F cross r"), where r is the position vector PR, and R is the point of application of the force. Its direction is perpendicular to both L and the line PR (and PQ). And its magnitude is F times PQ.​

    (See new Library entry on Moments at https://www.physicsforums.com/library.php?do=view_item&itemid=64)
     
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