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    kristo's Avatar
    kristo Posts: 41, Reputation: 0
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    #1

    Jan 23, 2007, 11:21 AM
    Balancing a bar
    On one end of a 6m bar there is a 600kg weight, on the other end there's a 200kg weight, the bar itself weighs 20kg.
    Where shall the point of support be so that the bar is balanced?

    I hope you understand it, English isn't my first language.. :(
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #2

    Jan 23, 2007, 11:34 AM
    Hi, here you need to balance the moments being supplied by each weight.

    So if we define x as the distance from the 600kg weight to the support

    Using:
    Moment = Force * Perpendicular distance (from fulcrum)
    and:
    Counterclockwise moment = Clockwise moment
    we find:



    this describes the moment felt in the clockwise and anticlockwise direction.
    600*x is the moment from the weight
    20(x/6) is the weight of the bar felt in this direction
    x/2 is the distance that the weight of the bar acts from the fulcrum

    then the same is written for the other direction

    The first step is obviously to cancel g, this is why I put it in brackets.

    You can solve this equation for x to find the distance that the fulcrum must be from the 600kg weight

    If you have any problem with understanding, please ask further!
    good luck!

    PS: i did this very quickly and have tried to catch any errors, I make no claims to correctness and you should check everything yourself.
    kristo's Avatar
    kristo Posts: 41, Reputation: 0
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    #3

    Jan 23, 2007, 01:22 PM
    Thanks a lot for your help!
    I get this far, but don't really know what to do next, a little help, please?
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #4

    Jan 23, 2007, 01:32 PM
    Expand the brackets, group like terms, solve the quadratic, that's about all there is to it.

    Hopefully you get a meaningful answer! :D
    kristo's Avatar
    kristo Posts: 41, Reputation: 0
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    #5

    Jan 24, 2007, 06:31 AM
    So, x is about 1.5 meters, right?
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #6

    Jan 24, 2007, 06:34 AM
    No idea but that sounds about right! :)
    if the bar was weightless then it would be exactly 1.5, so nearly 1.5 taking into account the weight of the bar sounds spot on.
    Make sure you understand how I made the equation in my first post, that's the most important thing to learn.
    kristo's Avatar
    kristo Posts: 41, Reputation: 0
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    #7

    Jan 24, 2007, 07:01 AM
    I understand it for most of the part, but could you explain the x/2 and x/6 part a bit?
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #8

    Jan 24, 2007, 07:07 AM
    x/6 is the fraction of the bar that is on that side of the fulcrum, so by multiplying it by 20g we get the weight of the bar on that side of the fulcrum.

    x/2 is the distance that this weight acts from the fulcrum. You understand that a solid object can be simplified to a point at the centre of mass of the object? Same principle here.

    so by multiplying them we find the moment from the bar in the anticlockwise direction.
    kristo's Avatar
    kristo Posts: 41, Reputation: 0
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    #9

    Jan 24, 2007, 07:19 AM
    I think I got it now, thank you :).
    Sorry about the half comment, accidentally pressed F5 while writing it and can't edit.
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #10

    Jan 24, 2007, 07:22 AM
    Heh no problem! Thanks for excercising my moment knowledge!

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