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How far from the left end should the fulcrum be placed?

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data

    A 0.105-, 48.7--long uniform bar has a small 0.055- mass glued to its left end and a small 0.150- mass glued to the other end. You want to balance this system horizontally on a fulcrum placed just under its center of gravity.

    How far from the left end should the fulcrum be placed?


    2. Relevant equations



    3. The attempt at a solution

    What I did was used the static equilibrium equation and set the pivot to be at the far right in order to cancel out those forces.
    m2=mass of the bar
    m1= mass of the ball to the far left
    x the distance to the left from the Fulton
    l= length of the bar
    I set clockwise to be the positive direction

    Ʃτ=0→ m1gx+m2g(l/2) solved for x and got

    x=-m2(l/2)/m1

    The answer I got was incorrect the correct answer is 31.8 cm and they used a different approach from mine.

    Cheers
     
  2. jcsd
  3. Apr 15, 2012 #2

    PeterO

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    Homework Helper

    I would use centres of mass.

    The Centre of Mass for the Bar is clearly at its centre, and its mass is given.

    You can find where the Centre of Mass of the two masses can also be found in the usual way.

    You then find the Centre of Mass of those two Centres of Mass.
     
  4. Apr 15, 2012 #3

    PeterO

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    Homework Helper

    When doing the rotations as you did, you account for mass of Bar - at a distance half way out, the little mass all the way out [both anti-clockwise] and the fulcrum force equal to sum of both masses plus the bar. I think you have left some weight forces out.
     
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