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Repositioning to take more weight

  1. Feb 15, 2010 #1
    1.Two men lift the ends of a 10m beam onto their shoulders, so that it is horizontal. The stronger of the two suggests he reposition himself to take 50% more of the weight than his colleague. Where should he put his shoulder



    2.clockwise moments=anti-clockwise moments



    3.50w+25w=25w
    Where w=weight
    so the man should be in the center of the beam because he has 50% of the beam to his left and 25% to his right. Leaving 25% to the weaker man

    Am I on target here please?
     
    Last edited: Feb 15, 2010
  2. jcsd
  3. Feb 17, 2010 #2
    I'm finding this difficult to know how to do this. Maybe like this:
    The ratio we need to be 50% more is 3:1
    (Fstrongxdistance%)=(Fweakxdistance%)
    3:1
    This is 75:25

    So now I need to take moments around either the center of the beam or around each of the men to get this ratio. But none of my workings are giving this ratio. Not sure:rolleyes:

    So, for example: Taking moments from the center of the beam:

    75% of 5m = 3.75
    Fstrongx3.5m=Fweakx5m
    Fstrong=Fweak5/3.5
    =1.4

    Fweak=Fstrong3.5/5
    =0.7

    This is a 2:1 ratio. A 100% difference - not what i want
    But is this the right technique?
     
    Last edited: Feb 17, 2010
  4. Feb 17, 2010 #3
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    Last edited: Feb 17, 2010
  5. Feb 17, 2010 #4
    50% more does not mean a ratio 3:1.
    If one force is 1, the other is 1+0.5=1.5 so the ratio is 1.5:1 or 3:2.

    Just write the equations for the balance of the forces and the balance of the torques.
     
  6. Feb 17, 2010 #5
    Why would I consider torque, as this is about turning effects. In this problem there is no net force acting in any direction, right? And there is no net turning effect about any point, as the ladder is still in equilibrium, and is still being carried by the two men. Confused.
    When you say balance of the forces and torques, do you mean put them to zero, 0?
     
  7. Feb 17, 2010 #6
    Yes, that's why there is no rotation: the sum of the torques is equal to zero.
    and the sum of the forces is also zero
     
  8. Feb 17, 2010 #7
    How can I take the sum of something when I don't know where he is gonna stand?
     
  9. Feb 17, 2010 #8
    How can I take the sum of something when I don't know where he is gonna stand?
    How are these equations gonna look?
     
  10. Feb 17, 2010 #9
    How can I take the sum of something when I don't know where he is gonna stand?
     
  11. Feb 18, 2010 #10
    For the forces:
    F1+F2-W=0
    where W is the weight and F1, F2 are the two forces.

    For the torques, assume that we use the end of the beam as the pivot point and that one of the guys is at this end and the other one is at a distance x from the end.

    W*(L/2)-F2*x=0

    where L is the length of the beam.
    You can also use the middle of the beam as pivot and we have
    F1*(L/2)-F2*(x-L/2)=0

    In any case, the force that acts right at the pivot point have zero torque.
     
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