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How to calculate counterweight at a different height?

  1. Mar 10, 2015 #1
    1. The problem statement, all variables and given/known data
    m2=3kg, m1=?
    the diagram has been attached
    2. Relevant equations
    I know that we find counterweights by equating equation mass x distance frm pivot but how do we find if it is at different height
    3. The attempt at a solution
    Is it m1yh1=m2xh2
    is counterweight at all dependent on height?
     

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  2. jcsd
  3. Mar 10, 2015 #2

    BvU

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    Does the full problem statement mention that the two beams to which the masses are attached are horizontal ?

    If yes, then your attempt looks good.
    Mass x g is force
    force x distance from pivot is torque
    No net torque means no angular acceleration => balance. g divides out.

    Advice: draw the forces in the diagram.
     
  4. Mar 10, 2015 #3
    That is basically m1y=m2x (in this case)
    But what about the effects of h1 & h2 ?
    The two beams are at different heights so will the approach change?
     
  5. Mar 10, 2015 #4

    BvU

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    Pity you don't show any equations. Your "Is it m1yh1=m2xh2" doesn't make sense to me. What does it represent ? A wild guess ? Is it a multiplication ? If so, I have no idea what y and x stand for. And the dimensions I don't recognize.

    Time to read the guidelines and make better use of the template . . .

    And completing the problem statement might be a good idea too: m1 can be anything if there are no restrictions. A restriction could be that the stuff is in equilibrium, but if you don't tell, nobody knows !

    [edit] I am a bit too strict. Your "That is basically m1y=m2x (in this case)" is correct for equilibrium. It doesn't feature h1 or h2.
     
    Last edited: Mar 10, 2015
  6. Mar 10, 2015 #5

    haruspex

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    To be precise (and this seems to be the crucial point here), the distance in that formula is the distance from the pivot to the line of action of the force, not to the point of application of the force.
     
  7. Mar 10, 2015 #6
    Sorry for the inconvenience.
    m1 & m2 are in equilibrium , y= 300mm & x=900 mm and hence m1y=m2x.
    So, if m1 & m2 are in equilibrium, will different heights have any effect? h1=900mm & h2=913mm.
     
  8. Mar 11, 2015 #7

    BvU

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    "That is basically m1y=m2x (in this case)" is correct for equilibrium.

    h1 and h2 are not present in the expression, so they have no effect !
     
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