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Static Equilibrium - board problem

  1. Mar 26, 2008 #1
    1. The problem statement, all variables and given/known data

    A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.0 g stacked over the 12.0 cm mark, the stick is found to balance at the 45.5 cm mark. What is the mass of the meter stick?

    2. Relevant equations

    netF = 0
    netTorque = 0
    The answer is 74 g

    3. The attempt at a solution

    I know there is an upward force exerted by the knife, a downward force exerted by the weight of the coins, and another downward force exerted by gravity acting on the meter stick itself, which I originally thought was at the .5 mark.

    Does there need to be a second upward force at the .445 m mark, or is that where I put my pivot point?

    This shouldn't be a hard problem, but I don't think I really understand how the force diagram should look.
    Last edited: Mar 26, 2008
  2. jcsd
  3. Mar 27, 2008 #2


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    Science Advisor
    Homework Helper

    Hi Seraph! :smile:

    I think the question is slightly misleading.

    It intends you to assume that there is only one knife-edge, and it is moved to the .445 m mark.

    So the only force you need worry about at the .5 m mark is the weight of the stick. :smile:
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