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Torque equilibrium. Where would the masses need to be placed?

  1. Oct 18, 2014 #1
    • Warning: Posting template must be used for homework questions.
    A triple beam balance, like the scales we use every week in lab, works using rotational equilibrium. If a 253 g mass is placed on the pan, 1 cm from the pivot, and there are three masses, 1 g, 10 g, and 50 g masses on the other side can be slid back and forth between 0 cm and 10 cm from the pivot, where would the masses need to be placed? Full credit only if you provide values at whole-integers of centimeters for the three masses.

    So far I've got: .253-.05r-.01(r+x)-.001(r+x+y)=0

    but I can't figure out how to solve for all the variables. any help is appreciated!
     
  2. jcsd
  3. Oct 18, 2014 #2

    billy_joule

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    Science Advisor

    Have you included the entire problem statement?
    Your equation implies that maybe you haven't. You want your variables to be distance from the pivot, not from the previous mass, In your eqn 'x' is the distance from the 50g mass to the 10g mass, which isn't really relevant. So instead of your eqn:

    0.253 -.005r - 0.01x - 0.001y = 0, where 0 ≤ r,x,y ≤ 10 and integers

    This can be solved by iteration pretty quickly. There is one solution where they are three consecutive integers...

    Or are you expected to be more mathematically rigorous?
     
  4. Oct 18, 2014 #3
    This is the whole problem statement. I apologize, I'm not sure what iteration is. I'm a liberal arts major out of my comfort zone, taking a college algebra-based physics course.
     
  5. Oct 18, 2014 #4

    gneill

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    Staff: Mentor

    Since you'll be comparing moments about a pivot, so long as all the units are the same (grams for mass, cm for distances) you won't need to convert to kg and meters. Just write the sum of the torques for each side of the pivot as an equation. Place the torque for the 253 gram mass on one side of the equals and the sum of the other torques on the other side. You really don't even need to convert the masses to weights since that's just multiplying every mass on both sides by a constant (g), and thus they all cancel out.

    The problem can then be solved quite easily by inspection.
     
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