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Finding mass of a meter stick using torque

  1. Feb 15, 2015 #1
    1. The problem statement, all variables and given/known data
    I did a lab and am trying to find the mass of a meter stick. Our fulcrum was at the 60cm mark at the meter stick. A 100 g mass was placed 30 cm away from the fulcrum on the shorter side. A 50 g mass was placed 43 cm away from the fulcrum on the longer side. How do I find the mass of the meter stick?

    2. Relevant equations
    Torque=rfsintheta

    3. The attempt at a solution
    Not sure how to calculate this
     
  2. jcsd
  3. Feb 15, 2015 #2

    Nathanael

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    I assume when you placed the masses there, the meter stick was balanced? What does this say about the net torque? Is anything other than the 100g and 50g masses causing any torque?
     
  4. Feb 15, 2015 #3
    The net torque is 0 and the meter stick causes torque. When I find try to calculate the torque of the meter stick i get 230 which would make the mass 230 g. The actual measured mass of the stick is 97 so this can't be right I don't know what I did wrong
     
  5. Feb 15, 2015 #4

    Nathanael

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    Okay, please show your work.
     
  6. Feb 15, 2015 #5
     

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  7. Feb 15, 2015 #6

    Nathanael

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    Well firstly, the masses are in grams not kilograms! So the weight is not 980 and 490 Newtons, it is 0.98 and 0.49 Newtons. I would suggest not even calculating the weights, just call it 0.1g and 0.05g because the g ends up canceling so why not save a bit of effort.
    (It is very often good to not plug in numbers till the end!)

    Anyway, you wrote "980*0.4" and "490*0.33" and I have no idea where you got these numbers from...

    What is the torque due to the 100g mass? What is the torque due to the 50g mass?
     
  8. Feb 15, 2015 #7
    I got the .4m and .33m because those are the distances from the center of gravity of the meter stick.
    The torque of the 100g would be .4m X .98 newtons = .392 Nm.
    The torque of the 50g would be .33m X .49 newtons = .1617 Nm.
    Right? Am I supposed to use the distances form fulcrum of the center of gravity?
    If this is correct I would do .392Nm - .1617Nm = .2303Nm which would be the torque of the meter stick. And since it is .1 meters away from center of gravity wouldn't that make the meter stick .2303 kg?
     
  9. Feb 15, 2015 #8

    Nathanael

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    You can take the torque about any point, but if you choose to take the torque about any point other than the fulcrum, then you will have to account for the torque due to the normal force from the fulcrum. You don't know what this normal force is, so you ought to take the torque about the fulcrum (that way the normal force has a distance zero and thus has zero torque).

    And since what is 0.1 meters from the center of gravity...? Certainly that is not the distance that the force of gravity acts?
     
  10. Feb 15, 2015 #9
    The fulcrum is .1 m from the center of gravity so I'm not sure how to account for this and how to set up the equations
     
  11. Feb 15, 2015 #10

    Nathanael

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    There's no point in taking the torque about the center of gravity because the thing you're trying to solve for (gravity) causes zero torque! So there's no way to solve for it...

    Like I said, you can take the torque about any point you want to. (There must be zero torque about every point!) So around what point should you take the torque?
     
  12. Feb 15, 2015 #11
    The fulcrum
    So should I find the torque of each mass by doing .3m X .98N = .294Nm and .43m X .49N = .2107Nm? Then would I subtract those to find the torque of the fulcrum
     
  13. Feb 15, 2015 #12

    Nathanael

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    And what do you get?
     
  14. Feb 15, 2015 #13
    .0833Nm then how do I find the mass of the meter stick from this number
     
  15. Feb 15, 2015 #14

    Nathanael

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    That is the torque due to gravity, right? Which equals...?
     
  16. Feb 15, 2015 #15
    I'm not sure what you mean. How would you go from .0833Nm to just grams if there's no m? What would the m be?
     
  17. Feb 15, 2015 #16

    Nathanael

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    Forget about the 0.0833Nm.... Does gravity cause a torque or not? What is the torque from gravity? Is it zero?
     
  18. Feb 15, 2015 #17
    Yes I'm not sure what to do next to get the mass of the meter stick
     
  19. Feb 15, 2015 #18

    Nathanael

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    Ok.. but what is the torque from gravity? Can you write an expression for it?

    Write out your torque equation and include the torque from gravity, and set the net torque equal to zero.
     
  20. Feb 15, 2015 #19
    Would it be torque of mass 1 + torque of mass 2 + torque of gravity = 0
     
  21. Feb 15, 2015 #20

    Nathanael

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    Yes. Now what is the value of these torques? Specifically, the torque from gravity... At what distance does the force of gravity act from the fulcrum?
    Put it all in to your torque equation.
     
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