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Using torque to find mass of meter stick

  1. Jun 10, 2013 #1
    1. The problem statement, all variables and given/known data

    22) A 1-kg rock is suspended from the tip of a meter stick at the 0- cm mark so that the meter stick balances like a see-saw when the fulcrum is at the 25-cm mark. From this information, what is the mass of the meter
    stick?
    A) 1/4 kg
    B) 1/2 kg
    C) 3/4 kg
    D) 1 kg
    E) more than 1 kg
    Answer: D

    2. Relevant equations

    Torque== length of lever arm * Force

    3. The attempt at a solution

    I keep on receiving B.....

    Counter Clockwise Torque:
    0.25*F1
    • F1==Force between 0 and 25 cm==g(1+.25MS)
    • MS== mass of stick

    Clockwise Torque:
    0.75*F2
    • F2==Force between 25 and 100 cm==g(0.75MS)


    Counter Clockwise Torque==Clockwise Torque (because it balances)

    0.25*g(1+.25MS)==0.75*g(0.75MS)
    0.25g+0.0625gMS==0.5625gMS
    0.25g==0.5625gMS-0.0625gMS
    2.5==5.0MS
    MS==0.5
     
  2. jcsd
  3. Jun 11, 2013 #2

    haruspex

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    You're not calculating the moment of that portion of stick correctly. How far is that mass from the fulcrum?
     
  4. Jun 11, 2013 #3
    Isn't it 25 cm away, since mass is at 0 and fulcrum at 25 cm? Or is it zero since the portion of the ruler accounted starts right after the fulcrum?
     
  5. Jun 11, 2013 #4
    Your meter is "split" in two by the fulcrum. Where is the weight of the each portion acting at, at which distance?
     
  6. Jun 11, 2013 #5
    The 25 cm long part weighs 1\4th of the total and 75 cm weighs 3\4ths, doesn't it?
     
  7. Jun 11, 2013 #6
    Hello, yes, that would be correct, but that is not what I was pointing at.
    If you look at the first post on this topic, you wrote following:

    You are correct about the magnitude of the weights (0.25MS & 0.75MS), but you wrote a wrong position from which those forces act.

    Could you tell where are 0,25MS and 0,75MS located?
     
  8. Jun 11, 2013 #7
    Do you know where the center of mass of the meter stick is located?
     
  9. Jun 11, 2013 #8
    at 25cm?
     
  10. Jun 11, 2013 #9

    Do they both act from the center of gravity?
     
  11. Jun 11, 2013 #10
    Or is it half way?
     
  12. Jun 11, 2013 #11
    Yes. It is located at 0.5 meters from either end. This is where the total mass of the meter stick is taken to be concentrated (at least in a static situation like this). So the center of mass of the meter stick is situated 0.25 meters from the fulcrum.
     
  13. Jun 11, 2013 #12
    I got it! Thank you!

    .25mg==25g
    m==1!
     
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