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Torque and weight of a meterstick lab

  1. Jan 13, 2007 #1
    So, i did a lab where we find the center of mass of a meter stick and then move the pivot point away from the CM and then add masses to bring it to equilibrium again. By finding the torque of the hanging mass, using the formula "Torque of hanging mass = Fh (Dist hanger to pivot)" we also find the torque of the center of mass since they are equal and opposite. We then use to formula "Fe(Dist from CM to pivot) = m(g)(dist from CM to pivot)" we can find the mass of the meterstick.

    One of the analysis questions is "how can the weight of the meterstick be cancelled out in future experiments?" Does anyone know the answer?
     
  2. jcsd
  3. Jan 13, 2007 #2

    cristo

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    Well, what do you think the answer is? (note we cannot give answers to homework questions, as per PF rules!)

    Hint: Suppose an object has a force applied to it at point A. Where would you place the pivot to ensure that the torque due to this force is zero?
     
  4. Jan 13, 2007 #3
    the way the question is worded is odd. The weight is the force, and i'm basically using three equations to get to the mass of the meter stick. However, i don't need to use the middle equation because it's just extra work and that is the equation that has the force of earth on the meterstick. Is it considered "cancelling out the weight of the meterstick" if i just don't use that equation to get to the mass of the meter stick?
     
  5. Jan 13, 2007 #4

    cristo

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    Hold on, I'm not sure whether I'm understanding your question here. Here's how I read it: You have conducted experiments and managed to calculate the mass of the metre-stick.

    Now, i don't really know what the next bit means. Does it mean "state how the weight of the metre-stick can be cancelled out in future experiments determining the mass of the metre-stick" or does it mean "...generally in future experiments using the metre-stick?"

    Since it says "future" experiments, I don't see why you would have to find the mass of the metre-stick in these future experiments, since you now know it!

    Please post an exact quotation of the question, and it may become clearer to me!
     
  6. Jan 13, 2007 #5
    This was an exact quotation: "how can the weight of the meterstick be cancelled out in future experiments?" By future experiments, i think she means doing the experiment again with a different meter stick that would have a different mass.


    tBut, still the question doesn't really make sense. The whole purpose of the lab was to find the mass of the meterstick, but in order to do that, we found the weight first. What i'm getting from the question, is that she's saying that in a future experiment, how can we find the mass without finding the weight. But, i still dont know if that makes any sense.
     
  7. Jan 13, 2007 #6

    cristo

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    I don't see how it makes sense either! I cannot think of a way to calculate the mass of the metre stick, using torques, without finding out the weight of the stick! I think the only way this question makes sense is in the sense of doing "other" experiments with this metre stick, in which case its weight would be "cancelled out" by taking torques about the center of mass.

    The question, as it stands, is very ambiguous, and, as such, I don't know what answer she wants!

    If anyone else is reading, and can correct/back me up, please do!
     
  8. Jan 13, 2007 #7
    thanks for your help. at least i know that it's the question and not just me.
     
  9. Oct 10, 2011 #8
    I have a similar question, only much simpler. How do you find the mass of the meter stick given this information: The fulcrum supports a uniform meter stick. A block is hung at different distances on the stick and the stick balances.
     
  10. Apr 1, 2012 #9
    If you put the fulcrum at the center of gravity of the meter stick, then the mass of the stick will not matter. It could weigh 8000lbs or 1g. That's how you cancel the mass.
     
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