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What is wrong with our assumptions?

  1. Oct 30, 2016 #1
    [Moderator note: Thread moved from technical forum so no template shown]

    I am trying to do a simple experiment that estimates the mass of an object based on how an Atwood machine reacts when it is added to one side.

    For this experiment we have a pulley with 200-gram masses on either side of the rope. We then set it up so that the right mass was exactly one meter from the floor. Then we would attach the unknown mass to the other time, and measure the time in seconds it took for it to hit the floor.

    From this data we calculated that the acceleration was (1 meter / (seconds to fall)^2). We then put it into the following formula.
    usVFbxt.png
    I really dont know where we got this formula from. I also dont know wether the a on one side is expressing acceleration of the system, or just one of the masses.

    Since m1 and m2 are both 200 grams we were able to simplify the formula. While simplifying we considered that m2 would be (m1 + mu) (mu being the unknown mass). We then got this formula:
    um0Sxp9.png

    We tried out our first mass and it took it 3.3 seconds to travel one meter.
    We plugged everyhing into the formula and got

    Unfortunately this did not work out for us. We converted grams to kilograms and substituted everything in. We got a calculated mass that was way too low, the objects actual weight is 12 grams. What did we mess up?
     
  2. jcsd
  3. Oct 30, 2016 #2

    Student100

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    First, if you don't know where a formula comes from you should see if you can derive it.

    Why don't you draw a force diagram for the masses and try to come up with it, it'll clear up what it's actually trying to say.

    Edit: Okay I reread it, and think you're saying that the .2kg mass is still there. in which case, your simplification is okay.

    What did you get for acceleration? And the unknown mass?

    I'm going to report this post, as I think it would be more appropriately placed in homework.
     
    Last edited: Oct 30, 2016
  4. Oct 31, 2016 #3

    James R

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    You're missing a factor of 2 there. If the distance to fall is ##h##, the time taken is ##t## and the acceleration is ##a##, then ##h=\frac{1}{2} at^2##, which gives ##a=2h/t^2##.

    The two masses are connected by a string that doesn't stretch, so a is the magnitude of the acceleration of either one of the masses. One accelerates downwards while the other accelerates upwards at the same rate.

    Fine.

    Got what? Is something missing there?

    Are you trying to solve for ##m_u##, then, using the measured value of ##a##? Re-arranging your second equation gives $$m_u = \frac{2 m_1 a}{g-a}$$. Does that help?
     
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