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Pulley Problem

  1. Jan 17, 2008 #1
    1. The problem statement, all variables and given/known data
    A pulley system is hanging from a newton meter as shown below. If we assume that gravitational acceleration is [tex]10 m/s^{2}[/tex], what will the reading be on the newton meter as the masses accelerate?

    a) 3 Kg
    b) 13 N
    c) 27 N
    d) 30 N

    [​IMG]

    2. Relevant equations
    [tex]F_{net}=ma[/tex]
    [tex]F_{g}=mg[/tex]


    3. The attempt at a solution
    I think that the answer is d), because if you add the masses of the two objects, which turns out to be 3.0 Kg, and multiply that by the gravitational acceleration, you would get 30 N. Am I right?
     
  2. jcsd
  3. Jan 17, 2008 #2

    mgb_phys

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    If they were stationary - yes, but the weights are falling!
     
  4. Jan 17, 2008 #3
    So how would I get the correct answer?
     
  5. Jan 17, 2008 #4
    Draw a free body diagram and apply F=ma for each mass.
     
  6. Jan 17, 2008 #5
    I was playing around with some of the numbers and I got 27 N. However, I might have gone wrong somewhere.

    [​IMG]
     
  7. Jan 17, 2008 #6

    Shooting Star

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    No.

    I presume it's a frictionless pulley. Let T be the tension in the string. Let 'a' be the accn of the two masses, toward the heavier mass.

    For the 2 kg mass, 2*a = 2*g - T.
    For the 1 kg mass, 1*a = T - 1*g.

    The total reaction on the meter is 2T. Find that.
     
  8. Jan 17, 2008 #7
    2*a = 2*g -T
    2*3.333... = 2*10 - T
    6.666... = 20 - T
    T = 13.333...

    2T = 2*13.333...
    T = 26.666...
    T = 27 N

    Thanks Shooting Star.
     
  9. Jan 17, 2008 #8
    holy crap i have a problem on my homework just like that. sweet! thanks. except i need the acceleration. my blocks are 5.0kg (block A) and 2.0kg (block B)....help w/ acceleration please!
     
  10. Jan 18, 2008 #9

    Shooting Star

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    There are two eqns with a and T. What do you think you should do?
     
  11. Jan 22, 2008 #10
    The problem that I posted at the very beginning of this thread was off my memory. Here is how it was actually worded.

    [​IMG]

    Is the answer still 27N, because when I asked my teacher, she said that the answer was 30N.
     
  12. Jan 22, 2008 #11
    Can anyone help?
     
  13. Jan 23, 2008 #12

    Shooting Star

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    EDIT: I became a bit upset to see that you had written T= 26.6 N in post #7, but I guess you meant 2T. So , the answer of 27 N still stands.

    Ask your teacher again. The answer cannot be 30 N.
     
    Last edited: Jan 23, 2008
  14. Mar 6, 2008 #13

    Shooting Star

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  15. Mar 6, 2008 #14
    If the masses exert a force of 30N downwards, and the net force is 10 Newtons downwards, shouldnt the reaction force at the pulley be about 20N upwards? What is wrong with my answer?

    Ok I udnerstand where I was wrong now! haha =X
     
    Last edited: Mar 6, 2008
  16. Mar 6, 2008 #15

    Chi Meson

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    I agree that the answer is 27 N.

    I solved it without looking at anyone's answers in the thread, without even looking at the choices for the answers.

    The acceleration of the system is 3.33 m/s^2.

    The net force on the 2kg mass is ma=6.67 N. This is equal to mg -T, so T = 20 N - 6.67 N= 13.33 N. The scale pulls up on 2*T = 27 N.

    Your teacher is incorrect. Point out that the "30N" violates Newton's 3rd Law as well as the Law of conservation of momentum, as well as the Conservation of Energy principle.
     
  17. Mar 6, 2008 #16
    Ah, but if the masses are held, then released to accelerate, the reading will show 30N at that instant, then decrease to 26.6666 and oscillate about that point (since its a spring system). It takes time for the reading to move from 30N to 26.66N.

    *at least if I was a teacher who always wanted to be right, I would say this :)
     
  18. Mar 6, 2008 #17

    Chi Meson

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    As a teacher, I would not waste time splitting hairs with "gotcha" scenarios. It is a good idea to be reminded of what things really do in the real world, but no one is going to be able to calculate anything if we did take all real considerations into account.

    And even with real spring scales, there is just enough damping to produce quick and accurate demonstrations of just this problem. I just did it right now with a standard large-display spring scale. It works quite well. I think the OP should ask for a demo!
     
  19. Mar 6, 2008 #18

    Shooting Star

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    We are not talking about the Physics any more, nor about the rationale behind the answer given by the OP's teacher, the logic of which is nonsense. I have expressed my real concern in the other thread, viz., the fate of future students made to learn something wrong.
     
  20. Mar 6, 2008 #19
    This is just standard Atwood's machine calculation, but the multiple-guess answers are interesting. It seems obvious that the person who constructed the question knew the answer to be 27 N since that is the only reasonable way to have any discrimination in the choice between B and C. Did she "lift" the question from another prof without bothering to get the answer key?
     
  21. Mar 6, 2008 #20
    I think your teacher may not notice the question.
    Since the system is accelerated.
     
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