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Homework Help: Force applied on pulley

  1. Dec 13, 2013 #1
    1. The problem statement, all variables and given/known data
    2vhy77p.jpg
    A force of 25 N is applied on the pulley. M1 = 1.5 kg, M2 = 2.5 kg, light frictionless strings and pulley.

    a) What is the tension in the strings?
    b) What is the acceleration of the masses?
    c) What is the minimum Force to apply on the pulley so that M2 comes off the ground?


    2. Relevant equations
    ƩF = ma


    3. The attempt at a solution
    I am not totally sure how to do this question because of the extra force applied. I mean since there is another acceleration, I am thinking in my head that it won't be just mg but it will be something like m(g+ 25/9.8) but I am not sure. Also it seems like there won't be an acceleration in mass 2, but will be in mass 1. Can anyone point me in the right direction?
     
  2. jcsd
  3. Dec 13, 2013 #2

    Zondrina

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    Treat the two masses separately and draw free body diagrams for each. Don't forget to define a reference frame for each one separately.

    The tension in the string will be uniform.
     
  4. Dec 13, 2013 #3
    What do you mean defining a reference frame for each one separately?
     
  5. Dec 13, 2013 #4

    Zondrina

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    Notice mass 2 is heavier than mass 1. This means gravity will pull mass 2 downwards.

    It could be safe to say then that you should define down as positive in the case of mass 2.

    Does it make sense to define it as such for mass 1?
     
  6. Dec 13, 2013 #5
    Yea I know that, Like for the fbd for mass 2 is m2g - T = 0, and for mass 1 its T - m1g = m1a, that is what I wasn't sure about, because the second mass doesn't accelerate but the first mass does...because of the force
     
  7. Dec 13, 2013 #6

    Zondrina

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    What? Both masses are accelerating at the same rate.

    If you broke things down into the FBDs you would get two equations:

    ##F_{net_1} =F_T - F_G##

    ##F_{net_2} =F_G - F_T##

    Now you have some information to work with. Namely gravity and the actual values of mass.
     
  8. Dec 13, 2013 #7
    I could have solved that part on my own, its just then T - m1g = m1a and m2g - T = m2a. But my real problem is incorporating the force on the pulley.
     
  9. Dec 13, 2013 #8

    Zondrina

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    Okay very good. You have two linear equations now.

    What happens if you.. per say sub one into the other?
     
  10. Dec 13, 2013 #9
    Yea I know two equations two unknowns, and I can solve for both T and a, a = g(m2 - m1)/(m2+m1) = 9.8/4
    T = m1g +m1a = (1.5*9.8 + 1.5*9.8/4) N
     
  11. Dec 13, 2013 #10

    Zondrina

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    Yup looks good.
     
  12. Dec 13, 2013 #11
    But that doesn't solve the whole problem...the pulley is still accelerating up
     
  13. Dec 13, 2013 #12

    Zondrina

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    For ##m_2## to come off the ground, there's two options. Suppose we choose the option to introduce an applied force ##F_A##, which pulls up on the second mass (imagine pulling on the string with your hand).

    What happens now? Think in terms of an FBD.
     
  14. Dec 13, 2013 #13
    you mean like pulling down on m1? Then its just an extra F in the m1 in the m1 FBD, and a -F in the m2 FBD.
     
  15. Dec 13, 2013 #14

    Zondrina

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    From part a) and b) you have lots of information at your disposal. You know ##F_T ≈ 18.4N## and ##F_G ≈ 24.5## for ##m_2##.

    Keep it simple, apply the force to the second mass. The difference between the tension force and the force of gravity should give you the force you would need to apply to get the box to stop accelerating downwards.

    Then applying even 0.1N of extra force afterwards would cause the box to accelerate upwards.
     
  16. Dec 13, 2013 #15
    But that tension and acceleration was taken as if the pulley had 0 acceleration
     
  17. Dec 13, 2013 #16

    Zondrina

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    Yes exactly. You have to assume the limiting case of ##F_{net_2} = 0## so that the applied force is at a minimum. If you find the point where the mass stops accelerating, then you can conjecture how much force you would need to get it to accelerate upwards (based on the significant figures, which If I've been following there should only be 2).
     
  18. Dec 13, 2013 #17
    Maybe you didnt understand the question. The pulley is being pulled up with 25 N of force..wouldn't the tension and acceleration be different because of this fact?
     
  19. Dec 13, 2013 #18

    Zondrina

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    Ohh I didn't see that. I'm sorry about that. That would change the answer numerically a bit yes.

    I'm not sure exactly what they're pulling on the pulley for, but it's simply another force to consider in the FBD and equation.

    The wording of the question isn't the best IMO.
     
  20. Dec 13, 2013 #19
    But where would that force go? The pulley is massless
     
  21. Dec 13, 2013 #20

    Zondrina

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    If the pulley is being pulled on, and the strings are attached to the pulley, then the strings will also be pulled on by the same force.

    Since the masses are attached to the strings, they will follow suit.
     
  22. Dec 13, 2013 #21
    So F and T are in the same direction?
     
  23. Dec 13, 2013 #22

    Zondrina

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    What I don't understand from your question is why the 25N force is being applied or when it is being applied. From what I can gather they want you to exert a force on the pulley in question c), but they've given you a 25N force already. This doesn't seem to make sense.

    If you consider this 25N force for part a) and b), you would get the acceleration is up and the tension force is down. This obviously doesn't make sense to either of us because ##m_2## is going to the ground for part c).

    Could you perhaps provide the original question?
     
  24. Dec 13, 2013 #23
    Well the original question was on my final exam.. this was a bonus question
     
  25. Dec 13, 2013 #24
    For a and b you use thr 25 N force and for c) its the minimum force to get m2 off the ground
     
  26. Dec 13, 2013 #25

    Zondrina

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    I think I understand the context of the question now. For part a and b use the 25N force they've given you.

    For part c), treat it completely independently of part a) and b). You need to find that applied force yourself in this case.

    The methods to do this are no different than what has been done on the first page.
     
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