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Two Blocks and a Wheel

  1. Aug 2, 2012 #1
    1. The problem statement, all variables and given/known data

    (Sorry if I used any wrong term, English isn't my first language)

    Be A (mass M) and B (mass m) two blocks connected to each other through an ideal wire passing through a wheel (caster), each block stays at one of the side of the wheel. If we apply a force F pointing up into the center of the wheel, which will be the acceleration of the wheel in terms of M, n, g and F? (Neglect the air resistance and the frictions).

    Please, see the diagram:
    imagem.png

    2. Relevant equations



    3. The attempt at a solution

    It is wrong, I know, but I don't know why!

    Be M > m
    We have
    Mg - T = M.a
    T - mg = m.a
    Thus, T = 2Mmg/(M+m)
    So, the acceleration γ of the wheel should be F - 2T = (M+m).γ
    γ = (F - 4Mmg)/(M+m)^2
    Which is wrong...
     
  2. jcsd
  3. Aug 2, 2012 #2

    PeterO

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    Might this be simply that the downward force on the system is the weight of the two masses [I assume the pulley is taken as massless, as the mass is not given], while the upward force is F.

    From that you calculate a net force, with a total mass of M+m and so you get acceleration?
     
    Last edited: Aug 2, 2012
  4. Aug 2, 2012 #3

    TSny

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    Hi, Tiba. Welcome to PhysicsForums.

    Does the wheel have a mass?

    Note that in this problem the two blocks will not have the same magnitude of acceleration, so you don't want to use the same symbol "a" stand for both blocks.
     
  5. Aug 2, 2012 #4

    TSny

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    Also, note that F - 2T is the net force on the wheel. So, what mass should you be using in the equation F - 2T = mass * acceleration?

    [EDIT: If the wheel has a mass, then the net force on the wheel would not be F - 2T since you would have to include the weight of the wheel.]

    [EDIT 2: They must be assuming a massless wheel. Otherwise, you would need to know the moment of inertia of the wheel, which is not mentioned. (Moreover, with a massive wheel the tension would not be the same on each side.)]
     
    Last edited: Aug 2, 2012
  6. Aug 2, 2012 #5

    PeterO

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    To get a feel for this: consider the following scenarios:

    Lets have both masses sitting of the ground to begin with, and in fact "glue" mass M to the ground.

    When you apply and upward force, gradually increasing, at first nothing moves, since the weight of the mass m may be more than enough to resist a small applied force F.

    If F is big enough, the pulley and mass m will both accelerate up [remember I am considering M to be glued to the floor]. the pulley system means the acceleration of m is twice the acceleration of the pulley.

    As F is increased, the acceleration of m increases meaning the Tension in the wire increases. That increase will be limitless, since the mass m is still glued to the ground at this point.

    Eventually your calculations will produce a tension which happens to be greater than the weight of M.

    Now consider if mass M was not "glued" to the ground.
    Once the tension you calculate above exceeds the weight of mass M, both masses will leave the ground, and you have a whole new set of conditions applying.

    The Tension in the wire will still be the same on both sides, but now you have to consider both masses in your equation.

    Since these masses do not start on the ground, that second more complicated situation is the only one you have to analyse.

    Remember, if F < Mg + mg, the pulley will be accelerating down!
     
  7. Aug 2, 2012 #6

    TSny

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    Hi, PeterO. Actually, if you work it out I think you'll find that it's possible for F < Mg + mg and still have the pulley accelerate upward. (Think of an extreme case where M >> m, like a bowling ball and a feather.) The condition F < Mg + mg will only imply that the center of mass of the system accelerates downward, but the pulley could still accelerate upward as long as F is not too small.
     
  8. Aug 3, 2012 #7
    Hello,

    Thank you for your answers, but I'm still a little confused.
    Just to clear out the things:
    I) The Wheel is massless.
    II) M=2,0kg, m=1,0kg, F=30,0N.
    III) The correct answer given in the textbook is γ=1,250m/s²

    Can someone please show me the calculation for this?

    Also, this isn't the case where the blocks have an apparent weight [being Pa=m(γ+g)] rather than mg?

    Thank you!
     
  9. Aug 3, 2012 #8
    @PeterO

    If I understood you, than the system should be F - (M.g + m.g) = L.a
    Being L the mass of the wheel, but as the wheel has no mass and putting the values I wrote in the above post, 30 - 30 = 0.a, which keeps me in the dark.
     
  10. Aug 3, 2012 #9

    TSny

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    I would recommend that for this problem you take upward as the positive direction for all forces and accelerations.
    How would you then write the above two formulas? Keep in mind that the accelerations of M and m will not have the same magnitudes, so you'll need to let a1, say, be the acceleration of M and a2 the acceleration of m.

    F - 2T is the net force on the wheel. So, you have the wrong mass on the right side of the equation.

    Note that you have four unknowns, a1, a2, T, and γ. Once you fix up the above equations you will have 3 equations for these four unknowns. So, you'll need a fourth equation. You can construct a fourth equation from the constraint that the two masses are connected by a string of fixed length.
     
  11. Aug 3, 2012 #10
    Thanks, TSny, I think I got it!

    Let:
    M = 2, m = 1, F = 30, g = 10.

    T - M.g = 2,0a
    T - m.g = 1,0b
    F = 2T

    Then

    a = -2,50
    b = 5,0

    Finally:
    γ=(a+b)/2
    γ=1,25m/s²

    Right?
     
  12. Aug 3, 2012 #11

    TSny

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    That looks very good.:smile: ....Just wondering, how did you justify the relation γ = (a+b)/2?
     
  13. Aug 3, 2012 #12
    In relation to the acceleration of the wheel, a = -b.
    So, to an outside observer, a - γ = - (b - γ)
    Giving γ=(a+b)/2

    Right? :)
     
  14. Aug 3, 2012 #13

    TSny

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    Yes. Very nice.
     
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