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Math Help - Mechanics problem

  1. #1
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    Mechanics problem

    A 2kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.15m, to a hanging book with mass 3kg. The system is released from rest, and the books are observed to move 1.2m in 0.8s.

    My solution so far:
    using constant acceleration formula
    s = ut+\frac{1}{2}at^2
    1.2 = 0 +\frac{1}{2}(a)(0.8)^2
    a = 3.75

    For the hanging part:
    3g - T_1 = ma
    T_1 = 3g-ma
    T_1 = 3g-3(3.75) = 18.2(rounded) This agrees with my textbook

    however for the horizontal tension in the cord
    I make it
    T_1 - T_2 = ma
    T_2 = T_1 - ma
    T_2 = 18.2 - 2(3.75)
    T_2 = 10.7N

    however the answer in my textbook is 7.5 which would suggest
    T_2=2(3.75)

    where has the T_1 force gone?
    Have I missed out a force
    Also how can you have different tensions in the cord surely it would snap?
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  2. #2
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    If you draw the free body diagram for the 2kg book. T_1 = (2kg)a

    The equation T_1 - T_2 = ma is incorrect.
    If your pulley has no mass or is frictionless, then T_1 = T_2.
    However, this is not the case, so there is some inertia in the pulley.
    (T_2 - T_1)(0.15m/2) is the torque on the pulley.
    Last edited by snowtea; December 19th 2010 at 09:11 AM.
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  3. #3
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    why wouldnt the cord snap if it has different tensions in it?
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  4. #4
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    Tension is just a force, it can be balanced by other forces such as friction.

    Think about dragging a really long heavy chain on the ground.
    The front link of the chain has the most tension and friction.
    The back link of the chain may have only a little tension and a little friction.
    However, you are right, the net acceleration at each link must be the same (if it has fixed length).
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