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Pulley Problem

  1. Jul 12, 2006 #1
    I am having a hard time with this problem:

    "A 9.00-kg hanging weight is connected by a string over a pulley to a 5.00-kg block that is sliding on a flat table. If the coefficient of kinetic friction is 0.200, find the tension in the string."

    I started the problem by drawing two free-body diagrams, one for the weight and one for the block. By doing that, I was able to come up with my components:

    Note: M1 = 5.00-kg
    M2 = 9.00-kg
    so when I am referring to "1," (such as Fx1) I am referring to the object with mass 5.00 kg.

    Fx1 = -f + T = m(ax)
    Fy1 = n - mg = 0

    Fx2 = 0
    Fy2 = T - mg = M(ay)

    From here, I realized that n = mg and f = u(n). We know what u is (0.200) and we know what n is, so I plugged those values in.

    f came out to be -9.8 N.

    From this point, I am stuck. I tried solving for T in the second y component equation, and just ended up with T = 88.2 + 9ay

    Any help on how to solve this problem is greatly appreciated. Thanks.
     
  2. jcsd
  3. Jul 12, 2006 #2
    Assume both masses are connected by an inextensible string and that the string remains taut the whole time. With this, how are the magnitudes of [tex]a_{x}[/tex] and [tex]a_{y}[/tex] related?

    Also, you may want to check your working for the vertical forces acting on the 9.00-kg mass. In what direction is the mass accelerating?
     
    Last edited: Jul 12, 2006
  4. Jul 12, 2006 #3
    ax and ay are equal to each other because they're attached to the same string. Correct?

    The mass is moving down. That is why mg is negative in the Fy2 equation.
     
  5. Jul 12, 2006 #4

    Hootenanny

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    In this case, yes.
    However, Mg is acting in the same direction as the velocity.
     
    Last edited: Jul 12, 2006
  6. Jul 12, 2006 #5
    You are right in saying that the magnitudes of [tex]a_{x}[/tex] and [tex]a_{y}[/tex] are equal.

    With regard to your next statement, doesn't the weight (Mg, note that M is used for the heavier mass and m for the lighter) act in a downward direction too?
     
  7. Jul 12, 2006 #6
    From this point, I try solving for T, which is T = 9ay - 88.2. Because the accelerations are equal, I dropped the "x" and "y" sub labels and just called it "a."

    After solving for T, I substituted T into my Fx1 equation so I could solve for acceleration. What I had in mind was to solve for acceleration, and then substitute acceleration back into the T equation and solve for T. Am I thinking correctly?

    Thanks.
     
  8. Jul 12, 2006 #7

    Hootenanny

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    Sounds good to me. :smile:
     
  9. Jul 12, 2006 #8
    "T = 9ay - 88.2" implies that the tensional force on the 9.00-kg mass and its weight act in the same direction. Based on my understanding of the problem, this does not seem the case...

    You also mention "...I substituted T into my Fx1 equation so I could solve for acceleration. What I had in mind was to solve for acceleration, and then substitute acceleration back into the T equation and solve for T. Am I thinking correctly?" Alternatively, you can make a (acceleration) the subject of both equations and eliminate it to solve for T (tension).
     
    Last edited: Jul 12, 2006
  10. Jul 12, 2006 #9

    Hootenanny

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    I apologise indeed. Pizzasky is correct, your two terms should be the other way round; T = 88.2 - 9a.
     
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