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Mass of a block via kinetic friction

  1. Feb 15, 2008 #1
    [SOLVED] Mass of a block via kinetic friction

    1. The problem statement, all variables and given/known data

    In Fig, two blocks are connected over a pulley. The mass of block A is 5.9 kg and the coefficient of kinetic friction between A and the incline is 0.36. Angle θ of the incline is 43°. Block A slides down the incline at constant speed. What is the mass of block B?


    2. Relevant equations

    3. The attempt at a solution

    OK so I'm unsure of how how to compute this question. I'm going to need to find the tension in this problem in order to find the mass of block two correct?

    So for m1:
    Fy = N-mg+(Tsin43-(Tsin133 [90+43 for the Fk force]*.36))

    How do I solve for T here if I don't have a second equation to help me solve? There isn't a normal force on block B seeing as it is in the air so I couldn't impose the two to get rid of N right?

    Fx = Tcos(43)-Tcos(133)*.36

    Is this correct?

    For m2:
    Fy = T-mg
    Fx = No movement

    T>mg or block A would not be in kinetic friction.

  2. jcsd
  3. Feb 15, 2008 #2

    Doc Al

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    Staff: Mentor

    This will be a bit easier to analyze using axes that are parallel and perpendicular to the plane, instead of horizontal and vertical.

    If T > mg, the blocks would accelerate.
  4. Feb 15, 2008 #3
    Well thats sort of what I'm trying to do. I'm making the friction and tension south west and north east respectively. I kind of figured the way I set that up was incorrect but I haven't had to do two forces at the same time before.
  5. Feb 15, 2008 #4

    Doc Al

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    Staff: Mentor

    For mass A, choose your x-axis as parallel to the incline and your y-axis as perpendicular. Then only the weight will need to be broken into components.
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