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Homework Help: Finding Acceleration Of A Bucket Moving Downwards With A Pulley

  1. Sep 26, 2007 #1
    1. The problem statement, all variables and given/known data
    A 26.5 kg block is connected to an empty 1.00 kg bucket by a cord running over a frictionless pulley (Fig. 4-57). The coefficient of static friction between the table and the block is 0.435 and the coefficient of kinetic friction between the table and the block is 0.320. Sand is gradually added to the bucket until the system just begins to move.

    2. Relevant equations
    So far I have solved the mass of the sand added to the bucket, which is 10.528kg, ignoring the mass of the bucket. I am still unsure of how to setup the equation, because I am confused on what I use for tension. The tension I originally used should have changed due to the mass of the sand now added to the bucket. The first time I solved for tension, I didn't need acceleration, as the bucket was at rest. Every equation I try to set up comes out needing two variables, neither of which I have any idea how to solve for.

    3. The attempt at a solution
    When I try to find tension, my equation comes out to one of these two (Sorry if I write these wrong, still figuring out how to type out an equation).

    For the Tension in the block, i get T=[tex]\mu[/tex]N + ma, which requires the use of either tension or acceleration to solve. I got this equation by using a free-body diagram, which had weight going down the y-axis, normal force going up the y-axis, tension to the right on the x-axis, and Friction going left on the x-axis. The sum of my forces would then be Tension minus friction equals mass times acceleration, so I moved friction over, and converted it to mu times normal force.

    My Other equation is for the bucket, which is T=mg +ma. For this one, ny free body has two forces acting on it, both on the y-axis. Tension going up, weight moving it down. My sum of forces came out to be Tension minus weight equals mass times acceleration.

    What am I missing here, or am I over complicating things? :confused:
  2. jcsd
  3. Sep 26, 2007 #2
    Ok, If I'm understanding what it is you're trying to solve for, it is the tension in the line between the two blocks. You have already correctly calculated the additional sand mass that must be added into the bucket to overcome the static friction, so we are now working with a kinetic friction acting on the block located on the table.

    To get to the tension, you first want to solve for what the acceleration of the two bodies is together (this can be done without breaking it down into separate free body diagrams knowing that the tension in the line must be constant). You can then use that acceleration and go back to the free body diagram of either block to solve for the tension in the cord.

    This is my first post on here and I'm not too sure of how much information is normally given (I dont want to just hand the solution over). I hope this can at least send you in the right direction. Let me know if you need more help.
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