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A Man Raises A Platform

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data
    A 27.5 kg person stands on a 55 kg platform. He pulls on the rope that is attached to the platform via the frictionless lower-right pulley. He pulls the rope at an angle of 29◦ to the horizontal, as shown in the figure below. Assume: g = 9.8 m/s2 . Ignore friction. The platform remains level.

    If he pulls the platform up at a steady rate, how much force is he pulling on the rope?

    2. Relevant equations

    3. The attempt at a solution
    I don't know where to start. I don't understand what my free body diagram would look like. Would it be the weight of the person and platform combined, with Force up and w down? I'm just really confused.
  2. jcsd
  3. Sep 21, 2010 #2
    well its not a complicated FBD, the man standing with the combination of the platform contribute a FG downwards, but the platform contributes a Normal forces to keep the system in the Y direction in equilibrium (before he pulls on the rope).

    initially (not pulling on rope)

    Fy = N - Fg = 0; therefor Fg = N

    when pulling the rope, you break down your y component of your force

    Fy = N - Fg + Fpsin(t) = ma; Fp is the force of you pulling on the rope, but you only want the y component.


    Fx = Fpcos(t) = 0 ; but the guys trying to go up, not side to side; therefor its zero

    hope that helps

    Attached Files:

    • FBD.png
      File size:
      4 KB
    Last edited: Sep 21, 2010
  4. Sep 22, 2010 #3
    All of that makes sense but then I get that:

    I don't have acceleration to find out what the force should be.

    I have answer choices:

    1.F = 269.5 N
    2. F = 323.4 N
    3. Cannot be determined.
    4. F = 808.5 N
    5. F = 404.25 N
    6. F = 202.125 N
    7. F = 539 N
    8. F = 606.375 N
    9. F = 485.1 N
  5. Sep 22, 2010 #4
    The point is that there is no acceleration as the man is pulling at a steady rate
  6. Sep 22, 2010 #5
    Oh! My bad.

    But wouldn't it be
    Fy = Fpsin(t) = 0

    Meaning F = 0?

    But I know that's not right.
  7. Sep 22, 2010 #6
    So, if he's pulling himself up at a constant rate, acceleration is zero. But what do I do with that in the situation of Force? The force isn't zero because he is exerting a force.
  8. Sep 22, 2010 #7
    I'm lost as what to do from here, then:

    If I don't have acceleration, what can I do with this formula?
  9. Sep 22, 2010 #8
    Forces in Y = -(M1+M2)g+Fsin(t) = 0; //M1 = mass of man, M2 = mass of plank

    solve for F
  10. Sep 22, 2010 #9
    -(M1+M2)g+Fsin(t) = 0; //M1 = mass of man, M2 = mass of plank

    Okay, so:
    F=1667.664927 N

    Which isn't an answer choice.
  11. Sep 23, 2010 #10
    Anyone? I have to finish this tonight.
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