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A 2 Block Swing

  1. Aug 2, 2013 #1
    1. The problem statement, all variables and given/known data
    A box of mass M is at rest on a horizontal table. The coefficient of static friction between the box and the table is k. The box is connected by a taut string to a block of unknown mass m , initially held horizontally as shown and then released. The box begins to slide at the instant the block reaches the bottom point of its swing. Find the mass m of the block.


    2. Relevant equations



    3. The attempt at a solution
    I took the force of gravity, mg, and set it equal to the friction force between the box and the table, kMg, getting m=k*M.
     

    Attached Files:

  2. jcsd
  3. Aug 2, 2013 #2

    gneill

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    Gravity and friction are not the only forces involved. The block m is following a circular path...
     
  4. Aug 2, 2013 #3
    Is the third force angular acceleration?
     
  5. Aug 2, 2013 #4

    gneill

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    Acceleration is not a force. What force is invoked to cause a body to move in a circular path?
     
  6. Aug 2, 2013 #5
    Is it a centripetal force?
     
  7. Aug 2, 2013 #6
    Yes, the mass must experience a centripetal force in order to move in a circular path. So how is this force provided?
     
  8. Aug 3, 2013 #7
    The force is provided by the circular swinging of the rope and is equal to mv^2/r. Is that right?
     
  9. Aug 3, 2013 #8

    gneill

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    The rope is the means by which the centripetal force is applied to the moving mass, yes. In addition, the rope also must supply a force that counteracts the weight of the mass at the bottom of the swing. Both forces are then acting in the same direction. What's the total?
     
  10. Aug 3, 2013 #9
    The total is mg+mv^2/r. Is that correct?
     
  11. Aug 3, 2013 #10

    gneill

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    Yes. You have enough information to continue. So rather than inching along one confirmation at a time, how about you take a stab at the full solution now?
     
  12. Aug 3, 2013 #11
    Ok, I got m=2gkMr/(2^5/6*v^2+2*g*r). Is that right?
     
  13. Aug 3, 2013 #12

    gneill

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    Show your work. You've introduced new variables...

    What is v? You should be able to determine v from the given information (I assume that r is given, although you didn't say so, right?)
     
  14. Aug 3, 2013 #13
    r is not given in the problem and I can't figure out what v is.
     
  15. Aug 3, 2013 #14

    gneill

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    If r is not given, then you have no choice but to introduce it as a variable. No problem. But you should be able to determine v, after all, it's just a mass falling though some distance in a gravitational field. Think about conservation rules.
     
  16. Aug 3, 2013 #15
    Is v equal to pi*r*g/2, giving a final answer of m=k*M/(g*r)?
     
  17. Aug 3, 2013 #16

    gneill

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    No. There's no pi involved; you want the change in height (in the gravitational field) in order to find the change in potential energy. What energy is it "traded" with, and what expression defines it?
     
  18. Aug 3, 2013 #17
    Ok so is v=sqrt(2*g*r)?
     
  19. Aug 3, 2013 #18

    gneill

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    Yup. Carry on.
     
  20. Aug 3, 2013 #19
    Ok so the final answer is 1/3*k*M. Is that right?
     
  21. Aug 3, 2013 #20

    gneill

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    Looks good! :smile:
     
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