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Acceleration of mass with friction

  1. Aug 24, 2015 #1

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

    At the instant shown the block is moving down the slope with P= 49 N, β= 38°, θ= 33° and M= 33 kg.


    2. Relevant equations

    What is the acceleration of the block up the slope if the kinetic coefficient of friction is 0.5?
    3. The attempt at a solution
    So I split all the forces into the normal direction and parallel to the plane

    Forces in the normal = 0 = N - mgcos33 + 49sin38
    N = 241.34

    Forces parallel to the plane = 49cos38 - mgsin33 + 0.5N
    = -17.036
    F/m = a
    a = -17.036/33 = -0.516
    But I am not getting the right answer, please help
     
  2. jcsd
  3. Aug 24, 2015 #2
    Draw the FBD for mass M.
     
  4. Aug 24, 2015 #3
    Photo on 25-08-2015 at 12.00 pm.jpg
     
  5. Aug 24, 2015 #4
    Now what, I added up all the forces in each direction. Not sure where I went wrong
     
  6. Aug 24, 2015 #5
    Oh, my earlier post didn't load.

    First of, does friction oppose or support the motion of the block ?
    And second, is the force on M being exerted by P, or the pulley ?

    Hope this helps.
     
  7. Aug 25, 2015 #6

    haruspex

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  8. Aug 25, 2015 #7

    haruspex

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    We are told that acceleration is up the slope but the present velocity is down.
     
  9. Aug 25, 2015 #8
    I think the force I am missing is tension in the pulley,
    Not sure but is it mgsin(theta)
     
  10. Aug 25, 2015 #9

    haruspex

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    Not sure what you mean by tension "in" the pulley. The rope runs around the pulley. It will be more convenient to think of it as two separate ropes, but you need know the tension in each (which will be?)
     
  11. Aug 25, 2015 #10
    Would it be p + mgsin(theta)
     
  12. Aug 25, 2015 #11

    haruspex

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    What is the tension in the upper part of the rope?
     
  13. Aug 25, 2015 #12
    Is it 49N ?
     
  14. Aug 25, 2015 #13
    Oops ! Missed the down.
     
  15. Aug 25, 2015 #14
    so P + mgsin(theta) + mgcos(theta)*friction
     
    Last edited: Aug 25, 2015
  16. Aug 25, 2015 #15
    Sorry my bad..
     
    Last edited: Aug 25, 2015
  17. Aug 25, 2015 #16

    haruspex

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    Yes, it's P, which is given as 49N.
    It doesn't say, but you should take the pulley as massless and having no axle friction. If the tension in the upper part of the rope is P, what must it be in the lower part? Hint: think about the net torque on the pulley.
     
  18. Aug 25, 2015 #17

    haruspex

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    Please do not provide answers. The system on the homework forums is that we provide hints, explain misconceptions, point out algebraic errrors, etc.
     
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