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Incline, two boxes, determine acceleration

  1. Jan 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Two masses M1 = 2.5kg and M2 = 4.0kg are on inclines and are connected together by a string . The coefficient of kinetic friction between each mass and its incline is Uk = 0.30.

    If moves up, and moves down, determine their acceleration.

    [​IMG]


    2. Relevant equations



    3. The attempt at a solution

    M1:

    F=ma
    x:T-0.30N-(2.5)(9.8)SIN51 = 2.5a
    y:N-(2.5)(9.8)cos51=0
    N=(2.5)(9.8)cos51

    T - 0.30((2.5)(9.8)cos51) - (2.5)(9.8)sin51 = 2.5a
    T= 2.5a + 0.30((2.5)(9.8)cos51) + (2.5)(9.8)sin51

    M2:

    x: (4.0)(9.8)sin21 - T - 0.30N = 4.0a
    y: N-(4.0)(9.8)cos21 = 0

    (4.0)(9.8)sin21 - T - 0.30((4.0)(9.8)cos21) = 4.0a
    -T = 4.0a - (4.0)(9.8)sin21 + 0.30((4.0)(9.8)cos21)
    T = -4.0a + (4.0)(9.8)sin21 - 0.30((4.0)(9.8)cos21)

    therefore

    2.5a + 0.30(m1gcos51) + m1gsin51 = -4.0a + m2gsin21 - 0.30(m2gcos21)

    6.0a = -0.30((2.5)(9.8)cos51) - (2.5)(9.8)sin51 + (4.0)(9.8)sin21 - 0.30((4.0)(9.8)cos21)

    a = -0.31


    This answer is negative and I think it should be positive, but I am not even sure if I have taken the appropriate steps.

    Thanks for any help!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 29, 2009 #2

    Delphi51

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    I fear you are off to a bad start. A lot of it is right in the details, but I think you are missing the whole idea of the thing. This
    x:T-0.30N
    suggests you are thinking in terms of x and y forces (not correct) and that the 0.30 is a force when in fact it is the coefficient of friction (with no units).

    You must resolve the mg force on each mass into a component that goes parallel to the ramp and a component that goes perpendicular to the ramp. The part perpendicular to the ramp is a "normal" force that causes a friction force UP the ramp, opposing the component parallel to the ramp. So two more calcs for the friction forces using that coefficient. Then you will know the net force down the ramp on each side.
     
  4. Jan 29, 2009 #3
    You must resolve forces parallel and perpendicular to the plane. The friction is the coefficient of friction multiplied by the 'normal' force.

    EDIT: beaten to it
     
  5. Jan 29, 2009 #4
    Isnt the normal force just mg(cos51) for the first box?'

    If not please explain why.
     
  6. Jan 29, 2009 #5

    Delphi51

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    Yes.
     
  7. Jan 29, 2009 #6
    Thats what I used in my original work.

    I took the mgcos51 and found that friction = 0.30(mgcos51)

    then I used that when determining

    F=MA

    in the x direction:

    T - (friction force) - x component of mg = ma

    T - 0.30(N) - mgsin51 = 2.5a

    I then solved for T with the other box.

    Please show me where my misstep is
     
  8. Jan 29, 2009 #7

    Delphi51

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    Ah, I guess I just had trouble understanding your solution. The "x component" is certainly confusing, suggesting a horizontal component and you haven't written the formula for the friction force anywhere . . . have mercy on your marker or you will receive no mercy in return!

    You must have an incorrect sign here since the friction force and the parallel component are in opposite directions.
     
  9. Jan 29, 2009 #8
    I do not understand.

    T - 0.30(N) - mgsin51 = 2.5a

    T is in the direction of the pulley so it is positive

    Friction is in the opposite direction of the acceleration so it is negative

    mg according to my coordinate system would also have negative x,y components.

    I am not sure which sign is incorrect.
     
  10. Jan 29, 2009 #9

    Delphi51

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    Oops, I forgot that the mass might be accelerating upward!
    I don't know whether the .3FN has the same sign as the mg sin(51) or not!
    Won't know that until you find the down-the-ramp forces on both sides.
    Then I would make the larger one positive, subtract the smaller one and subtract both friction forces to get ma. And watch out for the possibility of there being no acceleration because the net pulling force is less than the total friction force.

    You probably have a more sophisticated method than I do with your tension formula.
    I'll work out the numbers and see what I get.
     
  11. Jan 29, 2009 #10

    Delphi51

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    Okay, I'm getting about 19 N down the ramp on the left and about 14 N down the ramp on the right. That is a net of about 5 N. Nowhere near enough to overcome the combined friction forces on both sides. So the acceleration will be zero.
     
  12. Jan 29, 2009 #11
    Unfortunately that answer is not accepted by my online workbook. Thanks for your efforts, I'll take this one in to the prof to find out more. I don't expect u to work so hard on one questions.

    Thanks!
     
  13. Jan 29, 2009 #12

    Delphi51

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    Thanks to you - I think I learned how to do this kind of problem.
     
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