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Block on inclined plane with pulleys and friction

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data
    nE4Vv.png
    coefficient of kinetic friction = 0.15

    Find the acceleration of block a.

    2. Relevant equations
    see below

    3. The attempt at a solution
    http://www.wolframalpha.com/input/?i=-200sin30%2B0.15*200cos30%2B2x%3D-200%2F32.2*a%3Bx-40%3D40%2F32.2*b%3Ba%3D0.5*b

    I used x for tension and a and b for the acceleration of the blocks.

    I assumed that the acceleration of block a is down the plane and so block b would go up, but the answer I got was negative! I tried assuming that block a goes up and block b goes down, but I got a negative answer again! Are the equations that I used wrong? or the blocks don't move at all?
     
  2. jcsd
  3. Feb 19, 2012 #2

    tiny-tim

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    hi phlstr! :smile:

    your geometric constraint b = 0.5a is wrong, it should be b = -0.5a

    (look at the diagram! :wink:)
     
  4. Feb 19, 2012 #3
    Hi :). But I already made the net mass*acceleration negative for A and the net force for B positive. I used the geometric constraint just for the magnitude..

    and I think you mean a=0.5b. :shy:
     
    Last edited: Feb 19, 2012
  5. Feb 19, 2012 #4

    tiny-tim

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    no, both equations are measuring up as positive

    (we can see that eg from the fact that your x (tension) is positive both cases)

    so if a increases, b decreases :wink:
     
  6. Feb 19, 2012 #5
    What do you mean by measuring up as positive?
     
  7. Feb 19, 2012 #6

    tiny-tim

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    i mean uphill for A, and directly up for B
     
  8. Feb 19, 2012 #7
    So these are my equations:

    [itex]-200sin\left(30\right)+(0.15)(200)cos\left(30\right)+2T=-\frac{200}{32.2}a[/itex]
    I use negative values for forces pointing downhill, and positive for pointing up. Since I assumed that the net force for block A is going down, I made the right side negative..

    [itex]T-40=\frac{40}{32.2}b[/itex]
    Negative is down and positive is up. Since block A is going downhill, then block B is accelerating upwards and the net force is upwards.

    [itex]a=0.5b[/itex]
    The magnitude of the acceleration of block A is half of the acceleration of block B.

    This is how I understand the equations that I'm using.. Can you point where it's wrong?
     
  9. Feb 19, 2012 #8

    tiny-tim

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    ah, now that i see it on the same page, it does looks right :smile:

    ok, i get positive times a on the LHS = a positive number on the RHS,

    what did you get?
     
  10. Feb 19, 2012 #9
    I plugged the equations in wolfram alpha and got
    a=-0.534946 b=-1.06989, T=38.6709

    I also tried using substitution and got the same for A...
    and it's negative??
     
  11. Feb 19, 2012 #10

    tiny-tim

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    let's see :smile:
     
  12. Feb 19, 2012 #11
    [itex]-200sin\left(30\right)+(0.15)(200)cos\left(30\right)+2T=-\frac{200}{32.2}a[/itex]
    [itex]T-40=\frac{40}{32.2}b[/itex]
    [itex]a=0.5b[/itex]

    [itex]T=\frac{40}{32.2}b+40[/itex]
    [itex]2a=b[/itex]
    [itex]T=\frac{40}{32.2}2a+40[/itex]

    [itex]-200sin\left(30\right)+(0.15)(200)cos\left(30\right)+2\left(\frac{40}{32.2}2a+40\right)=-\frac{200}{32.2}a[/itex]

    [itex]-200sin\left(30\right)+(0.15)(200)cos\left(30\right)+\frac{160}{32.2}a+80=-\frac{200}{32.2}a[/itex]

    [itex]-200sin\left(30\right)+(0.15)(200)cos\left(30\right)+80=-\frac{200}{32.2}a-\frac{160}{32.2}a[/itex]

    [itex]5.980762114=-11.18012422a[/itex]

    [itex]a=-0.5349459[/itex]




    :smile:
     
  13. Feb 19, 2012 #12

    tiny-tim

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    oh … very small and negative!

    ie uphill

    so if the kinetic friction is uphill, the motion would have to be uphill also, which is impossible

    and if we make the kinetic friction downhill, then that +25 becomes -25, and the motion is also downhill, again impossible

    so the blocks don't move

    you could find the friction force F by solving the same pair of equations, with 0 on the RHS and F in the middle of the LHS :wink:
     
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