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3 pulleys - 2 masses on incline plane

  1. Dec 17, 2015 #1
    1. The problem statement, all variables and given/known data
    http://tinypic.com/r/qs9q4o/9 http://tinypic.com/r/qs9q4o/9
    Verify that the log will move up the ramp under the given conditions
    2. Relevant equations


    3. The attempt at a solution

    m1 = 220

    m2 = 130

    mu = 0.4

    theta = 25

    g = 9.81

    N = (m1*g)*cos(theta)


    1) m2g – T = m2a Forces acting on Block B

    2) 2T – m1g*sin(theta) – mu*N = m1a Forces acting on log


    1) T = m2g – m2a

    2) 2 (m2g – m2a) – (m1g*sin(theta)) – mu*N = m1a


    2m2g – 2m2a – (m1g*sin25) – (0.4 * N ) = m1a

    (220*130*9.81) – (220*9.81*sin25) – (0.4*1956) = m1a + 2m2a

    (2550.6) – (912.1) – (782.4) = a(m1+2m2)

    856.1 / (m1+2m2) = a

    856.1 / (220+260) = a


    a = 1.78m/s^2



    Unsure about the forces acting on the double pulley. I am right to assume it is 2T? Does this give me the correct value of 'a' to assume the log is accelerating up the slope?
     
  2. jcsd
  3. Dec 17, 2015 #2
    Yes. Is that incorrect?
     
  4. Dec 17, 2015 #3

    haruspex

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    At the same time as you replied, I deleted my post because I noticed you later had 2a terms, so I thought I should check whether you had defined a as the acceleration of one and were using 2a for the acceleration of the other. But having checked, no, it seems you did use a for both.
    The rope is constant length. As the weight descends a distance y, what happens to the three sections of rope?
     
  5. Dec 17, 2015 #4
    The rope carrying Block B increases as the block lowers. The two lengths of rope carrying the log stay equal to each other and shorten. Is that what you were asking?
     
  6. Dec 17, 2015 #5

    haruspex

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    Yes, but a bit more detail... if block B descends a distance y, how much do the two lengths holding the log shorten by?
    What does that tell you about accelerations?
     
  7. Dec 17, 2015 #6
    If block B decends by 1m, then the log will move 2m due to the two pulleys?
     
    Last edited: Dec 17, 2015
  8. Dec 17, 2015 #7

    haruspex

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    No. One step at a time, no wild guesses please.
    You wrote, correctly, that the two ropes parallel to the slope remain the same length as each other. We also know the total length of the rope is constant. Write that as an equation relating the lengths of the sections.
     
  9. Dec 17, 2015 #8
    Is the rope pulling the log two times the length of the rope supporting the block?
     
  10. Dec 17, 2015 #9
    I got nearly the same the answer (rounding error) so that could be a good sign. I treated the force on the double pulley as 2x like you, im pretty sure thats the way to do it, I could be wrong though. A little tip, I like to break down each force into each direction, form a sum of forces in each direction then form a triangle of the resultant force (being the hypotenuse), just the way my teacher taught me. It makes it nice and simple for figuring out which force is going where.
     
  11. Dec 17, 2015 #10

    haruspex

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    I said no wild guesses.
    Let the total length of the rope be L.
    If the vertical length is y, how long are the other two sections?
     
  12. Dec 17, 2015 #11
    I believe the other two sections would both be y also?
     
  13. Dec 17, 2015 #12
    If we are saying that force from the block is doubled due to the set up of the double pulley then something has to compensate... the distance moved.
     
  14. Dec 17, 2015 #13

    haruspex

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    That would make the total length 3y, but, since the total length is constant, y would be constant, so the weight cannot move?
    Please, please, please try writing equations to represent the clear facts:
    The total length of the string is constant;
    The two lengths parallel to the slope are equal in length.
    I will not respond further until you show an attempt to do that.
     
  15. Dec 17, 2015 #14
    Yes i agree. Would this affect the acceleration? Would it make the acceleration 2*a at the log?
     
  16. Dec 17, 2015 #15
    Previously you stated the total length of the rope was L and vertical was y. In my equations i have labelled the forces acting on the ropes as 2T at the log and T acting against the block. I don't understand what you're asking?
     
  17. Dec 17, 2015 #16

    haruspex

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    It is not (directly) related to the forces. It is just a matter of the way the lengths of the sections relate to each other.
    If each of the sections parallel to the slope is length x, and the vertical section is length y, what equation relates x, y and the total length L?
     
  18. Dec 17, 2015 #17
    Is it as simple as 2x+y=L?
     
  19. Dec 17, 2015 #18
    Think about the equation of acceleration, distance/time squared, if the log is moving less distance in the same time then....
     
  20. Dec 17, 2015 #19
    It is accelerating slower?
     
  21. Dec 17, 2015 #20

    haruspex

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    Yes!
    Now, what does that tell you about the accelerations? What do you do to a distance variable to turn it into an acceleration variable?
     
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