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Pulleys, tension, and blocks oh my!

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data
    In the figure, the pulleys and the cords are light, all surfaces are frictionless, and the cords do not stretch.
    (a) How does the acceleration of block 1 compare with the acceleration of block 2? (Use a_2 for a2.)
    (b) The mass of block 2 is 1.22 kg. Find its acceleration as it depends on the mass
    m1 of block 1. (Use m_1 for m1.)
    (c) Evaluate your answer for m1 = 0.537 kg. Suggestion: You may find it easier to do part (c) before part (b).
    (d) What does the result of part (b) predict if m1 is very much less than 1.22 kg?
    (e) What does the result of part (b) predict if m1 approaches infinity?
    (f) What is the tension in the long cord in this last case?
    (g) Could you anticipate the answers (d), (e), and (f) without first doing part (b)? Explain your answer.


    2. Relevant equations
    F = ma


    3. The attempt at a solution

    Alright.. so here's my force diagram.

    http://binary-snobbery.com/pics/physics09072009.jpg

    I labelled all of the tensions the same, because... I don't know. I just figured they'd all be the same... with the exception of the 2T which is just T+T.

    A/B) a1 = 2a2. I don't really know why, but it was my best guess, because m2 has 2T acting as an opposing force to its natural tendency toward motion.

    C) I tried this:
    m1:
    [tex]\Sigma[/tex] Fx = T = m1a1

    m2:
    [tex]\Sigma[/tex] Fy = 2T - m2g = m2a2

    So then I used the identity found in part A to solve the system for a2.

    2(m1a1) - m2g = m2a2
    2(m1(2a2)) - m2g = m2a2
    4m1a2 - m2g = m2a2

    Do a little algebra and you end up with:
    a2 = m2g / 4(m1 - m2)

    I plugged in the given values, but I got the wrong answer. That's basically where I gave up. I am missing something fundamental, I guess.
     
    Last edited: Sep 8, 2009
  2. jcsd
  3. Sep 8, 2009 #2

    rl.bhat

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    For m2 it should be
    Fy = m2*g - 2T.
     
  4. Sep 8, 2009 #3
    Why? That's completely counterintuitive.

    If you think of going UP as positive acceleration on the y axis and going DOWN as negative acceleration (as I think most of the physics world does), then g is -9.8m/s2 and not 9.8m/s2. Hence,

    Fy = 2*T - m2g
     
  5. Sep 8, 2009 #4

    rl.bhat

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    In that case Fy should be negative, because it is in the direction of g
     
  6. Sep 8, 2009 #5
    Well, I just thought about how "Direction" in the case of acceleration is whether or not the velocity is decreasing or increasing... Not so much the "direction" of the acceleration. I guess you could say that if something is slowing down, it's accelerating in the opposite direction that it's headed. Anyway.. I think really we're splitting hairs.

    I guess what I'm trying to say is that... I'm not sure what you're trying to say.
     
  7. Sep 8, 2009 #6

    rl.bhat

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    When the system is released, in which direction m2 will move?
    If it moves in the down ward direction what will be its acceleration?
    When you drop a body, what happens to it velocity? Increases or decreases?
    If it increases what is the sign of g?
     
  8. Sep 8, 2009 #7

    ideasrule

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    The problem is that Fy is a negative number if you define it as 2T-m2g instead of m2g-2T. That means a2 would be negative, which means a1 doesn't equal 2*a2; it's equal to -2*a2.

    I usually like to keep all forces positive; it's much more intuitive that way.
     
  9. Sep 8, 2009 #8
    all this discussion can someone answer part A?
     
  10. Sep 8, 2009 #9

    ideasrule

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    The original poster already gave the right answer for part A:

    a1=2a2

    He made a mistake with the signs later on, but there's no problem with that equation.
     
  11. Sep 8, 2009 #10
    But forces are vectors :(

    I looked at the problem some more tonight, and I'm just lost. I setup an appointment with a tutor tomorrow (fortunately, our department has free personal tutoring). I'm hoping they can help shed some light on it.
     
  12. Sep 8, 2009 #11
    yea i read it after i posted my bad
    but how do you figure part b
     
  13. Sep 8, 2009 #12
    what school do you go to
     
  14. Sep 8, 2009 #13
    Well.. If I'm reading the responses, so far, correctly... I was on the right track. If you know that a1 = 2*a2, then you can use that identity to get an equation out of the force equations for each of the masses. I still don't understand where that identity comes from, though.
     
  15. Sep 8, 2009 #14
    i still can't figure out b.
    if anyone can help please email me OSearcy4@aol.com
     
  16. Sep 8, 2009 #15

    rl.bhat

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    The acceleration in m2 is given by
    a2 = (m2*g - 2T)/m2
    But T also equal to m1*a1
    Substitute this value in the above equation and put a1 = 2*a2, find the value a2.
     
  17. Sep 9, 2009 #16
    After thinking about it for a while today, I figured out part a..

    The distance that m2 travels downward as m1 travels to the right is 1/2 the distance travelled by m1, i.e.
    x_1 = 1/2 x_2
    because of the pulley situated above m2 and its connection to a stationary object on the other side of the gap.

    They each move this distance (x, and .5x respectively in the same amount of time). This means that their velocities would be of the form:
    v1 = xt and v2 = x/2 * t

    If you take the derivative of these velocity functions wrt to time, you arrive at:
    a1 = x and a2 = x/2

    Solve a2's equation for acceleration for x, and you are presented with the relationship between a1 and a2:
    a1 = 2a2.

    What I'm stuck on is the tension in the rope...

    We're assuming that the tensions on either side of the lower pulley (above m2) are equal to each other. Why can we make that assumption?
     
  18. Sep 9, 2009 #17
    Also, I've finished the problem... but I don't understand why the tensions are equal.
     
  19. Sep 9, 2009 #18

    ideasrule

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    It's because the rope is assumed to be massless. Consider the first cm of the rope; because ma=0, the tension pulling it forwards must be equal to the tension pulling it backwards. The same argument applies to the 2nd cm, and the 3rd, and so on. It follows that tension is the same throughout a massless rope. Real ropes are amazingly light for their strength, so this approximation works well in the real world.
     
  20. Sep 9, 2009 #19
    Oh. I'd sort of intuited that, but I was pretty doubtful of myself. I was imagining the two lengths of the rope to the right and left of the lower pulley, but I couldn't decide what I thought about the tension. Thinking about it again, though... it makes perfect sense.

    Thanks for all the help!
     
  21. Sep 13, 2009 #20
    I think it should be :

    displacement of m1 = x
    displacement of m2 = 0.5x

    then,

    V1 = x1 / t => xt-1
    dv/dt = -1xt-2
    a1 = -xt-2

    V2 = 0.5x / t => 0.5xt-1
    dv/dt = -0.5xt-2
    a2 = -0.5xt-2

    then a1 / a2 = 2/1
    then a1 = 2a2

    please correct me if I am wrong.

    by the way, I also have no idea how to figure out the displacement of m1 is equal to half of m2. Hope someone could kindly explain to me =)
     
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