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Atwood Incline Pulley problem with two pulleys

  1. May 21, 2015 #1
    The problem statement, all variables and given/known data

    Given that a weight m1 is attached to a string going over a perfect pulley attached to another pulley suspending weight m2, and that weight m1 is on an incline with an angle of 37 degrees, find acceleration, tension, which mass hits the ground first, time needed to hit the ground, and max possible static friction. Note that m2 < m1. m1 is 2m from the bottom of the incline and m2 is 1m off the ground. Note that my teacher didn't actually write out a problem for us, so I'm trying to write this out as coherently as possible based on the drawing of a system we were given. So if something about this doesn't make sense please tell me.

    Okay, so for a simple Atwood incline problem I kinda know what to do, but what about with double pulleys? It shouldn't be that confusing because I just have to use 2T for the weight accelerating upwards (m2) since it's attached to a pulley and not just the string attached to m1, right? I'll attach an image to see if that clears up what I mean. I've put down my work for most of the problem and I'm wondering if someone could help me check it or tell me how I can confirm I've done a problem like this right and self-check it. Also I completely don't know how to find the max possible static friction for a system to still work, can anyone explain this to me?

    rh7h9j.jpg

    The attempt at a solution

    T = m2*g/2
    T = m2*g/2 = m1*g*sin 37°+m1*a
    g cross
    m2 = 1.2m1+2m1a
    a1 = (m2-1.2m1)/2m1
    2ℓ = 4.00 = a*t^2
    t = √4/a = √8m1/(m2-1.2m1)
    Vf1 = a1*t
    a2 = 2h/t^2 =2(m2-1.2m1)/8m1 = (m2-1.2m1)/4m1 = a1/2

    Where m1 = 10 kg and m2 = 15 kg,

    a1 = (m2-1.2m1)/2m1 = (15-12)/20 = 3/20 = 0.15 m/sec^2
    t = √8m1/(m2-1.2m1) = √8*10/(15-12) = 5.164 sec
    Vf1 = a1*t = 0.15*5.164 = 0.775 m/sec
    a2 = a1/2 = 0.15/2 = 0.075 m/sec^2
    Vf2 = a2*t = Vf1/2 = 0.387 m/sec
     
  2. jcsd
  3. May 21, 2015 #2

    haruspex

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    You're overlooking m2's acceleration. Consider F=ma for m2.
     
  4. May 21, 2015 #3
    Yup, I did. I went back and re-resolved F-tension for both parts.

    For F-tens. (m2) I have (m2a+m2g)/2

    For F-tens. (m1) I have (m1g sin(37))-(u m1g cos(37))-(m1a)

    Does this look better? I think I've got it now but I'm just checking.

    2a4wj92.jpg
     
  5. May 21, 2015 #4

    haruspex

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    Yes, those tension equations look right. (Haven't checked the image.)
     
  6. May 21, 2015 #5
    Okay, I'm a little stuck. I got an acceleration a1 of 2.72, thus a2 is 1.36, t then equals 1.213, v1f = 3.3 m/s and v2f = 1.65 m/s. m1 will hit the ground first. However, I still don't understand how to calculate max possible friction for the system to still move, I just looked online for help but I'm not really getting it.
     
  7. May 21, 2015 #6

    haruspex

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    I don't understand how you got a value for the acceleration without substituting a value for u. I don't see any value given. At the max value of u (or arbitrarily close to it), the acceleration becomes arbitrarily close to 0 and the time goes off to infinity.
     
  8. May 21, 2015 #7
    Oh, wow. I forgot to put up the givens for my problem. I was waiting to substitute them in, sorry. u=.05, m1=15kg,m2=10kg, g=10m/s^2
     
  9. May 21, 2015 #8

    haruspex

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    The acceleration you quote in post #5 seems too much. Please post your working.
     
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