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Pulley System Where String Breaks

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data

    Two particles A and B, of masses 2m kg and m kg respectively, are connected by a light inextensible string which passes over a fixed smooth pulley. The system is RELEASED from REST, with both portions of the string vertical and taut, while A and B are at the SAME HEIGHT.

    1. Find the magnitude of the acceleration of the particles and the tension in the string.

    The string breaks when the speed of each particle is u m/s. Find, in terms of u, the difference in height between the particles A and B:

    2. When the string breaks
    3. When B reaches its highest point, assuming that A has not reached the ground and B has not reached the pulley.

    Find the speed of A when B reaches its highest point.


    2. Relevant equations

    F = ma
    g = 10 m/s^2


    3. The attempt at a solution
    (2m)(a) = (2m)(10 m/s^2) - T -- (1)
    (m)(a) = T - (m)(10) -- (2)

    (2m)(10) - (2m)(a) = (m)(a) + (m)(10)... Rearranging this in order to solve for a and I got
    a = 3.33 m/^2 -- (3)

    Plugging (3) in (1) or (2) in order to solve for T and I got 13.33


    Now can anyone help me with the rest? I'm lost with #2 and #3.
     
  2. jcsd
  3. Apr 3, 2013 #2

    Doc Al

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    Staff: Mentor

    Well, if you've found the acceleration then make use of it. How long does it take to reach a speed of u?

    Once the string breaks, what's the acceleration of the masses?
     
  4. Apr 3, 2013 #3
    I would use the kinematic equation vf = vi + at with my vi = 0 (since the system was released from rest) and my vf to be u. Solving for time, t, I'd get t = u/3.33

    Wouldn't the acceleration the moment the string breaks still be 3.33 m/s^2 ??
     
  5. Apr 3, 2013 #4

    Doc Al

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    Good. Keep going. How far has each moved during that time?

    Just after the string breaks, what forces act on the masses?
     
  6. Apr 3, 2013 #5
    Okay. So I can get the respective displacements of each particle by using df = di + vt + (1/2)(a)(t^2), where v = u, and t = u/3.33. What I got for A was df = di - (u^2)/3.33 + (u^2)/2 [I subtracted them since A would be going down] while for B df = di - (u^2)/3.33 + (u^2)/2 [Added them since B would be going up]. Since the problem is asking for their difference, I subtracted these 2, leaving me with 2[(u^2)/3.33 + (u^2)/2]. Is that right?

    I'm still not sure for #3 though. Still thinking it through.
     
    Last edited: Apr 3, 2013
  7. Apr 3, 2013 #6

    Doc Al

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    Careful. u is the final velocity, not the initial. In that formula, v stands for the initial velocity.
     
  8. Apr 3, 2013 #7
    Yeah, but at the moment the string breaks the initial velocity would be u, right?
     
  9. Apr 3, 2013 #8

    Doc Al

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    At the moment the string breaks, the velocity is u. That's given.

    Whether that's the final or initial velocity depends on whether you are solving question 2 or question 3.
     
  10. Apr 3, 2013 #9
    Wouldn't u be the initial velocity to be used when solving for #2 and #3? Once the string breaks you would have a new system (free fall) where in A would have an initial downward velocity of u while B would have an initial upward velocity of u as well. Is my logic correct?

    For #3, what I'm thinking is still use u as my initial velocity and set my final velocity to be 0 since it's asking for the moment B reaches its highest point. I would then solve the distance traveled by B. I'd also solve for time, t, and then use t that to solve for the distance travel by A.
     
  11. Apr 4, 2013 #10

    Doc Al

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    That logic is fine for #3, but not for #2. To find the distance they move from starting position to where the string breaks, u is the final velocity.

    That's good.
     
  12. Apr 4, 2013 #11
    So what would be my initial velocity for #2? 0?
     
  13. Apr 4, 2013 #12

    Doc Al

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    Right. The system is released from rest.
     
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