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Tension and Blocks

  1. Sep 14, 2009 #1
    3 blocks are hanging on pulleys:

    O
    | \
    | \
    M3 \
    \
    O
    | \
    | \
    M1 M2

    o= pulleys


    the masses of 3 blocks are M1 = 12 kg, M2 = 18 kg, and M3 = 30 kg. The pulleys and strings are massless, there is no friction, and gravity points downward. The whole system is held fixed, then released at rest. What is the acceleration of the 30 kg block? Enter a positive answer if it goes up, a negative answer if it goes down, or 0 if the acceleration is zero.

    so pretty much the 30 kg mass is attatched to a pulley which is attached to another pulley which is held by the the 2 masses of 12 kg and 18 kg.

    As far as doing this problems I know that I need to use the formula:
    T-mg=ma

    and to figure out the acceleration of the M1 and M2 I used this:

    a= (m2 −m1)g/ (m1 + m2)

    which i then got a= 1.96 m/s^2
    which (plugging into the above equation using M1) i got tension to be 141.12 N
    Also noted by this equation:

    T= 2(M1M2)g / (m1 +m2)

    Now this is where i get stuck. I dont know where to go from there. I got the tension for the 2 pulleys and also the acceleration but how do I apply that to the next pulley which is attatched to M3 (30 kg) which will get my acceleration of block M3.
     
  2. jcsd
  3. Sep 14, 2009 #2
    Block M1 and M2 are the counter weights to M3 so they are not actually attached to block M3. My picture got screwed up.
     
  4. Sep 15, 2009 #3
    here is the picture
     

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  5. Sep 15, 2009 #4

    Doc Al

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    Staff: Mentor

    You'll definitely need this. (It's just Newton's 2nd law applied to the mass.)

    That assumes that the pulley is fixed, which is not the case here.

    Start by applying Newton's 2nd law to each mass.
     
  6. Sep 15, 2009 #5
    ok so here is what i get:

    T3= M3 (g+a3)

    T2= M2 (g+a2)
    T1= M1 (g+a2)

    T2= T1... so now what?

    I have 2 unknown accelerations (a3, a2) and 2 unknown tensions. But somehow T3 has to equal T2 and T1...i dont know.
     
  7. Sep 15, 2009 #6

    Doc Al

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    The first thing to do is assign an arbitrary direction to the accelerations. For example, I will assume M3 accelerates downward with magnitude a3. (If you guess wrong, no big deal--a3 will be negative.)

    You can't assume that M1 and M2 have the same acceleration. Do this: Assume M2 accelerates down with magnitude a with respect to the moving pulley. What then is M2's acceleration with respect to the floor, which is what counts for applying Newton's laws?

    Do the same for M1. (How do the motions of M1 and M2 relate?)

    It's true that T2 = T1. Good.

    To relate T3, T2, and T1, analyze the forces acting on the moving pulley.
     
  8. Sep 15, 2009 #7
    hey, sorry. I am a little bit confused with the 2nd part that you answered. I understand that the accelerations aren't the same (same tensions, different masses= different acceleration).

    I got lost with the accelerating with respect to the pulley and floor part.
     
  9. Sep 15, 2009 #8
    are you indicating that the acceleration from M1 and M2 has an affect on the acceleration of M3?

    such that if if M1 pulls down with a 4 m/s2 and M2 gets pulled up with 2 m/s2, that when M3 pulls the pulley containing M1 and M2, the acceleration M2 plays a role with the acceleration of M3?
     
  10. Sep 15, 2009 #9

    Doc Al

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    Good. Be sure to reflect that in your equations by using different symbols: a1 & a2.

    It's tricky. You know that if M1 moves 1 m toward the pulley, then M2 must move 1 m away from the pulley. So their accelerations with respect to the pulley are equal but opposite. But the pulley is accelerating, so you need to add the two accelerations together to get the total acceleration of each mass. (I said with respect to the floor, which is an inertial frame, since Newton's laws only apply in inertial frames.)

    Definitely. They're all connected and affect one another.
     
  11. Sep 15, 2009 #10
    Ah! so your saying if the pulley is accelerating 1 m/s2 downwards and the M1 is lets say 4 m/s2 upward..its really just 3 m/s2.

    Vice versa, if M2 is accelerating downward lets say 5 m/s2 its really 6 m/s2 (because of the pulley)

    So its very much a vector problem.
    So how do I get the pulleys acceleration?
     
  12. Sep 15, 2009 #11
    would the pulleys acceleration be the acceleration of M3?
     
  13. Sep 15, 2009 #12

    Doc Al

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    Exactly, assuming you mean this: The pulley accelerates 1 m/s^2 downward with respect to the floor; M1 accelerates 4 m/s^2 upward with respect to the pulley. Thus M1 accelerates 3 m/s^2 upward with respect to the floor--that's the actual acceleration that you'd use in Newton's 2nd law.

    Right.
    You'll set up the equations and solve for it.

    Yes, but in the opposite direction.
     
  14. Sep 15, 2009 #13
    like i said i got the tension to be 141.12 N

    So acceleration for M1 (12)= 1.96N
    And accleration for M2 (18)= -1.96N

    M3 as said before is T-F=ma
    T=m(g-a).

    Now tension (using the forumla earlier) would it be T= 2(M3*(M1+M2))g / (m3 +m2+ m1)?

    or do I know apply M1= (T-F)a instead of adding the 2 masses together as shown above?

    so it would be T=2(M3* (T-F)a)g/ (M3+M1)
     
  15. Sep 15, 2009 #14

    Doc Al

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    How did you get this? Show the equations that you solved to get it.
     
  16. Sep 15, 2009 #15
    T= 2(M1*M2)g/(M1+M2)

    M1= 12
    M2= 18
     
  17. Sep 16, 2009 #16

    Doc Al

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    Show how you derived this result.
     
  18. Sep 16, 2009 #17
    i got this from my book. is it wrong?
     
  19. Sep 17, 2009 #18

    Doc Al

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    That equation is a solution to a particular problem: Two masses hanging from a pulley that is at rest. That doesn't apply here.

    If you understand how that formula was derived, apply the same principles to this problem and derive a new formula. It's almost always a better idea to start from the basic principles--Newton's 2nd law--and derive the correct formula yourself.
     
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