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An inclined plane, a pulley, and three masses

  1. Dec 10, 2012 #1
    1. The problem statement, all variables and given/known data
    A system comprising blocks, a light frictionless pulley, a frictionless incline, and connecting ropes is shown. The 9 kg block accelerates downward when the system is released from rest. The acceleration of the system is closest to:

    A.) 1.9 [itex]m/s^2[/itex]
    B.) 2.1 [itex]m/s^2[/itex]
    C.) 1.7 [itex]m/s^2[/itex]
    D.) 1.5 [itex]m/s^2[/itex]
    E.) 2.3 [itex]m/s^2[/itex]

    x5gNC.jpg

    2. Relevant equations

    [itex]F = ma[/itex]


    3. The attempt at a solution
    1. First

      [itex]m_{1} = 6 kg[/itex]
      [itex]m_{2} = 4 kg[/itex]
      [itex]m_{3} = 9 kg[/itex]

      [itex]\theta = 30°[/itex]

      [itex]a = a_{x} = a_{y}[/itex]

    2. Then, the sum of forces on the three masses

      [itex]\sum F_{x1} = T_{2}-m_{1}g\sin\theta = m_{1}a[/itex]
      [itex]\sum F_{y1} = 0[/itex]

      [itex]\sum F_{x2} = T_{1}-T_{2}-m_{2}g\sin\theta = m_{2}a[/itex]
      [itex]\sum F_{y2} = 0[/itex]

      [itex]\sum F_{x3} = 0[/itex]
      [itex]\sum F_{y3} = T_{1}-m_{3}g = m_{3}a[/itex]

    3. Combine [itex]F_{x1}[/itex], [itex]F_{x2}[/itex], & [itex]F_{y3}[/itex] and isolate [itex]a[/itex]...

      [itex]a = \frac{2T_{1} - g (m_{1}\sin\theta + m_{2}\sin\theta + m_{3})}{(m_{1} + m_{2} + m_{3})}[/itex]

    4. Solve for [itex]T_{1}[/itex]

      [itex]\sum F_{y3} = T_{1}-m_{3}g = m_{3}a[/itex]
      [itex]\sum F_{y3} = T_{1}-m_{3}g = 0[/itex]
      [itex]T_{1} = m_{3}g = (9 kg)(9.81 m/s^2) = 88.29 N[/itex]

    5. Plug [itex]T_{1}[/itex] into [itex]a[/itex] and solve

      [itex]a = 2.1 m/s^2[/itex]
      Or, answer B.


    Is this the correct solution and answer? Did I solve correctly for T_1? I don't think it's right because the tension should not equal weight of m_3 because technically the block IS accelerating downward at this instant, isn't it?
     
    Last edited: Dec 11, 2012
  2. jcsd
  3. Dec 11, 2012 #2
    smashd, ja if the tension in the rope was the same as the weight no acceleration will take place: Very important note: Define your axis for the entire system, I see you take the x axis for your first 2 masses as parallel with the surface so the y axis should be perpendicular to this surface for the entire system you cannot change the axis for the 3rd mass... After defining your axis, draw the 3 FBD's for the masses, you are on more or less the right track, lets see if we can get to solution here... Do the FBD first...
     
  4. Dec 11, 2012 #3
    Thanks for the input, WillemBouwer.

    So [itex]\sum F_{3y}[/itex] should be:

    [itex]\sum F_{3y} = m_{3}g - T_{1} = m_{3}a[/itex]

    Then [itex]a[/itex] would become after combining the forces on the system:

    [itex]\frac{m_{3}g - m_{2}g\sin\theta - m_{1}g\sin\theta}{m_{1} + m_{2} + m_{3}}[/itex]


    Which is still 2.1 m/s^2, but this is the proper solution to the problem?
     
  5. Dec 11, 2012 #4
    yes, that looks better, ja as you take the acceleration as positive downward the weight force should be positive aswell... and it cant be 0 as you stated...
     
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