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Three blocks, two suspended, one on an incline

  1. Sep 24, 2006 #1
    Hi. I'm having a dickens of a time coming up with the correct answer for this.

    I shall describe the question as it includes an image. There are three blocks altogether, b_1, b_2, b_3, each with a mass m_1, m_2, and m_3, respectively, and the blocks are all connected together by two ropes, T_1 between b_1 and b_2, and T_2 between b_2 and b_3. Blocks b_1 and b_3 are hanging from a pulley at the end an incline that has an angle of theta degrees (going up from left to right, b_1 on the left, b_3 on the right) and on this incline is b_2 which has a friction coefficient mu.

    I've set it up a bunch of different ways but have not found the correct solution. Here is the last version:

    m_1 g - T_1 = m_1 a
    T_1 + sin(theta) m_2 g - mu cos(theta) m_2 g - T_2 = m_2 a
    T_2 - m_3 g = - m_3 a (*)

    For *, This was the last attempt to I made the acceleration negative (I tried it positive before).

    Any help would be greatly appreciated!! Thanx, lizzy
     
    Last edited: Sep 24, 2006
  2. jcsd
  3. Sep 24, 2006 #2
    let @ = incline angle (theta)
    u = friction coefficient

    (m1)g - T1 = (m1)a
    T1 + (m2)gsin@ - u(m2)gcos@ - T2 = (m2)a
    T2 - (m3)g = (m3)a

    sum the three equations ...

    (m1)g + (m2)g(sin@ - ucos@) - (m3)g = a(m1 + m2 + m3)

    solve for "a" now?
     
  4. Sep 24, 2006 #3
    thankx

    Thank you for your response.

    For m_1 = 7.5, m_2 = 2.7, m_3 = 2.5, @ = 24, u = 0.15, I came up with 3.91561. I entered it into the homework server but it said my answer was wrong and that was my last chance at the solution.

    There are two other questions related to it though which I might be able to get right with your help:

    What is the tension in the cord connected to the 2.5 kg block (b_3)?

    * I answered 34.289 but that was wrong.

    What is the tension in the cord connected to the 7.5 kg block (b_1)?

    * I have not answered this one at all yet (but I have more than one shot of an answer, I think 7 altogether)

    Thank you! :-)
     
  5. Sep 24, 2006 #4
    using your data, I get a = 8.3 m/s^2
     
  6. Sep 24, 2006 #5
    (m1)g + (m2)g(sin@ - ucos@) - (m3)g = a(m1 + m2 + m3)

    ==>
    (m1)g + (m2)g(sin@ - ucos@) - (m3)g
    a = -------------------------------------
    (m1 + m2 + m3)

    7.5 * 9.8 + 2.7 * 9.8(sin(24) - .15 cos(24)) - 2.5 * 9.8
    = ------------------------------------------------------
    ( 7.5 + 2.7 + 2.5)

    56.1364
    = ------------ = 4.42019
    12.7

    does that not look correct?

    T2 = (m3)a + (m3)g = 2.5 * 4.42019 + 2.5 * 9.8 = 35.5505

    ** That's right according to the homework server **

    T1 = (m1)g - (m1)a = 7.5 * 9.8 - 7.5 * 4.42019 = 40.3486

    ** right again! **

    When I first tried to solve for a, I forgot to put in the g next to m2.

    My original equations were just like yours but I tried to solve it by solving for T1 and T2 and then substituting those in - that must have been where I messed up I suppose - is there something special to summing the equations?

    Thank you for all your help.
     
    Last edited: Sep 24, 2006
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