3 Sliding Masses; find speed

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data
    The three blocks shown are relased at t=0 from the position shown in the figure. Assume that there is no friction between the table and M2, and that the two pulleys are massless and frictionless. The masses are: M1 = 1.0 kg, M2 = 7.0 kg, M3 = 3.0 kg.
    Calculate the speed of M2 at a time 1.45 s after the system is released from rest.
    (M1 is on left side of box, M2 is on the top of the box, M3 is on the right side of the box. Corners of box are pulleys.)


    2. Relevant equations
    None.



    3. The attempt at a solution
    I have ZERO clue.
     
  2. jcsd
  3. Sep 20, 2011 #2
    These problems are always very interesting ones.

    First, you have to draw a FBD for all three masses.
     
  4. Sep 20, 2011 #3
    Where is the figure? :confused:
     
  5. Sep 20, 2011 #4
  6. Sep 20, 2011 #5
  7. Sep 20, 2011 #6
    So basically, there are two pulleys attached to the box.

    There is a single, massless string run through the pulley system.

    M1 is hanging on the left side of the box and M3 is hanging on the right side of the box (attached to string I presume).

    M2 has a string connected on both sides? Yes?
     
  8. Sep 20, 2011 #7
    The image should be 3boxes on a pulley system.
     
  9. Sep 20, 2011 #8
    Yes, Aggression, that is right.
     
  10. Sep 20, 2011 #9
    Is that a university website or something like that?
    If so, i think i can't access that until i am a student of it.
     
  11. Sep 20, 2011 #10
    Alright, now we can get this problem started.

    So, you'll have 3 separate FBDs.

    M1's FBD will have a T force going up and a Weight Force going down.

    M2's FBD will have tension forces on both sides, a normal force going up that is equal to a weight force going down.

    M3's FBD will have a T force going up and a weight force going down.

    Someone correct me if I'm wrong.
     
  12. Sep 20, 2011 #11
    Image should be attached.
     

    Attached Files:

  13. Sep 20, 2011 #12

    PeterO

    User Avatar
    Homework Helper

    consider the 3 masses plus string as a single sytem, with the pulleys just adjusting a couple of directions.

    The Force trying to move the system clockwise is the weight of the right hand mass.
    The force trying to move the system in the anti-clockwise direction is the weight of the left hand mass.
    One of those is bigger and "wins".
    The net force [one weight minus the other] acts on the whole system. F = ma is then used.
     
  14. Sep 20, 2011 #13
    But i don't have the acceleration of the force...
     
  15. Sep 20, 2011 #14
    PeterO, please don't directly come on F=ma relation. It would be better if you help the OP make equations for each body. :smile:
     
  16. Sep 20, 2011 #15
    Okay, can you give me a hint please?
    I'm sorry, but i'm really bad at physics and I'm stressed and Idk where to go.
     
  17. Sep 20, 2011 #16
    Well, after you draw your FBDs, you will have a system of 3 equations.

    So, sum the forces in each situation.

    We will denote T[itex]_{1,2}[/itex] as the tension force between the M1 and M2 object.
    We will denote T[itex]_{2,3}[/itex] as the tension force between the M2 and M3 object.
    T[itex]_{2,1}[/itex] = T[itex]_{1,2}[/itex]
    T[itex]_{3,2}[/itex] = T[itex]_{2,3}[/itex] By Newton's 3rd law.


    Now, sum of the forces for M1....

    [itex]\Sigma[/itex]F = T[itex]_{1,2}[/itex] - m[itex]_{1}[/itex]g = m[itex]_{1}[/itex]a

    Sum of the forces for M2.....

    [itex]\Sigma[/itex]F = - T[itex]_{2,1}[/itex] + m[itex]_{2}[/itex]g - m[itex]_{2}[/itex]g + T[itex]_{2,3}[/itex] = m[itex]_{2}[/itex]a

    Sum of the forces for M3....

    [itex]\Sigma[/itex]F = -T[itex]_{3,2}[/itex]+m[itex]_{3}[/itex]g = m[itex]_{3}[/itex]a


    I'm not 100% sure that these are correct. Someone double check me.
     
  18. Sep 20, 2011 #17
    MEH! I have no clue!
     
  19. Sep 20, 2011 #18
    Let's assume they are. haha

    Now, add up all the equations. The tension forces will cancel leaving you with.....

    [itex]\Sigma[/itex]F = m[itex]_{3}[/itex]g - m[itex]_{1}[/itex]g = m[itex]_{1}[/itex]a + m[itex]_{2}[/itex]a + m[itex]_{3}[/itex]a

    So, factor out a g from the left side, and an a from the right side.....

    g(m[itex]_{3}[/itex]-m[itex]_{1}[/itex]) = a(m[itex]_{1}[/itex] + m[itex]_{2}[/itex] + m[itex]_{3}[/itex])

    Solve for a and that will give you the acceleration of each block.

    m1 = 1.0 kg
    m2 = 7.0 kg
    m3 = 3.0 kg
     
  20. Sep 20, 2011 #19
    where did m2g go?
     
  21. Sep 20, 2011 #20
    m2g canceled out because the equation for M2 included.....

    m2g - m2g = 0

    So, that's why they canceled out.
     
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