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Three blocks suspended on 2 pulleys to form a M shape

  1. Sep 18, 2012 #1
    1. The problem statement, all variables and given/known data
    A block with a mass of 25 kg is hung midway between two pulleys, with a rope connecting a hook at the top of the block to each of the two pulleys. On the other side of the pulleys, both ropes are connected to blocks with masses of 15 kg. The 25 kg mass in the middle sags downward until the 3 blocks and 2 pulleys form the shape of an “M”. If the pulleys are 45 cm apart, how far below the level of the pulleys does the top of the 25 kg mass need to be in order for the system to be in equilibrium?


    2. Relevant equations

    Sum of Forces =0

    3. The attempt at a solution
    I have calculated the tensions in one rope to be 147N using 15kg*9.8m/sec^2 and the Fg of the middle block is 245N I know I need to find the angles of the rope to the ceiling by splitting the tension of the rope in x and y directions, but I can not figure out how to get those values.
     
  2. jcsd
  3. Sep 18, 2012 #2
    The magnitude of vectors A and B:
    A=25g N
    B&C=15g N

    [itex]\vec{A}=\vec{B}+\vec{C}[/itex]

    From above equation you can find the direction of B and C vectors for equilibrium.
     
  4. Sep 18, 2012 #3
    If I come up with the forces add them 147N+147N≠245N So What am I doing Wrong?
     
  5. Sep 18, 2012 #4
    Yes, it's wrong.
    You have to think in vectors, which has magnitude and direction.
    If the vectors are in same direction, ordinary arithmetic can be performed on them.

    Another point is that the vector can be resolved/replaced by its components.
    The best directions of vector components is in X(horizontal) and Y(vertical)direction.
     
  6. Sep 18, 2012 #5
    I know that vectors are comprised of a magnitude and direction or can be written in component form. I just do not know the direction of the tension on the middle block so I do not know how to put it in component form without direction. What I am haing the most trouble with is how to separate that tension from the rope onto the middle block which is 147N into its' components. Am I trying to figure this out the wrong way? Could you please tell me which direction the tension is in or how to figure the direction out?
     
  7. Sep 18, 2012 #6
    Let F be the tension.
    There are 2 components that you can do normal arithmetics on identical direction.
    Please note that there is another tensin on mirror side.
    The weight vector is in Y direction.

    http://imageshack.us/a/img805/2684/20946647.jpg [Broken][/QUOTE]
     
    Last edited by a moderator: May 6, 2017
  8. Sep 18, 2012 #7
    Is weight vector is in the negative y direction? I have drawn all this out, but I do not have angle theta. Theta is not given in the problem, thats something I need to figure out. If the weight vector is in the negative direction it would have an equal and opposite force in the positive direction which would be 245 N correct? But you cannot have a triangle with side larger than its' hypotenuse so would force in the positive y direction half of 245N since it is beeing supported by two ropes?
     
  9. Sep 18, 2012 #8
    If the force in the positive y direction is half, then the angle would equal 56.4 degrees and the height would be 22.5*tan(theta)= 33.9cm from the ceiling. Is that right?
     
  10. Sep 18, 2012 #9
    To make the body in equilibrium, the sum of vectors with negative Y direction is equal to sum of vectors in positive Y direction.

    Likewise in X direction.

    Add: As in post #6, nothing is said about tangent.
     
    Last edited: Sep 18, 2012
  11. Sep 18, 2012 #10
    so is this right <-245j>= 147(xi+yj)+147(-xi+yj) ? Now how am I supposed to solve for the components?
     
  12. Sep 18, 2012 #11
    the tangent is to solve for the height of the block to the ceiling and the distance between the pulleys is 45cm so once I have theta I can solve for the distance using trig and half that distance
     
  13. Sep 18, 2012 #12
    What is the vertical component of the tension of one of the cables?
     
  14. Sep 18, 2012 #13
    is it half of the force from the middle weight
     
  15. Sep 18, 2012 #14

    rl.bhat

    User Avatar
    Homework Helper

    You can find θ using the parallelogram law of Forces.
    R2 = P2 + Q2 + 2PQcosθ.
     
  16. Sep 18, 2012 #15
    There are 2 components forces upward each of 15g N tension.
    Total positive upward force is - 2 x 15g x Sinθ
    Downward force = 25g

    You can get θ.
    From this you can get height which equal to L/2(Tanθ)

    Sorry your post#8 is correct.
     
  17. Sep 18, 2012 #16
    That is alright thanks for the help. I needed it.
     
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