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Pulley system in equilibrium

  1. Oct 7, 2007 #1
    The problem involves a picture so i posted it so it can be seen and easier or you guys to help me


    i know that the total sum of the tensions in this need to equal 37.4N + 61.4N

    but i get confused when angles get brought into play

    link to problem

    http://i199.photobucket.com/albums/aa314/anglum/help.jpg
     
  2. jcsd
  3. Oct 7, 2007 #2

    Doc Al

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    Staff: Mentor

    Examine the first "knot". What must be true about the forces acting on that knot? Hint: Consider vertical and horizontal components separately.
     
  4. Oct 7, 2007 #3
    i am so lost on this problem... would the forces actin on the first knot need to equal 37.4N???
     
  5. Oct 7, 2007 #4

    Doc Al

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    Staff: Mentor

    Since the system is in equilibrium, the total force must be zero.
     
  6. Oct 7, 2007 #5
    ok so if the total force on the first knot is 0 then i can assume that

    37.4N + T1+T2 = 0??
     
  7. Oct 7, 2007 #6
    i also need to find the other angles value in degrees as a part 4 of the question
     
  8. Oct 7, 2007 #7

    Doc Al

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    Staff: Mentor

    No, since forces are vectors and must be added as vectors. As I said earlier: Consider vertical and horizontal components separately.
     
  9. Oct 7, 2007 #8
    i am totally lost right now... i dont see y the forces of t1 t2 and the weight dont equal to zero
     
  10. Oct 7, 2007 #9
    the vertical componentei s37.4 and the horizontal is t2 ????
     
  11. Oct 7, 2007 #10

    Doc Al

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    The vectors add to zero, but not just the numbers.

    For each of those three forces list the following:
    (1) x-component
    (2) y-component

    Some of the components will be zero.
     
  12. Oct 7, 2007 #11
    ok for the wegith hangin down the x component is 0 and the y component is 37.4N

    for T2 the horizontal is WHAT while the vertical is 0

    and for T1 the vertical is sin41 = T1/X

    and for T1 the horizontal is cos 41 = T1/X
     
  13. Oct 7, 2007 #12
    ahhhh i figured out the first 3 parts

    im stuck on the angle
     
  14. Oct 7, 2007 #13

    Doc Al

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    Staff: Mentor

    Since it's downward, call it -37.4 N.

    Call the horizontal component T2.

    The vertical component is T1 sin41.
    The horizontal component is -T1 cos41.

    Now add up the vertical components and set to zero. Do the same for the horizontal components.
     
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