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Find the tension help

  1. Oct 31, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the tension in each cord in the figure, if the weight of the suspended object is w.

    [​IMG]

    1)Find the tension of the cord A in the figure (a)
    2)Find the tension of the cord B in the figure (a).
    3)Find the tension of the cord C in the figure (a).
    4)Find the tension of the cord A in the figure (b).
    5)Find the tension of the cord B in the figure (b).
    6)Find the tension of the cord C in the figure (b).




    3. The attempt at a solution

    I got the solutions, but I want to learn how to do this problem.


    What I have done is broken down the tension into x and y components.

    as so: [​IMG]

    if they are in equilibrium then the total force is 0.

    ok I now understand part c and f, because the only tension acting on that string is the weight of the object pulling down.

    update:

    I progressed some more (hopefully what I did is in the right track):

    Sum of Fx =

    Acos30 = Ax
    Bcos45 = Bx

    Sum of Fy =

    Asin30 = Ay
    Bsin45 = By
    w = Cy

    Solutions: 1).73205w 2) .896575w 3) w 4) 2.732w 5) 3.346w 6) w
     
    Last edited: Oct 31, 2007
  2. jcsd
  3. Oct 31, 2007 #2

    Doc Al

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    Staff: Mentor

    Signs matter. I'd say:
    Ax = -Acos30 (because it points to the left)
    Bx = +Bcos45

    Sum of Fx = 0, so:
    Ax + Bx = 0
    -Acos30 + Bcos45 = 0

    That's one equation.


    Correct the signs and write the vertical equilibrium equation. That's your second equation. You can solve them together to get A and B.
     
  4. Oct 31, 2007 #3
    so it would be:

    Sum of Fx::: -Acos30 + Bcos45 = 0
    Sum of Fy::: Asin30 + Bsin45 -w = 0

    now, the A's cancel out, and it would be 2Bcos45sin45-w=0....b=w? can't be. :(
     
  5. Oct 31, 2007 #4

    Doc Al

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    Staff: Mentor

    Good.

    What do you mean they cancel out?

    Write one variable in terms of the other (from one equation) and substitute (into the other equation).

    -Acos30 + Bcos45 = 0
    Acos30 = Bcos45
    A = Bcos45/cos30

    etc...
     
  6. Oct 31, 2007 #5
    ok so it would be

    Bcos45/cos30 + B sin 45 -w = 0

    .81649658B + .707106781B - w = 0

    1.523603361B = w

    B = w/ 1.523603361

    ..
     
  7. Oct 31, 2007 #6

    Doc Al

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    Staff: Mentor

    You forgot the sin30 in the first term.
     
  8. Oct 31, 2007 #7
    B(cos45/cos30)(sin30) + B(sin 45) -w = 0

    B (.81649658)(.5) + B (.707106781) -w

    .40824829B + .707106781B - w = 0

    1.115355071B = w

    B = w/1.115355071
     
  9. Oct 31, 2007 #8

    Doc Al

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    Staff: Mentor

    Looks OK.
     
  10. Oct 31, 2007 #9
    but the answer for the tension of B is "T_B =.896575w" o_O

    but if I try doing it like this:

    ((Bcos45)/cos30)(sin30) + B(sin 45) -w = 0

    instead of this

    B(cos45/cos30)(sin30) + B(sin 45) -w = 0

    I get .8164962w

    good enough?
     
  11. Oct 31, 2007 #10

    Doc Al

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    Staff: Mentor

    Realize that B/1.115355071 = (1/1.115355071)B

    Calculate 1/1.115355071
     
  12. Oct 31, 2007 #11
    :O

    wow, forgot about that 1 in the numerator.

    thank you for your help. :)
     
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