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Homework Help: Tensions and Weights!

  1. Jan 27, 2010 #1
    1. The problem statement, all variables and given/known data
    See figure.


    2. Relevant equations
    [tex]\sum F = ma[/tex]


    3. The attempt at a solution
    By just looking at the picture I was able to generate 2 equations but I seem to have 3 unknowns so I'm sorta stuck right now. Here are my equations:

    [tex] 50 = T_{ba}sin(130) + T_{bc}sin(\alpha) [/tex] and,

    [tex] 40 + T_{bc}sin(180+\alpha) = T_{cd}sin(50)[/tex]

    Any ideas of anything I'm missing? Am I taking the correct approach to this problem? Is there a simplier way!?

    Thanks,
     

    Attached Files:

  2. jcsd
  3. Jan 27, 2010 #2

    tiny-tim

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    Hi jegues! :smile:

    hmm … two vertical equations :rolleyes:

    now i wonder where you could get a third equation from? :wink:
     
  4. Jan 27, 2010 #3
    I think I may have figured out how to generate the extra equations I need.

    I know the the rings where all the cables are attached are in equilibrium in the x and y directions.

    If I apply [tex]\sum F_{x} = 0[/tex] I should be able to generate enough equations to solve this problem.

    Is this the correct approach?

    EDIT: [tex] T_{bc}cos(\alpha) + T_{ba}cos(130) = 0[/tex]
     
  5. Jan 27, 2010 #4

    tiny-tim

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    That's it! :smile:

    (though it might be easier in this case to use Tba and Tcd instead. :wink:)
     
  6. Jan 27, 2010 #5
    So we just "ignore" the section in the middle? (BC) It is all in equilibrium so it shouldn't affect our answers anyways correct?

    So,

    [tex] T_{ba}cos(130) = T_{cd}cos(50) [/tex],

    Correct?
     
  7. Jan 27, 2010 #6

    tiny-tim

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    That's right! :smile:

    (except for the sign … why do you persist in using angles > 90º ? …

    you can make mistakes doing that! :wink:)
     
  8. Jan 27, 2010 #7
    I run into sign errors whenever I've done it any other way. I think our professor showed us a way to correct it either using sin or cos for the given angle but I've always found it easier to simply use cosines for the x components and sines for the y component.

    I'm a little lost at the best way to solve the 3 equation 3 variable system, it's always easier when I don't have to solve for an angle.... ( [tex] \alpha[/tex] in this case)

    Any more tips!?
     
  9. Jan 27, 2010 #8
    I'm left with the following three equations:

    (1) : [tex] T_{ba}sin(130) + T_{bc}sin(\alpha) = 50[/tex]

    (2) : [tex] T_{bc}sin(180+\alpha)-T_{cd}sin(50) = -40[/tex]

    (3) : [tex] T_{ba}cos(130) - T_{cd}cos(50) = 0 [/tex]

    I'm getting a headache trying to solve them :s

    EDIT: Doh! I just reliazed I had 4 unknowns. I'm going to need another equation aren't I?
     
  10. Jan 27, 2010 #9
    So solving for 4 unknowns,

    (1) : [tex] T_{ba}sin(130) + T_{bc}sin(\alpha) = 50[/tex]

    (2) : [tex] T_{bc}sin(180+\alpha)-T_{cd}sin(50) = -40[/tex]

    (3) : [tex] T_{ba}cos(130) - T_{cd}cos(50) = 0 [/tex]

    (4) : [tex] T_{bc}cos(\alpha) + T_{ba}cos(130) = 0 [/tex]

    EDIT: Is this really what I'm stuck with doing? This is going to be an algebraic nightmare...
     
  11. Jan 27, 2010 #10

    tiny-tim

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    eek! :yuck: sorry, but i'm with your professor on this …

    don't be so inflexible, choose cos or sin to be appropriate for the diagram, not to make you feel at home!
    hmm … I've just noticed you have four unknowns, not three …

    looks like you're going to need that Tbc and Tba equation anyway.

    Tips? No, just use common-sense, and keep going. :smile:

    EDIT: if you used 50 instead of 130, it wouldn't look so intimidating. And sin(180º + α) = -sinα.
     
  12. Jan 27, 2010 #11
    How is this done? He went over it really fast in class so I couldn't really understand how he was determining to either place a cos or a sin for the given angle.

    Thank again, and am I actually stuck with that algebraic nightmare!? :yuck:
     
  13. Jan 27, 2010 #12

    tiny-tim

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    It's always cos of the angle between the force and the direction.

    It's never sin (except, of course, that sometimes the angle marked in the diagram is the "wrong" angle … so if they tell you that the angle to the horizontal is α, but you need cos of the angle to the vertical, then since it's always cos, that means it's cos(90º - α), which is sinα :wink:)
     
  14. Jan 27, 2010 #13
    instead of making things complicated using T[c] why dont u just name them T1.T2,T3? easier that way so u dont confuse them
     
  15. Jan 27, 2010 #14
    So if I'm looking at angles based from the horizontal axis I use cos(a) and if its from the vertical sin(a)? How do you know which sign to assign to these?

    Also,

    ?
     
  16. Jan 27, 2010 #15
    maybe instead of the horizon u should use a bearing instead?
     
  17. Jan 27, 2010 #16
    Still wondering if I'm actually stuck with solving a system of 4 equations and 4 unknowns... :yuck: ???
     
  18. Jan 28, 2010 #17

    tiny-tim

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    (just got up :zzz: …)
    If the force is in (approximately) the same direction as the positive axis, then it's +, if it's the other way, its -.

    And I repeat, always use cos.
    Well, for a start, it's only 3, because you immediately know that Tab = Tcd.

    Get on with it! :smile:
     
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