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Cable equilibrium homework

  1. Jun 17, 2009 #1
    peeee.jpg
    This question BUGS me

    I've tried it a whole lot of times, my teacher explained it to me at least 2 times but he came up with the wrong answer.

    I had a couple class mates explain it but it was an Extremely long/condusing way of doing it.

    I know the x components are equal so there is no x force

    so you are finding the Y component only.

    I do all the trig to find the y components and its the wrong answer.





    Anyone have a good way to solve this?

    thanks
     
  2. jcsd
  3. Jun 17, 2009 #2

    cepheid

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    Re: Equalibrium

    Easy. You are assuming a static situation, i.e. point x is in equilibrium. This means that the net force on it is zero. That condition gives you two equations (one for the x components, and one for the y components). You also have two unknowns: the force exerted by the post on point x (Fpost), and the force exerted by the angled cable on point x (Fcable).

    Two equations, two unknowns, all you have to do is solve:

    [tex] \sum F_x = 2200\textrm{N} - F_{\textrm{post}}\cos(75^{\circ}) - F_{\textrm{cable}}\cos(58^{\circ}) = 0 [/tex]


    [tex] \sum F_y = F_{\textrm{post}}\sin(75^{\circ}) - F_{\textrm{cable}}\sin(58^{\circ}) = 0 [/tex]

    Edit: I got 2551 N, which is closest to 2600 N. But I did it hastily and might have made algebra mistakes. What's the answer?
     
    Last edited: Jun 17, 2009
  4. Jun 17, 2009 #3
    Re: Equalibrium

    Yes its 2600

    I am still kind of confused.

    so you find the ratio between Fpost and F cable In one of the equations and plug it into the other?

    then what does the force on x =? post -/+ cable?
     
  5. Jun 17, 2009 #4

    LowlyPion

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    Re: Equalibrium

    Yes.

    Doesn't resolving the second equation with the first substitution yield the right result?
     
  6. Jun 17, 2009 #5
    Re: Equilibrium

    I have Fc=2895 and Fp=2541 F_p is the answer correct?
     
  7. Jun 17, 2009 #6

    cepheid

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    Re: Equalibrium

    Yes. One way of solving a system of two equations in two unknowns is to use the first equation to eliminate one variable by expressing it in terms of the other, and then substitute that relation into the second equation to solve for the second variable.

    What do you mean?

    The net force on point x is zero, because it is not moving! We used this fact to solve the problem.

    The force on point x DUE TO the post (which is what you are trying to solve for) is just Fpost, by definition.
     
  8. Jun 17, 2009 #7

    cepheid

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    Re: Equilibrium

    Well I already told you that I got Fp = 2551 N, so I guess our answers differ by 10 newtons.
     
  9. Jun 17, 2009 #8
    Re: Equilibrium

    Thank you.
     
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