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Static equlibrium problem

  1. Jun 18, 2008 #1
    1. The problem statement, all variables and given/known data
    A vertical load P is applied at end B of rod BC. The constant of the spring is k and the spring is unstretched when [tex]\theta[/tex] = 0. (a) Neglecting the weight of the rod, express the angle [tex]\theta[/tex] corresponding to the equilibrium position in terms of P, k, and l. (b) determine the value of [tex]\theta[/tex] corresponding to equilibrium if P = 2kl.

    http://img229.imageshack.us/img229/3746/77369190ws9.jpg [Broken]

    2. Relevant equations

    [tex]\sum F_{x}=0[/tex]

    [tex]\sum F_{y}=0[/tex]

    Force of Spring = ks

    3. The attempt at a solution

    There are three forces acting on the spring: P, [tex]F_{B}[/tex],[tex]F_{C}[/tex]. The force at C has an x component pointing to the left, and a y component pointing up. The force at B is equal to a force of a spring, ks.

    I guess where I am running in to trouble is when I am trying to solve for the components of the Force at B.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jun 18, 2008 #2


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    By using Lami's theorem you can find the relation between P, ks and theta. By using the properties of triangle you can find the relation between s, l and theta. Try.
  4. Jun 19, 2008 #3
    Do I need to know the original unstretched length of the spring?

    and btw: the correct answers are supposed to be

    a) [tex]tan^{-1} \frac{P}{kl}[/tex]
    b) 63.4 degs
    Last edited: Jun 19, 2008
  5. Jun 19, 2008 #4


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    Along AB force ks acts. So according to Lami's theorem P/sin(angleABC) = ks/sin(angleCBP). Now from properties of triangle s/sin(angleACB) = l/sin(angleBAC). Solve these equations. You will get the result.
  6. Jun 19, 2008 #5
    Last edited: Jun 19, 2008
  7. Jun 19, 2008 #6
    So I am assuming you want me to solve these equations for [tex]\theta[/tex], right?

    Well, this is what I have:

    [tex]\frac{s}{sin( \theta)} = \frac{l}{sin( \angle BAC)}[/tex]

    [tex]s = \frac {l sin (\theta)}{sin( \angle BAC)}[/tex]

    then plug s into:

    [tex]\frac{P}{sin( \angle ABC)} = \frac{ks}{sin( \angle CBP)}[/tex]

    to get:

    [tex]\frac{P}{sin( \angle ABC)} = \frac{klsin(\theta)}{sin( \angle BAC) sin( \angle CBP)}[/tex]

    and so

    [tex]P sin( \angle BAC) = klsin(\theta)[/tex]

    [tex]\frac{P sin( \angle BAC)}{kl} = sin(\theta)[/tex]

    and so, I solve for theta:

    [tex]\theta = sin^{-1} \frac{P sin( \angle BAC)}{kl} [/tex]

    which isn't the right answer. Am I doing something wrong, or am I not understanding what you wrote clearly?
    Last edited: Jun 19, 2008
  8. Jun 19, 2008 #7


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    Angle ABC = 90- theta/2, Angle CBP = 90 + theta, Angle ACB = theta and Angle CAB = 90 - theta/2. Remember that in triangleABC AC = BC. Substitute these values in the equations.
  9. Jun 19, 2008 #8


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    In triangle ABC angle ABC =Angle BAC, and Angle CBP = 90 + theta. SO sin(angle CBP) = sin( 90 + theta) = cos(theta). Substitute these values in your fourth step in 6th mail, you will get your answer.
  10. Jun 19, 2008 #9
    are you sure that sin(90+theta) = cos(theta)? It's very close, but I don't think that's a correct statement.

    edit...nevermind. I'm assuming you meant pi/2 instead of 90.

    THANKS!!! I cannot tell you how much of a help you've been.
    Last edited: Jun 19, 2008
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