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Stability of a given structure composed of three rods and a string

  1. Apr 17, 2005 #1
    Three rods with lenght l and mass m are arranged like the figure in the attached file. There connections are without friction.
    One rod is horizontal and the other two parallell and in the same plane.
    An elastic string connects point A and B.
    The tension in the string is proportional to its length.

    Determine the equillibrium angle a and investigate the stability.

    Any ideas on how to solve this problem?

    Thank you.
     

    Attached Files:

  2. jcsd
  3. Apr 17, 2005 #2
    I dont even know how to start on this one.
    Maybe someone can give some hints?
     
  4. Apr 18, 2005 #3
    Is the elastic coefficient k of the string also given as data in the problem?
     
  5. Apr 18, 2005 #4
    only that

    tension in the string = k * length of string
     
  6. Apr 18, 2005 #5

    OlderDan

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    There must be something more to this. What is holding these rods together? Are they hinged? Is the system subject to gravity? Is some part of this aparatus being held in place so it cannot free fall?
     
  7. Apr 18, 2005 #6

    DaveC426913

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    Wait. I think I get it.

    See enhanced diagram.

    "What is holding these rods together? Are they hinged?"
    Yes, as stated: "...[their] connections are without friction..."
     
    Last edited: Nov 28, 2006
  8. Apr 18, 2005 #7

    OlderDan

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    I think you've got it!!

    Let's try this for starters

    For equilibrium, neglect the thickness of the rods. Reference the zero of gravitational potential energy U to the top mount. Then

    [tex] U = -2mg \l cos a [/tex]

    The string is stretched to a length x given by the law of cosines

    [tex] x^2 = \l^2 + \l^2 -2 \l^2cos(90 - a) [/tex]

    [tex] x^2 = \l^2 + \l^2 -2 \l^2sina [/tex]

    [tex] x^2 = 2 \l^2 (1 - sina) [/tex]

    The potential energy of the string equals the lost gravitational potential energy

    [tex] 2mg \l cos a = \frac{1}{2}kx^2 = k \l^2 (1 - sina) [/tex]

    [tex] \frac{1 - sina}{cos a} = \frac{2mg}{k \l} [/tex]
     
    Last edited: Apr 18, 2005
  9. Apr 19, 2005 #8
    You are using an energy conservation approach. Remember that we don't know the initial height. Even if we did, we would only find the point at which the system will temporarily stop (still no equilibrium). What we are asked is to find the equilibrium angle.

    Wrong.

    JohanL. Use the equilibrium condition with forces that apply on the bottom rod.
     
  10. Apr 19, 2005 #9

    OlderDan

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    You are quite right. So my erroneous calculation, if done correctly, should be the minimum angle if the system were released from rest with an angle of a = 90, assuming the elastic constant of the string is sufficient to keep the thing from going past vertical. That assumption is of course not justified- My bad.
     
  11. Apr 19, 2005 #10
    Also another hint: the absence of friction at the hinges makes the reaction force of a hinge colinear with the rod to which it joins (see why). With this, it should be very easy to set up the equilibrium conditions for the horizontal and vertical directions and express the angle a in terms of m, l, and k.
     
  12. Apr 19, 2005 #11

    OlderDan

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    Let me see if I can salvage a bit of dignity after making my earlier sophomoric error. An energy approach is valid if done correctly, so let's see if I can do it right. Again, referencing gravitational potential energy to zero at a = 90

    [tex] U = -2mg \l cos a [/tex]

    The string is stretched to a length x given by the law of cosines

    [tex] x^2 = \l^2 + \l^2 -2 \l^2cos(90 - a) [/tex]

    [tex] x^2 = \l^2 + \l^2 -2 \l^2sina [/tex]

    [tex] x^2 = 2 \l^2 (1 - sina) [/tex]

    The total potential energy combining gravity and the spring is the sum of the two

    [tex] U_{total} = -2mg \l cos a + \frac{1}{2}kx^2 [/tex]

    [tex] U_{total} = -2mg \l cos a + k \l^2 (1 - sina) [/tex]

    Equilibrium is at the minimum of the potential energy (not the zero, DOH!!!! as Homer would say). The zero of the potential energy will give the limiting value for a. As I think more one it, there will be a limiting value somewhere, though it might well be at a negative angle between zero and 90.

    [tex] \frac {dU_{total}}{da} = 2mg \l sin a + -k \l^2 cosa = 0 [/tex]

    [tex] 2mg sin a = k \l cosa [/tex]

    [tex] tan a = \frac{k \l}{2mg } [/tex]

    Working out the result directly in terms of the forces involved as recommended by ramollari is a worthwhile exercise. If I have done my work right, the two results should agree.
     
  13. Apr 19, 2005 #12
    Yeah, thats the right answer.
    Thank you!

    The answer for the stability is
    stable if 0 < a < pi/2
    unstable if pi < a < 3pi/2

    And its unstable in that range because the potential energy dont have a minimum there.

    I have another problem i havent been able to solve. Maybe you can give me some hints?

    If you have two prisms on top of each other, one smaller than the other, and the one on top glids down to the ground the lower prisms moves to the left.
    The problem is to find how far the lower prism moves.

    There is no friction between the prisms or between the lower prism and the ground.

    If you choose an orthogonal coordinate system with the x-axis parallell to the ground and if Tpi is the kinetic energy for the i:th prism you get for the lagrangian

    L = Tp1 + Tp2 - mgh

    where h is the height to the center of mass of prism 2, the prism on top of the other.
    Then i guess you need a constraint. Here im not sure about how to set up the equation for the constraint. The prisms are in contact all the time but i cant see an obvious relation between their center of masses or something like that.
    Im not very good at mechanics!
     

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  14. Apr 19, 2005 #13

    Doc Al

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    Hint: What's the net horizontal force on the system (both prisms)?
     
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