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2D force equilibrium - harder than you might think

  1. Sep 27, 2007 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    [tex]\Sigma F = \Sigma F_{x} + \Sigma F_y [/tex]
    [tex] F_{spring} = -kx [/tex]

    the Pythagorean identities

    Not sure what else

    3. The attempt at a solution

    No idea how to start this. WE usually resolve each force into its [tex] F_x [/tex] and [tex] F_y [/tex] components, but I don't see how to do that here.

    As a note, or professor told us to work this out to a certain point, where we are down to a single unknown, and to then just guess and check to find an approximate answer.

    So what I need help with is to figure out how to leave myself with an equation containing no unknown other than [tex] \theta [/tex]
  2. jcsd
  3. Sep 27, 2007 #2


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    I'm a bit rusty on statics, but aren't you forgetting a relevant equation? Something about the moments about each point add to zero? I think you need to treat this as a standard truss problem with unknown final length of the spring. As you work through the equations, it will probably work down to an equation that can't be solved analytically, probably something like f1(theta) = f2(theta) and you need to iteratively guess at theta to find the solution.
  4. Sep 27, 2007 #3
    Yeah, there shouldn't be an unbalanced force anywhere in the equation. What I don't get is how do you break the force acting along the spring into an x and y component?

    For the cord, it is [tex] F_x_{AB} = F_{AB} * cos(\theta) [/tex] and [tex] F_y_{AB} = F_{AB} * sin(\theta) [/tex]. But, again, I don't see how to do this for the spring. But, still;

    [tex] F_x = F_x_{AB} + F_x_{AC} = 0 [/tex]
    [tex] F_y = F_y_{AB} + F_y_{AC} + P = 0 [/tex]
  5. Sep 28, 2007 #4


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    It took some head scratching, but I figured it out. I was able to work it to [itex]f_1(\theta) = f_2(\theta)[/itex] just like I thought would be the case. This problem is what's called a statically indeterminate problem. It's due to the 2 pinned joints. If you sum the forces and moments as usual, the horizontal reaction forces drop out and there is no way to determine what they are. Another equation is needed. The key is recognizing that the resultant reaction forces must act along the member. Hint: what does that tell you about the ratio of the y-reaction force to x-reaction force at B?
    Last edited: Sep 28, 2007
  6. Sep 28, 2007 #5
    You dont need to find the moments of the forces here. If you just study the equilibrium position, and then use Newtons' Laws of Motion to resolve the forces in the x and y direction, you should get your answer. What forces do you think are acting on the system at that point of time?
  7. Sep 28, 2007 #6
    Maybe I was wrong... I cant get past [tex]Tsin\theta +\sqrt{k^2x^2-T^2cos^2\theta}=10[/tex]. Cant figure out how to eliminate T (tension in the string). You either need one more equation or perhaps you can use the sine rule to find a relation between the lengths, which would give you x...
  8. Sep 28, 2007 #7
    I am not able to understand one thing.Is the spring remaning unstrecthed even after the force is applied?

    Then we may proceed.
  9. Sep 28, 2007 #8
    If you assume that angle ACB at equilibrium is [tex]\phi[/tex], using the sine rule, this gives you: [tex]\frac{l+x}{sin\theta}=\frac{l}{sin\phi}[/tex]. Using Newton's Laws, you can get another two equations by equating forces along the x and y axes. There are, however, four variables, and three equations. Any ideas on a fourth equation?
  10. Sep 28, 2007 #9
    No. The spring gets stretched by a distance x (assume). This causes a restoring force along the direction of the string. When we did problems like these, the situation at equilibrium was such that the third angle, BAC, would be 90 degrees. Then the sine rule would be able to give you a fourth equation. As it is, you're always one equation short.
    Last edited: Sep 28, 2007
  11. Sep 28, 2007 #10
    I am getting two equations but they are different from yours

    [tex]T_1\cos\theta\ = (T_2 + kx)\cos\phi[/tex]

    [tex]T_1\sin\theta\ + (T_2 + kx)\sin\phi[/tex] = P

    I am trying to figure out the other two.
  12. Sep 28, 2007 #11


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    I solved it using the method described in my last post. If there is another way, it would be interesting to know what it is.

    No too much detail though, since this is homework for Vidatu!
  13. Sep 28, 2007 #12
    Well, I got it. Or rather, I've gotten rid of all unknowns other than [tex] theta [/tex]. Once I've written it up nice, and found the real answer, I'll post my notes here.
  14. Sep 28, 2007 #13
    By which way did you get the answer by hotvette's way or by the way everlasting and i were discussing.
  15. Sep 29, 2007 #14
    I got these two equations coupled with the one I posted using sine rule. This leaves us one equation short. Now the problem here is there are multiple solutions to the problem for different values of [tex]\theta[/tex] and [tex]\phi[/tex]. If one of these were a constant/known, the problem would be solved.
  16. Sep 29, 2007 #15
    Ok iam online so we can discuss about the probelm

    Dont you think that by adding the sine rule we are just adding an extra variable L ie the lenght of the rope.

    We will have to use the data that bc= 0.4 metres
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