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Equilibrium and displacement of a Particle

  1. May 28, 2013 #1
    1. The problem statement, all variables and given/known data

    The springs AB and BC have stiffness k and an unstretched length of l. Determine the
    displacement d of the cord from the wall when a force F is applied to the cord. See Picture attached.

    l = 3 m
    k = 600 N/m
    F = 200 N

    2. Relevant equations
    Fspring= ks

    s= l - l0


    3. The attempt at a solution
    The force of the spring is 600(s) and I know that s = length of spring stretched - length unstretched. So for the sum of forces in the x direction I get 2(600)(s)[d/√(d2+1.52)] = 200... I divide both sides by 2 and get (600)(s)[d/√(d2+1.52)] = 100...

    from here I know that I can get s from equation above thus it becomes 600[√(d2+1.52) - 1.5][d/√(d2+1.52)] = 100... but I don't know what else to do from here... any replies would be greatly appreciated.

    Attached Files:

  2. jcsd
  3. May 29, 2013 #2
  4. May 29, 2013 #3
    Do you know what the answer is supposed to be? When I work it out, I get d=0.16667m.

    I think you are addressing the forces in the x direction incorrectly. If you look at the force exerted on the spring, you can think about it as stretching the spring a distance d in the x direction while also stretching the spring l/2 in the y direction.

    When you look at it like this, then the force due to one spring in the x direction is Fs=(k)(d). And since you have two springs the forces in the x direction should look like this:


    Solve for d.

    I don't know why you're using the term [d/√(d2+1.52)]

    (someone please correct me if I'm way off on this)
    Last edited: May 29, 2013
  5. May 29, 2013 #4
    jldibble is right

    I think you got your idea wrong when you said the force of the spring is 600.
  6. May 29, 2013 #5
    the force of the spring is k(s)... k=600 is given, and I know s = l - l0; therefore Fspring=600(l-l0).

    l0 is also given and equals 1.5m so the equation becomes 600(l - 1.5). Looking at the triangle the two springs create with the wall we can deduce that l (length of the stretched spring) is going to be equal to √(d2+1.52) if we split it into 2 right triangles and from the Pythagorean theorem.

    So putting all this together Fspring=600[√(d2+1.52) - 1.5].

    Now if we were to break it down into its x and y components:
    ƩFx=0: Fspring[d/√(d2+1.52) + Fspring[d/√(d2+1.52) - 200 = 0

    the √(d2+1.52) would be the equivalent of cosθ.

    ƩFy=0: Fspring[1.5/√(d2+1.52) - Fspring[1.5/√(d2+1.52).

    My issue is when I solve for d.... I guess I might have to do trial and error.
  7. May 29, 2013 #6
    sorry it should be s=*lstretched-l0...
  8. May 29, 2013 #7
  9. May 29, 2013 #8
  10. May 29, 2013 #9
    right I know how to set up the problem... but once I try to solve for d the algebra just throws me off.
  11. May 30, 2013 #10


    User Avatar

    Staff: Mentor

    The original post said that the unstretched length of a spring is l (lower case L). But l is given as 3m. Why are you using l/2 for the unstretched length?
  12. May 30, 2013 #11
    I think it's a typo (or just very poorly worded) since the attached drawing, which seems to have been copied directly from the text, shows their combined unstretched length as l and it's consistent with the solution given.

    I doubt you're supposed to solve it analytically. Approximate the solution by making an inital guess and iterate.
  13. May 30, 2013 #12


    User Avatar

    Staff: Mentor

    I'd have to side with it being a typo, since the diagram does not actually imply that the unstretched length is l/2 (although it is tempting to believe so for convenience sake!). The initial configuration could have the unstretched springs form an equilateral triangle with the wall.
  14. May 30, 2013 #13
    Yes, you're right. I assumed too much based on the appearance of the sketch.
  15. Jun 8, 2013 #14
    yeah sorry i worded the problem wrong...still, thank you for all the replies...ill do a better job next time.
  16. Jun 9, 2013 #15
    original length of each spring = l0/2

    extended length of each spring = [itex]\sqrt{d^2+(l_0/2)^2}[/itex]

    tensile force in each spring = [itex]k(\sqrt{d^2+(l_0/2)^2}-(l_0/2))[/itex]

    horizontal component of tensile force in each spring = [itex]k(\sqrt{d^2+(l_0/2)^2}-(l_0/2))\frac{d}{\sqrt{d^2+(l_0/2)^2}}=kd(1-\frac{(l_0/2)}{\sqrt{d^2+(l_0/2)^2}})[/itex]

    sum of horizontal components of tensile force from springs = [itex]2kd(1-\frac{(l_0/2)}{\sqrt{d^2+(l_0/2)^2}})[/itex]

    [itex]F=2kd(1-\frac{(l_0/2)}{\sqrt{d^2+(l_0/2)^2}})=2kd(1-\frac{1}{\sqrt{1+(\frac{2d}{l_0})^2}} )[/itex]

    The solution to this equation for d is 0.997 m.
    Last edited: Jun 9, 2013
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