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Springs, Restoring Force, and Hooke's Law

  1. Feb 13, 2006 #1
    I'm having some trouble with the following question:


    Obviously I know to use Hooke's equation somewhere in this problem: [itex]F = -kx[/itex]

    In the above equation, F = the force, caused by gravity, of the board pulling the spring horizontally, and k = 176 N/m as given. So, I need to find x, the length at which the spring is stretched.

    What I'm having trouble with is finding the component of the gravitational force that is pulling the spring horizontally. My first instinct told me to try [itex]mg cos(50.0^o)[/itex], but I can see now that that's very wrong.

    The answer in the back of the book is 0.236 m, but I can't seem to get that no matter what I do. Could anyone help?
  2. jcsd
  3. Feb 13, 2006 #2
    This situation is a static equilibrium, so apply the corresponding law for the torques (ie their sum is zero).

    Put the origin of the board on the bottom left side at the angle of 50°.

    Call the length oof the board l (you don't need to know l)

    Applying the law for static equilibrium yields :

    [tex]\frac{l}{2}10.1*9.81*sin(40) = l*F*sin(50)[/tex]

    Solve for F and F = 176x...You will find the x you need in meters.

  4. Feb 13, 2006 #3
    Torques hadn't even crossed my mind. :rofl:

    Thanks so much, Marlon. I completely understand it now.
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