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2 hanging masses with springs, find spring constant

  1. May 21, 2012 #1
    1. The problem statement, all variables and given/known data
    http://i.imgur.com/0gexJ.png



    2. Relevant equations
    F=kx (Hooke's Law), ƩFy=0



    3. The attempt at a solution
    I drew a free-body diagram for each block. For block A, I get the equilibrium equation: Fa = (10kg)(9.81ms^-2) = 98.1N = kx_a
    x_a = 300mm-250mm = 50mm
    k = 1.962kN/m

    For block B I get the equilibrium equation of Fb = 98.1 - Wa = kx_b but it seems as I don't need this equation.

    Is this correct, and if not, should I draw a "combined" FBD? If so, how? Thanks.
     
  2. jcsd
  3. May 21, 2012 #2

    Doc Al

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    Staff: Mentor

    Your equation is not obtained from analyzing block A alone, but by treating A and B as a combined object. So you answered your own question, even if you didn't realize it.

    If you analyzed A alone, you'd have to include the force from each spring and only the weight of A.
     
  4. May 21, 2012 #3
    Oh I see. From drawing seperate FBD's, I get the equations:
    kxa = kxb + mag (block A)
    kxb = mbg (block B)
    Substitute kxb back into the first equation to get kxa = (ma + mb)g, where xa = 300mm - 250mm, thus k = 1.962kN/m.

    Is this correct? Thank you.
     
  5. May 21, 2012 #4

    Doc Al

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    Staff: Mentor

    Perfectly correct. Note that your final equation is equivalent to what you did earlier.
     
  6. May 21, 2012 #5
    Thank you for you time and help =)
     
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