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Derive the force from different spring configurations

  1. Oct 28, 2005 #1
    Does anyone know a website that will derive the force from different spring configurations, i.e. two springs in series with different spring constants. I would just like to be able to understand and work with any combination of springs: two in series, connected to 3 in parallel connected to one in series...etc. I want to be able to understand from the general case. Halliday and Resnik does not go into detail at all, and I didn't have any luck googling. Thanks. -Brent
     
  2. jcsd
  3. Oct 29, 2005 #2
    ^bump

    I'd be satisfied with a decent explanation of two springs in series with different spring constants.
     
  4. Oct 29, 2005 #3

    Integral

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    For 2 springs in series you have a common force, each spring has a force of mg acting (neglecting the mass of the springs). So:

    [tex]- x_1 k_1= mg [/tex]
    and
    [tex] - x_2 K_2 = mg [/tex]

    For the pair of springs you can write:
    [tex] -x_T K_T = mg [/tex]


    But we must have:
    [tex] x_T = x_1 +x_2 [/tex]

    For the total we now can write:
    [tex] -(x_1 + x_2) K_T =mg [/tex]

    Observe that from our first relationships we have :

    [tex] -x_n = \frac {mg} {k_n} [/tex]

    n= 1,2

    so:
    (now I can eliminate the pesky negative sign)
    [tex] (\frac {mg} {k_1} + \frac {mg} {k_2})K_T = mg[/tex]

    Finally we get :

    [tex] \frac 1 {K_t} = \frac 1 { k_1} + \frac 1 {k_2} [/tex]

    The key to the parallel case is that the displacement of the springs are equal. I'll let you do that one.
     
    Last edited: Oct 29, 2005
  5. Oct 29, 2005 #4

    Q_Goest

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    The force on a spring equals the spring constant times the change in length.

    (1) dF=ktotal dxtotal

    and for a series of springs, dF is the same for each, so

    (2) dF = k1 dx1 + k2 dx2 + k3 dx3 etc...
    or dF = sum (kn dxn)

    and the total distance x is
    (3) dxtotal = sum (dF/kn)

    Now substitute (3) into (1)

    dF = ktotal dxtotal = ktotal sum(dF/kn)

    Divide by dF

    1 = ktotal sum (1/kn)

    Put ktotal on the opposite side

    ktotal = 1/ (sum (1/kn))

    (sorry for the lousey presentation here.)

    So the total spring constant for a bunch of springs in series is:

    ktotal = 1 / ( sum (1/kn) )

    Where
    ktotal = total or equivalent spring constant of the springs in series
    kn = one of the springs from 1 to n

    Example:

    For 3 springs with k's
    1. 10 (lb/in)
    2. 24 (lb/in)
    3. 40 (lb/in)

    The equivalent spring constant is

    k = 1/ (1/10 + 1/24 + 1/40 )

    k = 1/ (.1 + .0416667 + .025)

    k = 6 (lb/in)

    If you use the same basic logic, you should also be able to determine a more generic formula for springs in series and parallel.

    Edit: I see Integral beat me to it! lol
     
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