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Initial Tension Of Parallel Spring System

  1. Mar 23, 2015 #1
    1. The problem statement, all variables and given/known data
    How to calculate initial tension or pretension in a system of two parallel springs.
    F=5.33213N
    K1+K2=53.0597
    x=0.095

    Individual springs have initial tension of 0.25N so I would expect an answer of 0.5N but I can't get there.

    2. Relevant equations
    F=(k1+k2)x

    3. The attempt at a solution
    Total force applied - Initial tension = (k1+k2)x
    Total force applied - (k1+k2)x = Initial tension = 0.29?
     
  2. jcsd
  3. Mar 23, 2015 #2
    Could you please restate the problem exactly as it reads? And provide a drawing perhaps? Also, I have never seen the word pretension used that way. Ever.
     
  4. Mar 23, 2015 #3
    Those pretentious springs! Such over-inflated senses of self-worth!
     
  5. Mar 23, 2015 #4
    Calculate the initial tension of two springs in parallel by treating them as a single spring. Initial tension as in the force needed to begin extending the spring. Ktotal is the sum of individual spring constants and total force is shared between both springs.
     

    Attached Files:

  6. Mar 23, 2015 #5
    I would say that (k1 + k2)x - F = initial spring tension, not F - (k1 + k2)x.
     
  7. Mar 23, 2015 #6
    Think about the initial state of the springs. The tension in them can be expressed as (k1 + k2)[x][/0] and then the force F further stretches the springs to the value (k1 + k2)x. Thus, (k1 - k2)x - F = initial tension.
     
  8. Mar 23, 2015 #7
    But Fafter equilibrium=ktotal*x
    where Ftotal is F required to overcome initial tension+Fafter equilibrium
    so Ftotal-Initial Tension=ktotal*x
     
  9. Mar 23, 2015 #8
    Yeah that looks good. Thank you for clearly and concisely representing the quantities.
     
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