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Combining Spring Constants

  1. Feb 8, 2015 #1
    1. The problem statement, all variables and given/known data
    For the following arrangement of two springs, determine the effective spring constant, keff. This is the force constant of a single spring that would produce the same force on the block as the pair of springs shown for this case.

    nyckyf.jpg
    (Spring 1 is attached to Spring 2 which is attached to a block of mass m)

    Solution Provided by Book:

    Imagine that the block was displaced a distance x to the right of its equilibrium position. Let x1 be the distance that the first spring is stretched and let x2 be the distance that the second spring is stretched. Then [itex] x = x_1+ x_2 [/itex]. But [itex] x_1 = \frac {-F}{k_1} [/itex] and [itex]x_2 = \frac{-F}{k_2} [/itex], so

    [itex]\frac{-F}{k_1}+ \frac{-F}{k_2} = x[/itex]

    [itex]-F(\frac{1}{k_1}+ \frac{1}{k_2}) = x[/itex]

    [itex]F = \frac{1}{\frac{1}{k_1} + \frac{1}{k_2}}x[/itex]

    [itex]F = -\frac{k_1k_2}{k_1+k_2}x[/itex]

    Therefore,

    keff = [itex]\frac{k_1k_2}{k_1+k_2}[/itex]

    2ciiw78.jpg




    2. Relevant equations

    F=-kx

    3. The attempt at a solution

    Basically I'm trying to understand how the solution they gave works. I can follow the logic up until the last line of the solution, keff = [itex]\frac{k_1k_2}{k_1+k_2}[/itex].
    I understand that each spring exerts the same amount of Force. This is because as x, in F=-kx, increases, k compensates by decreasing.


    When I began I thought

    Fnew = keff * x

    I can solve for x with

    [itex]x = x_1 + x_2[/itex]

    [itex]x_1 = \frac{-F_1}{k_1}[/itex]

    [itex]x_2 = \frac{-F_2}{k_2}[/itex]

    This leads me to
    [itex]\frac{-F_1}{k_1}+ \frac{-F_2}{k_2} = x[/itex]

    and if you factor out the F(which you can do because the springs exert the same force) it equals

    [itex]-F(\frac{1}{k_1}+ \frac{1}{k_2}) = x[/itex].

    Moving it to the other side yields

    [itex]F = -\frac{1}{\frac{1}{k_1} + \frac{1}{k_2}}x[/itex]

    And if you add the fractions they equal

    [itex]F = -\frac{k_1k_2}{k_1+k_2}x[/itex]

    but then they set [itex]F = -\frac{k_1k_2}{k_1+k_2}x[/itex] equal to Fnew = keff * x and come up with the final answer

    keff = [itex]\frac{k_1k_2}{k_1+k_2}[/itex]

    Fnew does not equal F(The force of one spring). How can they set the equations equal to each other?
     
    Last edited: Feb 8, 2015
  2. jcsd
  3. Feb 8, 2015 #2

    TSny

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    Consider the original system of the two springs connected together. If you pulled on the free end to stretch the system by some distance, how does your force compare to the force that each spring exerts separately? How does your force compare to the force you would exert to stretch the equivalent spring (with keff) the same amount?
     
  4. Feb 8, 2015 #3
    I reformatted my post and tried to elaborate a little more because I realized that not even I could understand what I was asking.


    To answer your questions

    My Force = Force of Spring 1 + Force of Spring 2
    Force of Spring 1 = Force of Spring 2
    My Force = 2F
    My Force = Equivalent Spring Force
    Equivalent Spring Force = 2F?

    But 2F = keff * x cannot be set equal to [itex]F = -\frac{k_1k_2}{k_1+k_2}x[/itex]

    because the 2nd equation yields the value of the force of only one spring, doesn't it?
     
    Last edited: Feb 8, 2015
  5. Feb 8, 2015 #4

    TSny

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    This isn't correct. Suppose you are holding the free end of the two-spring system so the system is stretched at some distance x. Draw a free body diagram for each spring.
     
    Last edited: Feb 8, 2015
  6. Feb 8, 2015 #5
    I'm having trouble visualizing the forces in the spot between the springs. I know the Force my hand has would be distributed throughout the whole system.
     
  7. Feb 8, 2015 #6

    TSny

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    In the figure below I have shown the two springs being pulled by your applied force Fap. Below that, I have drawn the two springs separately for the free body diagrams. Can you complete the diagrams by describing what other forces should be included in the free body diagrams?
     

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  8. Feb 8, 2015 #7

    Stephen Tashi

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    A spring doesn't exert "a force" (singular). Only when you consider one end of the spring pulling on a free body can you define the force due to the spring. When you use a symbols like "F" without explanation it isn't clear which force you are talking about. If "F" represents the magnitude of several different forces, you should explain that too.

    To define what your symbols represent, draw (or describe) some free body diagrams.
     
  9. Feb 8, 2015 #8
    I have this so far. I'm not sure if I am missing anything. I know I've already said this, but this situation is very difficult for me to visualize for some reason.
     

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  10. Feb 8, 2015 #9

    TSny

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    Good. You are only missing one force that acts on spring #1.
     
  11. Feb 8, 2015 #10
    I've scratched my brain but I don't have any idea. Does it have something to do with the opposite force of the applied force?
     
  12. Feb 8, 2015 #11

    TSny

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    The left end of the the first spring is attached to the wall. The wall must exert a force on spring #1 as shown below.

    Now considering spring #1, what can you say about the magnitudes of the force from the wall and the force from spring #2?
     

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  13. Feb 8, 2015 #12
    Are they equal?
     
  14. Feb 8, 2015 #13

    TSny

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    Yes. If an object is at rest and remains at rest, then the net force acting on the object is zero.

    What can you say about the two forces acting on the ends of spring #2?
     
  15. Feb 8, 2015 #14
    They are also equal?
     
  16. Feb 8, 2015 #15

    TSny

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    Yes. Think about how the applied force is related to the force given by the formula F2 = k2 x2.

    It might help to think about what's going on in the simple case where you stretch a spring a distance x with your left hand holding one end of the spring and your right hand holding the other end. Both hands pull on the spring. How does the force of one of your hands compare to the force given by the formula F = kx?
     
  17. Feb 8, 2015 #16
    The amount my hand moves is equal to the x in F=kx.
     
  18. Feb 8, 2015 #17

    TSny

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    Yes. The important thing is that whenever a spring with spring constant k is stretched by a distance x, each end of the spring is being pulled with a force given by kx.

    So, if you look at the free body diagram for spring #2, what is the magnitude of the applied force Fap?
    Is it equal to k2 x2 or is it equal to k1 x1 + k2 x2?
     
  19. Feb 8, 2015 #18
    It would be just k2x2. The applied force does not exert a force on the first spring. Only the 2nd spring exerts a force on the 1st spring which is equal to the applied force.
     
  20. Feb 8, 2015 #19

    TSny

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    Right. (And the wall also exerts a force on #1 spring). Good.
     
  21. Feb 8, 2015 #20
    So basically does that mean that where ever you are in the system, the force is going to be constant??? Is this why they can refer to the force without any subscript, because they don't have to specify since it's the same everywhere?
     
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