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How do I find the reduced force on an object due to springs?

  1. Jul 20, 2010 #1
    1. The problem statement, all variables and given/known data
    This question is beyond my level, I am doing this for bonus marks. I am unaware of how to do it, I am merely requesting a formula, or a series of steps to help me. I have not yet done calculus.

    Imagine that a 1 kg box is falling from a height of 5.2 m, with 4 springs attached to its bottom.
    What is the reduced force on the box within the first instant of impact due to the springs?

    Please make up various needed variables, such as the spring amplitude, if needed to aid in explanation.

    2. Relevant equations

    Unaware of EQn's.

    3. The attempt at a solution

    Lack thereof explained above.
     
  2. jcsd
  3. Jul 20, 2010 #2
    I guess my first question is what level of physics are you in? You honestly have no clue of what equations to use?
     
  4. Jul 20, 2010 #3
    I'm in grade 11... I've tried multiple other forums (such as XKCD's) and sites (such as Answers.Yahoo.com), but either there was insufficient knowledge or bashing. I figure I'd give this forum a try.

    As for the actual math, I am aware of its complexity and the fact that calculus may be necessary, but I still want to try.
    So, to recap, I am trying to figure out how to go from a box dropping with springs on the bottom (energy equations, in this case mechanical energy eqns), to the diminished force in the first instant of impact (engineering equations, like Hooke's law).

    So, please explain the best you can, unless you feel like you're above being taxed by an angsty teen.
     
  5. Jul 20, 2010 #4
    Ah ok. Well, I can tell you right now that you can solve this problem with just algebra if you use pre-determined equations (ie you don't derived equations).

    Have you covered Hooke's Law and momentum in your physics class yet (I'm assuming you've done Newton's Laws)?
     
  6. Jul 20, 2010 #5
    Unfortunately we will not be covering Hooke's law nor other momentum questions.
    So (based on your link), are you suggesting that I use the F = dp/dt equation in order to find the reduced force on the box due to the springs?

    BTW, is the force itself reduced, or just spread out over a dimension of time?
     
  7. Jul 20, 2010 #6

    rock.freak667

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    Do you have any information about the spring? Such as the spring constant?
     
  8. Jul 20, 2010 #7
    Because this is an actual physical assignment (an egg drop report), I am given no external data. How would I go about finding the spring constant?
     
  9. Jul 20, 2010 #8

    rock.freak667

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    For a spring, F=kx. So just place a force on the spring (such as a known weight) and measure the deflection 'x'.
     
  10. Jul 20, 2010 #9
    x is the displacement of the end of the spring from its equilibrium position. So, if I were to press down on the springs with the same force it experiences when hitting the ground, would I measure from where the top of the spring was to where it is now, or from the middle to where it is now? 1

    And once I found x, would I divide negative F with x (-F / x) to find the spring constant? 2

    FINALLY, once I found the spring constant, what would I do from there?3

    Thanks for helping!
     
  11. Jul 20, 2010 #10
    Ok, let's take these equations and see if you can solve it.
    F=ma
    F=kx
    F=m(Δv)/(Δt)

    where Δv is the change inl velocity Δt is the change time (thanks for the symbols rock.freak667!!!! :D).

    So, from here, what would you do next?
     
    Last edited: Jul 20, 2010
  12. Jul 20, 2010 #11

    rock.freak667

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    You don't need to use the same amount of force. Just use standard weights, 1 kg, 2 kg, etc. So just rest the 1 kg weight on one spring.


    Where the spring is naturally, that will be your equilibrium position, when the weight is on it, measure from that point to where it is now.



    From here now, you can use conservation of energy, initially it has gravitational potential energy, which would be converted to elastic potential energy (remember you have 4 springs).
     
  13. Jul 20, 2010 #12
    Assuming that:
    Mass = 1 kg
    Acceleration = 9.8 m/s^2
    I know that F = 1 N.
    Thus, if F = kx, and x is an assumed 0.02m, then:
    k = 50.

    so... now what? (and what unit is k measured in?)
     
  14. Jul 20, 2010 #13

    rock.freak667

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    Well you need to use the energy equation.

    gravitation pe = elastic pe in 4 springs.
     
  15. Jul 20, 2010 #14
    oh, using the equation (1/2)kx^2, I can figure out the elastic potential per spring, if I discovered x using a 1kg weight on 1 spring. And from elastic potential to...
     
  16. Jul 20, 2010 #15
    Is the elastic potential equal to the mech energy halfway, or gravitational at the top?
     
  17. Jul 20, 2010 #16

    rock.freak667

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    The placing of the 1 kg on the spring was to get k.

    Now you are dropping the mass, so you need to get new value for 'x' due it being dropped. If you are dropping it from 5.2m, then it has gravitational potential energy. (How do you find gravitation potential energy?)

    EDIT:



    At the top there is gpe, at the bottom there is only elastic. So equate them.
     
  18. Jul 20, 2010 #17
    Alrighty, so:

    h = 5.2m
    m = 1 kg
    g = 9.8

    Eg = mgh
    Eg = 50.96
    50.96 = (1/2)kx2
    101.92 = 50x2
    x = sqrt(5096).
    x = 71.39

    Now, what do I do? Do I work my way back to force, and find the difference?
     
  19. Jul 20, 2010 #18

    PhanthomJay

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    At the instant of impact, the spring deflection is 0. So what force does the spring exert on the mass at the instant of impact?
     
  20. Jul 20, 2010 #19

    rock.freak667

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    Right well you have 4 springs, so the bold line should be 4*(1/2)kx2.

    That will give you the extension (well compression) all the springs undergo. Then if you just use Hooke's law, you will get the force each spring takes. Then multiply that by 4.

    Though you are dropping it from 5.2 meters, which is quite high, so I am not sure if your springs will break or deform if the actual spring constant is too small.
     
  21. Jul 20, 2010 #20
    I will be using a parachute and crumple zone which will reduce the force dramatically, I just wanted to use a simple example to help introduce myself to this new concept.

    So, just to clarify, will the force I find be less than if I just dropped the box?
     
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