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Homework Help: Problem with springs

  1. Jun 30, 2010 #1
    1. The problem statement, all variables and given/known data
    We have a string with property D (I don't know the name in english, it's dimension is N/m, the force required to stretch it 1m), which is small enough so that if we put the spring on the ground, it'll be shorter than it's standard height. The question is, when will this spring stretch more; if we hang it down from somewhere, or if we hold it in a way that the two ends each are 45 degrees to the vertical (sort of like a smiley, \/, a chain that you hold on the two ends)


    2. Relevant equations
    G=mg
    F(spring)=Dx


    3. The attempt at a solution
    My idea was the following, but I heard it isn't correct:
    If the spring hangs from somewhere, F=G is pulling it from the top and keeping it stationary (SummaF=0), while G is pulling down the middle part. This can be thought of as if we were pulling each end with F/2=G/2 force, which means a force of mg/2 is stretching the spring.
    If the spring is positioned the other way around, we have G pulling it in the middle, and two F's on each ends, and since the vertical components of the forces must equal each other, 2*(F*sinalpha)=mg, where alpha = 45, from which F=mg/(2sqr(2)/2)=mg/sqr(2) is stretching the spring. And since mg/sqr(2) > mg/2, the spring should be stretched more in the second position.

    Thanks for any ideas! If my attempt wasn't clear I'll be happy to try and clarify it.
    -Tusike
     
  2. jcsd
  3. Jun 30, 2010 #2
    The value you have labeled "D" is called the spring constant. The traditional symbol is "k" as in F = kx.

    When the spring is suspended, the force that causes the spring to stretch is the tension from the support. This force supports the whole weight of the spring mg not just mg/2.

    Your analysis of the second situation is okay.

    Continue the question to the extreme: what if alpha = 90? Then the two halves are suspended next to each other. Will this cause more or less stretch than the original configuration? It should be less (in fact, half) -- but according to your reasoning they would be equal.
     
  4. Jul 1, 2010 #3
    I see. So when hanging, it get's stretched by mg.
    Thanks!
     
  5. Jul 1, 2010 #4

    PhanthomJay

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    No, that can't be correct. If a massless spring (no weight) with a spring constant k was vertically suspended from the ceiling with a mass of M attached to the free end, then the force in the spring is a constant force of Mg, and its extension would be Mg/k.
    But if a non-massless spring is hanging under its own weight Mg, the force in the spring is not constant, it is a variable force which is Mg at the ceiling and 0 at the free end. The average force in the spring is thus Mg/2, so its extension under its self weight is Mg/2k, which I think is what you said in the beginning.
     
  6. Jul 1, 2010 #5
    OK then, I seem to be getting nowhere:) Then there must be a problem with my reasoning in the second situation, when it's hanging with it's end 45 degrees to the vertical (\/).Although I really don't see the problem with that either. It's just that I went to a camp, and we had this problem, worth 4 points; and the solution I described in my first post was worth 2 points. I would really love to believe I was correct, but It's hard to imagine the people organizing the camp made a mistake...
     
  7. Jul 1, 2010 #6
    Okay. I've been thinking about this on and off for a bit.

    Suppose we have a massless spring hanging from the ceiling with a weight mg. It gets pulled down a certain amount = x. Imagine the spring is cut in half: two springs in series holding up the weight. Nothing changes. Both halves have to support the full weight.

    Now imagine that half the weight (mg/2) is transported to the upper spring. There is no change in the upper spring: it still has to support mg. But the lower spring only has to support mg/2. So this half is stretched only half of its original value. Total extension = (3/4)x.

    Continue the process: keep subdividing the springs and moving the weight. In the limit, this is a model of the massive spring. And the total extension will be (1/2)x.

    So now I agree with PhantomJay. :smile:

    However, this idea also applies to the second case where the massive spring is suspended by two 45 degree forces. Each side of the V has to balance an average force of mg/4, not mg/2.
     
  8. Jul 2, 2010 #7
    OK, I think we can assume that for the first situation, the stretch x = mg/(2k) is then correct.
    However, I really don't see how you got x=mg/(4k), I'm not really sure it's correct.
     
  9. Jul 2, 2010 #8

    PhanthomJay

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    It's even less than that. The vertical component of the average force is mg/4 in each leg of the "V". But the spring constant of each spring leg is now 2k, because by folding it back, you've halved the unextended original length of the spring, therefore doubling its stiffness.
     
  10. Jul 3, 2010 #9
    OK, I see how each half of the spring would have a spring constant k'=2k, but I still don't know how you guys got that mg/4 idea. If the vertical components of the force the spring is mg/4, that'd mean whatever's holding it is extracting the same force on the spring. So, for the vertical direction, the forces acting on the spring are mg pulling it down, and two mg/4's holding it up. The spring should start to fall down... I drew what I think you said, if it isn't the way you imagined, please draw it like I did.

    -Tusike
     

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  11. Jul 4, 2010 #10

    PhanthomJay

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    Oh, I wasn't envisioning you had the catenary curve in mind, but yes, a wire attached to a support at each end and hanging under its own weight will take the shape of a catenary curve. But don't confuse the reactions at the supports with the tension force in the spring (elastic cord). While the support reactions must be mg/2, the force (tension) in the cord at the support, from equilibrium considerations, is mg(sq rt 2)/2 at a 45 degree angle, whereas the horizontal tension at the low point is mg/2......the tension force varies along its length.
     
  12. Jul 4, 2010 #11
    OK, so this is what I arrived to:
    What I said in the beginning was that if the spring is simply hanged, the stretch would be x=mg/(2k). We found that this indeed is correct.
    As for when hanged in a way that the ends are alpha=45 degrees to the vertical, I said the force acting on both ends would be F=mgsqr(2)/2, so the stretch would be x = mgsqr(2)/(2k). This would mean that the spring is more stretched in this situation, sqr(2) times more than in the first one.
    We found that this isn't correct, because contrary to the first case, in the second case we have to calculate with a k'=2k spring constant, which would mean that x=mgsqr(2)/(4k), which is sqr(2)/2 times the stretch we calculated in the first situation. So the spring is more stretched in the first situation (sqr(2)/2 < 1). That would explain why I got 2 points instead of 4, and it those seem logical and correct, so thanks everyone! :)

    EDIT: Nope, sorry, there's still no difference between the solution and what I said in the beginning. I forgot that it doesn't really matter if we count with k or k'=2k, because if we count with the latter, we need to multiply what we get by 2, since the string has two parts, which leaves us with what we had when we counted with k, x=2*mgsqr(2)/(4k) = mgsqr(2)/2...
     
    Last edited: Jul 4, 2010
  13. Jul 4, 2010 #12
    I think this is a tough question. I haven't come up with a reasonable and precise qualitative explanation yet, so I did some calculations to see how it would be. I found that the extension in case 2 is greater. It is about 0.57mg/k (the precise result is not nice at all, so I didn't bother remembering it :tongue2:).

    Though I'm not sure if my calculation is right or not, I think I can roughly explain this in some sense (Oh yes, my intuition told me that the extension in case 2 should be greater).
    - The average tension in case 1 is mg/2.
    - In case 2, the tension varies along the spring, and it is symmetrical. The tension at one end is T1 = mg/sqrt(2), and the tension at the lowest point of the spring is T2 = mg/2. So we may roughly estimate the average tension = (T1 + T2)/2 = 0.6mg.
    The average tension in case 2 is greater, which means that the extension along the spring should be greater, and thus, the overall extension in case 2 is greater.
     
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