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Elastic Force

  1. Jan 18, 2014 #1
    1. The problem statement, all variables and given/known data

    A block of mass m, is suspended by a spring with a spring constant k and stretches a length x . If two blocks of mass m/2 are suspended by a spring with the same spring constant k (ii), how much will the spring stretch?

    http://imagizer.imageshack.us/v2/800x600q90/12/sidr.png [Broken]


    2. Relevant equations

    Fg=mg

    Fe=-kx

    3. The attempt at a solution

    I think each mass of m/2 stretches the spring x/2, so together would stretch the same as the block of mass m alone.
     
    Last edited by a moderator: May 6, 2017
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  3. Jan 18, 2014 #2

    hilbert2

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    Suppose that we have case (i) and the spring and mass are at equilibrium, with the spring being stretched a distance x. The mass is pulling the spring with force mg, but doesn't the support the spring is hanging from have to "pull" the spring from opposite end with the opposite force -mg? (otherwise it would not be a static situation)

    In case (ii) the spring is pulled from one end with force mg/2 and from the other end with force -mg/2.
     
  4. Jan 18, 2014 #3
    But the question is how much does the spring stretch in (ii), not how is (i) a static situation, you suppose (i) as a static situation of course, but it's not what the problem is about.
     
  5. Jan 18, 2014 #4

    hilbert2

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    I'm just pointing out that in case (i) the spring is stretched from both ends with force mg and in the case (ii) it's stretched from both ends with only half that force...
     
  6. Jan 18, 2014 #5
    Are you supposing that the spring has mass? Because in the problem they say nothing about the mass of the spring, so I suppose it with no-mass. Otherwise I don't understand you very well.
     
  7. Jan 18, 2014 #6

    tiny-tim

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    hi thonwer! :smile:

    you should consider the tension (the force) on each side of the spring in both cases

    as hilbert2 :smile: says …
    and i don't understand what you are saying here about "static" …
    … they are both in static equilibrium :confused:

    (and mass has nothing to do with it)
     
  8. Jan 18, 2014 #7
    I still don't get it :( I use Newton's Second Law in (i) Fe+T=mg
    in (ii) for each block Fe=mg/2 ?
     
  9. Jan 18, 2014 #8

    tiny-tim

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    i don't understand what you're doing here

    what is T?

    on which body are those the external forces? :confused:
     
  10. Jan 18, 2014 #9
    T=tension

    The external forces are on the blocks.

    I'm a little bit lost I think. Which would be the free body diagram?
     
  11. Jan 18, 2014 #10

    tiny-tim

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    Fe (i assume that's the spring force) does not act on the blocks, only the tension (and the weight) acts on the blocks

    (the spring is not connected to the blocks: the string, with its tension, is between them)
    you should do the free body diagram for the spring

    what external forces are acting on the spring? :smile:
     
  12. Jan 18, 2014 #11
    What I've done is this:

    http://imagizer.imageshack.us/v2/800x600q90/202/7fbl.jpg [Broken]

    I don't get anywhere :S
     
    Last edited by a moderator: May 6, 2017
  13. Jan 18, 2014 #12
    Ok, I got to the point in which in (i) x=mg/k and in (ii) x'=mg/2k so, the answer is, the string stretches in (ii) x/2 ? or do I have to add the two blocks in (ii) and the string stretches x?
     
  14. Jan 18, 2014 #13

    PhanthomJay

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    you've got to the right point and one of those answers is correct, but how you go it there is a mystery becasuse your free body diagram is not only incorrect, it is not even a free body diagram when you show every internal force in one uncut section and with force directions shown without regard to Newton 3.
    In part 1, the force in the spring is mg and it elongation is x. In part 2, the force in the spring is ---?--- and per Hookes law its elongation is----?---? I would guess 50% of beginners would guess wrong. Which group are you in?
     
  15. Jan 19, 2014 #14
    In part 1, the force in the spring is mg and it elongation is x. In part 2, the force in the spring is ---?--- and per Hookes law its elongation is----?---? I would guess 50% of beginners would guess wrong.

    In part 2, the forces in the spring are the tension and the elastic force, in the blocks are mg/2 and the tension. As the tension is the same in the spring and blocks tension=mg/2. In the spring we have tension=elasticforce => mg/2=kx' => x'=mg/2k. This is half the x of part 1.
     
  16. Jan 19, 2014 #15

    tiny-tim

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    thonwer, you keep talking about the "elastic force" (which you also write as Fe)

    where have you got this expression from? :confused:

    your free body diagram, or your force equation, must only use external forces

    whatever the "elastic force" is, it isn't external, is it?

    do the first spring again, and include the tension in the other string, between the spring and the ceiling :smile:
     
  17. Jan 19, 2014 #16
    With elastic force I mean Hook's Law, F=-kx, it's a translation which I see I've made wrong.
     
  18. Jan 19, 2014 #17

    tiny-tim

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    hi thonwer! :smile:
    it isn't the translation that's wrong, it's the way you're using it

    the F in Hooke's law is the applied force, the external force …

    in this case it is the tension in the string

    your F and T are the same thing!

    as i said, start again with the first spring … what are the external forces on it? :wink:
     
  19. Jan 19, 2014 #18
    I don't understand, I'm sorry. In (i) the spring suffers from the mg of the block and the tension, the block suffers from mg and kx from the spring. Am I right?

    In (ii) the spring suffers from tension of both strings, each block suffers from mg/2 and the tension. Is this right? If so, where do I have to put kx ?
     
  20. Jan 19, 2014 #19
    Oh, so the F in Hooke's Law is the net force?
     
  21. Jan 19, 2014 #20

    PhanthomJay

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    Yes, correct. But you need to get a better handle on free body diagrams and the difference between internal and external forces. The only forces external to the system are the weights of the blocks. When you look at one of the hanging blocks, you isolate it in a free body diagram to identify the forces acting on it. Its weight acts down and the tension in the cord acts up, and since the block is in equilbrium , you conclude from Newtons first law that the cord tension is mg/2. Note that the tension and weight act on the block , not in the block. (The force in the block is the internal tension mg/2). Same principles apply for the other block.
    Now look the free body diagram of the spring. The 2 forces acting on it are the cord tensions mg/2 from each side. The elastic force in the spring is determined by cutting a section thru the spring, and determining from equilibrium that the internal spring force is mg/2.
     
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