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Statics Springs Design Problem

  1. Mar 4, 2013 #1
    Statics Springs "Design Problem"

    1. The problem statement, all variables and given/known data

    Here's a problem I've been very determined to figure out, but I can't. I attached the picture. This single problem is 5% of my grade, so I'd appreciate any help!

    A plate storage system for a self-serve salad bar in a restaurant is shown. As plates are added to or withdrawn from the stack, the spring force and stiffness are such that the plates always protrude above the tabletop by about 60 mm. If each plate has 0.509 kg mass, and if the support A also has 0.509 kg mass, determine the stiffness k and unstretched length L0 of the spring. Assume the spring can be compressed by a maximum of 40% of its initial unstretched length before its coils begin to touch. Also specify the number of plates that can be stored. Assume the system has guides or other mechanisms so the support A is always horizontal.


    2. Relevant equations

    Spring force=stiffness*(L-Lo) L=final elongation and Lo=initial elongation

    3. The attempt at a solution
    There's 3 parts to the spring equation and I can only get one by a random number of plates (since you cant tell how many exactly their are in the picture) and taking their weight force. By summing the forces in the y direction i get Force of spring-mg=0, so Fs=mg. That's what I tried and got me...nowhere. I'm guessing I should find the stiffness first since it's listed first in the problem statement, but i'm not sure how to, or what to use for elongation.
     

    Attached Files:

  2. jcsd
  3. Mar 5, 2013 #2

    CWatters

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    I haven't solved it but I'd try writing two equations based on the one you cited:

    1) An equation with just one plate.
    2) An equation with nPlates where n is unknown but causes the spring to be compressed by 40%

    See if thats enough equations to solve for the unknowns. If not then pick another number of plates and write another equation?

    Something like that anyway :-)

    Basically for each plate you add you know the change in the displacement and the change in mass. You might be able plot a graph and extrapolate to zero plates?
     
  4. Mar 5, 2013 #3
    Thanks I know what you mean. So the spring goes down 15mm every time a plate is added. But do you think the unstretched length of the spring is 660mm (so then when theres 4 plates on top protruding 60mm it will be level with the table top)? Because, thats what I assumed, so i took 40% of 660mm and got like 396mm. So that would me the max compressed. then i took 396/15 and got max of 24 plates. So i added up the weight of the plates to get spring force and solved for stiffness 247=-k(396-660) so k≈0.4 N/mm.

    Could this be right!? I really hope so. I think it's similar to what you told me to do with 2 equations, but i just used one with the max spring lengths and trays assuming the spring unstretched is about 660mm.

    Thanks
     
  5. Mar 5, 2013 #4

    haruspex

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    Yes, and you know the weight of each plate, so that tells you k.
    Not quite. You're overlooking that the bottom plate effectively occupies 30mm height. So what does that make the platform height with no plates?
     
  6. Mar 5, 2013 #5
    Okay k stiffness changes after each plate is added then?

    And okay would I add 15mm to compensate for the first plate height making it 375mm unstretched length?
     
  7. Mar 5, 2013 #6

    haruspex

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    No, the stiffness (modulus) is a constant.
    The OP is wrong in referring to 'unstretched' length. The fully relaxed state of the spring would be its uncompressed length, and that must exceed 600mm. And that's not simply the length with no plates - the platform A also weighs as much as a plate. So it's like the length with -1 plates.
     
  8. Mar 5, 2013 #7
    Okay because when I was adding one plate on and subtracting 15mm for length and then solving for k several times, I was getting different answers. This must be because I had the wrong uncompressed length?
     
  9. Mar 6, 2013 #8

    haruspex

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    Possibly, but getting k is much simpler than that - just take the weight of one plate and divide by the change in length of the spring its addition causes: ΔF/Δx.
     
  10. Mar 6, 2013 #9
    k=0.33 Everything makes sense now. Thanks a lot
     
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