Statics Springs Design Problem

AI Thread Summary
The discussion revolves around solving a statics problem involving a spring mechanism for a salad bar plate storage system. The key challenge is determining the spring stiffness (k) and the unstretched length (L0) based on the mass of the plates and the spring's compression characteristics. Participants suggest using equations that relate the spring force to the weight of the plates and the resulting displacement to find k. There is clarification on the concept of unstretched length versus uncompressed length, emphasizing that the spring's length must account for the weight of the support as well. Ultimately, the correct approach simplifies to calculating k by dividing the weight of a plate by the change in spring length it causes.
jklops686
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Statics Springs "Design Problem"

Homework Statement



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 As plates are added to or withdrawn from the stack, the spring force and stiffness are such that the plates always protrude above the tabletop by about 60 mm.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.

Homework Equations



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

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 can't 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.
 

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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 that's 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?
 
CWatters said:
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 that's 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?

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 there's 4 plates on top protruding 60mm it will be level with the table top)? Because, that's 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
 
jklops686 said:
So the spring goes down 15mm every time a plate is added.
Yes, and you know the weight of each plate, so that tells you k.
when there's 4 plates on top protruding 60mm it will be level with the table top)?
Not quite. You're overlooking that the bottom plate effectively occupies 30mm height. So what does that make the platform height with no plates?
 
haruspex said:
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?

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?
 
jklops686 said:
Okay k stiffness changes after each plate is added then?
No, the stiffness (modulus) is a constant.
And okay would I add 15mm to compensate for the first plate height making it 375mm unstretched length?
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.
 
haruspex said:
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.

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?
 
jklops686 said:
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?
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.
 
haruspex said:
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.

k=0.33 Everything makes sense now. Thanks a lot
 
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