Supporting spring, object, top spring problem.

  • Thread starter mechdesignron
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In summary: Fw=(spring force 1)+spring force 2. K1=spring constant for spring 1, K2=spring constant for spring 2. D1 and D2 are preload on the springs. K1*D1 and K2*D2 are the same because they are in parallel. When you change Fw and keep the other constants the same, it affects the distance the object drops (delta) because Fw is the force applied to the object. In summary: If you have two springs in parallel, and want to know how an object will fall, just use this equation: FW+K2N2(D2-Delta)
  • #1
mechdesignron
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FW+(Spring Force 2)K2N2(D2-Delta)

(spring force 1)K1N1(D1+Delta)

I have a spring set(springs in parallel) on top of an object and a spring set supporting the object. The weight of the object is represented by FW. The bottom springs are F1 and the top springs F2. There is preload on the springs, represented by D1 & D2 repectively. I want to know how far the object will fall (Delta) in the given system.

I came up with this formula.

FW+K2N2(D2-delta)=K1N1(D1+delta)

Some magic algebra and Presto:

delta=(FW+K2*N2*D2-K1*N1*D1)/(K1*N1+K2*N2)

OK my problem. It doesn't seem to work. I plugged this into my handy-dandy spreadsheet and when I change the K value-keeping them both the same- it works fine. The higher the K the less the object falls. but when I change the compressed distance -keeping them both the same-, (which should, by hooks Law make the spring stiffer), does not change the distance the object moves. Please help.

Thank you
 
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  • #2
Clarify a couple of points, please.

What are N1 and N2?

What do you mean by "fall"?

What is delta?

How did you arrive at your equation? It might be a good idea if you would put your full development process in a post so that I can see how you got here.
 
  • #3
mechdesignron: Assuming no coils touch, increasing the compressed distance of a linear elastic spring does not increase its stiffness. Instead, by Hooke's law, the force on the spring increases, not its stiffness.

I will call the spring constants of your two spring sets k1 and k2. Although you do not say what N1 and N2 are, we see from your formulation that k1 = K1*N1, and k2 = K2*N2. The equation in the middle of your post is correct. However, we know from equilibrium before Fw is added, k1*D1 = k2*D2. Substituting this into your middle equation, the equation simplifies to Fw - k2*delta = k1*delta. Solving for delta gives delta = Fw/(k1 + k2), which is exactly the result expected. In other words, no amount of preload compression (if your spring coils do not touch) changes the distance an added object drops (delta). The only way to change the distance the object drops is by changing k1, k2, and/or Fw. Your observed results are correct.
 
  • #4
Thank You nvn

It looks like nvn has awnsered my question but I will clarify to both clean up the post.
The top "equation" is an attempt at a free body diagram with the underline representing the supported object. N=number of springs, by fall I mean delta (the amount an object moves).
 

1. How does the spring affect the object in a top spring problem?

In a top spring problem, the spring supports the weight of the object and provides a restoring force to keep it in equilibrium. The spring stretches or compresses in response to the weight of the object, and the magnitude of the restoring force is directly proportional to the displacement of the spring from its equilibrium position.

2. What factors affect the stability of the object in a top spring problem?

The stability of the object in a top spring problem is affected by the weight of the object, the stiffness of the spring, and the length of the spring. A heavier object will require a stronger spring to maintain equilibrium, while a longer spring will provide more stability due to its larger displacement capacity.

3. How do you calculate the spring constant in a top spring problem?

The spring constant, also known as the force constant, can be calculated by dividing the force applied to the spring by the displacement of the spring from its equilibrium position. This value is represented by the letter "k" in the equation F = kx, where F is the force and x is the displacement.

4. What happens to the spring when the object is in motion in a top spring problem?

When the object is in motion in a top spring problem, the spring will also be in motion due to the forces acting upon it. As the object moves up or down, the spring will stretch or compress accordingly. However, the restoring force of the spring will always act in the opposite direction of the displacement, keeping the object in equilibrium.

5. Can a top spring problem be solved using Hooke's Law?

Yes, a top spring problem can be solved using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. This law is often used to calculate the spring constant and solve for unknown variables in spring problems.

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