Understanding Springs: What Happens with Applying Force?

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Discussion Overview

The discussion revolves around the behavior of a spring system when a force is applied to one of the masses connected by the spring. Participants explore the implications of Hooke's Law, the role of reaction forces, and the dynamics of the system in a frictionless environment.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether the spring would contract when a force is applied to one of the spheres, suggesting that without an opposing force, the spring would not compress.
  • Another participant argues that if a force is applied to the masses, the system would accelerate, leading to a compression of the spring, and provides a mathematical derivation to support this view.
  • Some participants emphasize the importance of reaction forces, stating that the inertia of the spheres would result in a force that causes the spring to compress.
  • A participant introduces a second-order differential equation to describe the motion of the spheres and inquires about its application, including the potential addition of damping effects.

Areas of Agreement / Disagreement

Participants express disagreement regarding the conditions under which the spring would compress. Some assert that a reaction force is necessary for compression, while others argue that the applied force leads to acceleration and subsequent compression of the spring. The discussion remains unresolved with multiple competing views.

Contextual Notes

Participants reference concepts such as Hooke's Law, inertia, and differential equations without reaching a consensus on the specific dynamics of the spring system under the described conditions.

jacobsmith
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I am rather confused about springs. This may be an obvious question, so just bear with me.

Now, say you had two spheres of any mass, a spring (that obeys Hooke's Law) connecting them, and the whole system was without external forces or friction (we can say, suspended in space). Now, what would happen if you applied a force to one of the spheres, in the direction of the other sphere? How far would the spring contract - would it even contract in the first place?
 
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jacobsmith said:
would it even contract in the first place?

No. If there is no opposing force the spring won't compress. There must be a reaction force on the opposite end of the spring for it to compress. Otherwise, it will just move in space.

CS
 
I disagree. Let's say we have two masses attached to a spring, m1 and m2. The spring constant is known and is equal to k. Let's say we push the masses with force F until the entire system is accelerating at a. We know, from free-body diagrams, that F-Fspring = m1*a and Fspring = m2*a. Eliminating a, we have (F - Fspring)/m1 = Fspring/m2. Solving for Fsrping we have (F*m2)/((m1+m2)) = Fspring = kx. Thus the spring will compress until x = (F*m2)/(k*(m1+m2))
 
stewartcs said:
No. If there is no opposing force the spring won't compress. There must be a reaction force on the opposite end of the spring for it to compress. Otherwise, it will just move in space.

CS
The reaction force is due to the inertia of the spheres via f=ma. A force on one causes both to accelerate and a force between them (it'll be half the force applied to the first).
 
russ_watters said:
The reaction force is due to the inertia of the spheres via f=ma. A force on one causes both to accelerate and a force between them (it'll be half the force applied to the first).

Opps! Forgot about inertia...you can go ahead and smack me now. :redface:

CS
 
You can decribe the response of the spheres with respect to each other with a simple second order DE.

mx'' - kx = 0
 
Topher925 said:
You can decribe the response of the spheres with respect to each other with a simple second order DE.

mx'' - kx = 0

How would I apply this equation? Would I use it for both spheres?

And if I wanted to add in damping, the equation would be this: mx'' - cx' - kx = 0 where c is the damping coefficient?

Thank you all for your help.
 

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