Is F=Kx Always Necessary for Spring Forces?

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Discussion Overview

The discussion centers on the relationship between the applied force \( F \) and the spring force \( kx \) in the context of spring mechanics. Participants explore whether \( F \) must always equal \( kx \), considering various scenarios including stationary masses and constant velocity motion.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants question whether \( F \) must equal \( kx \) and suggest that \( F \) can be a separate force applied to the mass.
  • There is a discussion about the implications of Newton's laws, particularly whether they apply in this context.
  • One participant notes that if the mass is stationary or moving at constant velocity, then \( a \) in the force balance equation \( F - kx = ma \) would be zero.
  • Another participant suggests that \( F \) can take any value, indicating that \( F \neq kx \) is possible under certain conditions.
  • Participants explore conditions under which \( F \) could equal \( kx \), particularly in equilibrium situations.

Areas of Agreement / Disagreement

Participants express differing views on whether \( F \) must equal \( kx \), with some asserting it can vary while others explore conditions where they might be equal. The discussion remains unresolved with multiple competing perspectives.

Contextual Notes

Participants highlight the importance of the mass's state (stationary or in motion) and the nature of the applied force in determining the relationship between \( F \) and \( kx \). There are unresolved assumptions regarding the definitions of forces and the specific conditions being considered.

navneet9431
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Is it necessary that the applied force F is equal to the spring force 'kx'?
 

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navneet9431 said:
Is it necessary that the applied force F is equal to the spring force 'kx'?
Are you trying to invoke Newton's 2nd law or Newton's 3rd law? What, if anything, do either have to say about the situation?
 
I am not trying to invoke any law
jbriggs444 said:
Are you trying to invoke Newton's 2nd law or Newton's 3rd law? What, if anything, do either have to say about the situation?
 
Is the mass stationary?
 
No
sophiecentaur said:
Is the mass stationary?
 
navneet9431 said:
View attachment 232217
Is it necessary that the applied force F is equal to the spring force 'kx'?
The way you've drawn the picture, it looks like F is a separate force applied to the mass, not the spring force. You'll have to tell us what the drawing means - you drew it! (presumably).
 
navneet9431 said:
No
Make the spring as weak as you like. Would F be zero?
 
Yes,F(on right) is a separate force applied to mass m other than the spring force acting on the left side.
russ_watters said:
The way you've drawn the picture, it looks like F is a separate force applied to the mass, not the spring force. You'll have to tell us what the drawing means - you drew it! (presumably).
 
Please write down for us your Newton's 2d law force balance equation on the mass M.
 
  • #10
F-kx=ma
Chestermiller said:
Please write down for us your Newton's 2d law force balance equation on the mass M.
 
  • #11
navneet9431 said:
F-kx=ma
Perfect. Does that answer your question?
 
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  • #12
navneet9431 said:
Yes,F(on right) is a separate force applied to mass m other than the spring force acting on the left side.
In that case it can have literally any value you choose to give it.
 
  • #13
So it means that F=/=kx
Chestermiller said:
Perfect. Does that answer your question?
 
  • #14
But is it possible that F=kx in any condition?
 
  • #15
navneet9431 said:
But is it possible that F=kx in any condition?

If the mass is stationary, or if F is applied in such a way that the mass is moving at a constant velocity. Look at the equation you wrote. What is "a" under these two situations?

Zz.
 
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  • #16
Thanks
ZapperZ said:
If the mass is stationary, or if F is applied in such a way that the mass is moving at a constant velocity.

Zz.
 
  • #17
navneet9431 said:
But is it possible that F=kx in any condition?

YOU will choose that.

If you choose |F|>|kx|, then m won't be at rest, but will have a constant acceleration rightwards.
If |F| = |kx|, then you have achieved an equilibrium, nothing moves. (In lab, this is how we calculate spring constant, where F=mg:biggrin:)

If |F| < |kx|, then you will have the spring pulling m leftwards.
 
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