Springs connected to a mass in series?

AI Thread Summary
The discussion focuses on understanding the mechanics of springs connected in series, particularly how forces and displacements interact. Participants explore the application of Newton's laws to derive equations for the system, emphasizing that the total displacement of the mass is the sum of the displacements of each spring. They clarify that while the force applied to each spring is the same, the relationship between the spring constants must be established to find an effective spring constant for the series configuration. The conversation highlights the importance of recognizing that the restoring forces of the springs must equal the applied gravitational force. Ultimately, the goal is to derive an effective spring constant that accurately represents the behavior of the series spring system.
  • #51
jack action said:
Yes, there are.

Doing the free body diagrams (FBD) for each component (the ceiling ##c##, the spring ##k_1##, the spring ##k_2## and the mass ##m##):

By definition, from one FBD to the next:
$$F_c = F_{k1}$$
$$F_{k1} = F_{k2}$$
$$F_{k2} = F_m$$
mass FBD:
$$F_m = mg$$
Therefore:
$$F_c = F_{k1} =F_{k2} = F_m = mg$$
The force ##mg## goes through every component.
No, that can't be right. F K1 would be directed upward, not downwards, because the spring will be stretched downwards, and so FK1 will apply a restoring force upwards.
 
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  • #52
flinnbella said:
No, that can't be right. F K1 would be directed upward, not downwards, because the spring will be stretched downwards, and so FK1 will apply a restoring force upwards.
On the thing that is attached to below ( i.e. spring 2 ) it is applying a restoring force ##F_{k2}##, which is equal and opposite the force that is applied to spring 1. (##F_{k1} = - F_{k2}##)
 
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  • #53
jack action said:
The force mg goes through every component.
This is true but that means they all have magnitude mg but they are not all mutually action-reaction pairs. There are three such pairs but the others are known equal and opposite because each spring is not experiencing acceleration or deformation.
 
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  • #54
hutchphd said:
This is true but that means they all have magnitude mg but they are not all mutually action-reaction pairs. There are three such pairs but the others are known equal and opposite because each spring is not experiencing acceleration or deformation.
And in the case of the oscillator( which is @flinnbella original problem) the forces on opposite ends of each spring are equal and opposite because the springs are “massless”.
 
  • #55
Yes I had forgotten the Original Post!
 
  • #56
hutchphd said:
Yes I had forgotten the Original Post!
Yeah… we have been walking around a little bit in hopes to get a flag in the ground…somewhere.
 
  • #57
flinnbella said:
No, that can't be right. F K1 would be directed upward, not downwards, because the spring will be stretched downwards, and so FK1 will apply a restoring force upwards.
Those are only the magnitudes. You can see the direction of the vectors on the drawing where I already assumed that the forces acting on each component are equal and opposite.
 
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