Two bars connected by a spring on a floor with friction

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Homework Help Overview

The problem involves two bars of masses connected by a spring on a frictional surface, with the objective of determining the minimum force required to move one of the bars. The discussion centers around the forces acting on the blocks, the role of friction, and the energy considerations involved in the system.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equations of motion for both blocks, questioning the role of the force equations and the assumptions regarding friction coefficients. There are inquiries about the behavior of the spring and the direction of forces acting on the blocks.

Discussion Status

Participants are actively exploring different interpretations of the problem, particularly regarding the conditions under which the blocks move and the implications of the spring's behavior. Some guidance has been offered about focusing on free body diagrams and the relationship between forces and motion.

Contextual Notes

There are indications of potential errors or ambiguities in the problem statement, particularly concerning the coefficients of friction and the definitions of forces involved. Participants are also considering the implications of static versus kinetic friction in their analysis.

  • #31
PeroK said:
You need to think about what happens after the equilibrium point.
Is this the equation we should be concerned with?
##Fx - μm_2gx - 1/2kx^2 = -1/2m_2v^2##
 
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  • #32
Null_Void said:
Is this the equation we should be concerned with?
##Fx - μm_2gx - 1/2kx^2 = -1/2m_2v^2##
No. That's no use, as you yourself have repeatedly complained about the ##v^2## term being a nuisance.
 
  • #33
PeroK said:
No. That's no use, as you yourself have repeatedly complained about the ##v^2## term being a nuisance.
Will it perform simple harmonic motion about the equilibrium point?
 
  • #34
Null_Void said:
Will it perform simple harmonic motion about the equilibrium point?
No.
 
  • #35
Null_Void said:
But an extra velocity term ##v## is still present? What value should ##v## take in this case
Suppose there is still a velocity term at that point. What would happen if you were to make F just a tiny bit less?
 
  • #36
PeroK said:
No.
## Fx - 1/2kx^2 - μm_2gx= 0##
Is this the right equation?
 
  • #37
Null_Void said:
## Fx - 1/2kx^2 - μm_2gx= 0##
Is this the right equation?
It is. It's useful because that's the point at which ##m_2## stops. And that's the point at which the force in the spring is at its maximum.
 
  • #38
PeroK said:
It is. It's useful because that's the point at which ##m_2## stops. And that's the point at which the force in the spring is at its maximum.
Ah I get it now. Combining this with the equation @haruspex pointed out, I'll get
##F = μ(m_1g/2 + m_2g)## which is the correct answer, Thanks for your help everyone!
 
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