Two bars connected by a spring on a floor with friction

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SUMMARY

The discussion centers on a physics problem involving two blocks, ##m_1## and ##m_2##, connected by a spring on a frictional surface. The key equations derived include the force equations for each block: ##F - kx - μm_1g = m_1a## for block 1 and ##kx = μm_2g## for block 2. The minimum force ##F## required to move block ##m_1## is determined to be ##F = μ(m_1g/2 + m_2g)##, where ##μ## represents the coefficient of friction and ##k## is the spring constant. The discussion emphasizes the importance of free body diagrams and energy considerations in solving the problem.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with free body diagrams (FBD)
  • Knowledge of spring mechanics and Hooke's law
  • Concept of kinetic and static friction coefficients
NEXT STEPS
  • Study the principles of energy conservation in mechanical systems
  • Learn about the dynamics of systems involving springs and friction
  • Explore advanced applications of free body diagrams in complex systems
  • Investigate the differences between static and kinetic friction in practical scenarios
USEFUL FOR

Students of physics, mechanical engineers, and anyone interested in understanding dynamics involving springs and frictional forces in mechanical systems.

  • #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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