Understanding Driving Force in the m2 Equation: Explained by Experts

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SUMMARY

The discussion centers on the m2 equation in classical dynamics, specifically the term b1(x'2 - x'1), which is identified as a negative term despite being associated with the motion of m2. Participants clarify that a driving force is an external force that initiates motion, while the frictional force is an internal force that does not drive the system but rather depends on its motion. The conversation also highlights the importance of understanding the relationship between the forces acting on m1 and m2, particularly through Newton's third law, which states that forces between two bodies are equal in magnitude and opposite in direction.

PREREQUISITES
  • Understanding of classical dynamics principles
  • Familiarity with Newton's laws of motion
  • Knowledge of force interactions, specifically frictional forces
  • Ability to interpret equations of motion
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  • Study the concept of driving forces in classical mechanics
  • Learn about Newton's third law and its applications in dynamics
  • Explore the implications of frictional forces in multi-body systems
  • Analyze specific case studies involving the m2 equation and its terms
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Students of physics, particularly those studying classical mechanics, educators explaining dynamics concepts, and anyone interested in understanding force interactions in multi-body systems.

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Homework Statement
(Problem 3.26; Classical Dynamics of Particles and Systems)
Figure 3-B illustrates a mass m1 driven by a sinusoidal force whose frequency is w. The mass m1 is attached to a rigid support by a spring of force constant k and slides on a second mass m2. The frictional force between m1 and m2 is represented by the damping parameter b1 and the friction force between m2 and the support is represented by b2. Construct the electrical analog of the system, and calculate the impedance.
Relevant Equations
Inserted image of equations in question. These are from the solution manual.
565-3-26p-i10.png
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I am having trouble understanding why the second term in the m2 equation, b1(x'2 - x'1), is a negative term. Given that this force is the reason why m2 is moving in the first place, why is it not considered a driving force? I think that I don't have a clear understanding of what driving force means.
 
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A driving force is an external force imposed on the system that drives its motion. The frictional force is not driving because it is an internal force between different parts of your system that is not forcing the motion of the system but instead its form depends on the motion of the system.

Since it is the frictional force from 1 on 2, it is the 3rd law pair of the frictional force from 2 on 1. It must be equal in magnitude and opposite in direction.
 
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What happens if the value of b1 is so big that v1=v2?
 
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Orodruin said:
A driving force is an external force imposed on the system that drives its motion. The frictional force is not driving because it is an internal force between different parts of your system that is not forcing the motion of the system but instead its form depends on the motion of the system.

Since it is the frictional force from 1 on 2, it is the 3rd law pair of the frictional force from 2 on 1. It must be equal in magnitude and opposite in direction.
Got it. I understand. Thank you!
 
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ANAli said:
Homework Statement:: (Problem 3.26; Classical Dynamics of Particles and Systems)

I am having trouble understanding why the second term in the m2 equation, b1(x'2 - x'1), is a negative term. Given that this force is the reason why m2 is moving in the first place, why is it not considered a driving force? I think that I don't have a clear understanding of what driving force means.
One way to see if the signs make sense is to consider specific cases. For example, if ##m_1## is moving to the right (##\dot x_1>0##) and ##m_2## is at rest (##\dot x_2=0##), the top mass will push the bottom mass to the right so you would expect ##\ddot x_2>0##. In this case, the second equation reduces to ##m_2 \ddot x_2 = + b_1 \dot x_1 > 0##, which is in agreement. Similarly, if you consider the case where ##\dot x_2>0## and ##\dot x_1 = 0##, you'll see the signs of the other two terms are correct as well.
 

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