Understanding Driving Force in the m2 Equation: Explained by Experts

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Homework Help Overview

The discussion revolves around understanding the m2 equation in the context of classical dynamics, specifically focusing on the interpretation of terms related to driving forces and frictional forces within a system. Participants are exploring the nature of these forces and their implications on motion.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the definition of driving force and its distinction from internal forces, particularly friction. There is an inquiry into why a specific term in the m2 equation is negative and how it relates to the motion of the masses involved. Some participants suggest examining specific cases to clarify the signs of the terms in the equation.

Discussion Status

The discussion is active, with participants providing insights into the nature of driving forces and friction. Some have expressed understanding after receiving explanations, while others continue to explore the implications of the terms in the equation and their physical meanings.

Contextual Notes

The original poster references a specific problem from a textbook, indicating that the discussion is framed within the constraints of a homework assignment. There is an emphasis on understanding the dynamics of the system rather than providing direct solutions.

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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.
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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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