Vibrations of a Hanging Chain: Modeling Tension with PDEs

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SUMMARY

The discussion focuses on modeling the vibrations of a hanging chain using partial differential equations (PDEs). Specifically, it addresses the equilibrium tension in the chain, defined as $\tau(x)=\rho \cdot g \cdot x$, where $\rho$ is the mass density and $g$ is the gravitational acceleration. The equilibrium condition is established by balancing the upward tension against the downward gravitational force acting on the chain segment. Participants are encouraged to derive the tension equation based on these principles.

PREREQUISITES
  • Understanding of partial differential equations (PDEs)
  • Knowledge of basic mechanics, particularly tension and gravitational forces
  • Familiarity with the concepts of mass density and equilibrium in physical systems
  • Ability to interpret mathematical expressions and equations related to physics
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  • Study the derivation of PDEs for vibrating strings and chains
  • Explore the applications of tension in various physical systems
  • Learn about the implications of mass density in dynamic systems
  • Investigate the role of gravitational acceleration in equilibrium conditions
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Students and professionals in physics, particularly those focusing on mechanics and wave phenomena, as well as mathematicians interested in the application of PDEs in real-world scenarios.

cbarker1
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Dear Everyone,

I am having trouble with how to start with one part of the question:
"In this exercise, we derived the PDE that models the vibrations of a hanging chain of length $L$. For convenience, the x-axis placed vertically with the positive direction pointing upward, and the fixed end of the chain is fastened at $x=L$. Let $u(x,y)$ denote the deflection of the chain, we assume is taking place in $(x,u)$-plane, as in the figure, and let $\rho$ denote its mass density (mass per unit length).
(Here is where I have trouble starting)
Part a:
Show that, in the equilibrium position, the tension at a point $x$ is $\tau(x)=\rho \cdot g \cdot x$, where g is the gravitational acceleration.

Thanks,
Cbarker1
 
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Hi Cbarker1,

In the equilibrium position the chain is hanging straight down. At a point $x$ along the chain there are two forces we must consider: (i) the weight of the chain pulling the point down via gravity and (ii) the tension at $x$ pulling up. The equilibrium equation is then $$\tau(x)-mg=0,$$ where $m$ is the mass of the portion of the chain pulling down at the point $x$. Can you continue from here?
 

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