# PhysicsEquilibrium

#### Cbarker1

##### Active member
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

#### GJA

##### Well-known member
MHB Math Scholar
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?