A uniform flexible chain of linear mass density and fixed length hangs at rest under…

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Mechanics & Dynamical Systems · Lagrangian mechanics

Action and Symmetry, Ch Multipliers, non-holonomic constraints and forces outside the potential

Problem

A uniform flexible chain of linear mass density

\rho = 0.35\,\mathrm{kg\,m^{-1}}

and fixed length

\ell

hangs at rest under gravity between two fixed supports. In equilibrium it takes the shape

y(x)

that makes its total gravitational potential energy

V[y] = \rho g\int_{x_1}^{x_2} y\sqrt{1+y'^2}\,\mathrm{d}x

stationary subject to its length,

\int_{x_1}^{x_2}\sqrt{1+y'^2}\,\mathrm{d}x = \ell,

being fixed, where

x

is horizontal,

y

is measured vertically upward from the chain's lowest point, and

g = 9.8\,\mathrm{m\,s^{-2}}

. The chain is perfectly flexible (no bending stiffness) and air resistance is ignored. ((a)) Introduce a Lagrange multiplier

\lambda_1

for the length constraint, entering the integrand as

-\lambda_1\sqrt{1+y'^2}

so that

-\lambda_1

will come out positive. Using the Beltrami first integral on the augmented integrand (which has no explicit

x

), show that the shape satisfies

\rho g y - \lambda_1 = \rho g c\sqrt{1+y'^2}

for some constant

c

of dimension length, and hence that

y(x) = \lambda_1/(\rho g) + c\cosh\!\left((x-x_0)/c\right)

for constants

c

x_0

. ((b)) At the chain's lowest point

y' = 0

and, by the choice of origin,

y = 0

there. Use this to express

\lambda_1

in terms of

\rho

g

and

c

, and identify

-\lambda_1

with the constant horizontal component of tension

T_0

carried along the whole chain (the standard statics result for a catenary). ((c)) For this chain the catenary parameter is

c = 1.2\,\mathrm{m}

. Give the numerical values of

\lambda_1

and

T_0