ORDINARY DIFFERENTIAL EQUATIONS POWERTOOL

Unit 30 -- Application: Suspended Wire

Outline of Unit 30

Initialization

> restart;

> with( DEtools ):

> with( plots ):

> with( linalg ):

Warning, the name changecoords has been redefined

Warning, the name adjoint has been redefined

Warning, the protected names norm and trace have been redefined and unprotected

>

A model for the shape of a homogeneous flexible wire that is hanging in its equilibrium position is derived from the physical requirement that the total force on any segment of the cable must be in balance. Consider the segment of the curve AP from the lowest point A = ( 0, ) to an arbitrary point P = = ( , ) on the curve. The forces acting to hold the cable in equilibrium are the tensions at the endpoints and the weight of the segment of the cable. Because the unit tangent at A is - i = < -1, 0 >, the tension at A is = i where is the magnitude of the tension at A. Likewise, the tension at P is the tangential force . Assuming the cable has density kg/m, the total mass of the cable is

> mass := m(t) = delta * Int( sqrt( 1 + diff( y(tau), tau )^2 ), tau=0..t );

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These forces are in balance when

0 = j

= i - j

Equating the horizontal and vertical components of the forces yields

> Fhoriz := 0 = T / sqrt( 1 + diff( y(t), t )^2 ) - H;

> Fvert := 0 = T * diff( y(t), t ) / sqrt( 1 + diff( y(t), t )^2 ) - g * delta * Int( sqrt( 1 + diff( y(tau), tau )^2 ), tau=0..t );

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These two equations can be combined to form the integro-differential equation

> q1 := eval( Fvert, isolate( Fhoriz, T ) );

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To convert this to a differential equation, differentiate with respect to

> q2 := isolate( diff( q1, t ), diff( y(t), t$2 ) );

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or

> ode1 := diff( y(t), t$2 ) = alpha * sqrt( 1 + diff( y(t), t )^2 );

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where .