An inextensible heavy chain
Lies on a smooth horizontal plane,
An impulsive force is applied at A,
Required the initial motion of K.
Let ds be the infinitesimal link,
Of which for the present we’ve only to think;
Let T be the tension, and T + dT
The same for the end that is nearest to B.
Let a be put, by a common convention,
For the angle at M ‘twixt OX and the tension;
Let Vt and Vn be ds‘s velocities,
Of which Vt along and Vn across it is;
Then Vn/Vt the tangent will equal,
Of the angle of starting worked out in the sequel.
In working the problem the first thing of course is
To equate the impressed and effectual forces.
K is tugged by two tensions, whose difference dT
Must equal the element’s mass into Vt.
Vn must be due to the force perpendicular
To ds‘s direction, which shows the particular
Advantage of using da to serve at your
Pleasure to estimate ds‘s curvature.
For Vn into mass of a unit of chain
Must equal the curvature into the strain.
Thus managing cause and effect to discriminate,
The student must fruitlessly try to eliminate,
And painfully learn, that in order to do it, he
Must find the Equation of Continuity.
The...