626 JAMES CLERK MAXWELL. 



Let T be the tension, and T + eft 



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 V t and V w be ds's velocities, 



Of which V 4 along and V n across it is ; 



y 

 Then -^ the tangent will equal, 



v * 

 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 



(1) Must equal the element's mass into Vj. 

 V n 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 V w into mass of a unit of chain 



(2) 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 reason is this, that the tough little element, 



Which the force of impulsion to beat to a jelly meant, 



Was endowed with a property incomprehensible, 



And was " given," in the language of Shop, " inexten- 



sible." 



It therefore with such pertinacity odd defied 

 The force which the length of the chain would have 



modified, 



That its stubborn example may possibly yet recall 

 These overgrown rhymes to their prosody metrical. 

 The condition is got by resolving again, 

 According to axes assumed in the plane. 

 If then you reduce to the tangent and normal, 



(3) You will find the equation more neat tho' less formal. 



