122.] PARALLEL FORCES. 7! 



the vertices of the polygon will lie on a parabola whose axis 

 is vertical. 



To find its equation, let us suppose, for the sake of simplicity, 

 that one side of the polygon, say the >th, is horizontal so that 

 a k =o. Taking this side as axis of x, its middle point O as origin, 

 the co-ordinates of the vertex P k are ^a, o, if a be the length of 

 the horizontal side and hence also that of the horizontal projec- 

 tion of every side. 



Putting W/H=r, we have tan<* A = o, tan A+1 = r, tan a k+2 

 = 2T, ; hence the co-ordinates of P k+1 are x=^a, y = ar ; those 

 of P k+2 are x = ^a, y = aT J r2ar = ^ar\ those of P k+3 are x=*^a, 

 =6<2T, etc. ; those of the /zth vertex after/^are 



v 



Eliminating n, we find the equation 



which represents a parabola whose axis is the axis of y, and 

 whose vertex lies at the distance \ar = \aW/H below the 

 origin O. 



122. Let the number of sides be increased indefinitely, the 

 length a and the weight W approaching the limit o, but so 

 that the quotient a/W remains finite, say lim (a/ W)= i/w. 

 Then lim (a/r} H/w, lim (#T)=O; so that the equation of the 

 parabola becomes 



w 



where w is evidently the weight of the cord, or chain, per unit 

 length. 



The parabola is, therefore, the form of equilibrium of a cord 

 suspended from two points when the weight of the cord is uni- 

 formly distributed over its horizontal projection. This is, for 



