STATICS. 



[121.. 



i+l 



(i+ i)th sides intersect ; let 4 , a. +l be the angles at which these 

 sides are inclined to the horizon, and W t the weight suspended 

 from the vertex P t (Fig. 27). 



Cutting the cord on both sides of P if and introducing the 



tensions T t and 7 (+1 , the condi- 

 tions of equilibrium of the point 

 P t are found by resolving the 

 three forces W it T it T i+1 horizon- 

 tally and vertically (Art. 100) : 



T t+1 cos a i+1 = T i cos a it (i) 



T i+1 sin a i+1 = T ( sin a t + W t . (2) 



The former of these equations 

 shows that, whatever the weights 

 W and the lengths and inclina- 

 tions of the sides, the horizontal 



components of the tensions T are all equal. Denoting this con- 



, - 



stant value by //, we have 



Fig. 27. 



7*! cos ! = T 2 cos 2 = = T i cos a.i='" = 



(3) 



Substituting the values of T t and T i+l as obtained from these 

 relations, into (2), this equation becomes 



W. 



tan a i+l = tan , + ' 

 ri 



(4) 



which shows that as soon as all the weights and the inclination 

 and tension of any one side are given, the inclinations and ten- 

 sions of all the other sides can be found. 



121. Let us now assume that the weights W are all equal. 

 Then the values of tan a i+l given by (4) form an arithmetical 

 progression. If, in addition, we assume that the sides of the 

 polygon are such as to have equal horizontal projections, i.e. 

 if we assume the weights to be equally spaced horizontally, 



