24() MIS' i.oUS THEOKI 



Let P "be t .Itant normal action, and >-uppo>,- 



( J \\ ith the vrrlieal : then the 

 . 'ii will i .ml its direction will ma! 



angle - - Q with the vertical on the other side of it 1 1 



supposing the string to become rigid, and resolving h- 

 tally, 



\\e- 



Again, resolving vertically, and denoting by IT the 

 of the weights of the system which hangs over the cyl: 

 we have 



Pcos + p.P cos (^ - 0} - TF= 0. 



\2 / 



Hence we obtain tan 6 = /*, 



W 



VI. Suppose a heavy string which is not of mi: 

 density and thickness to be suspended from two fixed points, 

 and to be in equilibrium. Let t be the tension at any point, 

 6 the angle which th- t at that point makes with the 



horizon; then t cos 9 will be constant. For ima-ine any por- 

 tion of the string to become riiri'l, then the only horizontal 

 forces which act on it are the resolved parts of the tensions at 

 end; and these must therefore be equal in ma ;::,itul'' : 

 fore 



t cos = constant = r suppose (1). 



Let w be the weight of the portion of the string contained 

 between any iixe.l point and the variable, point considered. 

 i by resolving the forces vertically we obtain in a similar 

 manner 



t sin 6 w = constant ; 

 fore w = r tan -f constant (2). 



in, proceeding as in Art. 103, that is resolving tin- 

 forces which act on an element along the normal, we find 



( -ymcos0 = (3), 



