SOLUTION ; The procedure used is identical to that of the preceding problem. 

 Calculate 



d 30.0 



= 0.031 



2 2 



gT (9.8 (10) 



enter Figure 7-75 to the breaking-limit curve and read 



H 



b 



= 0.0205 



2 

 gT 



Therefore 



H, = 0.0205 gT^ = 0.0205 (9.8) (10)^ = 20.1 m (65.9 ft) 



and 



iL = 1^ = 15 



H, 20.1 ^'^^ 

 D 



o 

 From Figures 7-71 and 1-11, using d/gT = 0.031 and interpolating between 



H « and H = 1/4 H, for H = 0.15 H^ , 

 K. = 0.44 



From equation (7-37), 



F . = C, og -!^ HK . 



2 



F. =2.0 (1025.2) (9.8) ''^?'^^ (3) (0.44) = 1,875 N (422 lb) 

 ^m 4 



and from equation (7-38), 



F = 1.2 (0.5) (1025.2) (9.8) (0.3) (3)^ (0.20) = 3,255 N (732 lb) 



Dm 



Compute W from equation (7-41), 



_ V _ 2.0(0.3) _ 

 ^ ^H 1.2 (3) ^'^' 



7-130 



