5. 



W^ = 1000 lb 

 b 







^b 1000 

 °s 2 400 

 ^1 



2 

 ^ lb-sec 

 2.5 ^^, 





Assume c = 





6. 



Let At = .01 sec 





7. 



a. F = W, - D V 



- c (o)A (o: 



= 1000 - 2.5 (20)^ - (20) (.5892) 

 = -11.784 lb 



b. a = '-^I'll^ = -.00212 ft/sec^ 



5576 



c. V = 20 - (.00212) (.01) 



= 19.999981 ft/sec 



x^^= + (20) (.01) 



= .2 ft 



This completes the first iteration. In preparation for the second 

 iteration, it is noted that the embedment depth after the first iteration 

 is .2 feet. Since this is within the increment, AJl = 0.5 feet, centered 

 about £. = 0, the summation portion of the force equation still contains 

 only one multiplication term. The calculation continues as follows: 



d. F = +1000 - 2.5 (19.99998)^ - (20) (.5892) = -13.5 lb 



e. a = Zl^'J' = .00242 ft/sec^ 



JJ/D 



f. v^, = 19.99998 - (.00242) (.01) 



'2At " 19.99995 ft/sec 



n o M / 19. 999981 + 19.99995 , . ^^ . 

 ^2At = °-2 + ( 2 ^ ^-^^^ 



= .40 ft 



24 



