1. a. F^ = W, - W, 



Wj = 7sVs = 30 X 212 = 6,360 1b 



Wb = 50,000 1b 



Fq = 50,000 - 6,360 = 43,640 lb 



b. D/B = 0.393 



from Figure 3 (or Equation 5) 



Fib/Fq = 0-670 



c. Fib = 0-670 X 43,640 = 29,240 lb 



F- Fgib = F,t, + Wb - Wj = 29,240 + 50,000 - 6,360 



= 72,880 lb 



or, with a safety factor of 1 .5, 



Fgjb = (1.5 X 29,240) + 50,000 - 6,360 = 87,500 1b 



To ensure immediate breakout, a line force of 87,500 pounds 

 should be applied to the object. 



G. It is determined that a force of 87,500 pounds cannot be applied 

 with the existing equipment. 



H. A force of 60,000 pounds, however, can be applied. The amount 

 of time required for breakout with this force Is desired. 



16,3601b 



Fb 



= Fg, - W, + W, 



Fcb 



= 60,000 lb 



Fb 



= 60,000 - 50,000 + 6,360 = 



Fb/Fib 



= 16,360/29,240 = 0.56 



K. From Figure 5 (or Equation 7) 



T = 1.41 X lO^db-min/ft^^l 



Ik = 



TD^ /D^2 

 P \B, 



fb ^ 16,360 

 A 90 



1 82 psf 



tu = 



1.41 X 105 X (2.36)2 /2.36V „^ .^2 ■ 



—:^] = 6.7 X 10-^ mm 



182 \ 6 / 



1.1 hr 



31 



