Case -1
To bring the ship upright the stbd LM must be equal to port side LM.
LM = (Weight x Distance)
3173.88 tm = ( W x 12)
Hence W = (3173.88/ 12)
= 264.49 t
Case -2
Volume of the tank = (L x B x D)
= (12 x 12 x 9)
= 1296m3
Now, Weight of FW on stbd tank = (Volume x Density)
= (1296 x 1)
= 1296 t.
Now, calculating 1/3 of mass of FW
= (1296 x 1/3)
= 432t.
| Ship’s wt | KG | VM | LM |
| 10000 | 9.0 | 90000 | 3173.88 S |
| 432 | 6t | (-) 2592 | 5184.00 P |
Final W = 10000 Final VM = 87408 tm 2010.12 P
Final KG = (Final VM/Final W)
Final KG = (87408/10000)
= 8.741m
Solid GM = (KM – KG)
= (10.8 – 8.741)
= 2.059m
We know that :
FSC = (LB3 x di) / (12 x W)
= (12 x 123 x 1) /(12 x 10000)
= 0.3456m
Fluid GM = (2.059 – 0.3456)
tanθ = (2010.12/(10,000 x 1.7134)
=6.69 degree P
Question no.19. Sub question 3
Kg will change when being shifted from derrick to lower hold 2m above kg so GM will also change
Gm= 0.58 and list will be 1.60degree to starboard
Q.7
Final listing moment is 700(p), not starboard
For case 2; since, both the tanks are slack, therefore, Total FSC = 0.3456m x 2 = 0.6912m.
Good luck
Ex -11 List Ques No. 18, Part 2 have a mistake in total Vertical movements…..it should be 87000 not 87200 which gives us a final answer as follows
List = 3* 51′ or 3.85 degrees
Q.7 listing moment is 700 port not stbd