# Stability – I : Chapter 11

###### Solution :-

Given :
Cargo shifted (w)   = 200t,
Distance  = 10m,
d = 5m from CL to stbd
W = 10000 t ,
KG = 7.0m &
KM = 7.4m

 Ship’s wt KG VM 10000 t 7.0m 70000 tm 200t (shift) 10m (-) 2000 tm

Final W =10000t                                             Final VM= 68000 tm

We know that:
Final KG = (final VM)/ (Final W)
Final KG = (68000/10000)
= 6.8m

Final GM= (KM – KG)
= (7.4 – 6.8)
= 0.6m

Listing moment (LM) = (weight x distance )
= (200 x 5)
= 1000tm

tanθ =  LM/(W x GM)
= 1000 /(10000 x 0.6)
= 9.46 degree

### About the author

#### Vikrant_sharma

• Ankit says:

Q.7

Final listing moment is 700(p), not starboard

• Aba says:

For case 2; since, both the tanks are slack, therefore, Total FSC = 0.3456m x 2 = 0.6912m.
Good luck

• Anupam says:

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

• Ankit says:

Q.7 listing moment is 700 port not stbd