# Stability – I : Chapter 11

###### Solution:

W = 5000t, GM = 0.3m,
Weight shifted (w) = 20t & d= 5m

Listing moment = (20 x 5 )
= 100tm

We know that :
Tanθ  = ( Listing Moment) /(W x GM)
= 100/(5000 x 0.3)
= 3.8 degree.

1. ###### On a ship of W 8000t, GM 2.0m, If the following transverse shifting were done , find the list:
• 200cargo shifted 4m to stbd
• 100t cargo shifted 2 m to port
• 100t cargo shifted 4m to port
• 50t stores shifted 20m to stbd

Solution:-

We can calculate Listing moment due to shifting of cargo
LM(1)  = ( weight x distance )
=   (200 x 4)
= 800 tm (S)

Again,   LM (2) = (Weight x distance )
= (100 x 2)
= 200 tm (P)

LM(3)  = ( weight x distance )
=   (100 x 4)
= 400 tm (P)

LM(4)  = ( weight x distance )
= (50 x 20)
= 1000 tm (S)

Final listing moment (LM) = LM(1) + LM(2) +LM(3)+ LM(4)
= 800(S) + 200(P) + 400(P) + 1000(S)
=1800(S) + 600(P)

So, final listing moments = (1800 – 600)
= 1200 tm (S)

Now, Tanθ  = (Final LM )/(W x GM)
= (1200/(8000 x 2 )
= 4.29 degree (S) #### 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