# Stability – I : Chapter 3

###### Solution :

Volume of rectangular log = (L x b x h)
= (8m x 2m x 2m)
= 32m3

Weight = (U/W volume ) x (Density of displaced water)
Weight   = (8 x 2 x1.6) x(1)
= 25. 6t

RD = (mass / volume)
= 25.6 / 32
= 0.8t/m3 .

###### Solution :

Volume of rectangular log = (L x B X H )
= (5m x 1.6m x 1.0m)

Weight of  log  = 6t
SW RD = 1.025

Weight = (u/w volume )x (Density of water displaced)
6   =   (5 x 1.6 x D) x (1.025)
D  =  (6 /( 5 x 1.6 x1.025)
= 0.73m
Hence draft = 0.73m

As we know that
Density  = (mass /volume)
= (6 / (5 x 1.6 x 1 )
=0.75 t/m3

###### Solution :

Area of rectangular log = ( B X H ) = 3m x 2m
RD of log  = 0.7 t/m3
Draft in water of RD 1.01 can be calculated  as;
Density= (mass/volume)
0.7 =  mass/ volume
Mass = volume x 0.7
Mass   = (L x 3 x2) x ( 0.7)
= 4.2L t

Now mass = (U/w volume)  at depth of  D m x (1.01)
4.2L   =( L x 3 x D )x( 1.01)
D   = ( 4.2 / 3 x 1.01 )
= 1.386m.

Hence draft in water of RD 1.01  is 1.386m

#### Amit Sharma

Graduated from M.E.R.I. Mumbai (Mumbai University), After a brief sailing founded this website with the idea to bring the maritime education online which must be free and available for all at all times and to find basic solutions that are of extreme importance to a seafarer by our innovative ideas.

• sailoramit.2085 says:

PROBLEM NUMBER 4 AND 5 ARE COMPLETELY WRONG, HOW CAN ONE USE FORMULA FOR A VOLUME OF A RECTANGLE TO DERIVE VOLUME OF A TRIANGLE

• Pulkit malik says:

Question 4 and 5 are wrong.

• Pulkit malik says:

Questions 4 and 5 are wrong.

• Hardik says:

Question 4&5 are wrong .

• Shivam kumar says: