Stability – I : Chapter 3

  1. A rectangular log of wood 8m long, 2m wide and 2m high floats in FW at a draft of 1.6m with one face of horizontal. Find its mass and RD.
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 .

  1. A rectangular log of wood 5m x 1.6m x1.0m weight 6t and floats with it largest face horizontal. Find its draft in SW and its RD.
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

  1. A rectangular log 3m broad and 2m high floats with its breadth horizontal. If the density of the log is 0.7 tm-3, find its draft In  water of RD 1.01.
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

About the author

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.

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