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

  1. A cylinder 2m in diameter and 10 m log floats in FW, with its axis horizontal, at draft of 0.6m. Find its mass.
Solution :

Diameter of cylinder=  D = 2m,
Radius = ( D / 2 )
= 1m
length = 10m, depth = 0.6m

Mass =  (u/w volume) x (density of displaced water)

Weight   =     ( πr2h x density )
= (3.1416 x 1 x 1 x 0.6) x (1)
= 1.88t.

  1. A barge of triangular cross section is 20m long, 12m wide and 6m deep. It floats in SW at a draft of 4m. Find its displacement.
    Solution :

Volume of triangular cross section = (L x B X H )
= (20 x 12 x6)

Depth  = 4m

Here Displacement means =  displacement by triangular cross section.
In triangle ABC, CG is perpendicular  to  AB
So triangle GBC and EFC are similar,
So by law of similar angle triangle
GB / EF =   GC / EC
6 / EF   = 6/ 4
EF = (6x 4)/6
=  4m

Now, DF = (2EF)
= 4×2
= 8m

Displacement  = ( u/w volume ) x (1.025)
= (20 x 8 x 4) x (1.025)
= 656 t .

  1. A cylindrical drum of 1.2m diameter and 2m height floats with it axis vertical in water of RD 1.016 at a draft of 1.4m. Find the maximum mass of lead shots that can be put in it with sinking it.
Solution:

Radius = (d/2)
=(1.2m/2)
= 0.6m

H = 2m, D =1.4m , RD = 1.016

Mass of the cylinder =  (r2 h) x(density)
= (3.1416 x 0.6 x 0.6 x 2) x (1.016)
= 2.298t

Mass of the cylinder at 1.4m = 1.4 x  r2 h)  x (density)
= 1.4 x (3.1416 x 0.6 x0 .6) x (1.016)
= 1.608 t

Hence ,the maximum mass of lead shots that can be put in it with sinking  is  (2.298 t – 1.608 t) =  0.69 t

  1. A rectangular barge of 10 m long and 5m wide, floating in SW at a draft of 3m, is being lifted out of the water by a heavy-lift crane. Find the load on the crane when the draft has reduced to 1m.
    Solution:

Area of rectangular barge = ( L x B)
= 10m x 5m,
RD = 1.025

Mass when draft 3m = (u/w volume at 3m) x (density)
= (10 x 5 x 3) x (1.025)
= 153.75t.

Mass after being lifted by the heavy crane =(u/w volume at 1m depth )x (density)
= (10 x5 x 1) x (1.0250
= 51.25 t

Hence ,Load taken by the crane = (153.75 – 51.25)
=  102.5 t

  1. A rectangular box 2.4 m long, 1.2m wide and 0.8m high, floats in water of RD 1.012m at an even keel draft of 0.2m. find maximum mass of SW that can be poured in to it without sinking it.
Solution :

Given :
L = 2.4m, B = 1.2m, H = 0.8m
RD =  1.012 , Depth = 0.2m

Mass of rectangular box = (L x B x H )x (1.012)
= (2.4 x 1.2 x 0.8) x (1.012)
= 2.33t

Mass of box even keel when draft  is 0.2m
= (2.4 x 1.2 x 0.2) x(1.012)
= 0.583

Hence, Salt water that  can poured  is (2.33 – 0.58)
= 1.75 t.

  1. A box – shaped vessel of 18450 t displacement in 150m long and 20 m wide. Find its draft in SW.
     Solution:

We know that:

Displacement =   (U/w volume)x  (density)
18450  =    ( 150 x 20x D) x (1.025)
D       =    ( 18450 ) / ( 150 x 20 x 1.025)
Draft  = 6m

  1. A box-shaped vessel 120m long and 15m wide is floating in DW of RD 1.005 at a draft of 5m. if her maximum permissible draft in SW in 6m , find how much cargo she can now load .
Solution:

Area of box shape vessel = (Lx b)
= 120 x 15m,
RD = 1.005, Draft =5m
Mass     = (u/w volume ) x (density)
= (120 x 15 x 5) x (1.005)
= 9045t .

Maximum cargo she can load = (120 x 15 x 6) x (1.025)= 11070 t

Cargo can be load = (11070 – 9045)
= 2025t.

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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