Stability – I : Chapter 3

  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

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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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