Stability I – Chapter 3 Flotation Solutions

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 .

2. 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

3. 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

4. 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.

5. 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 .

6. 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