Stability – I : Chapter 3 |

# 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

###### 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.

###### 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 .

###### Solution:

=(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

###### 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

###### 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.

###### 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

###### 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. • Asir Muhammad says: