# Stability – I : Chapter 3

###### 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 #### 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.

• sailoramit.2085 says:

PROBLEM NUMBER 4 AND 5 ARE COMPLETELY WRONG, HOW CAN ONE USE FORMULA FOR A VOLUME OF A RECTANGLE TO DERIVE VOLUME OF A TRIANGLE

• Pulkit malik says:

Question 4 and 5 are wrong.

• Pulkit malik says:

Questions 4 and 5 are wrong.

• Hardik says:

Question 4&5 are wrong .

• Shivam kumar says:

• Asir Muhammad says: