Stability – I : Chapter 3

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