Stability – I : Chapter 8

5. A homogeneous log of wood 3m x 0.75m x 0.75m floats in SW with one face horizontal .If the RD of the log is 0.8m, calculate the vertical distance between its COG and its COB .

Solution:

Volume of log of wood = (L x B x H)
= (3m x 0.75m x 0.75m)

RD of SW = 1.025
RD of the log = 0.8m

Mass of the log = (volume x density)
= (3 x 0.75 x 0.75 ) x (0.8)
= 1.35 t

Since,  log floats freely so
Mass = Displacement
1.35 = (u/w volume) x(density)
1.35 = (L x B x draft) x  (1.025)
1.35 = ( 3x 0.75 x draft) x (1.025)

So, draft = 1.35/(3 x 0.75 x 1.025)
Draft  = 0.585 m

Again, we know that:
COB = (1/2 x draft)
= (0.585/2)
= 0.2925m

COG = (1/2 x height)
= (0.75 / 2)
= 0.375m

Hence, Vertical distance between COB and COG
= (0.375 – 0.295)
= 0.083m