Stability – I : Chapter 8

6. A homogeneous log of wood of 0.5m square section floats in water of RD 1.005 at a draft 0.4m with one of its faces horizontal . Find the vertical distance between its COG and its COB in water of RD 1.020.

Solution:

RD = 1.005 & Draft  = 0.4m

We know that:
Displacement (W) = (volume ) x (density)
= (0.5 x 0.5 x 0.4) x(1.005)
= 0.1005t

Now ,calculating draft for the same displacement  at RD of 1.020.
W = ( Area x draft ) x( density)
0.1005 =(0.5 x 0.5 x draft) x( 1.020)

Draft = 0.1005/(0.5 x 0.5 x 1.020)
= 0.394m

Again , we know that :
COB = (draft /2)
= (0.395/2)
= 0.197m

COG = (Height/2)
= (0.5/2)
= 0.25m

Now, Vertical distance between COG and COB can be easily calculated as = (0.25 – 0.197)
= 0.053m