Stability – I : Chapter 10

Solution:

Given:
Total FSM = 2087.4 tm
FSM deducted from 2P = 684.0 tm
New FSM = 1403.4

FSC =(FSM/W)
= (1403.4 /1000)
= 0.14

GM solid = (KM – KG)
= (9.9 – 8.950)
= 0.95m

We know that:

GM fluid = (GM solid – FSC)
= (0.95 – 0.14)
= 0.81m.

12. Vessel in same condition as in question 10, transfers SW ballast from FP IN TO NO 3s and in to AP, such that NO3s becomes full while FP and AP remain partly full . Find the GM fluid if final KG is 8.880m.

Solution:

According to question,
Now, No 35 becomes full, so FSM deducted

Again, New FSM created by FP and AP
We know that:
FP = (10 x 1.025)
= 10.25 tm.

AP = (20 x 1.025)
= 20.5 tm.

So, FSM = (10.25 + 20.50) tm
= 30.75 tm

New, final FSM = (2087.4 – 246.0 + 30.75)
= 1872.15 tm

FSC = (1872.15 /10000)
= 0.187

New KG = 8.880m (given)

GM (Solid)   = (KM – KG)
= (9.9 – 8.880)
= 1.02m

GM (Fluid) = GM (solid) – FSC
= (1.02 – 0.187)
= 0.833m.