# Stability – I : Chapter 9

###### Solution:

GZ =Righting lever, GM = Metacentric height

We know that Righting moment = (W x GZ )
Where, W = displacement for all of keel, M = Transverse meta centre.
Displacement (W) = 10000t,
GM = 0.4m, heel = 5degree

We know that
RM ( righting moment or moment of statical stability)
= (W x GM x sinθ)
= (10000 x 0.4 x sin 50 )
= 348.62tm

###### Solution :

Displacement (W) = 12000t,
Heel  = 6degree
GZ = 0.1m  &  KM = 8.2m

We know that :
GZ = GM sinθ,
0.1 = GM sin 60.
GM = (0.1 / sin 60)
= 0.956m
KG = (KM – GM)
= (8.2 – 0.956)
= 7.44m

RM (righting moment or moment of statical stability)
= (W GM sinθ)
= (12000 x 0.956 x sin60 )
= 1199.15 tm.

###### Solution :

Displacement (W)= 14000t,
Heel = 8degree ,
RM = 400tm &  KG 7.3m

We know that :
RM (Righting moment or moment of statical stability)
= (W x GM x sinθ)
400 = (14000  x GM x sinθ)
GM = 400/ (1400 x sin80  )
GM = 2.053m

Now, KM = (KG + GM )
= (7.3 + 2.053 )
= 9.353m.

###### Solution :

Displacement (W) = 8000t,
KB = 3.5m &  KM = 6.5m ,
KG = 6m, heel= 20degree.

We can calculate:
GM = (KM – KG)
= (6.5 – 6)
= 0.5m

Since, GZ = Sinθ(GM + 1/2 BM tan2θ)
GZ = sin 200 (0.5 + 1/2 x 3 x tan2200 )
Again , GZ = 0.239 ,

RM (Righting moment or moment of statical stability)
= (W.GZ)
= 8000 x 0.239 = 1911.8 tm

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