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IF THE CENTER OF GRAVITY OF A SHIP IN UPRIGHT POSITION IS 10 m ABOVE THE CENTER OF
GRAVITY OF THE PORTION UNDER WATER (Bo), THE DISPLACEMENT BEING 1000 metric tons
(1 metric ton = 1000 kg) AND THE SHIP IS TIPPED 30 degrees CAUSING THE CENTER OF
BUOYANCY TO SHIFT SIDEWISE BY 8 m.
(1) FIND THE LOCATION OF METACENTER FROM Bo.
(2) FIND THE METACENTRIC HEIGHT
(3) WHAT IS THE VALUE OF THE RIGHTING OR OVERTURNING MOMENT IN kg-m.
SOLUTION :
z = MBosin
θ
8 m = MBosin(30°)
MB
0
= 16 m
M is located 16 m above Bo
MG = MB
0
− GB
0
MG = 16 m − 10 m
MG = 6. 000 m
RM = BFx̅
RM = ( 1000 metric ton) (
1000 kg
1 metric ton
) ( 6 sin(30°))
RM = 3 , 000 , 000 kg − m
A LOADED SCOW HAS DRAFT OF 1.8 m IN FRESH WATER, WHEN ERECT. THE SCOW IS 6 m WIDE,
12 m LONG AND 2.4 m HIGH. THE CENTER OF GRAVITY OF THE SCOW IS 1.8 m ABOVE THE
BOTTOM ALONG THE VERTICAL AXIS OF SYMMETRY
(1) DETERMINE THE INITIAL METACENTRIC HEIGHT
(2) WHAT IS THE MAX SINGLE WEIGHT THAT CAN BE MOVED TRANSVERSELY FROM THE
CENTER OF THE UNLOADED SCOW OVER THE SIDE WITHOUT SINKING THE SCOW.
SOLUTION :
MB
0
B
2
12D
tan
2
(θ)
MB
0
( 6 m)
2
12 ( 1. 8 m)
tan
2
MB
0
= 1. 667 m
MG = MB
0
− GB
0
MG = 1. 667 m − 0. 9 m
MG = 0. 767 m
tan(θ) =
- 6 m
3 m
θ = 11 .310°
MB
0
( 6 m)
2
12 ( 1. 8 m)
tan
2
MB
0
= 1. 7 m
∑ M
cg
= −BF
x̅
+ P
y
BF = ( 1. 8 m)( 6 m)( 12 m)( 9. 81 kN/m
3
BF = 1271. 376 kN
- 376 ( 0. 8 sin
) = P (
cos( 11. 310 °)
P = 65. 199 kN
