The immersed cross-sectional area of a ship 120m long, commencing from aft are 2,40,79,100,103,104,104,103,97,58 and 0 m2 calculate - (i) Displacement; (ii) Longitudinal position of the centre of buoyancy
If you are noticing some error in problems kindly comment below.Thanks
Given
L = 120 m
To find
i) Displacement
ii) Longitudinal position of the centre of buoyancy
Solution
If you are noticing some error in problems kindly comment below.Thanks
Given
L = 120 m
To find
i) Displacement
ii) Longitudinal position of the centre of buoyancy
Solution
Area
|
SM
|
Product for volume
|
Lever
|
Product for 1stmoment
|
2
|
1
|
2
|
+5
|
+10
|
40
|
4
|
160
|
+4
|
+640
|
79
|
2
|
158
|
+3
|
+474
|
100
|
4
|
400
|
+2
|
+800
|
103
|
2
|
206
|
+1
|
+206
|
104
|
4
|
416
|
0
|
Σma = +2130
|
104
|
2
|
208
|
-1
|
-208
|
103
|
4
|
412
|
-2
|
-824
|
97
|
2
|
194
|
-3
|
-582
|
58
|
4
|
232
|
-4
|
-928
|
0
|
1
|
0
|
-5
|
0
|
Σv = 2388
|
Σmf = -2542
|
It is given that L = 120 m and there are 10 equidistant spaces. So common interval
h = 120 / 10
= 12 m
i) We know that displacement = Density x Under water volume ----------------(1)
But Under water volume = (h/3) x Σv
= (12/3) x 2388
= 9552 m3
(1) -------------------------->
Displacement = 1.025 x 9552
= 9790.8 tonne
ii) Centre of buoyancy from midship = h x (Σma + Σmf) / Σv
= 12 x (2130 - 2542 ) / 2388
= 2.070 m forward
NOTE : If Σma is greater than Σmf , the buoyancy will be forward of midship.
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