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25 April 2019

42.NA April 2019 Q.6(b)

April 25, 2019 Posted by AK 6 comments
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

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.

6 comments:

  1. I think whatis mentioned is wrong.. If aft moment is greater than forward.. C. O. B will be aft of midship.. Please clarify my doubt..

    ReplyDelete
  2. I mean in note.. What is mentioned

    ReplyDelete
  3. ii) Centre of buoyancy from midship = h x (Σma + Σmf) / Σv
    = 12 x (2130 - 2542 ) / 2388
    = -2.070 m forward

    The answer will be in Negative.. so y it is not considered..? I saw Reed's book for the same, it is considered as +ve. Any reason..?

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  4. This comment has been removed by the author.

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