A body has the cylindrical upper portion of 4m diameter and 2.4 m deep. The lower portion, which is curved; displaces a volume of 800 litres of water and its centre of buoyancy is situated 2.6m below the top of cylinder. The centre of gravity of the whole body is 1.6 m below the top of the cylinder and the total displacement of water is 52 KN. Find the metacentric height of the body.

SOLUTION:

Given,

Volume of water displaced by the lower portion =800 L

`=0.8 m^3`

Also the total displacement of the water =52 KN

Thus the total volume of water displaced is`=52/9.81=5.3 m^3`

Hence the volume of water displaced by the upper cylindrical portion is`=5.3-0.8=4.5 m^3`

Let h be the distance between free surface of liquid from top of the body.

Volume of water d.... Show More

SOLUTION:

Given,

Volume of water displaced by the lower portion =800 L

`=0.8 m^3`

Also the total displacement of the water =52 KN

Thus the total volume of water displaced is`=52/9.81=5.3 m^3`

Hence the volume of water displaced by the upper cylindrical portion is`=5.3-0.8=4.5 m^3`

Let h be the distance between free surface of liquid from top of the body.

Volume of water displaced by upper portion`=pi/4**4^2**(2.4-h)`

or,`4.5=pi**4**(2.4-h)`

or,`h=2.04 m`

Now the distance of centre of buoyancy of the cylindrical portion from the top is,

`OB_2=h+(2.4-h)/2=2.22 m`

Let B be the centre of buoyancy of the whole body,

`OB=((Volume\ of\ Curved\ po rtion**OB_1)+(Volume\ of\ Cyl i ndrical\ po rtion**OB_2))/(Total\ Volume\ of\ water\ displaced)`

`=(0.8**2.6+4.5**2.22)/(0.8+4.5)`

`=2.277 m`

Now,

`BG=OB-OG`

`=2.277-1.6`

`=0.677 m`

Also,

`BM=I_(yy)/V`

Where,

`I_(yy)`=M.I of the plane of the body at the water surface.

or,`I_(yy)`=M.I of the cylindrical portion (top portion) about its c.g.

`=pi/64**4^2`

`=12.566 m^4`

And `V`=total volume of liquid displaced

or,`V=5.3 m^3`

Thus,

`BM=12.566/5.3`

`=2.37 m`

Hence the metacentric height is,

`GM=BM-BG`

`=2.37-0.677`

`=1.693 m`