Show that a cylindrical buoy of 1m diameter and 2m height weighing 7.848 KN will not float vertically in sea water of density 1030 kg/`m^3`. Find force necessary in a vertical chain attached at centre of base of buoy that will keep it vertical.

SOLUTION:

Let,

h be the depth of immersion 

Now, from law of Floatation,

Weight of cylinder = Buoyant force 

`7.848=(pi**1^2)/4**h**1030**9.81**1/1000`

`h=0.9899 m`

Now, the distance of c.g. from base is,`=2/2=1m`

Similarly, the distance of C.B. from base is,`=h/2=0.49479`

Now,

`BG=OG-OB`

`=0.50503`

Now the metacentric height is,

`GM=I _(yy)/V-BG`

Where, `I_(yy)`=M.I of the plan about y axis.

`=(pid^4)/64`

And V=volume of ....Show More

SOLUTION:

Let,

h be the depth of immersion 

Now, from law of Floatation,

Weight of cylinder = Buoyant force 

`7.848=(pi**1^2)/4**h**1030**9.81**1/1000`

`h=0.9899 m`

Now, the distance of c.g. from base is,`=2/2=1m`

Similarly, the distance of C.B. from base is,`=h/2=0.49479`

Now,

`BG=OG-OB`

`=0.50503`

Now the metacentric height is,

`GM=I _(yy)/V-BG`

Where, `I_(yy)`=M.I of the plan about y axis.

`=(pid^4)/64`

And V=volume of cylinder immersed

`=(pid^2)/4*h`

Thus,

`GM=d^2/(16h)-0.50503`

`=1^2/(16**0.9899)-0.50503`

`=-0.442`

Here the negative sign shows that the metacentre lies below the centre of gravity and hence the cylinder will be in unstable equilibrium and hence the cylinder will not float vertically in water.

Let T force is applied in a vertical chain attached at the centre of base of buoy to keep the buoy vertical.

Now, we have to find the combined position of c.g. and C.B.  For combined centre of buoyancy, let h' be the depth of immersion when force T is applied.

From the law of floatation,

Total downward force = weight of water displaced 

`7848+T=(pid^2)/4*rho_w*h'*g`

`7848+T=pi/4**1^2**1030**h'**9.81`

`h'=(7848+T)/7935.89`

Now the distance of new C.B is,

`C.B=AB=h'/2=(7848+T)/15871.8`

Now

Let,

`W_c`be the weight of cylinder

`x_1` be the distance of c.g (weight acting point) from  A

`x_2` be the distance of application of T force  from A.

Now, the combined c.g. due to weight of cylinder and due to tension T in the chain from A is,

`AG'=(W_c*x_1+T*x_2)/(W_c+T)`

`=((7848**2/2)+(T**0))/(7848+T)`

`=7848/(7848+T)`

Also,

`B'G'=AG' - AB'`

`=7848/(7848+T) - (7848+T)/15871.8`

Now the metacentric height is,

`GM=I/V-B'G'`

`=(pid^4)/(64**(pid^2)/4*h') -B'G'`

`=1/(16h')-B'G'`

`=1/(16 **(7848+T)/7935.89)-7848/(7848+T) +(7848+T)/15871.8` 

Now for stable equilibrium,

`GM >=0`

`=1/(16 **(7848+T)/7935.89)-7848/(7848+T) +(7848+T)/15871.8 >=0` 

`-117632/(16 (7848+T))+(7848+T)/(15871.8)>=0`

`(7848+T)/(15871.8)>= 117632/(16 (7848+T))`

`(7848+T)^2 >=(1080.23^2)`

`7848+T>=1080.23`

`T>=2954.3 N`

Thus the force in the chain must be at least equal to 2954.3 N so that the cylindrical buoy can be kept in vertical position.