# Question No.8

A cylindrical buoy 1.35 m in diameter and 1.8 m high has a mass of 770 kg. Show that it will not float with its axis vertical in sea water  ( S.G = 1.025 ). What minimum pull should be applied to a chain attached to the center of the base to keep it vertical ?

### Given :

• $S.Gseawater=1.025$

### Solution :

$First, we should calculate the height of the immersed part of the buoy h :$

$W=FB$

$770=1.025∗1000∗π4∗1.352∗h$

Second , check the stability of the floating body :

$MG=IminVimmersed−GB$

$Vimmersed=Ah$

$Since the body is uniform →$ $GB=H2−h2$

$Imin=π64D4$

$Since MG is negative, therefore the buoy is unstable !$

The buoy will not float with its axis vertical in sea water !

To find the minimum pull that should be applied to a chain attached to the center of the base to keep it vertical :

$W+T=FB$

$770+T=γseawaterAh'$

$770+T=1.025∗1000∗π4∗1.352∗h'$

$770+T=1466.43h'$ $→$ $1$

$MG'=IminVimmersed'−GB'=0$ $→$ $IminVimmersed'=GB'$

$Vimmersed'=Ah'$

$Vimmersed'=π4∗1.352∗h'=1.43h'$

$Imin=π64D4$ $→$

$W∗H2+T∗0=(W+T)∗x$

$770∗1.82=(770+T)∗x$

$x=693770+T$

$GB'=x−h'2$

$GB'=693770+T−h'2$

$IminVimmersed'=GB'$

$0.1631.43h'=693770+T−h'2$

From 1 :

$0.1631.43h'=6931466.43h'−h'2$ $∗h'$

$0.114=0.473−0.5h'2$ $→$ $h'2=0.718$

From 1 :

$770+T=1466.43∗0.847$ $→$ $T=472 kg$