Q & A – Fluid Mechanics – Buoyancy – Q.8

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Q & A – Fluid Mechanics – Buoyancy – Q.8

Q & A – Fluid Mechanics – Buoyancy – Q.8

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 :

  • D=1.35  m
  • H=1.8  m
  • m=770  kg
  • S.Gseawater=1.025

Solution :

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

W=FB

770  kg=γseawaterAh

770=1.0251000π41.352h

h=0.525   m

Second , check the stability of the floating body :

MG=IminVimmersedGB

Vimmersed=Ah

Vimmersed=π41.3520.525=0.751  m3

Since the body is uniform → GB=H2h2

GB=1.820.5252=0.6375  m

Imin=π64D4

Imin=π641.354=0.163  m4

MG=0.1630.7510.6375=0.42  m

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=1.0251000π41.352h'

770+T=1466.43h' 1

MG'=IminVimmersed'GB'=0 IminVimmersed'=GB'

Vimmersed'=Ah'

Vimmersed'=π41.352h'=1.43h'

Imin=π64D4 Imin=π641.354=0.163  m4

WH2+T0=(W+T)x

7701.82=(770+T)x

x=693770+T

GB'=xh'2

GB'=693770+Th'2

IminVimmersed'=GB'

0.1631.43h'=693770+Th'2

From 1 :

0.1631.43h'=6931466.43h'h'2 h'

0.114=0.4730.5h'2 h'2=0.718

h'=0.847  m

From 1 :

770+T=1466.430.847 T=472 kg

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