Lecture-4-Fluid Statics .ppt

mechanics

1. Lecture -4
Fluid Statics

2. Buoyancy & Stability
Examples of types of Buoyancy Problems:

3. Introduction:
Wheneverabodyisplacedoveraliquid,eitheritsinksdown
orfloatsontheliquid.
Twoforcesinvolveare:
1.GravitationalForce
2.Up-thrustoftheliquid
IfGravitationforceismorethanUpthrust,bodywillsink.
IfUpthrustismorethanGravitationforce,bodywillfloat.

4. Archimedes Principle:
“Wheneverabodyisimmersedwhollyorpartially
inafluid,itisbuoyedup(i.eliftedup)byaforce
equaltotheweightofthefluiddisplacedbythe
body.”

5. Buoyancy:
Abodyinafluid,whetherfloatingorsubmerged,isbuoyed
upbyaforceequaltotheweightofthefluiddisplaced.
“Thetendencyofafluidtoupliftasubmergedbody,because
oftheup-thrustofthefluid,isknownasforceofbuoyancyor
simplybuoyancy.”
Thebuoyantforceactsverticallyupwardthroughthecentroid
ofthedisplacedvolumeandcanbedefinedmathematicallyby
Archimedes’principleasfollows:fluid of volumeDisplaced
fluid of weight Specific
forceBuoyant



d
f
d
V
F
 dfd
VF

6. Center of Buoyancy:
It is defined as:
“The point, through which the force of buoyancy is supposed
to act.”
As the force of buoyancy is a vertical force and is equal to the
weight of the fluid displaced by the body, the centre of
buoyancy will be the centre of gravity of the fluid displaced.
In other words, the centre of buoyancy is the centre of area of
the immersed section.

7. 7
Problem-1
Find the volume of the water displaced and position of
centre of buoyancy for a wooden block of width 2.5 m and
of depth 1.5 m, when it floats horizontally in water. The
density of wooden block is 650 kg/m3 and its length 6.0 m.

8. Solution:

9. kN 4.9 W
19.62W14.72
thrustupward with concrete andblock of weight total theequating Now
kN 19.62 2 x 9.81
in water immersed completely isblock en the thrust wh Upward
2m displaced water of volume
in water, immersed completely isblock the that whenknow We
kN W 14.72 concrete andblock theof weight Total
kN 14.72 2 x 0.75 x 9.81 Weight its and
2m 0.5 x 1 x 4 block wooden of Volume
block.on wooden placed be torequired Concrete of weight thebeLet W
3
3







 Problem-2
Awoodenblockof4mx1mx0.5minsizeandofspecificgravity0.75
isfloatinginwater.Findtheweightofconcreteofspecificweight24k
kN/m3thatmaybeplacedontheblock,whichwillimmersethe
woodenblockcompletely.
Solution:

10. “Wheneverabody,floatinginaliquid,isgivenasmall
angulardisplacement,itstartsoscillatingaboutsomepoint.
Thispoint,aboutwhichthebodystartsoscillating,iscalled
metacentre.”
Metacentre:

11. Metacentric Height:
“Thedistancebetweencentreofgravityofafloatingbodyand
themetacentre(i.edistancebetweencgandminFig.)is
calledmetacentricheight.”
Metacentricheightofafloatingbodyisadirectmeasureofits
stability.
Morethemetacentricheightofafloatingbody,moreitwill
stableandviceversa.
Somevaluesofmetacentricheight:
MerchantShips=upto1.0m
SailingShips=upto1.5m
BattleShips=upto2.0m
RiverCraft=upto3.5m

12. Analytical Method for Metacentric Height:
Consideringashipfloatingfreelyinwater.Lettheshipbe
givenaclockwiserotationthroughasmallangleq(in
radians)asshowninFig.Theimmersedsectionhasnow
changedfromacdetoacd
1e
1.

13. Analytical Method for Metacentric Height:
TheoriginalcentreofbuoyancyBhasnowchangedtoanew
positionB
1.Itmaybenotedthatthetriangularwedgeocnhas
goneunderwater.Sincethevolumeofwaterdisplaced
remainsthesame,thereforethetwotriangularwedgesmust
haveequalareas.
Alittleconsiderationwillshow,thatasthetriangularwedge
oamhascomeoutofwater,thusdecreasingtheforceof
buoyancyontheleft,thereforeittendstorotatethevesselin
ananti-clockwisedirection.
Similarly,asthetriangularwedgeocnhasgoneunderwater,
thusincreasingtheforceofbuoyancyontheright,thereforeit
againtendstorotatethevesselinananticlockwisedirection.

14. Analytical Method for Metacentric Height:
Itisthusobvious,thattheseforcesofbuoyancywillforma
couple,whichwilltendtorotatevesselinanticlockwise
directionaboutO.Iftheangle(q),throughwhichthebodyis
givenrotation,isextremelysmall,thentheshipmaybe
assumedtorotateaboutM(i.e.,metacentre).
Letl=lengthofship
b=breadthofship
q=Verysmallangle(inradian)throughwhichtheshipis
rotated
V=Volumeofwaterdisplacedbytheship

15. Analytical Method for Metacentric Height:
From the geometry of the figure, we find that
am=cn=bq/2
--Volume of wedge of water aom
= ½ (b/2 x am)xl
= ½ (b/2 x bq/2)l (am = bq/2)
= b
2
ql/8
--Weight of this wedge of water
= b
2
ql/8 (=sp. Wt. of water)
--And arm L.R. of the couple = 2/3 b
--Moment of the restoring couple
= (b
2
ql/8) x (2/3 b) = b
3
ql/12 …(i)

16. Analytical Method for Metacentric Height:
--Andmomentofthedisturbingforce
=.VxBB
1 …(ii)
--Equatingthesetwomoments(i&ii),
b
3
ql/12=xVxBB
1
--Substitutingvaluesof:
lb
3
/12=I(i.e.momentofinertiaoftheplanoftheship)and
BB
1=BMxqintheaboveequation,
.I.qxV(BMxq)
BM=I/V
BM=Momentofinertiaoftheplan/Volumeofwaterdisplaced

17. Analytical Method for Metacentric Height:
--Now metacentric height,
Note: +ve sign is to be used if G is lower than B and,
–ve sign is to be used if G is higher than B.BGBMGM 

18. Problems:
1.Arectangularblockof5mlong,3mwideand
1.2mdeepisimmersed0.8mintheseawater.If
thedensityofseawateris10kN/m2,findthe
metacentricheightofblock.
2.Asolidcylinderof2mdiameterand1mheightis
madeupofamaterialofSp.Gravity0.7andfloats
inwater.Finditsmetacentricheight.

19. Conditions of Equilibrium of a Floating
Body:
Abodyissaidtobeinequilibrium,whenitremains
insteadystate,Whilefloatinginaliquidfollowing
arethethreeconditionsofequilibriumofafloating
body:
1.StableEquilibrium
2.UnstableEquilibrium
3.NeutralEquilibrium

20. 1. Stable Equilibrium:
Abodyissaidtobeinastableequilibrium,ifit
returnsbacktoitsoriginalposition,whengivena
smallangulardisplacement.
Thishappenswhenmetacentre(M)ishigherthan
centreofgravity(G)ofthefloatingbody.

21. 2. Unstable Equilibrium:
AbodyissaidtobeinaUnstableequilibrium,ifit
doesnotreturnbacktoitsoriginalposition,when
givenasmallangulardisplacement.
Thishappenswhenmetacentre(M)islowerthan
centreofgravity(G)ofthefloatingbody.

22. 3. Neutral Equilibrium:
Abodyissaidtobeinaneutralequilibrium,ifit
occupiesanewpositionandremainsatrestinthis
newposition,whengivenasmallangular
displacement.
Thishappenswhenmetacentre(M)concideswith
centreofgravity(G)ofthefloatingbody.

23. Problems:
3.Arectangulartimberblock2mlong,1.8mwideand1.2m
deepisimmersedinwater.Ifthespecificgravityofthe
timberis0.65,provethatitisinstableequilibrium.
4.Acylindricalbuoyof3mdiameterand4mlongisweighing
150N.Showthatitcannotfloatverticallyinwater.
5.Asolidcylinderof360mmlongand80mmdiameterhasits
base10mmthickofspecificgravity7.Theremainingpartof
cylinderisofspecificgravity0.5.Determine,ifthecylinder
canfloatverticallyinwater.