Lecture 8 Students1
28
-
Upload
ijordanscribd -
Category
Documents
-
view
219 -
download
0
Transcript of Lecture 8 Students1
Recap……
We have looked last lecture at
• Hydrostatic force on a submerged curved surface
Hydrostatic force on a submerged curved surface• Horizontal component, FH = FX
• Vertical component, FV = FY + W (Vector addition) where W = gV: is the weight of the enclosed liquid block of volume
V and acts downward through the centroid of this volume.
The magnitude of the resultant hydrostatic force acting onthe curved surface, FR = Tan θ = FV / FH
The location of the line of action of the resultant force can be determined by taking moments about an appropriate point.
2
V
2
H FF
The altitude of a hot air balloon is controlled by the temperature difference between the air inside and outside the balloon, since warm air is less dense than cold air.
When the balloon is neither rising nor falling, the upward buoyant force exactly balances the downward weight.
Buoyancy Forces
Buoyancy forcesBuoyancy
– As the pressure in a fluid increases with depth, the fluid exerts a resultant upward force on any body immersed in it.
Consider a cylinder submerged in water:Buoyancy = total vertical force exerted on the body due to pressure
• Pressure force on the top surface
• Pressure force on the bottom surface
A
L
FBottom
FTOP h
Free surface
ghAAPF TOPTOP
A=areaALhgAPF BOTTOMBOTTOM )(
Buoyancy forces• Total pressure force or net upward force:
A
L
FBottom
FTOP h
Free surface
A = Area
gVρgLAρ
ghAρL)Ag(hρFF ForceBuoyant
f
f
ff
TOPBOTTOM
That is:
Or in wordsBuoyancy force = (density of fluid) x g x (Volume of subject body)
bodyfb gVρF
Buoyancy forcesBuoyancy
The formula derived for a cylinder is applicable for any shape:
C
E
D
FFCFD
FCEDbodyfb gVF
The buoyant force acting on a body is equal to the weight of the liquid displaced by the body.
The buoyant force is independent of:a) the distance of the body from the free surface;b) the density of the solid body.
“A body experiences an up thrust (force) equal to the weight of fluid displaced”
“The up thrust acts through the centroid of the displaced volume”
Archimedes Principle
Buoyancy forces
Maximum Buoyancy Force
Weight
Drum Floats:
Archimedes PrincipleFloats/Sinks?
drumfluid gV
gVgM drumdrumdrum
BouyancyWeight
fluiddrum
A Hot-Air Balloon has a volume of 1000m3, If the surrounding air has a density of 1.21 kg/m3, what is the buoyancy force?
1. I need more info!2. 11.87 kN3. 11.87 x 103 kN4. None of the above
Example
A Hot-Air Balloon has a volume of 1000m3, If the surrounding air has a density of 1.21 kg/m3, what is the buoyancy force?
Buoyancy forces (cont.)• What is the buoyancy force on a body only partly immersed in the
liquid?
gVFb
where V is the volume of the part ofthe body immersed in the liquid.
The weight of the entire body must be equal to the buoyant force. i.e. FB = W
Buoyancy forces (cont.)FB = W → ρf g Vsub = ρbody g Vtotal
The submerged volume fraction of afloating body is equal to the ratio ofthe average density of the body tothe density of the fluid.
→
ρρ
vv
f
body
total
sub
How deep in the water does the drum float?
Alternatively what is h?
Archimedes PrincipleExample Question
kgM drum 250
mH 1h
mD 8.0
31020 mkgsw
How deep in the sea water does the drum float?
1. 0.488 m2. 1 m3. 0.122 m4. 0.497 m5. It sinks!6. None of the above
Archimedes PrincipleExample Question
kgM drum 250
mH 1h
mD 8.0
Check your understanding• Consider two 5-cm-diameter sphere balls – one made of steel, the other
wood – submerged in water. Will the buoyancy force acting on these two balls be the same or different?
Why?• Will the buoyancy force change if the balls are submerged in oil?
SteelWood
Fb
W
Fb
W
Worked example• A steel ball of 10cm-in-diameter submerged in water is hanging with a string
from top. Calculate the tensile force in the string. (The density of steel is 7800kg/m3.)
water
FS
W Fb
343
3body m105.23
20.13.14
34πR
34V
Is it Stable?
Terminology:•Centre of Buoyancy•Centre of Gravity•Stable Equilibrium•Neutral Equilibrium•Unstable Equilibrium•Metacentre•Metacentric Height
c.g.
c.b.
Stability
Stable?
Restoring Moment = WxG below B
B
W
G G
B
W
x
Rotation
Stability of Submerged ObjectsStable Equilibrium
B @ G
G
B
W
G
B
W
Neutral Equilibrium
Rotation
Stability of Submerged Objects
Overturning Moment = WxG above B
G
B
W
G
B
W
x
Unstable Equilibrium
Rotation
Stability of Submerged Objects
“The Metacentre is the point where the line of action of the up-thrust intersects the original vertical line through the centre of buoyancy and the centre of gravity.”
The restoring or overturning moment is
GM = metacentric height
When θ is small: ..GMW
sin..GMW
Stability of Floating BodiesMetacentric Height
GGB
B
W
B’
B
W
M Above G therefore “Stable”
Metacentric Height M
GM
Stability of Floating Bodies
G
B
B
W
G
B
B
W
M Below G therefore
“Unstable”
Metacentric Height
Stability of Floating Bodies
M
c.g.
c.b.
Metacentre
Stability of Floating BodiesStable?
We have looked today at:
• Buoyancy force• Stability of submerged and floating bodies
Next lecture• Rigid body motion
Summary
