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Archimedes’ Principle
• An object immersed in a liquid has an upward buoyant force equal to the weight of the liquid displaced by the object.
• An object will float if the upward buoyant force is greater than the object’s weight.Archimedes
287 – 211 BC
(courtesy F. Remer)
Archimedes’ Principle
• ‘Square’ bubble of gas in a tank of water
(courtesy F. Remer)
Archimedes’ Principle
• Water pressure in tank increases with depth
• p = pressure
• h = depth
ph
(courtesy F. Remer)
Archimedes’ Principle
• Horizontal Pressure Differences Balance
(courtesy F. Remer)
Archimedes’ Principle
• Force on bottom of ‘bubble’
Fbottom
pbottom FbottomA
ApF bottombottom
Archimedes’ Principle
• Force on top of ‘bubble’
Fbottom
Ftop
A
Fp top
top
ApF toptop
(courtesy F. Remer)
Archimedes’ Principle
• Buoyancy Force
Fbottom
Ftop
FB (pbottom ptop ) A
FB
FB Fbottom Ftop
FB pA
(courtesy F. Remer)
FB Apbottom Aptop
Archimedes’ Principle
• Pressure Difference Between Top & Bottom
• Negative by conventionpbottom
ptop
h
ph
g
p gh
(courtesy F. Remer)
Archimedes’ Principle• Combine Equations
pbottom
ptop
h
p gh
B
B pA
FB ghA
FB gV
V hA
(courtesy F. Remer)
Note: Density (ρ) = density of liquid Volume = displaced volume of liquid = volume of object
Archimedes Principle• The net force on the object is thus the
difference between the object’s weight and the buoyancy force of the displaced fluid.
• If the buoyancy of an object exceeds its weight, it tends to rise. An object whose weight exceeds its buoyancy tends to sink.
Fnet mg VgWeight of object
Buoyant Force of displacedfluid
Archimedes’ Principle
Cartoon here??Archimedes
287 – 211 BC
At the moment of Archimedes’ famous discovery.
Example
• What is the buoyancy in seawater of a piece of wood that weighs 10,000 N & measures 3m x 1m x 2m?
• The weight of wood = 10,000 N
• The volume of wood = 3m x 2m x 1m= 6 m3
• The corresponding weight of an equal volume of seawater: 6 m3 x 10300N/m3 = 61,800N
weight of water displaced = upward force = 61,800N
weight of wood = downward force = 10,000N
Net Force = 51,800N up
Example 2• A fully suited diver weighs 200 pounds. This diver
displaces a volume of 3.0 cubic feet of seawater. Will the diver float or sink?
• weight of equal volume of seawater:
3.0 ft3 x 64 lb/ft3. = 192 lb.
Weight of diver = down force = 200 lbs
Displaced weight of sea water = up force = 192 lb net force = 8 lbs down
• The diver will sink. This diver weighs 8 pounds in
the water and is over-weighted. Removal of eight pounds will allow the diver to “hover”.
Buoyancy
• Similar to parcel of air in atmosphere
• At Equilibrium– Density of Parcel Same
as Density of Environment
pe
pe
(courtesy F. Remer)
Archimedes Principle• Translation:
– objects more dense than water will sink - negatively buoyant ;
– objects less dense than water will float - positively buoyant;
– objects of the same density will remain at the same level and neither sink nor float - neutrally buoyant.
Buoyancy
• Density Difference Results in Net Buoyancy Force
pe
pe
B
Buoyancy
• Density Difference Results in Net Buoyancy Force
pe
pe
B
Buoyancy
• Net Buoyancy Force
pe
BgV)(B pe
Archimedes’ Principle
• Water is in hydrostatic equilibrium
• ρ = density• g = acceleration of
gravity
B
ph
g
ph
g