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