Archimedes Principle

Learning Objectives

• Describe Archimedes’ Principle.

• Define density, buoyancy, and specific gravity.

• Correctly calculate the buoyancy of an object in either fresh or salt water.

• Correctly solve lifting problems.

• Correctly calculate surface air volume equivalents.

Main Points

• Density

• Buoyancy

• Specific gravity

• Archimedes’ Principle

• Surface Equivalent air volume

• Lifting problems

Density

• Definition– Mass per Unit Volume

• Density of air at sea level– .08 lbs. per cu. ft.

• Hydrostatic Density– Salt Water

• 64 lbs. per cu. ft.

– Fresh Water• 62.4 lbs. per cu. ft.

Buoyancy

• Force that allows an object to float.

Specific Gravity

Density of a substance vs. density of pure

water.

Archimedes’ Principle

• An object partially or wholly immersed in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the object.

• Buoyancy of an object = – Weight of the water displaced by the object - Weight of the object

When placed in seawater, what is the state of buoyancy for each of these objects?Where will they end up?

Positive_______________________________________________

Neutral

________________________________________________Negative_

32 lbs1 cu ft

64 lbs1 cu. Ft.

96 lbs1 cu. ft

States of Buoyancy

• Positive buoyancy– Specific Gravity of the object is less than that

of the fluid

• Neutral– Specific gravity of the object is equal to the

specific gravity of the fluid

• Negative– Specific gravity of the object is greater than

that of the fluid

Example 1

• What is the buoyancy of an anchor with a dry weight of 100 lbs., and a volume of .22 cu. ft., when it is dropped in the ocean?

Answer to Example 1

Displaced wt.=

.22 cu. ft. x 64 lbs. per cu. ft. 14.08 lbs.

-Dry wt.

100 lbs.

Buoyancy

- 86 lbs

Example 2

How many 50 lb. lift bags will it take to lift

an object with a volume of 3.1 cu. ft. and a

dry weight of 289 lbs.?

Each lift bag weighs 2 lbs. and the object is

in fresh water.

Answer to Example 2

Displaced weight =

3.1 cu. ft. x 62.4 lbs./ cu. ft. 193.4 lbs.

-Dry weight

289 lbs.

Buoyancy

- 95.6 lbs.

Lift capacity = 50 lbs - 2 lbs = 48 lbs of lift / bag.

Use how many bags?

2 bags.

Surface Equivalent Air Volume

• How much air must you bring down from the surface if the object in example 2 is located at a depth of 120 ffw?

Surface Equivalent Air Volume cont.

• Buoyancy of the object -95.6 lbs

• How much lifting force must be generated to lift the object to the surface?– 95.6 lbs

Surface Equivalent Air Volume cont.

• How much freshwater must be displaced to generate the required lifting force?

• How is this calculated?– Force required/density of fresh water

• Density of fresh water – 62.4 lbs. per cu. ft.

• 95.6 lbs/62.4 lbs. per cu. ft. =– 1.53 cu. Ft. of water must be displaced

Surface Equivalent Air Volume cont.

How much air must we bring down from the

surface to displace 1.53 ft3 of fresh water at

a depth of 120 ffw.?

• Calculate Pata at a depth of 120 ffw.?– {Depth + 34}/34 = atm– {120+34}/ 34 = 4.5 atm

• Multiply Pata x Vol h20 to be displaced

– 1.53 x 4.5 = 6.93 cu. ft. at the surface

Lifting problem

• You have been enlisted to salvage an outboard motor lost at sea. You locate the outboard, which displaces 2 ft3 of water and weighs 900 lbs in air, at a depth of 66 ft. How much air will you need to add to a lift bag to bring the outboard to the surface? How much air will be in the lift bag once at the surface?

Calculate the Buoyancy of the Object

Volume = 2 ft3

Weight of the water displaced = 2 ft3 x 64 lbs/ft3 = 128 lbs

Dry weight = 900 lbs

Buoyancy of the Object128 lbs – 900 lbs = -772 lbs

Calculate the Volume of Water to be Displaced

How much lifting force is necessary?

772 lbs

How much water must be displaced

772 lbs / 64 lbs/ft3 = 12.06 ft3

Calculate How Much Air You Need to Bring Down from the surface

Calculate Pata

(66 / 33) + 1 = 4 ata

Multiply P ata x volume H20 to be displaced

4 ata x 12.06 ft3 = 48.24 ft3

How much air will be in the bag at the

surface?

Example 3

When properly weighted for diving in the

ocean, a diver and his gear weigh 224 lbs.

How must the diver adjust the amount of

weight in his weight system to be properly

weighted in fresh water?

Answer to Example3The volume of the diver and his equipment will not change

SW displacement = 224 lbs./64 lbs. per cu. ft. = 3.5 cu.ft.

FW displacement = 3.5 cu. ft. x 62.4 lbs./cu. ft. = 218.4 lbs.

Wt. system Adjustment = 224 lbs.- 218.4 lbs.

Answer:Remove 5.6 lbs

Shortcut Adjust up or down by 2.5% of total diver weight.This is the difference in density between ocean water andfresh water

Have we covered:

• Density

• Buoyancy

• Specific gravity

• Archimedes’ Principle

• Surface Equivalent air volume

• Lifting problems

Can You

• Describe Archimedes’ Principle?

• Define density, buoyancy, and specific gravity?

• Correctly calculate the buoyancy of an object in either fresh or salt water?

• Correctly solve a lifting problem?

• Correctly calculate Surface Air Volume Equivalents?

Last Thoughts

• Understanding and applying Archimedes’ Principle enables you to weight yourself properly and to achieve and maintain the appropriate state of buoyancy.

• Combining Archimedes’ Principal with Boyle’s Law enables you to correctly calculate the volume of gas and number of lift bags you will need to bring from the surface to ensure you can lift and object off the bottom.