Archimedes Principle

Learning ObjectivesDescribe 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 PointsDensityBuoyancySpecific gravityArchimedes PrincipleSurface Equivalent air volumeLifting problems

DensityDefinitionMass per Unit VolumeDensity of air at sea level.08 lbs. per cu. ft.Hydrostatic DensitySalt Water64 lbs. per cu. ft.Fresh Water62.4 lbs. per cu. ft.

Buoyancy

Force that allows an object to float.

Specific Gravity

Density of a substance vs. density of purewater.

Archimedes PrincipleAn 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 ft64 lbs1 cu. Ft.96 lbs1 cu. ft

States of BuoyancyPositive buoyancySpecific Gravity of the object is less than that of the fluidNeutralSpecific gravity of the object is equal to the specific gravity of the fluidNegativeSpecific gravity of the object is greater than that of the fluid

Example 1What 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 1Displaced wt.= .22 cu. ft. x 64 lbs. per cu. ft. 14.08 lbs. -Dry wt. 100 lbs. Buoyancy - 86 lbs

Example 2How many 50 lb. lift bags will it take to liftan object with a volume of 3.1 cu. ft. and adry weight of 289 lbs.?

Each lift bag weighs 2 lbs. and the object isin 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 VolumeHow 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 lbsHow 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 waterDensity 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 thesurface to displace 1.53 ft3 of fresh water ata depth of 120 ffw.?Calculate Pata at a depth of 120 ffw.?{Depth + 34}/34 = atm{120+34}/ 34 = 4.5 atmMultiply Pata x Vol h20 to be displaced1.53 x 4.5 = 6.93 cu. ft. at the surface

Lifting problemYou 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 ObjectVolume = 2 ft3Weight of the water displaced = 2 ft3 x 64 lbs/ft3 = 128 lbsDry weight = 900 lbsBuoyancy of the Object128 lbs 900 lbs =-772 lbs

Calculate the Volume of Water to be DisplacedHow much lifting force is necessary? 772 lbs

How much water must be displaced772 lbs / 64 lbs/ft3 = 12.06 ft3

Calculate How Much Air You Need to Bring Down from the surfaceCalculate Pata(66 / 33) + 1 = 4 ataMultiply P ata x volume H20 to be displaced4 ata x 12.06 ft3 = 48.24 ft3 How much air will be in the bag at thesurface?

Example 3When properly weighted for diving in theocean, a diver and his gear weigh 224 lbs.How must the diver adjust the amount ofweight in his weight system to be properlyweighted 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 lbsShortcut 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:DensityBuoyancySpecific gravityArchimedes PrincipleSurface Equivalent air volumeLifting problems

Can YouDescribe 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 ThoughtsUnderstanding and applying Archimedes Principle enables you to weight yourself properly and to achieve and maintain the appropriate state of buoyancy.Combining Archimedes Principal with Boyles 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.

Consider: The weight of the object. The volume of water displaced when the object is completely submerged. The density of seawater and the weight of the water displaced.