EXPERIMENT (2) BUOYANCY & FLOTATION …

1 EXPERIMENT (2) BUOYANCY & FLOTATION ( metacentric height )By:Eng. Motasem M. Fedaa M. Fayyad. 1 ARCHIMEDES PRINCIPLE Archimedes Principlestatesthatthebuoyantforcehasama gnitudeequaltotheweightofthefluiddisplac edbythebodyandisdirectedverticallyupward . Buoyantforceisaforcethatresultsfromafloa tingorsubmergedbodyinafluid. is the weight of the shaded areaF1and F2are the forces on the plane surfacesFBis the buoyant force the body exerts on the fluid2 ARCHIMEDES PRINCIPLE Theforceofthefluidonthebodyisopposite,or verticallyupwardandisknownastheBuoyantFo rce. Theforceisequaltotheweightofthefluiditdi splaces. Thebuoyantforcesactsthroughthecentroidof thedisplacedvolumeThe location is known as the center of : SUBMERGEDOBJECTS table Equilibrium: if when displaced returns to equilibrium Equilibrium: if when displaced it returns to a new equilibrium Equilibrium:Unstable Equilibrium:C > CG, Higher C < CG, Lower 4 STABILITY: SUBMERGEDOBJECT IftheCentreofGravityisbelowthecentreofbu oyancythiswillbearightingmomentandthebod ywilltendtoreturntoitsequilibriumpositio n(Stable).

2 IftheCentreofGravityisabovethecentreofbu oyancy,anoverturningmomentisproducedandt hebodyis(unstable). Notethat,Asthebodyistotallysubmerged, : FLOATINGOBJECTS lightly more complicated as the location of the center BUOYANCY can change:6 METACENTREANDMETACENTRICHEIGHT Metacentre point (M): This point, about which the body starts oscillating. metacentric height : Is the distance between the centre of gravity of floating body and the IfMliesaboveGarightingmomentisproduced,e quilibriumisstableandGMisregardedasposit ive. IfMliesbelowGanoverturningmomentisproduc ed,equilibriumisunstableandGMisregardeda snegative. IfMcoincideswithG, : 2-Theoretically: MG = BM + OB (2)In WaterOB= : To determine the metacentric height of a flat bottomed vessel in two parts:PART (1) : for unloaded and for loaded (2) : when changing the center of gravity of the : Thesetupconsistsofasmallwatertankhavingt ransparentsidewallsinwhichasmallshipmode lisfloated, , (1)Determination of floatation characteristic for unloaded and for loaded , Repeat step 4 for different weights; 100, 150, & 200 g, and take the corresponding angle of Repeat the above procedure with increasing the bottom loading by 2000 gm and 4000 Record the results in the Calculate GM practically where , W has three Draw a relationship between (x-axis) and GM (y-axis), then obtain GM when equals Calculate GM WeightOff balance GMGM at =0 BMOBTheo.

3 GMWb (gm)P (gm) (degree)(mm)from graph(mm)(mm)(mm) = = measurement:-Pontoon dimension : Depth (D) = 170 mmLength (L) = 380 mm, Width (W) = 250 height of the center of gravity of the pontoon is OGvm= 125 mm fromouter surface of vessel balance weight is placed at x = 123 mm from pontoon center weight of the pontoon and the mast Wvm= 3000 gmPURPOSE: To determine the metacentric height of a flat bottomed vessel in two parts:PART (1) : for unloaded and for loaded (2) : when changing the center of gravity of the :-Pontoon dimension : Depth (D) = 170 mmLength (L) = 380 mm, Width (W) = 250 height of the center of gravity of the pontoon is OGvm= 125 mm from outer surface of vessel balance weight is placed at x = 123 mm from pontoon center weight of the pontoon and the mast Wvm= 3000 gmPROCEDURE PART (2) : when changing the center of gravity of the Replace the bilge weights by 4x 50 gm Apply a weight of 300gm on a height of 190 mm from the pontoon Apply weights of 40, 80 &120 gmson the bridge piece loading pin, then record the corresponding tilting Calculate GM practically where 5.

4 Draw a relationship between in degrees (x-axis) and GM Practical (y-axis), then obtain GM when equals zero. 19 .3500)123(PGM=PROCEDURE6. Move 50 gm bilge weight to the mast ahead, then repeat steps 3,4& Repeat step 6 moving 100, 150 & 200 gm bilge weight to the Determine the height of the center of gravity for each loading condition according to equation20 WLWmWbWbWvmOG ++++=)2790()190(1)35()125(213500)2790()3 5()190(300)125(3000 LWmWbOG++++=228. Calculate GM theoretically according to equationGM (Th.) = BM + OB OGNotice:BM & OB are constants for all loading conditions, since the dimensions & the weight of pontoon do not balance GMBMOGTheo. GMP (gm) (degree)(mm)(mm)(mm)(mm)Mast Weight = Weight = weight = Weight = weight = (2) \Part (2)2425 QUESTIONS