Problem *3.93[Difficulty: 2]Given:Geometry of steel cylinderFind:Volume of water displaced; number of 1 kg wts to make it sinkSolution:The data isFor waterρ999kgm3⋅=For steel (Table A.1)SG7.83=For the cylinderD100 mm⋅=H1m⋅=δ1mm⋅=The volume of the cylinder isVsteelδπD2⋅4πD⋅H⋅+⎛⎜⎝⎞⎟⎠⋅=Vsteel3.22104−×m3⋅=The weight of the cylinder isWSGρ⋅g⋅Vsteel⋅=W7.83999×kgm3⋅9.81×ms2⋅3.22×104−×m3⋅Ns2⋅kg m⋅×=W24.7N=At equilibium, the weight of fluid displaced is equal to the weight of the cylinderWdisplacedρg⋅Vdisplaced⋅=W=VdisplacedWρg⋅=24.7 N⋅m3999 kg⋅×s29.81 m⋅×kg m⋅2⋅×=Vdisplaced2.52L=To determine how many 1 kg wts will make it sink, we first need to find the extra volume that will need to be dsiplaced
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.