Problem 3.90 - Problem *3.90 [Difficulty: 3] Given: Data on...

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Problem *3.90 [Difficulty: 3] Given: Data on sphere and weight T F B W Find: SG of sphere; equilibrium position when freely floating Solution: Basic equation F B ρ g V = and Σ F z 0 = Σ F z 0 = TF B + W = where TM g = M1 0 k g = F B ρ g V 2 = WS G ρ g V = Hence Mg ρ g V 2 + SG ρ g V 0 = SG M ρ V 1 2 + = SG 10 kg m 3 1000 kg × 1 0.025 m 3 × 1 2 + = SG 0.9 = The specific weight is γ Weight Volume = SG ρ g V V = SG ρ g = γ 0.9 1000 × kg m 3 9.81 × m s 2 Ns 2 kg m × = γ 8829 N m 3 = For the equilibriul position when floating, we repeat the force balance with T = 0 F B W 0 = WF B = with F B ρ g V submerged = From references (trying Googling "partial sphere volume") V submerged π h 2 3 3R h () = where h is submerged depth and R is the sphere radius
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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.

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