Problem 3.93

# Problem 3.93 - Problem*3.93[Difficulty 2 Given Geometry of...

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Problem *3.93 [Difficulty: 2] Given: Geometry of steel cylinder Find: Volume of water displaced; number of 1 kg wts to make it sink Solution: The data is For water ρ 999 kg m 3 = For steel (Table A.1) SG 7.83 = For the cylinder D 100 mm = H1 m = δ 1mm = The volume of the cylinder is V steel δ π D 2 4 π D H + = V steel 3.22 10 4 × m 3 = The weight of the cylinder is WS G ρ g V steel = W 7.83 999 × kg m 3 9.81 × m s 2 3.22 × 10 4 × m 3 Ns 2 kg m × = W 24.7N = At equilibium, the weight of fluid displaced is equal to the weight of the cylinder W displaced ρ g V displaced = W = V displaced W ρ g = 24.7 N m 3 999 kg × s 2 9.81 m × kg m 2 × = V displaced 2.52L = 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.

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