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Problem 3.94 - W w W a ρ w g ⋅ V d ⋅ − = so the...

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Problem *3.94 [Difficulty: 2] Given: Experiment performed by Archimedes to identify the material conent of King Hiero's crown. The crown was weighed in air and in water. Find: Expression for the specific gravity of the crown as a function of the weights in water and air. Solution: We will apply the hydrostatics equations to this system. Governing Equations: F b ρ g V d = (Buoyant force is equal to weight of displaced fluid) F b Mg W w Assumptions: (1) Static fluid (2) Incompressible fluid Taking a free body diagram of the body: Σ F z 0 = W w Mg F b + 0 = W w is the weight of the crown in water. W w Mg F buoy = Mg ρ w g V d = However in air: W a Mg = Therefore the weight measured in water is:
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Unformatted text preview: W w W a ρ w g ⋅ V d ⋅ − = so the volume is: V d W a W w − ρ w g ⋅ = Now the density of the crown is: ρ c M V d = M ρ w ⋅ g ⋅ W a W w − = W a W a W w − ρ w ⋅ = Therefore, the specific gravity of the crown is: SG ρ c ρ w = W a W a W w − = SG W a W a W w − = Note: by definition specific gravity is the density of an object divided by the density of water at 4 degrees Celsius, so the measured temperature of the water in the experiment and the data from tables A.7 or A.8 may be used to correct for the variation in density of the water with temperature....
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