qz10sol_0132Hf10 - 0 = B b – W i- W b The buoyancy on the...

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TA: Tomoyuki Nakayama Monday, November 15, 2010 PHY 2048: Physic 1, Discussion Section 0132H Quiz 10 (Homework Set #12) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ A block of wood has a mass of 5.00 kg and a density of 600 kg/m 3 . It is to be loaded with iron (7.87 × 10 3 kg/m 3 ) so that it will float in water (1.00 × 10 3 kg/m 3 ) with 0.850 of its volume submerged. a) What mass of iron is needed if the iron is attached to the top of the wood? The volume of the block is V b = m b / ρ b = 8.33 × 10 -3 m 3 Three forces are exerted on the system, the gravitational forces on the block and iron load, and buoyancy on the block. Since the system is in equilibrium, we have
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Unformatted text preview: 0 = B b – W i- W b The buoyancy on the block is equal to the weight of water displaced by the block. Therefore, W b = 0.85 ρ w V b g Expressing the buoyancy and gravitational forces in terms of volume and density, we solve the first equation for the mass of the block. 0 = 0.85 ρ w V b g – ρ i V i g – ρ b V b g ⇒ m i = ρ i V i = (0.85 ρ w – ρ b )V b = 2.08 kg b) What mass of iron is needed if the iron is attached to the bottom of the wood? Since the iron load is submerged in water, a buoyant force is exerted on it. The balance of forces equation yields 0 =B i + B b – W i- W b 0 = ρ w V i g + 0.85 ρ w V b g – ρ i V i g – ρ b V b g V i = (0.85 ρ w – ρ b )V b /( ρ i - ρ w) = 3.03 × 10-4 m 3 The mass of the iron load is m i = ρ i V i =2.38 kg...
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This note was uploaded on 01/25/2011 for the course PHY 2048 taught by Professor Field during the Summer '08 term at University of Florida.

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