Unformatted text preview: cylinder? According to Archimedes’ principle, the buoyancy on the cylinder is equal to the weight of water displaced by the cylinder. Since 1.5 cm is above the surface of water, the rest 6.5 cm is under the surface of water. The buoyancy is B C = ρ W (6.5/8)(Ah)g = 1.59 N b) What is the radius r of the lead ball? Newton’s 2 nd law applied to the whole system yields B B + B C – W B – W C = 0 We express the buoyancies and gravitational forces in terms of volumes and densities and solve the equation for the volume of the ball. ρ W V B g + ρ W (6.5/8)(Ah)g – ρ B V B g – ρ Ahg = 0 ⇒ V B = [(6.5/8) ρ W – ρ C ]Ah/( ρ B – ρ W ) = 7.93 cm 3 The radius of the ball is V B = (4 π /3)r 3 r = 3 √ (3V B /(4 π )) = 1.24 cm...
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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.
 Summer '08
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