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Unformatted text preview: and volume Ah 1 , while the fluid displaced in the lower fluid has density 4 and volume Ah 2 . Hence, F buoy = Ah 1 + 4 Ah 2 Balancing forces, we have W = F buoy , so that 2 A ( h 1 + h 2 ) = Ah 1 + 4 Ah 2 = 2 h 1 + 2 h 2 = h 1 + 4 h 2 Hence, solving for h 1 , we find h 1 = 2 h 2 But, H = h 1 + h 2 , so that 2 h 2 + h 2 = H = h 2 = 1 3 H, h 1 = 2 3 H Therefore, 1/3 of the objects volume lies in the lower fluid....
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.
 Spring '06
 Phares

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