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p0380 - ρ and volume Ah 1 while the fluid displaced in the...

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338 CHAPTER 3. EFFECTS OF GRAVITY ON PRESSURE 3.80 Chapter 3, Problem 80 Let h 1 and h 2 denote the distance from the top and bottom, respectively, of the object to the interface between the two fluids. Also, let H denote the distance from the bottom to the top of the object as shown in Figure 3.60. Figure 3.60: Submerged object in a 2-layer fluid. The weight of the object, W ,is 2 ρ times its volume, which is AH . Thus, W =2 ρ AH =2 ρ A ( h 1 + h 2 ) The buoyancy force is equal to the weight of the displaced fluid. The fluid displaced in the upper fluid has density
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Unformatted text preview: ρ and volume Ah 1 , while the fluid displaced in the lower fluid has density 4 ρ and volume Ah 2 . Hence, F buoy = ρ Ah 1 + 4 ρ Ah 2 Balancing forces, we have W = F buoy , so that 2 ρ A ( h 1 + h 2 ) = ρ Ah 1 + 4 ρ 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 object’s volume lies in the lower fluid....
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