{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

p0380

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

This preview shows page 1. Sign up to view the full content.

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
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online