p0370 - = 16 ρ gH 2 h +10 ρ gH 3 = 16 ρ gH 2 w h + 5 8 H...

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3.70. CHAPTER 3, PROBLEM 70 327 3.70 Chapter 3, Problem 70 Figure 3.55: Equivalent problems for two fluids separated by an arc. To begin, use superposition as illustrated in Figure 3.55. For Problem A, the arc’s projection on a vertical plane is a rectangle of height H and width 8 H . So, the centroid is located at z = 3 2 H and A =8 H 2 . Also, we are given that the area of OAB is 5 8 H 2 . Thus, F x A = ρ g zA =12 ρ gH 3 and F z A = ρ gH 2 (8 H )+ 5 8 ρ gH 2 (8 H )=13 ρ gH 3 Similarly, for Problem B, the centroid is at z = h + 1 2 H and A =8 H 2 , wherefore F x B =2 ρ g zA =16 ρ g w h + 1 2 H W H 2 and F z B
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Unformatted text preview: = 16 ρ gH 2 h +10 ρ gH 3 = 16 ρ gH 2 w h + 5 8 H W So, the solution to the original problem is F x = F x A − F x B = 12 ρ gH 3 − 16 ρ gH 2 w h + 1 2 H W = 4 ρ gH 3 w 1 − 4 h H W F z = F z A − F z B = 13 ρ gH 3 − 16 ρ gH 2 w h + 5 8 H W = 3 ρ gH 3 w 1 − 16 3 h H W Therefore, the horizontal force component, F x , vanishes when h = 1 4 H For this depth, the vertical force on the arc is thus F z = − ρ gH 3...
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This note was uploaded on 10/12/2009 for the course AME 309 at USC.

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