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p0364

# p0364 - area above the i th step is ih 2/n 2 Thus the total...

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320 CHAPTER 3. EFFECTS OF GRAVITY ON PRESSURE 3.64 Chapter 3, Problem 64 Figure 3.49: Swimming-pool steps. The projection of the steps onto a vertical plane is a rectangle of height h and width 6 h . So, the centroid and area are z = 1 2 h, A =6 h 2 Thus, the horizontal component of the force on the steps is F x = ρ g zA = ρ g w 1 2 h W p 6 h 2 Q =3 ρ gh 3 Turning to the vertical component, the cross-sectional area of the fluid above the first step in the xz plane is h 2 /n 2 . The area above the second step is 2 h 2 /n 2 . Generalizing, the
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Unformatted text preview: area above the i th step is ih 2 /n 2 . Thus, the total cross-sectional area for the complete set of steps is A cs = h 2 n 2 n 3 i =1 i = h 2 n 2 n ( n + 1) 2 = 1 2 w 1 + 1 n W h 2 So, the volume of the fluid above the steps is V = 6 hA cs = 3 w 1 + 1 n W h 3 Therefore, the vertical force on the steps is F z = ρ gV = 3 ρ g w 1 + 1 n W h 3 Thus, in vector form, the force on the steps is F = 3 ρ gh 3 } i + w 1 + 1 n W k ]...
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