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

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

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 = gV = 3 g w 1 + 1 n W h 3 Thus, in vector form, the force on the steps is F = 3 gh 3 } i + w 1 + 1 n W k ]...
View Full Document

This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

Ask a homework question - tutors are online