p0356 - hinge, viz., ( z cp z ) F = RF s = F s = 2 gR 2 ( z...

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

View Full Document Right Arrow Icon
3.56. CHAPTER 3, PROBLEM 56 309 3.56 Chapter 3, Problem 56 Figure 3.41: Hinged gate. The hydrostatic force on the gate is F = ρ g zA For the circular gate, the centroid, area and moment of inertia are z =2 R, A = π R 2 ,I = π 4 R 4 Thus, the force is F = ρ g (2 R )( π R 2 )=2 πρ gR 3 Now, the hydrostatic force is balanced by the reaction forces at the hinge and the stop (see Figure 3.41). We can determine the force at the stop by taking moments about the hinge, viz.,
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: hinge, viz., ( z cp z ) F = RF s = F s = 2 gR 2 ( z cp z ) The center of pressure is z cp = z + I zA = z cp z = I zA Thus, z cp z = 4 R 4 (2 R ) ( R 2 ) = 1 8 R Therefore, the force on the stop is F s = 2 gR 2 w 1 8 R W = 4 gR 3...
View Full Document

Ask a homework question - tutors are online