p0356

# p0356 - hinge viz z cp − z F = RF s = ⇒ F s = 2 πρ gR...

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