Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_178

⚡ Mecademy AIENG정역학 · ch5  Problem Statement The viewing window in the tank is made of one-fourth of a hemispherical shell of radius . The surface of the fluid is a distance above the highest point of the window. Determine the horizontal and vertical forces exerted on the shell by the fluid. Problem 5/178 (a) Horizontal Forces and 1. Formula: The horizontal force exerted on a curved surface in a given direction is equal to the force exerted on the projection of the surface onto a vertical plane perpendicular to that direction: where is the specific weight of the fluid, is the projected area, and is the depth of the centroid of the projected area from the fluid surface. 2. Substitution: The shell is an octant of a sphere. Its projection onto both the -plane and the -plane is a quarter-circle of radius . Thus, . The highest point of the window is at . The fluid surface is at . The centroid of a quarter-circle of radius is at a distance from its base (the -plane). The depth of the centroid from the fluid surface is . By symmetry, . 3. Calculation: To find a common denominator: r hA F x F y F = h γA h ˉ p γA p h ˉ yz xzrA = p πr 4 1 2 Az=rz= r+h r=zˉ 3π 4r xy =h ˉ (r+h)− 3π 4r F = x F = y γr+h− ( 3π 4r )( 4 πr 2 ) F = x γ − ⋅ [ 4 πr(r+h) 2 4 πr 2 3π 4r ] F = x γ − [ 4 πr(r+h) 2 3 r 3 ] F = x [3 π (h+ 12 γr 2 r)−4r] 4. Result: ● Final Conclusion: The horizontal force components in the and directions are equal in magnitude: . (b) Vertical Force 1. Formula: The vertical force on a