Find: F [lbf] in order to keep the gate vertical air F 267 water 293 440 880 water gate hinge R = 2.5 [ft] Find the force, F, in pounds force. [pause] In this problem, --- R the gate is semi-circular hinge

Find: F [lbf] in order to keep the gate vertical air F 267 water 293 440 880 water gate hinge R = 2.5 [ft] a reservoir of water is held back by a gate. From the cross section below, --- R the gate is semi-circular hinge

Find: F [lbf] in order to keep the gate vertical air F 267 water 293 440 880 water gate hinge R = 2.5 [ft] we notice the gate is semi-circular, having a radius of R, equal to, --- R the gate is semi-circular hinge

Find: F [lbf] in order to keep the gate vertical air F 267 water 293 440 880 water gate hinge R = 2.5 [ft] 2.5 feet. [pause] Since the water surface reaches the top of the gate, --- R the gate is semi-circular hinge

Find: F [lbf] in order to keep the gate vertical air F 267 water 293 440 880 water gate d=2.5 [ft] hinge R = 2.5 [ft] we know the depth of water, d, also equals, ---- R the gate is semi-circular hinge

Find: F [lbf] in order to keep the gate vertical air F 267 water 293 440 880 water gate d=2.5 [ft] hinge R = 2.5 [ft] 2.5 feet, and at the base of the gate, --- R the gate is semi-circular hinge

Find: F [lbf] in order to keep the gate vertical air F 267 water 293 440 880 water gate d=2.5 [ft] hinge R = 2.5 [ft] there is a hinge. [pause] The problem asks us to find the force, F, --- R the gate is semi-circular hinge

Find: F [lbf] in order to keep the gate vertical air F 267 water 293 440 880 water gate d=2.5 [ft] hinge R = 2.5 [ft] which will keep the semi-circular gate positioned vertically, by resisting ---- R the gate is semi-circular hinge

Find: F [lbf] in order to keep the gate vertical air F 267 water 293 440 880 water gate d hinge R = 2.5 [ft] the hydrostatic forces of the water pushing against the gate. [pause] In this problem, we’ll define point A, --- R the gate is semi-circular hinge

Find: F [lbf] in order to keep the gate vertical air A F 267 water 293 440 880 water gate d hinge R = 2.5 [ft] at the top of the gate, and we’ll define point B, --- R the gate is semi-circular hinge

Find: F [lbf] in order to keep the gate vertical air A F 267 water 293 440 880 water gate d hinge B R = 2.5 [ft] at the hinge, at the bottom of the gate. [pause] If we set the sum of the moments about point B, --- R the gate is semi-circular hinge

Find: F [lbf] in order to keep the gate vertical air A F water gate d hinge R = 2.5 [ft] B ΣMB = 0 = F * R – FH * (R-y’) equal to zero, the moment in the counter-clockwise direction is the force F, ---

Find: F [lbf] in order to keep the gate vertical air A F water R d R = 2.5 [ft] B ΣMB = 0 = F * R – FH * (R-y’) times the radius of the gate, R, and the clockwise moment is the product of ---

Find: F [lbf] in order to keep the gate vertical air A F water R d FH R = 2.5 [ft] B ΣMB = 0 = F * R – FH * (R-y’) the magnitude of the hydrostatic force, F H, acting on the gate at a distance of, ---

Find: F [lbf] in order to keep the gate vertical air A F water y’ R d FH R-y’ R = 2.5 [ft] B ΣMB = 0 = F * R – FH * (R-y’) R minus y prime, from the hinge of the gate. [pause] If we solve for the force, F, ---

Find: F [lbf] in order to keep the gate vertical air A F water y’ R d FH R-y’ R = 2.5 [ft] B ΣMB = 0 = F * R – FH * (R-y’) we’re left with an equation involving the variables, F H, R, ---- FH * (R-y’) F= R

Find: F [lbf] in order to keep the gate vertical air A F water y’ R d FH R-y’ R = 2.5 [ft] B ΣMB = 0 = F * R – FH * (R-y’) and y prime. [pause] Since we already know the radius of the gate, ---- FH * (R-y’) F= R

Find: F [lbf] in order to keep the gate vertical air A F water y’ R d FH R-y’ R = 2.5 [ft] B ΣMB = 0 = F * R – FH * (R-y’) R, we only need to solve for variables, F H, ---- FH * (R-y’) F= R

Find: F [lbf] in order to keep the gate vertical air A F water y’ R d FH R-y’ R = 2.5 [ft] B ΣMB = 0 = F * R – FH * (R-y’) and y prime. [pause] The horizontal force caused by the water, F H, equals, ----- FH * (R-y’) F= R

Find: F [lbf] in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R-y’ R = 2.5 [ft] B FH = Pc * A the hydrostatic pressure at the centroid of the gate, P c, times, the area of the gate, A. [pause] This pressure, P c, ---

Find: F [lbf] in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R-y’ R = 2.5 [ft] B FH = Pc * A equals, rho times g, times d c, where d c equals, --- Pc = ρ * g * dc

Find: F [lbf] in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R-y’ R = 2.5 [ft] B FH = Pc * A depth to the depth to the centroid of the gate, which is a different value than, y prime. For a semi-circle, d c, equals, --- cenroid Pc = ρ * g * dc

Find: F [lbf] in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R-y’ R = 2.5 [ft] B FH = Pc * A depth to R, minus the quotient, 4 R, divided by 3 PI. [pause] The area of a semi-circle, equals, --- cenroid Pc = ρ * g * dc dc R * 4*R dc= R- 3*π

Find: F [lbf] in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R-y’ R = 2.5 [ft] B FH = Pc * A A = 0.5 * π * R2 one half, PI, R squared. [pause] After making these substitutions, --- Pc = ρ * g * dc dc R * 4*R dc= R- 3*π

Find: F [lbf] in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R-y’ R = 2.5 [ft] B FH = Pc * A A = 0.5 * π * R2 we’ll rewrite our equation for the hydrostatic force, F H. Since we know the radius, R, --- 4*R Pc = ρ * g * dc dc= R- 3*π 4*R FH = 0.5 * ρ * g *π * R2 R- * 3*π

Find: F [lbf] in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R-y’ R = 2.5 [ft] B FH = Pc * A A = 0.5 * π * R2 we can subsitute in that value. For variable rho, the density of water, --- 4*R Pc = ρ * g * dc dc= R- 3*π 4*R FH = 0.5 * ρ * g *π * R2 R- * 3*π

Find: F [lbf] in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R-y’ R = 2.5 [ft] B equals, 62.3 pounds mass per cubic foot. [pause] And on planet Earth, the gravitational acceleration constant, ---- ρ= 62.3 [lbm/ft3] 4*R FH = 0.5 * ρ * g *π * R2 R- * 3*π

Find: F [lbf] in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R-y’ R = 2.5 [ft] B g, equals, 32.2 feet per seconds squared. [pause] Also, we’ll multilply this value by, ---- g= 32.2 [ft/s2] ρ= 62.3 [lbm/ft3] 4*R FH = 0.5 * ρ * g *π * R2 R- * 3*π

Find: F [lbf] in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R-y’ R = 2.5 [ft] B unit a unit conversion. [pause] This make the hydrostatic force, equal to, --- conversion g= 32.2 [ft/s2] ρ= 62.3 [lbm/ft3] lbf* s2 4*R 1 FH = 0.5 * ρ * g *π * R2 R- * * 3*π 32.2 lbm*ft

Find: F [lbf] in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R-y’ R = 2.5 [ft] B FH = 880.1 [lbf] unit 880.1 pounds force. [pause] Now that we have variables F H, and --- conversion g= 32.2 [ft/s2] ρ= 62.3 [lbm/ft3] lbf* s2 4*R 1 FH = 0.5 * ρ * g *π * R2 R- * * 3*π 32.2 lbm*ft

Find: F [lbf] in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R-y’ R = 2.5 [ft] B FH = 880.1 [lbf] unit R, we are left to figure out the value of variable, --- conversion g= 32.2 [ft/s2] ρ= 62.3 [lbm/ft3] lbf* s2 4*R 1 FH = 0.5 * ρ * g *π * R2 R- * * 3*π 32.2 lbm*ft

Find: F [lbf] ? in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R-y’ R = 2.5 [ft] B FH = 880.1 [lbf] unit y prime. [pause] y prime refers to the distance from the top of the gate, --- conversion g= 32.2 [ft/s2] ρ= 62.3 [lbm/ft3] lbf* s2 4*R 1 FH = 0.5 * ρ * g *π * R2 R- * * 3*π 32.2 lbm*ft

Find: F [lbf] ? in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R-y’ R = 2.5 [ft] B FH = 880.1 [lbf] unit to the point where the resultant hydrostatic force F H, is acting upon the gate. [pause] We’ll write out --- conversion g= 32.2 [ft/s2] ρ= 62.3 [lbm/ft3] lbf* s2 4*R 1 FH = 0.5 * ρ * g *π * R2 R- * * 3*π 32.2 lbm*ft

Find: F [lbf] in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R = 2.5 [ft] R-y’ FH = 880.1 [lbf] B Ic,x the equation for y prime, which involves variables, --- y’ = (R-yc) + (R-yc)*A

Find: F [lbf] in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R = 2.5 [ft] R-y’ FH = 880.1 [lbf] B Ic,x R, y c, I c x, and A. [pause] For a semi-circle, the term y c, equals, --- y’ = (R-yc) + R-yc (R-yc)*A y’ * * yc R

Find: F [lbf] in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R = 2.5 [ft] R-y’ FH = 880.1 [lbf] B Ic,x 4 times the radius, divided by 3 PI. [pause] And the area of the semi-circle, A, ---- y’ = (R-yc) + R-yc (R-yc)*A y’ * 4*R * yc= yc R 3*π

Find: F [lbf] in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R = 2.5 [ft] R-y’ FH = 880.1 [lbf] B Ic,x equals, PI R squared, divided by 2. [pause] The area moment of intertia, ---- y’ = (R-yc) + R-yc (R-yc)*A y’ * 4*R * yc= yc R A=π*R2/2 3*π

Find: F [lbf] in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R = 2.5 [ft] R-y’ FH = 880.1 [lbf] B Ic,x taken about an axis parallel to the x-axis, and running through the centroid of the gate, --- y’ = (R-yc) + R-yc (R-yc)*A y’ * 4*R * yc= yc R A=π*R2/2 3*π

π Find: F [lbf] - in order to keep the gate vertical air A F FH * (R-y’) water F= y’ R R d FH R-y’ π 8 Ic,x= - R4 B 9*π * 8 Ic,x is also a function of the radius, R. [pause] After writing out a revised equation for variable y prime, --- y’ = (R-yc) + R-yc (R-yc)*A y’ * 4*R * yc= yc R A=π*R2/2 3*π

π π Find: F [lbf] - + - in order to keep the gate vertical 8 R4 9*π * y’ = R- + 3*π 4*R (π * R2/2) R- 3*π * π 8 Ic,x= - R4 9*π * R = 2.5 [ft] 8 Ic,x we can plug in the variable R, --- y’ = (R-yc) + R-yc (R-yc)*A y’ * 4*R * yc= yc R A=π*R2/2 3*π

π π Find: F [lbf] - + - in order to keep the gate vertical 8 R4 9*π * y’ = R- + 3*π 4*R (π * R2/2) R- 3*π * π 8 Ic,x= - R4 9*π * R = 2.5 [ft] 8 Ic,x and y prime, equals, --- y’ = (R-yc) + R-yc (R-yc)*A y’ * 4*R * yc= yc R A=π*R2/2 3*π

π Find: F [lbf] - + in order to keep the gate vertical 8 R4 9*π * 4*R y’ = R- + 3*π 4*R (π * R2/2) R- 3*π * R = 2.5 [ft] 1.743 feet. [pause] At this point, --- y’ = 1.743 [ft] R-yc y’ * * yc R

π Find: F [lbf] - + in order to keep the gate vertical 8 R4 9*π * 4*R y’ = R- + 3*π 4*R (π * R2/2) R- 3*π * R = 2.5 [ft] FH * (R-y’) F= we can solve for the force, F. After substituting in variables, y prime, --- y’ = 1.743 [ft] R R = 2.5 [ft] FH = 880.1 [lbf]

π Find: F [lbf] - + in order to keep the gate vertical 8 R4 9*π * 4*R y’ = R- + 3*π 4*R (π * R2/2) R- 3*π * R = 2.5 [ft] FH * (R-y’) F= R and F H, the force F, equals, --- y’ = 1.743 [ft] R R = 2.5 [ft] FH = 880.1 [lbf]

π Find: F [lbf] - + in order to keep the gate vertical 8 R4 9*π * 4*R y’ = R- + 3*π 4*R (π * R2/2) R- 3*π * R = 2.5 [ft] FH * (R-y’) F= 266.5 pounds force. [pause] y’ = 1.743 [ft] R R = 2.5 [ft] F=266.5 [lbf] FH = 880.1 [lbf]

π Find: F [lbf] - + in order to keep the gate vertical 8 R4 9*π * 4*R y’ = R- + 3*π 4*R (π * R2/2) R- 3*π * 267 293 440 880 R = 2.5 [ft] FH * (R-y’) F= When reviewing the possible solutions, ---- y’ = 1.743 [ft] R R = 2.5 [ft] F=266.5 [lbf] FH = 880.1 [lbf]

π Find: F [lbf] - + in order to keep the gate vertical 8 R4 9*π * 4*R y’ = R- + 3*π 4*R (π * R2/2) R- 3*π * 267 293 440 880 R = 2.5 [ft] FH * (R-y’) F= the correct answer is A. [fin] y’ = 1.743 [ft] R R = 2.5 [ft] F=266.5 [lbf] FH = 880.1 [lbf] answerA

* ΔP13=-ρA* g * (hAB-h1)+… depth to centroid kg*cm3 1,000 g * m3 a monometer contains 5 different fluids, which include, --- kg*cm3 1,000 * g * m3