Find: FR [N] R water d FR R = 20 [m] circular 1.62*107 gate 1.72*107 Find the reaction force, F R, in newtons. [pause] In this problem, --- 1.62*107 1.72*107 1.82*107 1.92*107 gate d = 10 [m] w = 35 [m]
Find: FR [N] R water d FR R = 20 [m] circular 1.62*107 gate 1.72*107 a circular gate holds back a reservoir of water. The reservoir of water has a depth of, -- 1.62*107 1.72*107 1.82*107 1.92*107 gate d = 10 [m] w = 35 [m]
Find: FR [N] R water d FR R = 20 [m] circular 1.62*107 gate 1.72*107 10 meters. The radius of the circular gate, R, equals, --- 1.62*107 1.72*107 1.82*107 1.92*107 gate d = 10 [m] w = 35 [m]
Find: FR [N] R water d FR R = 20 [m] circular 1.62*107 gate 1.72*107 20 meters. [pause] And the gate has a total width, w, --- 1.62*107 1.72*107 1.82*107 1.92*107 gate d = 10 [m] w = 35 [m]
Find: FR [N] R water d FR R = 20 [m] circular 1.62*107 gate 1.72*107 measuring 35 meters. [pause] The problem asks us to find the hydrostatic force, F R, --- 1.62*107 1.72*107 1.82*107 1.92*107 gate d = 10 [m] w = 35 [m]
Find: FR [N] R water d FR R = 20 [m] circular 1.62*107 gate 1.72*107 which is made up of a horiztonal component, F R H, --- 1.62*107 1.72*107 1.82*107 1.92*107 gate d = 10 [m] w = 35 [m]
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,H FR= FR,H + FR,V 2 2 and a vertical conponent, F R V. [pause] Where subscripts H and V identify, --- FR FR,V
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,H FR= FR,H + FR,V 2 2 the horizontal and vertical components. [pause] The reaction force in the horizontal direction, equals, --- FR FR,V horizontal vertical
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,H FR= FR,H + FR,V 2 2 P c, times A, where the P c, equals, --- FR FR,V FR,H = Pc *A
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,H FR= FR,H + FR,V 2 2 the pressure at the centroid of the gate wall, from a horiztonal perspective, and A represent, --- FR FR,V FR,H = Pc *A pressure at centroid
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,H FR= FR,H + FR,V 2 2 the area of the gate. [pause] We will further define the pressure term, P c, as, --- FR FR,V area FR,H = Pc *A pressure at centroid
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] Pc = ρ *g * dc FR= FR,H + FR,V 2 2 rho, g, d c. We plug in the density of water, which is, --- area FR,H = Pc *A pressure at centroid
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] Pc = ρ *g * dc FR= FR,H + FR,V 2 2 999 kilograms per meters cubed. The gravitational accelration, g, equals, --- ρ= 999 [kg/m3] FR,H = Pc *A (for water) pressure at centroid
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] Pc = ρ *g * dc FR= FR,H + FR,V 2 2 9.81 meters per second squared. [pause] And since the gate appears like a rectangle from the horizontal persepctive, --- ρ= 999 [kg/m3] FR,H = Pc *A g= 9.81 [m/s3] pressure at centroid
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] d Pc = ρ *g * dc FR= FR,H + FR,V 2 2 2 d c, is simply, half the depth of the water, d, which is, --- ρ= 999 [kg/m3] FR,H = Pc *A g= 9.81 [m/s3] pressure at centroid
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] d Pc = ρ *g * dc FR= FR,H + FR,V 2 2 2 half of 10 meters. [pause] This makes the centroidal pressure equal to, --- ρ= 999 [kg/m3] FR,H = Pc *A g= 9.81 [m/s3] pressure at centroid
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] d Pc = ρ *g * dc FR= FR,H + FR,V 2 2 2 49,001 kilogram per meter, seconds squared. [pause] The area of the gate, A, equals, --- ρ= 999 [kg/m3] FR,H = Pc *A g= 9.81 [m/s3] pressure at Pc = 49,001[kg/m*s2] centroid
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR= FR,H + FR,V 2 2 the depth of water, d, times the width of the gate, w, and after plugging in, --- A = d * w FR,H = Pc *A pressure at Pc = 49,001[kg/m*s2] centroid
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR= FR,H + FR,V 2 2 these known values, the area of the gate equals, --- A = d * w FR,H = Pc *A pressure at Pc = 49,001[kg/m*s2] centroid
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR= FR,H + FR,V 2 2 350 meters squared. [pause] Now we can solve for the --- A = d * w FR,H = Pc *A A = 350 [m2] pressure at Pc = 49,001[kg/m*s2] centroid
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR= FR,H + FR,V 2 2 the horizontal reaction force on the gate, F R H, which equals, --- A = d * w FR,H = Pc *A A = 350 [m2] Pc = 49,001[kg/m*s2]
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR= FR,H + FR,V 2 2 1.715 times 10 to the 7, kilogram meters per second squared. After applying a unit converstion,--- Pc = 49,001[kg/m*s2] FR,H = Pc *A A = 350 [m2] kg * m FR,H = 1.715 * 107 s2
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR= FR,H + FR,V 2 2 and cancelling out terms, the units of this force are in, --- Pc = 49,001[kg/m*s2] FR,H = Pc *A A = 350 [m2] kg * m N * s2 FR,H = 1.715 * 107 1 * s2 kg * m
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR= FR,H + FR,V 2 2 Newtons. [pause]. Next, we’ll calculate the --- Pc = 49,001[kg/m*s2] FR,H = Pc *A A = 350 [m2] kg * m N * s2 FR,H = 1.715 * 107 1 * s2 kg * m
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR= FR,H + FR,V 2 2 the vertical component of the hydrostatic force, F R V. This force equals, --- Pc = 49,001[kg/m*s2] FR,H = Pc *A A = 350 [m2] FR,H = 1.715 * 107 [N]
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,V = Py * A FR= FR,H + FR,V 2 2 the pressure in the y direction, times, the area of the gate. [pause] This time, since were dealing a vertical force, --- Pc = 49,001[kg/m*s2] FR,H = Pc *A A = 350 [m2] FR,H = 1.715 * 107 [N]
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,V FR= FR,H + FR,V 2 2 FR,V = Py * A on a rounded surface, we can’t use any short cut methods, such as how we used the centroidal pressure, --- Pc = 49,001[kg/m*s2] FR,H = Pc *A A = 350 [m2] FR,H = 1.715 * 107 [N]
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,V FR= FR,H + FR,V 2 2 FR,V = Py * A when computing the horizontal component. Therefore, we’ll have to integrate the product, --- Pc = 49,001[kg/m*s2] FR,H = Pc *A A = 350 [m2] FR,H = 1.715 * 107 [N]
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,V FR= FR,H + FR,V 2 2 FR,V = ∫Py * A P y, and A, to determine the vertical component of the hydrodstatic force. [pause] In this equation, P y, equals, --- Pc = 49,001[kg/m*s2] FR,H = Pc *A A = 350 [m2] FR,H = 1.715 * 107 [N]
Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,V FR= FR,H + FR,V 2 2 FR,V = ∫Py * A rho times g, times the depth, d. And we’ll define the depth, d, equal to, --- Py = ρ * g * d FR,H = Pc *A FR,H = 1.715 * 107 [N]
Find: FR [N] R water θ w = 35 [m] d FR d = 10 [m] FR,V FR= FR,H + FR,V 2 2 FR,V = ∫Py * A R times the sine of angle theta, where theta is defined as, --- Py = ρ * g * d 1.715 * 107 [N] d= R * sin(θ)
Find: FR [N] R water θ w = 35 [m] d FR d = 10 [m] FR,V FR= FR,H + FR,V 2 2 FR,V = ∫Py * A the angle inside the gate, as shown. [pause] The area, A, --- Py = ρ * g * d 1.715 * 107 [N] d= R * sin(θ)
Find: FR [N] R water θ w = 35 [m] d FR d = 10 [m] FR,V A = w * d FR= FR,H + FR,V 2 2 FR,V = ∫Py * A equals the width, w, times the depth, d, where again, --- Py = ρ * g * d 1.715 * 107 [N] d= R * sin(θ)
Find: FR [N] R water θ w = 35 [m] d FR d = 10 [m] FR,V A = w * d FR= FR,H + FR,V 2 2 FR,V = ∫Py * A d equals R times the sine of angle theta. [pause] Now our equation for the vertical reaction force is, --- Py = ρ * g * d 1.715 * 107 [N] d= R * sin(θ)
Find: FR [N] R water θ w = 35 [m] d FR d = 10 [m] FR,V A = w * d FR= FR,H + FR,V 2 2 FR,V = ∫Py * A rho, g, R squared, w, sine squared theta. And we’ll integrate with respect to theta, --- Py = ρ * g * d 1.715 * 107 [N] d= R * sin(θ) FR,V = ∫ρ * g * R2 * w * sin2(θ)
Find: FR [N] R water θ w = 35 [m] d FR d = 10 [m] FR,V A = w * d FR= FR,H + FR,V 2 2 FR,V = ∫Py * A where the limits of integration will span from 0 radians, to, --- Py = ρ * g * d 1.715 * 107 [N] R * sin(θ) FR,V = ρ * g * R2 * w * ∫sin2(θ)*dθ
Find: FR [N] R water θ R = 20 [m] d FR d = 10 [m] FR,V A = w * d FR= FR,H + FR,V 2 2 FR,V = ∫Py * A theta. We can solve for theta by taking the arc sine of --- Py = ρ * g * d 1.715 * 107 [N] θ R * sin(θ) FR,V = ρ * g * R2 * w * ∫sin2(θ)*dθ
Find: FR [N] R θ=sin-1(d/R) θ R = 20 [m] d d = 10 [m] A = w * d FR= FR,H + FR,V 2 2 FR,V = ∫Py * A the depth of water, divided by the radius of the gate, since the structure of the gate, --- Py = ρ * g * d 1.715 * 107 [N] θ R * sin(θ) FR,V = ρ * g * R2 * w * ∫sin2(θ)*dθ
Find: FR [N] θ=sin-1(d/R) θ R = 20 [m] d R d = 10 [m] A = w * d FR= FR,H + FR,V 2 2 FR,V = ∫Py * A forms a right triangle, as shown. After pluggin in d, --- Py = ρ * g * d 1.715 * 107 [N] θ R * sin(θ) FR,V = ρ * g * R2 * w * ∫sin2(θ)*dθ
Find: FR [N] θ=sin-1(d/R) θ R = 20 [m] d R d = 10 [m] A = w * d FR= FR,H + FR,V 2 2 FR,V = ∫Py * A and R, we calcualte theta, equals, --- Py = ρ * g * d 1.715 * 107 [N] θ R * sin(θ) FR,V = ρ * g * R2 * w * ∫sin2(θ)*dθ
Find: FR [N] θ=sin-1(d/R) θ θ=π/6 R = 20 [m] d R d = 10 [m] A = w * d FR= FR,H + FR,V 2 2 FR,V = ∫Py * A PI over 6 radians, which is the same as 30 degrees. [pause] Now it’s time to integrate, --- Py = ρ * g * d 1.715 * 107 [N] π/6 R * sin(θ) FR,V = ρ * g * R2 * w * ∫sin2(θ)*dθ
Find: FR [N] θ=sin-1(d/R) θ θ=π/6 R = 20 [m] d R w = 35 [m] π/6 FR,V = ρ * g * R2 * w * ∫sin2(θ)*dθ sin squared theta, d theta, which equals, ---
Find: FR [N] - θ=sin-1(d/R) θ θ=π/6 R = 20 [m] d R w = 35 [m] π/6 FR,V = ρ * g * R2 * w * ∫sin2(θ)*dθ π/6 one half theta, minus, one quarter of sin 2 theta. [pause] After plugging in the limits of integration, --- θ sin(2*θ) FR,V = ρ * g * R2 * w * - 2 4
Find: FR [N] - - - + θ=sin-1(d/R) θ θ=π/6 R = 20 [m] d R w = 35 [m] FR,V = ρ * g * R2 * w * - 2 4 PI over 6, and zero, our equation for the vertical component of the reaction force, reduces to, --- (π/6) sin(2*(π/6)) FR,V = ρ * g * R2 * w * - 2 4 (0) sin(2*(0)) - + 2 4
Find: FR [N] - - + θ=sin-1(d/R) θ θ=π/6 R = 20 [m] d R w = 35 [m] (π/6) sin(2*(π/6)) FR,V = ρ * g * R2 * w * - 2 4 0.04529 rho, g, R squared, w. [pause] From before, we know the values of, --- (0) sin(2*(0)) - + 2 4 FR,V = 0.04529 * ρ * g * R2 * w
Find: FR [N] - - + θ=sin-1(d/R) θ θ=π/6 R = 20 [m] d R w = 35 [m] (π/6) sin(2*(π/6)) FR,V = ρ * g * R2 * w * - 2 4 rho and g, and we also know the given radius and width of the gate, --- g= 9.81 [m/s3] (0) sin(2*(0)) - + ρ= 999 [kg/m3] 2 4 FR,V = 0.04529 * ρ * g * R2 * w
Find: FR [N] - - + θ=sin-1(d/R) θ θ=π/6 R = 20 [m] d R w = 35 [m] (π/6) sin(2*(π/6)) FR,V = ρ * g * R2 * w * - 2 4 variables R and w. This makes the force, F R V, equal to, --- g= 9.81 [m/s3] (0) sin(2*(0)) - + ρ= 999 [kg/m3] 2 4 FR,V = 0.04529 * ρ * g * R2 * w
Find: FR [N] θ=sin-1(d/R) θ θ=π/6 R = 20 [m] d R w = 35 [m] g= 9.81 [m/s3] ρ= 999 [kg/m3] 6.214 times 10 to the 6 kilogram, meters, per second squared. Which has the same units as, ---- FR,V = 0.04529 * ρ * g * R2 * w kg * m FR,V = 6.214 * 106 s2
Find: FR [N] θ=sin-1(d/R) θ θ=π/6 R = 20 [m] d R w = 35 [m] g= 9.81 [m/s3] ρ= 999 [kg/m3] Newtons. [pause] Now that we know the both the vertical, --- FR,V = 0.04529 * ρ * g * R2 * w kg * m N * s2 FR,V = 6.214 * 106 1 * kg * m s2
Find: FR [N] θ=sin-1(d/R) θ θ=π/6 R = 20 [m] d R w = 35 [m] FR= FR,H + FR,V 2 2 and horizontal components of the reaction force, we can solve for, --- FR,H = 1.715 * 107 [N] FR,V = 6.214 * 106 [N]
Find: FR [N] θ=sin-1(d/R) θ θ=π/6 R = 20 [m] d R w = 35 [m] FR= FR,H + FR,V 2 2 the total reaction force, F R, which equals, --- FR,H = 1.715 * 107 [N] FR,V = 6.214 * 106 [N]
Find: FR [N] θ=sin-1(d/R) θ θ=π/6 R = 20 [m] d R w = 35 [m] FR= FR,H + FR,V 2 2 1.823 times 10 to the 7, Newtons. [pause] FR,H = 1.715 * 107 [N] FR,V = 6.214 * 106 [N] FR=1.823 * 107 [N]
Find: FR [N] θ=sin-1(d/R) θ θ=π/6 R = 20 [m] d R w = 35 [m] FR= FR,H + FR,V 2 2 When reviewing the possible solutions, --- 1.62*107 1.72*107 1.82*107 1.92*107 FR,H = 1.715 * 107 [N] FR,V = 6.214 * 106 [N] FR=1.823 * 107 [N]
Find: FR [N] θ=sin-1(d/R) θ θ=π/6 R = 20 [m] d R w = 35 [m] answerC FR= FR,H + FR,V 2 2 the correct answer is C. [fin] 1.62*107 1.72*107 1.82*107 1.92*107 FR,H = 1.715 * 107 [N] FR,V = 6.214 * 106 [N] FR=1.823 * 107 [N]
* 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