1. 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]

2. 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]

3. 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]

4. 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]

5. 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]

6. 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]

7. Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,H FR= FR,H + FR,V2 2
and a vertical conponent, F R V. [pause] Where subscripts H and V identify, --- FR
FR,V

8. Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,H FR= FR,H + FR,V2 2
the horizontal and vertical components. [pause] The reaction force in the horizontal direction, equals, --- FR
FR,V
horizontal
vertical

9. Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,H FR= FR,H + FR,V2 2
P c, times A, where the P c, equals, --- FR
FR,V
FR,H = Pc *A

10. Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,H FR= FR,H + FR,V2 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

11. Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,H FR= FR,H + FR,V2 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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

18. Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR= FR,H + FR,V2 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

19. Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR= FR,H + FR,V2 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

20. Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR= FR,H + FR,V2 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

21. Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR= FR,H + FR,V2 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]

22. Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR= FR,H + FR,V2 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 = * 107 s2

23. Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR= FR,H + FR,V2 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 = * 107 1 * s2
kg * m

24. Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR= FR,H + FR,V2 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 = * 107 1 * s2
kg * m

25. Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR= FR,H + FR,V2 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 = * 107 [N]

26. 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 = * 107 [N]

27. Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,V FR= FR,H + FR,V2 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 = * 107 [N]

28. Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,V FR= FR,H + FR,V2 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 = * 107 [N]

29. Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,V FR= FR,H + FR,V2 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 = * 107 [N]

30. Find: FR [N] R water w = 35 [m] d FR d = 10 [m] FR,V FR= FR,H + FR,V2 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 = * 107 [N]

31. Find: FR [N] R water θ w = 35 [m] d FR d = 10 [m] FR,V FR= FR,H + FR,V2 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)

32. Find: FR [N] R water θ w = 35 [m] d FR d = 10 [m] FR,V FR= FR,H + FR,V2 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)

33. 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)

34. 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)

35. 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)

36. 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

37. 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

38. 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

39. 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

40. 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

41. 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

42. 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, ---

43. 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

44. 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

45. 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, g, R squared, w. [pause] From before, we know the values of, ---
(0)
sin(2*(0)) - + 2 4
FR,V = * ρ * g * R2 * w

46. 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 = * ρ * g * R2 * w

47. 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 = * ρ * g * R2 * w

48. 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 = * ρ * g * R2 * w
kg * m
FR,V = * 106 s2

49. 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 = * ρ * g * R2 * w
kg * m
N * s2
FR,V = * 106 1 *
kg * m s2

50. 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 = * 107 [N]
FR,V = * 106 [N]

51. 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 = * 107 [N]
FR,V = * 106 [N]

52. 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 = * 107 [N]
FR,V = * 106 [N]
FR=1.823 * 107 [N]

53. 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 = * 107 [N]
FR,V = * 106 [N]
FR=1.823 * 107 [N]

54. Find: FR [N] θ=sin-1(d/R) θ θ=π/6 R = 20 [m] d R w = 35 [m] answerC
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 = * 107 [N]
FR,V = * 106 [N]
FR=1.823 * 107 [N]

55. * 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