1. 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

2. 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

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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, ---

13. 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 ---

14. 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, ---

15. Find: F [lbf] in order to keep the gate vertical air A F water y’ R dFH
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, ---

16. Find: F [lbf] in order to keep the gate vertical air A F water y’ R dFH
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

17. Find: F [lbf] in order to keep the gate vertical air A F water y’ R dFH
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

18. Find: F [lbf] in order to keep the gate vertical air A F water y’ R dFH
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

19. Find: F [lbf] in order to keep the gate vertical air A F water y’ R dFH
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

20. 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, ---

21. 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

22. 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

23. 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

24. 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

25. 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

26. 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

27. 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

28. 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

29. 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

30. 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 = [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

31. 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 = [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

32. 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 = [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

33. 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 = [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

34. 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 = [lbf] B
Ic,x
the equation for y prime, which involves variables, ---
y’ = (R-yc) +
(R-yc)*A

35. 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 = [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

36. 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 = [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

37. 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 = [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

38. 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 = [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

39. 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

40. 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

41. 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

42. 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’ = [ft]
R-yc y’ * * yc R

43. 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’ = [ft] R
R = 2.5 [ft]
FH = [lbf]

44. 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’ = [ft] R
R = 2.5 [ft]
FH = [lbf]

45. 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’ = [ft] R
R = 2.5 [ft]
F= [lbf]
FH = [lbf]

46. 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’ = [ft] R
R = 2.5 [ft]
F= [lbf]
FH = [lbf]

47. 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’ = [ft] R
R = 2.5 [ft]
F= [lbf]
FH = [lbf]
answerA

48. * 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