Fluid Pressures

1. Fluid Pressures

2. We are to calculate the force on a flat plate
submerged in a fluid.
Fluid Pressures

3. We are to calculate the force on a flat plate
submerged in a fluid.
Let A = Area of the plate
h = the depth
ρ = the weight of the fluid A
h
ρ
Fluid Pressures

4. Let A = Area of the plate
h = the depth
ρ = the weight of the fluid
The simplest case is when
the plate is submerged
horizontally in a fluid.
A
h
ρ
Fluid Pressures
We are to calculate the force on a flat plate
submerged in a fluid.

5. Let A = Area of the plate
h = the depth
ρ = the weight of the fluid
The simplest case is when
the plate is submerged
horizontally in a fluid.
A
h
ρ
The force F on the plate is the weight of the volume
of the fluid that sits over the plate,
Fluid Pressures
We are to calculate the force on a flat plate
submerged in a fluid.

6. Let A = Area of the plate
h = the depth
ρ = the weight of the fluid
The simplest case is when
the plate is submerged
horizontally in a fluid.
A
h
ρ
The force F on the plate is the weight of the volume
of the fluid that sits over the plate, that is
F = ρAh
Fluid Pressures
We are to calculate the force on a flat plate
submerged in a fluid.

7. Let A = Area of the plate
h = the depth
ρ = the weight of the fluid
The simplest case is when
the plate is submerged
horizontally in a fluid.
A
h
ρ
The force F on the plate is the weight of the volume
of the fluid that sits over the plate, that is
F = ρAh
We are to calculate the force exerted on plates of
various shapes that sit vertically in a fluid.
Fluid Pressures
We are to calculate the force on a flat plate
submerged in a fluid.

8. Suppose a plate is submerged
in water vertically with the top
of the plate is at a distance h
to the surface.
h
Fluid Pressures

9. Suppose a plate is submerged
in water vertically with the top
of the plate is at a distance h
to the surface.
h
We select a direction and set
the measurement x, say from
the top x = 0 to the bottom x = b
of the plate.
Fluid Pressures

10. Suppose a plate is submerged
in water vertically with the top
of the plate is at a distance h
to the surface.
h
We select a direction and set
the measurement x, say from
the top x = 0 to the bottom x = b
of the plate.
x=0
x=b
xi-1
xi
Let the sequence {0=x0, x1, x2, .. , xn=b} be an equi-
partition of [0, b] and xi–1 and xi be two consecutive
points.
Fluid Pressures

11. Suppose a plate is submerged
in water vertically with the top
of the plate is at a distance h
to the surface.
h
We select a direction and set
the measurement x, say from
the top x = 0 to the bottom x = b
of the plate.
x=0
x=b
xi-1
xi
Let L(x) = cross-sectional length at x, so
L(xi) = cross-sectional length at xi.
Let the sequence {0=x0, x1, x2, .. , xn=b} be an equi-
partition of [0, b] and xi–1 and xi be two consecutive
points.
Fluid Pressures
L(xi)

12. Suppose a plate is submerged
in water vertically with the top
of the plate is at a distance h
to the surface.
h
We select a direction and set
the measurement x, say from
the top x = 0 to the bottom x = b
of the plate.
x=0
x=b
xi-1
xi
Let L(x) = cross-sectional length at x, so
L(xi) = cross-sectional length at xi.
Let the sequence {0=x0, x1, x2, .. , xn=b} be an equi-
partition of [0, b] and xi–1 and xi be two consecutive
points.
Let Δx = the length of each subinterval.
Fluid Pressures
L(xi)
Δx

13. h
x=0
x=b
xi-1
xi
L(xi)
The cross-sectional area from
xi-1 to xi is approximately L(xi)Δx.
Δx
Fluid Pressures

14. h
x=0
x=b
xi-1
xi
L(xi)
The cross-sectional area from
xi-1 to xi is approximately L(xi)Δx.
So the cross-sectional area from
xi-1 to xi is approximately L(xi)Δx. Δx
h+xi
Fluid Pressures

15. The cross-sectional area from
xi-1 to xi is approximately L(xi)Δx.
So the cross-sectional area from
xi-1 to xi is approximately L(xi)Δx.
The depth of this strip of area is
approximately (xi+h).
Fluid Pressures
h
x=0
x=b
xi-1
xi
L(xi)
Δx
h+xi

16. The cross-sectional area from
xi-1 to xi is approximately L(xi)Δx.
So the cross-sectional area from
xi-1 to xi is approximately L(xi)Δx.
So the fluid force exerted on
this strip of area is approximately ρL(xi)(xi+h)Δx.
The depth of this strip of area is
approximately (xi+h).
Fluid Pressures
h
x=0
x=b
xi-1
xi
L(xi)
Δx
h+xi

17. The cross-sectional area from
xi-1 to xi is approximately L(xi)Δx.
So the cross-sectional area from
xi-1 to xi is approximately L(xi)Δx.
So the fluid force exerted on
this strip of area is approximately ρL(xi)(xi+h)Δx.
The depth of this strip of area is
approximately (xi+h).
Hence the total fluid force F is:
∑ ρL(xi)(xi+h) ΔxF = limn∞
Fluid Pressures
h
x=0
x=b
xi-1
xi
L(xi)
Δx
h+xi

18. The cross-sectional area from
xi-1 to xi is approximately L(xi)Δx.
So the cross-sectional area from
xi-1 to xi is approximately L(xi)Δx.
So the fluid force exerted on
this strip of area is approximately ρL(xi)(xi+h)Δx.
The depth of this strip of area is
approximately (xi+h).
Hence the total fluid force F is:
∑ ρL(xi)(xi+h) ΔxF = lim By FTC, this is
the definite integral
from 0 to b,
n∞
Fluid Pressures
h
x=0
x=b
xi-1
xi
L(xi)
Δx
h+xi

19. The cross-sectional area from
xi-1 to xi is approximately L(xi)Δx.
So the cross-sectional area from
xi-1 to xi is approximately L(xi)Δx.
So the fluid force exerted on
this strip of area is approximately ρL(xi)(xi+h)Δx.
The depth of this strip of area is
approximately (xi+h).
Hence the total fluid force F is:
∑ ρL(xi)(xi+h) ΔxF = lim By FTC, this is
the definite integral
from 0 to b,= ∫x=0
b
ρ L(x)(x+h) dx
n∞
Fluid Pressures
h
x=0
x=b
xi-1
xi
L(xi)
Δx
h+xi

20. Stated in words:
∫x=0
b
ρ L(x)(x+h) dxF =
distance in x
length in x
Fluid Pressures

21. Stated in words:
∫x=0
b
ρ L(x)(x+h) dxF =
distance in x
length in x
∫
b
ρ (cross-sectional length)(distance to surface) dxF =
x=a
Fluid Pressures

22. Stated in words:
∫x=0
b
ρ L(x)(x+h) dxF =
∫
b
ρ (cross-sectional length)(distance to surface) dxF =
distance in x
length in x
x=a
Example A. Find the force
exerted on a square plate
submerged in the fluid as
shown.
5
6
5
Fluid Pressures

23. 5
Place the x-measurement as
shown.
6
5
x
0
5
Fluid Pressures

24. 5
Place the x-measurement as
shown.
6
5
x
0
5
At x, the cross-sectional length
L(x) = 5.
L(x) = 5
Fluid Pressures

25. 5
Place the x-measurement as
shown.
6
5
x
0
5
At x, the cross-sectional length
L(x) = 5.
The cross-section is (6 + x) to
the surface.
L(x) = 5
Fluid Pressures

26. 5
Place the x-measurement as
shown.
6
5
x
0
5
At x, the cross-sectional length
L(x) = 5.
The cross-section is (6 + x) to
the surface. So the force is:
L(x) = 5
ρ 5(6 + x) dx =∫x=0
5
Fluid Pressures

27. 5
Place the x-measurement as
shown.
6
5
x
0
5
At x, the cross-sectional length
L(x) = 5.
The cross-section is (6 + x) to
the surface. So the force is:
L(x) = 5
ρ 5(6 + x) dx = ρ(30x + 5x2/2)|∫x=0
5
x=0
5
Fluid Pressures

28. 5
Place the x-measurement as
shown.
6
5
x
0
5
At x, the cross-sectional length
L(x) = 5.
The cross-section is (6 + x) to
the surface. So the force is:
L(x) = 5
ρ 5(6 + x) dx = ρ(30x + 5x2/2)| = 212.5ρ∫x=0
5
x=0
5
Fluid Pressures

29. 5
Place the x-measurement as
shown.
6
5
x
0
5
At x, the cross-sectional length
L(x) = 5.
The cross-section is (6 + x) to
the surface. So the force is:
L(x) = 5
ρ 5(6 + x) dx = ρ(30x + 5x2/2)| = 212.5ρ∫x=0
5
x=0
5
Note: If we place the x differently as
shown,
6
5x
0
11
6
L(x) = 5
Fluid Pressures

30. 5
Place the x-measurement as
shown.
6
5
x
0
5
At x, the cross-sectional length
L(x) = 5.
The cross-section is (6 + x) to
the surface. So the force is:
L(x) = 5
ρ 5(6 + x) dx = ρ(30x + 5x2/2)| = 212.5ρ∫x=0
5
x=0
5
Note: If we place the x differently as
shown, the integral will be:
6
5x
0
11
6
ρ 5* x dx =∫x=6
11
L(x) = 5
Fluid Pressures

31. 5
Place the x-measurement as
shown.
6
5
x
0
5
At x, the cross-sectional length
L(x) = 5.
The cross-section is (6 + x) to
the surface. So the force is:
L(x) = 5
ρ 5(6 + x) dx = ρ(30x + 5x2/2)| = 212.5ρ∫x=0
5
x=0
5
Note: If we place the x differently as
shown, the integral will be:
6
5x
0
11
6
ρ 5* x dx = ρ5x2/2| = 212.5ρ∫x=6
11
x=6
11
L(x) = 5
Fluid Pressures

32. 9
Example B. Find the force exerted on a triangular
plate submerged in the fluid as shown.
10
12
0
Fluid Pressures

33. 9
x
Example B. Find the force exerted on a triangular
plate submerged in the fluid as shown.
Place the x-measurement from bottom to top.
10
12
0
x
L(x)=4x/3
Fluid Pressures

34. 9
x
Example B. Find the force exerted on a triangular
plate submerged in the fluid as shown.
Place the x-measurement from bottom to top.
10
12
0
x
L(x)=4x/3
The cross-sectional length L at x satisfies:
L
x =
12
9
Fluid Pressures

35. 9
x
Example B. Find the force exerted on a triangular
plate submerged in the fluid as shown.
Place the x-measurement from bottom to top.
10
12
0
x
L(x)=4x/3
The cross-sectional length L at x satisfies:
L
x =
12
9
so L = 4x
3
Fluid Pressures

36. 9
x
Example B. Find the force exerted on a triangular
plate submerged in the fluid as shown.
Place the x-measurement from bottom to top.
10
12
0
x
L(x)=4x/3
The cross-sectional length L at x satisfies:
L
x =
12
9
so L = 4x
3
It's distance to the top is (19 – x).
Fluid Pressures

37. 9
x
Example B. Find the force exerted on a triangular
plate submerged in the fluid as shown.
Place the x-measurement from bottom to top.
10
12
0
x
L(x)=4x/3
The cross-sectional length L at x satisfies:
L
x =
12
9
so L = 4x
3
It's distance to the top is (19 – x).
Hence the force on it is
ρ (19 – x) dx
= 702ρ
∫x=0
9
4x
3
Fluid Pressures

38. 9
x
Example B. Find the force exerted on a triangular
plate submerged in the fluid as shown.
Place the x-measurement from bottom to top.
10
12
0
x
L(x)=4x/3
The cross-sectional length L at x satisfies:
L
x =
12
9
so L = 4x
3
It's distance to the top is (19 – x).
Hence the force on it is
ρ (19 – x) dx
= 702ρ
∫x=0
9
4x
3
Fluid Pressures

39. Example C. Find the amount of water force on a
semi–disc plate with 10-ft radius as shown.
10
Fluid Pressures

40. Place the x-axis as shown.
0
x
10
Example C. Find the amount of water force on a
semi–disc plate with 10-ft radius as shown.
Fluid Pressures

41. Fluid Pressure
Place the x-axis as shown.
0
x
10
100 – x2
Example C. Find the amount of water force on a
semi–disc plate with 10-ft radius as shown.
Fluid Pressures

42. Place the x-axis as shown.
0
x
10
100 – x2
Example C. Find the amount of water force on a
semi–disc plate with 10-ft radius as shown.
So the cross-sectional length at x is 2100 – x2.
Fluid Pressures

43. Place the x-axis as shown.
0
x
10
100 – x2
It's distance to the surface is x.
Example C. Find the amount of water force on a
semi–disc plate with 10-ft radius as shown.
So the cross-sectional length at x is 2100 – x2.
Fluid Pressures

44. Place the x-axis as shown.
0
x
10
100 – x2
It's distance to the surface is x.
Therefore the integral for the
fluid force is:
Example C. Find the amount of water force on a
semi–disc plate with 10-ft radius as shown.
So the cross-sectional length at x is 2100 – x2.
Fluid Pressures

45. Place the x-axis as shown.
0
x
10
100 – x2
It's distance to the surface is x.
Therefore the integral for the
fluid force is:
ρ 2x100 – x2 dx∫x=0
10
Example C. Find the amount of water force on a
semi–disc plate with 10-ft radius as shown.
So the cross-sectional length at x is 2100 – x2.
Fluid Pressures

46. Place the x-axis as shown.
0
x
10
100 – x2
It's distance to the surface is x.
Therefore the integral for the
fluid force is:
ρ 2x100 – x2 dx∫x=0
10
Use substitution;
set u = (100 – x2)
Example C. Find the amount of water force on a
semi–disc plate with 10-ft radius as shown.
So the cross-sectional length at x is 2100 – x2.
Fluid Pressures

47. Place the x-axis as shown.
0
x
10
100 – x2
It's distance to the surface is x.
Therefore the integral for the
fluid force is:
ρ 2x100 – x2 dx
= – (100 – x2)3/2|
∫x=0
10
2ρ
3 x=0
10
Use substitution;
set u = (100 – x2)
Example C. Find the amount of water force on a
semi–disc plate with 10-ft radius as shown.
So the cross-sectional length at x is 2100 – x2.
Fluid Pressures

48. Place the x-axis as shown.
0
x
10
100 – x2
It's distance to the surface is x.
Therefore the integral for the
fluid force is:
ρ 2x100 – x2 dx
= – (100 – x2)3/2|
= 2000ρ/3 = 41,600 lb
∫x=0
10
2ρ
3 x=0
10
Use substitution;
set u = (100 – x2)
Example C. Find the amount of water force on a
semi–disc plate with 10-ft radius as shown.
So the cross-sectional length at x is 2100 – x2.
Fluid Pressures