10 fluid pressures x

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

Let A = Area of the plate
h = the depth
ρ = the weight of the fluid A
h
ρ
Fluid Pressures

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

h = the depth
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

h = the depth
A
h
ρ
of the fluid that sits over the plate, that is
F = ρAh
Fluid Pressures

h = the depth
A
h
ρ
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

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

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

to the surface.
h
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

to the surface.
h
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.
points.
Fluid Pressures
L(xi)

to the surface.
h
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.
points.
Let Δx = the length of each subinterval.
Fluid Pressures
L(xi)
Δx

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

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

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

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

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

∑ 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

∑ 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

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

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

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

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

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

5
shown.
6
5
x
0
5
L(x) = 5.
The cross-section is (6 + x) to
the surface.
L(x) = 5
Fluid Pressures

5
shown.
6
5
x
0
5
L(x) = 5.
the surface. So the force is:
L(x) = 5
ρ 5(6 + x) dx =∫x=0
5
Fluid Pressures

5
shown.
6
5
x
0
5
L(x) = 5.
L(x) = 5
ρ 5(6 + x) dx = ρ(30x + 5x2/2)|∫x=0
5
x=0
5
Fluid Pressures

5
shown.
6
5
x
0
5
L(x) = 5.
L(x) = 5
ρ 5(6 + x) dx = ρ(30x + 5x2/2)| = 212.5ρ∫x=0
5
x=0
5
Fluid Pressures

5
shown.
6
5
x
0
5
L(x) = 5.
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

5
shown.
6
5
x
0
5
L(x) = 5.
L(x) = 5
ρ 5(6 + x) dx = ρ(30x + 5x2/2)| = 212.5ρ∫x=0
5
x=0
5
shown, the integral will be:
6
5x
0
11
6
ρ 5* x dx =∫x=6
11
L(x) = 5
Fluid Pressures

5
shown.
6
5
x
0
5
L(x) = 5.
L(x) = 5
ρ 5(6 + x) dx = ρ(30x + 5x2/2)| = 212.5ρ∫x=0
5
x=0
5
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

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

9
x
Place the x-measurement from bottom to top.
10
12
0
x
L(x)=4x/3
Fluid Pressures

9
x
10
12
0
x
L(x)=4x/3
The cross-sectional length L at x satisfies:
L
x =
12
9
Fluid Pressures

9
x
10
12
0
x
L(x)=4x/3
L
x =
12
9
so L = 4x
3
Fluid Pressures

9
x
10
12
0
x
L(x)=4x/3
L
x =
12
9
so L = 4x
3
It's distance to the top is (19 – x).
Fluid Pressures

9
x
10
12
0
x
L(x)=4x/3
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

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

Place the x-axis as shown.
0
x
10
Fluid Pressures

Fluid Pressure
0
x
10
100 – x2
Fluid Pressures

0
x
10
100 – x2
So the cross-sectional length at x is 2100 – x2.
Fluid Pressures

0
x
10
100 – x2
It's distance to the surface is x.
Fluid Pressures

0
x
10
100 – x2
Therefore the integral for the
fluid force is:
Fluid Pressures

0
x
10
100 – x2
fluid force is:
ρ 2x100 – x2 dx∫x=0
10
Fluid Pressures

0
x
10
100 – x2
fluid force is:
ρ 2x100 – x2 dx∫x=0
10
Use substitution;
set u = (100 – x2)
Fluid Pressures

0
x
10
100 – x2
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)
Fluid Pressures

0
x
10
100 – x2
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)
Fluid Pressures

10 fluid pressures x

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