Calculating Fluid Force on a Circular Pressure Release Gate in a Dam

In summary, a circular "pressure release gate" is located in the vertical face of a dam. The radius of the gate is 2 metres and the gate is positioned so that its top lies 5 metres below the surface of the water held in the reservoir behind the dam. The total fluid force acting on the door of the gate, when it is in a fully closed position, is F=pg∫2\sqrt{4-x^2}(5+x)dyF.
  • #1
Nyasha
127
0

Homework Statement


A circular "pressure release gate" is located in the vertical face of a dam. The radius of the gate is 2 metres and the gate is positioned so that its top lies 5 metres below the surface of the water held in the reservoir behind the dam.Find the total fluid force acting on the door of the gate, when it is in a fully closed position.


Homework Equations


Area= [tex]2\sqrt{4-x^2}[/tex] dy

Pressure = pg(5+x)

The Attempt at a Solution


F= pg∫ [tex]2\sqrt{4-x^2}[/tex] (5+x) dy
 
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  • #2
Nyasha said:
F= pg∫ [tex]2\sqrt{4-x^2}[/tex] (5+x) dy

Hi Nyasha! :smile:

(have a square-root: √ and a rho: ρ :wink:)

almost right :smile:, but the area of the slice is 2√(4 - y2) dy, at depth 5+y :wink:
 
  • #3
I don't see your point, tiny-tim. There is no coordinate system given with the problem. It perfectly legitimate to set up a coordinate system in which x is vertical.

[tex]pg\int_0^2 2\sqrt{4- x^2}(5+x)dx[/tex] is exactly the same as [tex]pg\int_0^2 2\sqrt{4- y^2}(5+y)dy[/tex]
 
  • #4
HallsofIvy said:
I don't see your point, tiny-tim. There is no coordinate system given with the problem. It perfectly legitimate to set up a coordinate system in which x is vertical.

[tex]pg\int_0^2 2\sqrt{4- x^2}(5+x)dx[/tex] is exactly the same as [tex]pg\int_0^2 2\sqrt{4- y^2}(5+y)dy[/tex]

erm … the point is you can't use x and y at the same time! :wink:

(i assume that's why Nyasha was puzzled about how to integrate)
 
  • #5
HallsofIvy said:
I don't see your point, tiny-tim. There is no coordinate system given with the problem. It perfectly legitimate to set up a coordinate system in which x is vertical.

[tex]pg\int_0^2 2\sqrt{4- x^2}(5+x)dx[/tex] is exactly the same as [tex]pg\int_0^2 2\sqrt{4- y^2}(5+y)dy[/tex]

If my integration is correct then how do l further simplify this integration using the area of a disk of radius "R" where x=Rsinθ and dx=Rcosθ and a simple symmetry argument
 
  • #6
Nyasha said:
If my integration is correct then how do l further simplify this integration using the area of a disk of radius "R" where x=Rsinθ and dx=Rcosθ and a simple symmetry argument

Make the substitution and see!

What do you get? :smile:
 
  • #7
tiny-tim said:
Make the substitution and see!

What do you get? :smile:

I get :


[tex]
pg\int_0^2 2\sqrt{4- R^2sin\theta^2}(5+Rsin\theta)Rcosd\theta
[/tex]


From here l don't know how to further simplify this
 
  • #8
Nyasha said:
[tex]
pg\int_0^2 2\sqrt{4- R^2sin\theta^2}(5+Rsin\theta)Rcosd\theta
[/tex]

… and now put R = 2 … :rolleyes:

(oh, and you haven't substituted the limits)
 
  • #9
tiny-tim said:
… and now put R = 2 … :rolleyes:

(oh, and you haven't substituted the limits)


[tex]

pg\int_0^2 2\sqrt{4- 4sin\theta^2}(5+2sin\theta)2cosd\theta

[/tex]


I am not supposed to multiple this integral by two using the symmetry argument
 
  • #10
Nyasha said:
[tex]

pg\int_0^2 2\sqrt{4- 4sin\theta^2}(5+2sin\theta)2cosd\theta

[/tex]


I am not supposed to multiple this integral by two using the symmetry argument

Nyasha, you still haven't changed the limits of integration …

and what is √(4 - 4sin2θ) ?
 
  • #11
tiny-tim said:
Nyasha, you still haven't changed the limits of integration …

and what is √(4 - 4sin2θ) ?


[tex]


pg\int_0^4 2\sqrt{4- 4sin\theta^2}(5+2sin\theta)2cosd\theta


[/tex]


What am l supposed to do with √(4 - 4sin2θ) ? I'm l supposed to use some trig identity
 
  • #12
If x = 2sinθ, and x goes from 0 to 2, what does θ go from?

And you must learn your trignonometric identities … what is 1 - sin2θ?
 
  • #13
tiny-tim said:
If x = 2sinθ, and x goes from 0 to 2, what does θ go from?

And you must learn your trignonometric identities … what is 1 - sin2θ?


[tex]


pg\int_0^2 {4\cos\theta}(5+2sin\theta)2cosd\theta


[/tex]


The limit of the integral is from 0 to π/2. I don't know how to put it into the integral using Latex source code.


[tex]

pg\int_0^2 ({20\cos\theta +8\cos\theta\sin\theta})2cosd\theta



[/tex]

Is [tex] 8\cos\theta\sin\theta} [/tex] = [tex] sin(8\theta) ?
[/tex]
 
  • #14
nyasha1 said:
[tex]


pg\int_0^2 {4\cos\theta}(5+2sin\theta)2cosd\theta


[/tex]


The limit of the integral is from 0 to π/2. I don't know how to put it into the integral using Latex source code.


[tex]

pg\int_0^2 ({20\cos\theta +8\cos\theta\sin\theta})2cosd\theta



[/tex]

Is [tex] 8\cos\theta\sin\theta} [/tex] = [tex] sin(8\theta) ?
[/tex]

Hi nyasha! :smile:

(you can put π/2 into the integral with ^{\pi /2} :wink:)

You don't like undoing those brackets, do you? :rolleyes:

It's a combination of cos2θ and cos2θsinθ.

No, sin2θ = 2sinθcosθ, but it doesn't work for sin8θ.

btw, going back to the original integral, it would have been simpler to just integrate x√(4 - x2) and √(4 - x2) separately, without substituting …

and shouldn't the limits have been -2 to +2? :smile:
 
  • #15
tiny-tim said:
Hi nyasha! :smile:

(you can put π/2 into the integral with ^{\pi /2} :wink:)

You don't like undoing those brackets, do you? :rolleyes:

It's a combination of cos2θ and cos2θsinθ.

No, sin2θ = 2sinθcosθ, but it doesn't work for sin8θ.

btw, going back to the original integral, it would have been simpler to just integrate x√(4 - x2) and √(4 - x2) separately, without substituting …

and shouldn't the limits have been -2 to +2? :smile:

If we integrate x√(4 - x2) and √(4 - x2) separately then do we need x=Rsinθ. The question says in order to evaluate the integral l will need the use the formula of disk of radius "R" where x=Rsinθ.

How did you get your limits to be -2 to +2 ?

[tex] pg\int_0^{\pi /2} {4\cos\theta}(5+2sin\theta)2cosd\theta



[/tex]Is this integral correct ?
 
Last edited:
  • #16
FluidForce.jpg


According to the solutions manual when l evaluate one of my integrals l should get zero. What am l doing wrong on this question ? Please help.
 
  • #17
Hi Nyasha! :smile:

Your integral (from 0 to 2) is only over the top half of the gate …

you need to integrate from -2 to 2.
 
  • #18
tiny-tim said:
Hi Nyasha! :smile:

Your integral (from 0 to 2) is only over the top half of the gate …

you need to integrate from -2 to 2.


I used the symmetry argument so that l could integrate from 0 to 2 and then multiply it by two. Is that correct ?
 
  • #19
Well, it would be … if it wasn't for the (5 + y).

(in physical terms: the pressure on the bottom half of the gate is obviously more than on the top half)
 
  • #20
tiny-tim said:
Well, it would be … if it wasn't for the (5 + y).

(in physical terms: the pressure on the bottom half of the gate is obviously more than on the top half)


Thanks very much tiny-tim it now makes perfect sense.
 

Related to Calculating Fluid Force on a Circular Pressure Release Gate in a Dam

1. What is the definition of fluid force in Calculus 2?

In Calculus 2, fluid force refers to the force exerted by a fluid on a surface, which is determined by the density of the fluid, the area of the surface, and the depth of the fluid.

2. How is fluid force calculated in Calculus 2?

The formula for calculating fluid force in Calculus 2 is F = ρghA, where F is the fluid force, ρ is the density of the fluid, g is the acceleration due to gravity, h is the depth of the fluid, and A is the area of the surface.

3. What is the difference between static and dynamic fluid force in Calculus 2?

Static fluid force refers to the force exerted by a fluid on a stationary surface, while dynamic fluid force refers to the force exerted by a moving fluid on a surface.

4. How is fluid force applied in real-world scenarios?

Fluid force is applied in many real-world scenarios, such as calculating the pressure on a dam due to water, determining the force on a boat's hull due to water, and calculating the lift force on an airplane wing due to air.

5. What are some applications of fluid force in engineering?

Fluid force has many applications in engineering, including designing structures such as dams and bridges to withstand the force of water, creating efficient and aerodynamic shapes for vehicles, and understanding the force distribution on airplane wings to improve flight performance.

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