Force required to produce wave in tank

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The discussion revolves around calculating the force required to move a flap in a wave-producing tank using either a pneumatic system or a motor and cam system. The user has calculated the mass of displaced water and the height when the flap is extended, leading to a force calculation of 4602.4 kN. There are concerns about the accuracy of the calculations, particularly regarding the application of force at different heights and the use of relevant equations. The user is also considering the work-energy theorem and how to incorporate pressure calculations into their analysis. Clarifications and feedback on these calculations are sought to ensure a correct understanding of the physics involved.
Whatamiat
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Homework Statement


I am building a tank to produce waves.
I have two options to power the flap that will be produced to create the waves:

1. Pneumatic system
2. Motor and Cam system

Before I can order motors or pneumatic air cylinders I need to know what force will be required to move the flap the required distance to produce the waves.

Sketches of both systems are attached
Sketch of problem is attached

Homework Equations



Fxd = mgh

P= FxA ?

The Attempt at a Solution



From attached images:

Mass of Water = 7500kg

Finding Distance:
d = tan 18 x 1 = 32.5cm

Finding mass of displaced water:
d x h x w = .325 x 1 x .75 = 243.75kg
but we are dealing with triangle so
mass of displaced water =243.75/2 = 121.875kg of water displaced

Calculating Height of water when flap is extended:
Mass of water = length of tank x width of tank x height of tank
7.5 = (10-.1625) x .75 x h
h = 1.0165 cm

Calculating Force:
d = 32.5/2 = 16.25 (as we are dealing with triangle not rectangle)

F x 16.25 = 7500x9.81x1.0165

F = 4602.4kN

My Thoughts
Very sure I am missing something fundamental with this solution but not sure what!

I feel I have ignored the fact that the force is being applied at 1.25m not 1m but don't know how to incorporate this.

Im also unsure if I have used the wrong figures for my Fxd=mgh solution.

I feel P=F/A can be incorporated but not sure how?

Thanks for any feedback!
 

Attachments

  • wave generater pneumatic.JPG
    wave generater pneumatic.JPG
    16.1 KB · Views: 462
  • wave generater.JPG
    wave generater.JPG
    18.3 KB · Views: 406
  • detailed sketch.JPG
    detailed sketch.JPG
    15.6 KB · Views: 515
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Hi Whatamiat! :smile:

Are you using the work-energy theorem? (it would have been clearer if you'd said so)

If so, in the formula work done = force "dot" displacement, the displacement is the displacement of the point of application of the force. :wink:
 
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