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Homework Help: Need help in relating force needed to drive mass of water.

  1. Dec 7, 2011 #1
    1. Ok I have to build a real life wave-maker for my Final Year Project(FYP). My wave-maker design is as shown below:
    http://img696.imageshack.us/img696/5888/drawntank.png [Broken]
    The proffessor said I need to be able to calculate the torque
    of the motor before I order it. I start out by calculating the force needed to move the water using the flap paddle.The equation I have applied is:

    Water's Volume = length*breadth*height = (1.23)*(0.57)*(0.45) = 0.3155 meter^3

    The torque is
    t = r*f*sin(theta)

    r = 0.35m

    theta = 10 degree

    f = ??

    I know Force = m*a

    m = water's volume in this case = 0.3155 meter^3

    but what is the acceleration, "a" here though??

    The spring is there to keep the flap paddle in neutral positon when the water is quiet.

    So my question is: How would I determine the minimum force that I need to drive the paddle? And how would I go from there to determine the torque that my motor would need to move the paddle??

    Sorry if this seems a bit messy, first time poster:redface:
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Dec 7, 2011 #2
    When the flap rotates clockwise is it compressing the spring and has to counteract the force of the water on the flap? Where is the centroid on the flap? Do you know the spring constant?
  4. Dec 7, 2011 #3
  5. Dec 7, 2011 #4
    Hi R, sorry for the late reply, it's been q long day for me. Ok, let's get back to your questions. Yeah as you can see the motor will have to have to rotate to pull the string, which in turn pull the paddle in the purpple arrow direction, so it will have to go against the spring. The spring constant is 6094N/m. I don't get what you mean by the flap's centroid? If you mean the point of rotation, it's the tip of the paddle(the inverted green triangle) which is hinged to the brown block.

    As for the motor, I don't really know either because I haven't really worked out the power needed to push the flap paddle.
  6. Feb 20, 2012 #5
    Hey RTW69 so you have any idea??
  7. Feb 20, 2012 #6


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    Staff: Mentor

    It's a good question, and I'm none the wiser than you. But to a first approximation, you might be able to say that the effect of the nodding paddle is to lift that displaced volume of water into a rectangular (?) shape atop the mean water level.
  8. Feb 21, 2012 #7
    Well, in this case, the 0.315^3 = 315 litres of water, which means it has 315 kilograms (693 lbs), if we have to lift it vertically upwards, then we're gonna need at least 693 lbs??
  9. Feb 21, 2012 #8


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    Staff: Mentor

    It doesn't merely support that weight of water. It accelerates it up and to the right. The faster the paddle moves, the taller the wave.
  10. Feb 21, 2012 #9
    Please ignore this post...
    Last edited: Feb 21, 2012
  11. Feb 21, 2012 #10
    Although the paddle is being pulled in the upright direction, the paddle itself actually moves in more on the x-axis. Here's a video to help visualize the thing in motion:

    I remember doing system modelling in 2nd year, and I think the equations are much more complicated than those i listed above.

    Basic equations I've got so far is:
    Fp = Force required to push paddle (or paddle's pushing force)
    Fw = water force acting on the paddle (when it's still)
    k = spring constant
    x1 = spring displacement


    Fp = -kx + Fw

    One of the guy from last year told me the force acting on the paddle is about 60 kg (no idea how he got that) so let's just say Fw = 60kg

    Since Fp will require a motor to drive it, I'll have to use the equation:
    T(Torque) = J[d^2(theta)/dt^2]

    and J = ??

    Now we have:

    J[d^2(theta)/dt^2] = -kx + 60kg

    And I'm kinda lost here...like how would I get "J" ?? and what kind of spring constant I need??
    Last edited by a moderator: Sep 25, 2014
  12. Feb 21, 2012 #11


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    Staff: Mentor

    Does the spring contribute any to the shape of the wave? The trailing edge, maybe? Otherwise, it may do no more than just reset the paddle ready for the next wave. If you are happy for the pressure of water to push the paddle back to its rest position, then maybe you don't need a very strong spring.
  13. Feb 22, 2012 #12
    Sorry for the late reply man, I was bogged down by all the lab questions yesterday. Anyway back to the topic, I certainly hope that the spring wouldn't contribute to the wave since it's first purpose was just there to make sure the paddle doesn't fall all the way back since there's enough space for the water to push it all the way to the left (which i dun want). In still water level, the water would still have force acting on the paddle, but the spring would have to hold it in a vertical position, 90 degrees (more or less) as opposed to the water level.
  14. Feb 25, 2012 #13


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    Staff: Mentor

    Sounds dangerous to me! The spring is held partly extended, and as it ages and stretches the paddle will begin to lie over to the left? Or a reflected wave could push it to the left, and allow water to spill out? Why not just use a rope, then you could be certain that the paddle will return to the vertical. Or place a strong compression spring on the left side of the paddle to act as a resilient stop.
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