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Hydrostatic forces on a gate with no width?

  1. Aug 1, 2013 #1
    Hydrostatic forces on a gate with no width??

    1. The problem statement, all variables and given/known data
    A gate is placed at the bottom of a dam wall at a distance of "H" below the surface (measured from the top of the gate). The gate is hinged so that there is 1.2m above the pivot point and 0.8m below the pivot. A stop is placed at the bottom of the gate to stop it from rotating. As the height of the water increases the force on the upper section of the gate will increase and eventually open the gate. Calculate the size of "H" in order for the gate to open.

    *no gate width was given
    *fluid in question is water


    2. Relevant equations
    F(resultant)=γ.yc.A
    F(position)=I/(yc.A)+yc


    3. The attempt at a solution
    I know the solution for this question is 0.666m from the answers. So far I have used the resultant force equation to calculate the force for both the top and bottom section of the gate and then attempted to find the force position using the second equation before finally putting this information into a moment equation about the pivot point. My only problem is that the only examples i can find use a gate width and so far using this method I haven't been able to come to answer that matches. In both of the equations I have listed I have attempted to substitute length of the gate for area, however, obviously this doesn't work. Any help would be great, thanks!
     
  2. jcsd
  3. Aug 1, 2013 #2

    tiny-tim

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    hi veryc0nfused! :smile:
    just call the width "w", and write out all the equations …

    you'll find the w will cancel out at the end! :wink:

    show us what you get :smile:
     
  4. Aug 1, 2013 #3
    Okay, still not managing to get the correct answer no matter what I do, hoping someone can give me a hand?

    My current method involves breaking the top and bottom section of the gate into two areas and considering them separately. Therefore;

    Force (top) = 9810(H+0.6)(1.2w)
    Force (bot) = 9810(H+1.2)(0.8w)

    Meaning the position of these forces are;

    Position (top) = [0.144/(1.2H+0.72)]+(H+0.6)
    Position (bot) = [0.043/(0.8H+1.28)]+(H+1.6)

    I then attempted to substitute these equations into the moment about the pivot using the distance for each force as:

    Distance (top) = (H+1.2)-position
    Distance (bot) = (H+2)-position

    However, doing this created a huge equation and the final result did not match in the end. I'm assuming there must be an easier way of going about this but just can't seem to find it?
     
  5. Aug 1, 2013 #4

    tiny-tim

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    before we go any further …

    i don't understand how this gate works :confused:
     
  6. Aug 1, 2013 #5

    mfb

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    Don't forget the units. How did you get 0.6 (which is right)?
    The 1.2 here is wrong.

    Where do all those numbers come from? Why do you multiply H with something?
    It would be useful if you start with formulas, and then plug in numbers. This makes it easier to understand what you calculate.

    Hmm, force depends on height... do you know integrals?


    @tiny-tim: The upper part can go outwards, pushing the lower part inwards - if the torque on the upper part exceeds the torque on the lower part (in magnitude).
     
  7. Aug 1, 2013 #6
    Okay, realised I was going about it all wrong. I ended up solving the question (I think but very open to criticism) by considering the gate as one entire area. By doing this all i had to do was consider the position of the resultant force and make sure that it would be above the pivot point. Thoughts?
     
  8. Aug 1, 2013 #7

    tiny-tim

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    sorry, still don't understand :confused:

    since the gate is totally submerged, what difference does it make?
    yes, you can certainly do that :smile:
     
  9. Aug 1, 2013 #8

    mfb

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    That is a good approach.

    The gate is "in" the dam - one side is not in the water. The height-dependent pressure depends on the height above the water - as the areas are not equal, the total torque depends on the height of the water level.
     
  10. Aug 1, 2013 #9
    You need to do integrations to get the moments on the parts of the gate above and below the pivot. Let x be the distance measured above the pivot, and y be the distance measured down from the pivot. The pressure in the region above the pivot is:
    [tex]p=ρg(H+1.2-x)[/tex]
    The pressure in the region below the pivot is:
    [tex]p=ρg(H+1.2+y)[/tex]

    The differential force on the region of the gate above the pivot between x and x + dx is:
    [tex]dF=wρg(H+1.2-x)dx[/tex]
    where w is the width of the gate.
    The differential moment about the pivot in this same region is:
    [tex]dM=wρg(H+1.2-x)xdx[/tex]
    Similarly, for the region below the pivot,
    [tex]dM=wρg(H+1.2+y)ydy[/tex]
    The moment equation above the pivot has to be integrated between x = 0 and x = 1.2. The moment equation below the pivot has to be integrated between y = 0 and y = 0.8.
    You then set the moments for the two regions equal to one another.

    chet
     
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