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Hydrostatics question - angled triangular gate?

  1. Mar 24, 2013 #1
    1. The problem statement, all variables and given/known data
    The question asks to find the magnitude of the pressure force on the gate:

    3. The attempt at a solution
    Here is the part of the solution:

    I don't get why zc is 3.667m. Shouldn't it be 3 + (2/3)sin40 since you want the depth of the centroid from the surface to the gate?

    Also, another way to solve this would be to find the horizontal force and vertical forces. This would mean that I would have to for the weight of the fluid under the gate in order to find the vertical force right?
  2. jcsd
  3. Mar 24, 2013 #2


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    If you draw the location of the centroid of the triangular gate on the diagram and work out the trigonometry, you will see that the location of the centroid in the problem statement is correct. Remember, the length of the gate from A to B is 2 / sin 50. You then draw 2/3 of this distance from B on the Front View. Working back from this point to the surface, you will see zc is 3.667 m.
  4. Mar 24, 2013 #3
    Ah that makes a lot of sense, thank you.

    So I decided to try the other method I mentioned where you take the magnitude of the Fhorizontal and Fvertical. However, my answer is off by around 3000N (the answer key gives 39000N). Do you guys know what is wrong?

  5. Mar 24, 2013 #4
    Just check the vertical component of the hydrostatic-pressure force. I'm getting it near 25 kN.
  6. Mar 24, 2013 #5
    Hmm... First off, am I right about the fact that there are only two vertical force - weight of the water and the projected vertical upwards force?

    Also, I wasn't sure if I should include the weight of the gate in the vertical force calculations... Even if I did the answer isn't 25kN.
  7. Mar 24, 2013 #6
    Yes those are the only vertical forces.

    Now excluding the weight, net force due to hydrostatic pressure*sin40 is the vertical component.
  8. Mar 24, 2013 #7
    Oh wow that is a very smart way to do it. So essentially you don't even solve for the vertical component, you wrote it in terms of the resultant force?

    But in case I wanted to solve for it using the method I currently have written, how would I do it? Because if the gate was curve like a circle, your method will not work anymore right?
  9. Mar 24, 2013 #8
    My method is valid for any shape as long as you can calculate the area because net force is equal to pressure times area of the surface and is perpendicular to the surface.

    Could you please explain how you arrived at the vertical component as I did not follow what [itex]y_{bot}[/itex] meant.
  10. Mar 24, 2013 #9
    Right, but for a curved shape, you wouldn't know what the angle was so just finding the horizontal force would not be enough right? Because in this example, we know that the surface is flat so we can use Fvertical = Fresultant * sin40 for the vertical force and then

    Fresultant2 = Fhorizontal2 + Fvertical2

    To solve for the resultant force.

    The method I used comes from a sample question in the textbook:

    In the solution, for the part where they solve for the vertical force, we see that they take the pressure at the very bottom.
  11. Mar 25, 2013 #10
    Does anyone know why this method does not work :O
  12. Mar 25, 2013 #11
    Thanks for showing me this method. I am seeing this for the first time. I usually employ integration to solve in case the gate is curved because that method can be used irrespective of the shape ( usually the integrals are solvable). I still don't get why they take the pressure at the bottom to find the vertical component of force though:( Could you please show how these formulae were derived or could you give a link to a site which shows the derivation?
    Last edited: Mar 25, 2013
  13. Mar 25, 2013 #12
    Unfortunately there is no derivation in the book.

    As for the integration method, how can you use integration for the curved circular gate example from the textbook?
  14. Mar 25, 2013 #13
    Check out the attachment for the diagram.

    Now consider a small element of area along the lateral surface of the cylinder with subtends an angle dα at the center.

    Pressure at this depth is ρg(h+x) and the force on the cylinder per unit length at this depth is simply the product of area = Rdα*unit length and is normal to the area i.e. directed to the center. It's x and y components can be found and separately integrated over required limits to get the total x and y components of force.

    Though I cannot find any mistake in my argument, the integral misses the textbook's answer.
    the X component that I get is ρgR(h+R/4) while the textbook says ρgR(h+R/2). There are similar discrepancies with the Y component. I shall check my integrals again tomorrow and tell you if I get anything.

    Attached Files:

  15. Mar 26, 2013 #14
    I am no expert. The y_component of force when calculated using integrals doesn't match the book's answer but x-component of force does.

    The pressure at height h+Rsinα is P=ρg(h+Rsinα), the force acting on area dA is dF=PdA, where dA=Rdα. The direction of this force is radially towards the centre.
    [tex]dF_x=dF\cos \alpha=ρg(h+Rsinα)R\cos \alpha d\alpha[/tex]
    Integrating the expression within the limits 0 to pi/2
    [tex]F_x= ρgR\left(h+\frac{R}{2}\right)[/tex]
    which matches with the book's answer. However,
    [tex]dF_y=dF\sin \alpha=ρg(h+Rsinα)R\sin \alpha d\alpha[/tex]
    [tex]F_y=ρgR\left(h+\frac{\pi R}{4}\right)[/tex]

    I would like an expert to comment on this.
  16. Mar 26, 2013 #15
    My bad. Thanks Pranav-Arora I rechecked my integrals. It matches with what you obtained.
  17. Mar 27, 2013 #16


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    I'm no expert either, but your calculation looks correct to me. It agrees with the solution in the figure in post #9 where the vertical force on the cylinder would be ##F_V = F_y - W##. Here, ##F_y## is the vertical force on the horizontal bottom surface of the volume of water below the cylinder as shown in the figure and ##W## is the weight of that water.
  18. Mar 27, 2013 #17
    Thanks TSny,

    According to our calculations, the vertical force due to the water would be 37.892 kN while the textbook says 39.2 kN. Though the difference is small, the answers still vary because of the methods used to obtain them.

    So, my question is whether the textbook's method is an approximation. If so, how far is it valid or is there a limit after which the approximation fails?
  19. Mar 27, 2013 #18


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    I believe the 37.892 kN is the correct answer for the vertical force on the cylinder; whereas, the 39.2 kN is the correct answer for the vertical force on the horizontal bottom surface of the water that lies beneath the cylinder. The difference of these equals the weight of the volume of water that lies vertically beneath the cylinder.
  20. Mar 27, 2013 #19
    I have a few doubts,

    1. So are (referring to the diagram), Fv=37.892 kN and Fy=39.2 kN?
    2.To find the weight per unit length of the cylinder, shouldn't Fv=37.892 kN be used as it is the vertical force component on the cylinder?
  21. Mar 27, 2013 #20


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    In going to part (b) where you want to find the weight per unit length of the cylinder, you should think of a free body diagram for the cylinder. Note that there will be horizontal and vertical components of "reaction" force at the hinge acting on the cylinder. These would have to be included in summing the forces in the horizontal and vertical directions. I think torque would be the way to go for this part.
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