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Finding resultant force on a segment of submerged wall

  1. Aug 18, 2015 #1
    • Member warned about posting with no effort and no template
    Please see the attached image. I understand the area represents the resultant force per unit width. But I'm unsure what equation to come up with for the purple area. I'd appreciate a little patience here, it's been ages since I did this.

    Thanks
     

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  2. jcsd
  3. Aug 18, 2015 #2

    RUber

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    What are you using to determine the width of the triangle? Is that just some constant times the depth, essentially the pressure from the column of water?
    Does the column have volume or are you just working with 2D? I don't know what physical constants to use, so I will call it C.
    Your triangle is defined by one side which is 125mm high, and 125Cmm wide. The width at any given height h is C(125-h). You could either use calculus or geometry to find out the area contained between h = 0 and h= 10mm.
     
  4. Aug 18, 2015 #3
    Thanks for the response, sorry therre wasn't enough info. This is 3 dimensional. Needing to now how much force 2 litres of water will put on the highlighted wall of 10mm with a full water body height of 125mm.
     

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  5. Aug 18, 2015 #4

    SteamKing

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    According to Pascal's Law, hydrostatic pressure is proportional to the depth of the fluid:

    https://en.wikipedia.org/wiki/Pascal's_law

    or, mathematically speaking:

    p = ρ g h

    where

    p = pressure in pascals (1 pascal = 1 N / m2)

    ρ = density of the fluid (ρ water = 1000 kg / m3)

    g = acceleration due to gravity, 9.81 m/s2

    h = height of fluid above point of interest, in meters

    The little doo-dad shown in the second picture is probably circular, with a diameter of 10 mm. Since pressure changes with depth, you'll have to integrate p over the circular area of this plug.
     
  6. Aug 18, 2015 #5
    What do you mean with integrate over the circular area? It's been too long since I did integration. Let's say the area of that 10 mm wall was 3x10^-4 m^2, could you guide me on calculating the approximate force applied aganst it form the water? -I'd really appreciate the help. I'm trying to manage too much at the moment.

    Attached is a simplified model of the part.
     

    Attached Files:

    Last edited: Aug 18, 2015
  7. Aug 18, 2015 #6
    ...Is there anybody out there?
     
  8. Aug 18, 2015 #7

    RUber

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    I think your first diagram may have been sufficient. The question is, what are you going to use as your constant?
    If you have 2L = 2000 cm^3 of water, in something 125mm=12.5 cm high, you can assume that the cross-sectional area of the column is 160cm^2.

    As SteamKing mentioned, gravity is 9.81 m/sec^2 and the density of water is 1 g/cm^3, so you still return to
    P(h) = C (125-h), where C is the pressure per unit height of the column.

    You are looking for the pressure from h = 0 to h = 10.
    ##\int_0^{10} C(125-h) = 125(10) - \frac{10^2}{2}##
    Or, with geometry,
    The width of the triangle at h= 10 is 115C.
    The width of the triangle at h = 0 is 125C.
    The rectangle with dimensions 10mm x 115Cmm has area ...
    The triangle with height 10mm and base length 10C has area...
    Total area = area of rectangle + area of triangle.

    Both approaches lead to the same result. So...what is C?
     
  9. Aug 18, 2015 #8
    thanks for getting back to me, much appreciated! With regards to C how do I find this out?
     
  10. Aug 18, 2015 #9

    RUber

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    For the physical constant, check out what SteamKing posted. It will be based on the volume per unit height of the column, Gravity, and the density of water. You should have all those values.
     
  11. Aug 23, 2015 #10
    Why am I not understanding this?

    I've done an engineering degree and this isn't making sense. Could someone please just take me through to the answer step by step for the new dimensions attached. I'd greatly appreciate the help.
     

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  12. Aug 23, 2015 #11
    How can it be that nobody is clear on this. I've had other people look at this and they are telling me different. Please someone clarify this.
     
  13. Aug 23, 2015 #12

    SteamKing

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    Well, what are these other people telling you? How do you know they aren't wrong?

    I think the replies you have been given are clear, but as you yourself asked, "Why am I not understanding this?"

    We can only do so much in the HW forums without violating the rules. Providing a step-by-step solution is a Bright Red Violation of those rules.

    Now, if you were to provide your own attempt at calculating what the hydrostatic force is, then you'll get comments on whether you are taking the correct approach or not.
     
  14. Aug 23, 2015 #13
    Ah OK, I didn't know that was the rule. Confusion comes from being advised to just use the normal hydrostatic equation - density x gravity x height and multiplying it with the area for force. Then I'm looking online and seeing hydrostatic force being calculated differently (seemingly) taking a resultant force from a triangle.

    If someone can just use the image I posted and take me through it with and explanation and a solution, it will make this less confusing.
     
  15. Aug 23, 2015 #14

    RUber

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    The triangle analogy is based on the idea that the force changes proportionally to the depth.
    The approach where you multiply the force by an area used a constant mean pressure assumption, probably at the midpoint of the plate's height.
     
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