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Hydrostatic Force Problem

  1. May 19, 2015 #1
  2. jcsd
  3. May 19, 2015 #2

    SteamKing

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    Let's bring some things over from your post on the SE so it's easier to discuss:

    eRHNm.jpg

    It's not clear what you mean here. The triangle has a depth of 8 units, and the top 3 units are not immersed in liquid. Only the shaded bits are immersed.

    It's not clear why you are assuming that this problem involves work.

    To start with, you should be applying Pascal's Law to this problem.

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

    The pressure (not the force) in a fluid is proportional to the depth below the surface of the fluid, thus ΔP = ρg Δh.

    For the purposes of this problem, ρg = 62.4 lb / ft3, and we will specify that the pressure of the fluid P = 0 at the surface.

    Thus, P = ρg (y - 3), and P = 0 for 0 ≤ y ≤ 3, and y = 0 at the apex of the triangle.

    Now, the pressure P will be a constant across the width of the triangle at a given depth y for 3 ≤ y ≤ 8 feet.

    Since pressure P = force / area, then we can say the force due to the water pressure along a narrow strip is F = P * area of the strip or

    F = P * width of the strip * dy (where dy is the height of the strip) For the triangle, the width of the strip, x, can be done as a proportion, because when

    y = 0, x = 0 and when y = 8, x = 6

    So x = 6 * y / 8 = 3y/4

    Setting up the force calculation on the triangle,

    ##dF = P dA = (ρg * y) * x * dy = (ρg * (y - 3)) * (3y/4) * dy ##

    ##F = \int dF = \int^8_3 (ρg * (y - 3)) (3y/4) dy = (3/4) 62.4 \int^8_3 (y^2 - 3y) dy ##

    ##F = (3/4) 62.4 ((1/3)y^3 - (3/2)y^2)|^8_3 = 3705\; lbs.##
     
  4. May 20, 2015 #3
    Thank you so much this is a wonderful explanation. I just don't understand the importance of modeling height as y-3 instead of 5-y. Isn't it the same thing essentially? Also isn't the slope of the right side of the triangle negative 8/3 x? So I don't know why it would work to multiply it by two yielding 3/4
     
  5. May 20, 2015 #4

    SteamKing

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    It depends on how you want to set the problem up. I chose to work from the apex of the triangle down toward the base. The term (y - 3) is used because the water doesn't go all the way from the base of the triangle to the apex, but stops 3 feet below it. At (y-3), the pressure due to the water is zero and increases according to the depth past that point.

    The slope of the side of the RHS of the triangle is indeed 8/3, but that fact alone is not what is needed to solve this problem.

    What you do need to know instead is the width of a horizontal strip across the triangle when measured at a certain distance y below the apex. That's why I chose to use a proportion, relating the width of the base, which is 6 feet, to the height of the triangle, which is 8 feet. In other words, you know that at the apex, the width is 0, at 8 feet below the apex, the width is 6 feet, and there is a straight line connecting these two points which determines the width.

    Therefore, the proportion:

    ##\frac{width\;at\;y} {y} = \frac {6} {8} ##

    will give the width of this triangle for any value of y.
     
  6. May 20, 2015 #5
    Thank you for the explanations. I dislike the idea of accounting for empty space so I choose to look at the shape as a trapezoidal container of water. I tried integrating ∫05(pg*y)*x*dy where y is then (5-y) and x = 3y/4 and got a totally wrong answer. I thought the width will be the same if I modeled it as a trapezoid.. not sure where I went wrong.
     
  7. May 20, 2015 #6

    SteamKing

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    I don't know what you mean by this.

    I started my integration at a depth 3 feet below the apex of the triangle because there is no water present, hence no pressure on the triangle. If you don't account for the empty space in some manner, you will not calculate the correct amount of force the water exerts on the triangle. It's as simple as that.

    You're placing too much blind faith in trying to integrate simple expressions for the pressure, without making sure that these are the correct expressions.

    This problem can be solved assuming a trapezoid, if you do it properly.

    Let's look at what you're integrating. It appears you want the depth to be measured down from the surface of the water, hence d = (5 - y), which also assumes that the bottom of the trapezoid is located at y = 0 feet. Now, you also used the width of the trapezoid as x = 3y/4. Let's check a few points and see if they correlate with what we know about the trapezoid:

    When y = 0 feet, the depth of the water is (5 - 0) = 5 feet, and the width of the trapezoid is 6 feet, according to the diagram in the problem. Substituting y = 0 into x = 3y/4 means that x = 0 feet, which is not the width of the trapezoid. This is a big red flag that what you have assumed to be the width of the trapezoid is in fact not true. Well, let's check another point. When y = 5 feet, the depth is (5 - 5) = 0 feet, which is true according to the figure. Checking the width at this location gives x = 3*5/4 = 3.75 feet. It's a little harder to check this last result, but we know for a fact that the width x = 3y/4 is wrong when y = 0. so let's see if we can fix it.

    We can calculate the slope of the sides of the triangle using the dimensions given in the figure. The vertical axis drawn through the apex of the triangle perpendicular to the base is also an axis of symmetry, and we can use this fact to simplify calculations somewhat.

    If we look at the slope of the left side of the triangle, we can easily see that the rise is 8 feet and the run is 3 feet, so the formula fro the slope ##m = \frac{rise}{run}## gives us m = 8/3. The equation of the line which forms the left side of the trapezoid is y = 8x/3.

    Normally, the equation of a line is used to find a series of y values if we plug in a series of x values, but in this case, we wish to find the x values which correspond with the series of y values we are plugging in, so we must modify the equation of the line y = 8x/3 to give x = 3y/8.

    The width of the trapezoid is decreasing as we move up from its base, so it would seem that a good trial function for the width would be x = 3 - 3*y / 8. The value of x measures half the width of the trapezoid from the axis of symmetry, so the expression for x must be multiplied by 2, which means

    x = total width of the trapezoid = 2 * (3 - 3*y / 8) = 6 - 6y / 8 = 6 - 3y / 4.

    Let's check a few points to see if the width we are calculating with the formula for x matches the trapezoid as drawn.

    When y = 0 feet, we are at the base, and x = 6 - 3 * 0 / 4 = 6 feet, which is the distance shown on the figure. So far, so good.
    When y = 8 feet, we are at the tip of the triangle, from which the trapezoid is formed, and x = 6 - 3 * 8 / 4 = 6 - 24 / 4 = 6 - 6 = 0 feet. It looks like we have a winner!

    Now, it appears we have the key ingredients to make our integral for the hydrostatic pressure on the trapezoid:

    ##F=\int_0^5 ρgd * dA = \int_0^5 ρgd * width * dy = \int_0^5 ρg * (5 - y) * (6 - 3y/4) * dy##

    ρg = 62.4 lbs / ft3 is a constant, so

    ##F= ρg \int_0^5 (5 - y) * (6 - 3y/4) * dy##

    expanding the integrand,

    ##F= ρg \int_0^5 (30 + (-6 - 15/4)* y + 3y^2/4) dy = ρg \int_0^5 (30 - 39*y/4 + 3y^2/4) dy##

    Running through the integration:

    ##F= ρg * [(30y - 39*y^2/8 + 3y^3/12)]|_0^5##

    ##F= 62.4*[150 - 39*25/8 + 3*125/12] = 62.4 * 59.375 = 3705\; lbs.##

    which force matches the other calculation. :wink:
     
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