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Finding the force using calculus

  1. Oct 31, 2006 #1
    The Deligne Dam on the Cayley River is built so that the wall facing the water is shaped like the region above the curve y=0.5 x^2 and below the line y= 160. (Here, distances are measured in meters.) The water level is 22 meters below the top of the dam. Find the force (in Newtons) exerted on the dam by water pressure. (Water has a density of 1000 kg/m^3, and the acceleration of gravity is 9.8 m/sec^2 .)



    F(x) = density of fluid* Volume displaced *g = 1000*9.8*volume displaced

    so what is my volume displaced? is it simply .5 *x^2 -160 ??

    and what integral do I do? are my limits of integration from 0 to 22?


    thank you!!

    blumfeld0
     
  2. jcsd
  3. Nov 1, 2006 #2
    Woah, woah. This isn't a buoyancy problem. I don't think any fluid is in fact being displaced here. By the way, you didn't mention the overall depth of the water.

    Go back to your definitions. Pressure is Force per unit Area, P=F/A. You're being asked to compute the total force on the dam due to fluid pressure. At the simplest level, what you need to do is find the area of the face of the dam touching the water, then multiply that area by the "average" water pressure on the area. That will give you a very, very crude "guess"timate of the answer.

    OK, the trouble here is that not only does the water pressure vary with depth, increasing as you go downwards, the direction the dam faces in is also changing, meaning you have to compute the direction of the force as well as magnitude. To top it all off, the surface area may also be dependant on position, by which I mean the amount of surface area as you look "head on" at the dam really depends on how sloped it is at the point you're looking at.

    This is where your integral calculus comes in. You will need to find two things;

    1) A representation of the pressure as a function of the depth d. P(d)
    2) A representation of the surface area vector for the dam. dS(x,d), where d is the depth. This will give you the direction the surface faces in, and the infinitesimal area of the surface at a point.

    You then need to compute the two dimensional integral
    [tex]\iint_A P(d) d\mathbf{S}(x,d) dd dx[/tex]

    Integration here is really the advanced "multiplication" we need to compute the total force on the surface face. It's essentially analagous to the basic F= P*A in the simple case.
     
  4. Nov 1, 2006 #3

    HallsofIvy

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    No, you do not integrate "from 0 to 22" and, more importantly, you do not integrate with respect to x!

    x is measured horizontally and pressure does not change horizontally. You need to integrate vertically, with respect to y. The height of the water is 160- 22= 158 m. You will need to integrate, with respect to y, from 0 to 158.
    Of course, since y= (1/2)x2, [itex]x= \pm \sqrt{2y}[/itex] so the length of a thin horizontal strip is is [itex]2\sqrt{2y}[/itex] and its area is [itex]2\sqrt{2y}dy[/itex]. The force on that strip is the pressure due to a weight of water above that height, y, in a 1m by 1m column, times that area. The weight of that water is mgh= (9.81)(1000)(158- y) so your integral is
    [tex](2)9810\int_0^{158}(158-y)\sqrt{2y}dy[/tex]
     
  5. Nov 1, 2006 #4
    omg! im in calculus 2. we havent even learned 2-D integrals. thats not til next semester. ok thank you!! i will need to sit down and understand this. i have a feeling it will on the final!


    blumfeld0
     
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