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Force Exerted By a liquid integration problem

  1. Oct 28, 2012 #1

    B3NR4Y

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    Gold Member

    1. The problem statement, all variables and given/known data
    A rectangular swimming pool is 20 feet wide and 40 feet long. The depth of the water varie uniformly from 3 feet at one end to 9 feet at the other end. Find the total force exerted at the bottom of the pool.


    2. Relevant equations

    The force F exerted by a liquid of constant density "p", where the functions f and g are continuous on [c,d], is
    \begin{equation}
    F=\int_c^d \, p(k-y)[f(y)-g(y)]dy
    \end{equation}
    The equation of the line making up the bottom of the pool is
    \begin{equation}
    \frac{-3}{10}x+6=y
    \end{equation}

    3. The attempt at a solution
    The depth of any rectangle below the surface would be (9-y), I reasoned.
    I tried to do the integration
    \begin{equation}
    \begin{split}
    F&=62.5\int_0^6 (9-y)(\frac{10(y-6)}{-3})dy\\
    &=26250\\
    \end{split}
    \end{equation}
    The book says this answer is wrong, as I suspected. I can't seem to find a way to think of it. Ways I've tried:
    Finding the volume of water above the wedge caused by the incline + the wedge volume * 62.5
     
    Last edited: Oct 28, 2012
  2. jcsd
  3. Oct 29, 2012 #2
    You don't really need calculus for this problem. The total force is just the volume of the pool times the density of water. Notice that a cross section of the pool is a trapezoid.
     
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