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Section 8.3 in stewart calculus problem 8

  1. May 24, 2014 #1
    So I'm working On hydrostatic pressure. The problem #8 in stewart

    Is a trapezoid with the base above water.

    4

    \------------------------/ 1(above watee)
    \ / 1(in watee)
    \ /
    ---------------
    2

    So if the base was a number say
    50. N the bottom was 15.

    I COULD split the trapezoid in half.

    15 +10 base

    And bottom would be 15.

    Similar triangles for the rectangle slab would give me. A/(height of object minus height in water-Y)=10/20

    Then I'll solve for A.


    The problem here is that the number 4 does not break up as nicely as this does. How would I relate the sides?

    Thanks
     
  2. jcsd
  3. May 24, 2014 #2

    Simon Bridge

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    I'm afraid you are not making sense.
    You need to relate the numbers to dimensions in order to make sense.

    A trapezoidal trough may have length L, bottom-width B, top-width W, and overall-height H.
    It may be filled with water to height y.

    We'd expect W>B, and L>W, so it's like a pig-trough. But it does not have to be that way.

    w=B+(W-B)y/H would be the width of the trough at the water-height,
    and the volume of water in the trough would be V=y(B+w)L/2.

    Or it could be a trapezoidal boat, like a punt with a flat prow and stern.
    In that case, w is the width of the punt at the water-line and V is the volume of water displaced.

    Now... what are you talking about?
     
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