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Creating a berm with a 1:1 slope

  1. Aug 11, 2011 #1
    1. The problem statement, all variables and given/known data

    324,000 cubic feet of soil is to be placed in a rectangular area (constructed from hay bales) of 80 feet x 100 feet. Using a slope of 1:1 for the fill soil, how many bales high must the walls be stacked to accommodate the soil?
    Assume the hay bales have a height of 18"

    2. Relevant equations
    A = L x W
    V = A x h

    Volume of a pyramid...? 1/3(L x W x h) = V

    3. The attempt at a solution

    I started by calculating the required height (assuming the soil was filled without a slope):

    324,000 cu ft = 8000ft x h
    h = 40.5 ft

    I know from here I would simply divide the total height (h) by the height of a hay bale (18"), to get the number of bails. However, I am thinking that my calculated height is too much, since the soil will be slopped (1:1 slop) and extend above the walls. My question is, how do I figure out how high the soil can be slopped above the walls? and what the volume of soil extending above the walls would be. I was thinking this may have something to do with calculating the volume of a pyramid... but I'm a bit confused as to how to proceed. Any help or suggestions would be greatly appreciated.
    Last edited: Aug 11, 2011
  2. jcsd
  3. Aug 11, 2011 #2


    Staff: Mentor

    The problem says 32,400 cu ft.
  4. Aug 11, 2011 #3
    That was a typo. Thanks for pointing it out. I have corrected it in my original post.
  5. Aug 11, 2011 #4


    Staff: Mentor

    Do you have an idea of how the berm will look, with its sloped (not slopped) sides?
  6. Aug 11, 2011 #5
    When sloped, it should resemble a pyramid.
  7. Aug 11, 2011 #6


    Staff: Mentor

    The problem would be a bit easier if the base happened to be square, rather than rectangular.
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