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How do I locate the centroid and the center of gravity of a square pyramid?

  1. Sep 24, 2009 #1
    Height=104
    Base=220 x 220
    Density=2500
     
  2. jcsd
  3. Sep 25, 2009 #2
    Finding the center of gravity for this object is gonna take some grunt-work.

    I'll help you get started with the integral. First, let's look at a thin slice of the pyramid from a top-down view. It's in the shape of a square, whose center is pretty obvious.

    So now we know the center of gravity of a square slice (Let's call the length of the side [tex]x[/tex], and the width of the slice, [tex]dy[/tex], and we know its mass: [tex]dm=\rho\cdot dv[/tex] where [tex]\rho[/tex] is the density of the pyramid and [tex]dv[/tex] is the volume of the square slice)

    The best place to place our origin is at the top of the pyramid. Try and find a way to express [tex]x[/tex] through [tex]y[/tex], which we will define as the distance from our origin to the center of gravity of a square slice.

    From there on, it's just a question of plugging what you get into the definition of the center of gravity:
    First, it's obvious that the center of gravity will be somewhere on the pyramid's line of symmetry, since the individual contributions from each of the square slices will line up on this line (Imagine the pyramid reducing into a string of beads, each with the mass of the square slice it's come to replace).

    [tex]y_{CoM}=\frac{1}{M_{total}}\int y\cdot dm[/tex]
     
  4. Sep 25, 2009 #3
    Ok so I can calculate dm then. dm=25000*(length*base*height of square) = 2500*48400 dy = 121000000 dy. Am I on the right track there? And then I would use that formula for Ycom to find the y component of the center of gravity? So Ycom = (1/Mtotal)*integral of y*121000000 dy from y=0 to y=104. Is that correct or am I way off? Also, what is Mtotal? Mx plus My?
     
  5. Sep 25, 2009 #4
    You are way off. For your function, you should express everything using [tex]y[/tex] where [tex]y[/tex] is the distance of the square slice from the top of the pyramid.

    [tex]dm=\rho S\cdot dy[/tex] where [tex]S[/tex] is the area of the square slice and [tex]\rho[/tex] is the density.

    You need to find [tex]S[/tex] and express it in terms of [tex]y[/tex]
    To do this, remember that the total volume of a pyramid is: [tex]\tfrac{1}{3}AH[/tex] where [tex]A[/tex] is the area of the base, and [tex]H[/tex] is the height of the pyramid.

    As for what [tex]M_{total}[/tex] is:
    Mass is a scalar, not a vector. There's no such thing as [tex]M_x[/tex] or [tex]M_y[/tex] or [tex]\vec M[/tex].

    The total mass is just that. The TOTAL mass of the pyramid. [tex]M=\rho\cdot V[/tex] where [tex]V[/tex] is the total volume of the pyramid.
     
  6. Sep 25, 2009 #5
    Thank you very much, I think I am on the right track now. I solved for Volume and came up with an answer of 1.7x10^6 m^3. Using that Volume, I then solved for the Total Mass of the pyramid, which came out to be 4.2x10^9 kg.

    Now, I am still confused as to what to do next. I know that the next step is going to be to somehow plug everything into the integral of (density x area of square slice) dy. I am not sure on how to find the area of the square slice though, which you expressed as S. I know that it needs to be in terms of y and usually when I solve for this in other problems they give us two equations that we can manipulate in a way so that they are in terms of y. However I don't know how to do that for this problem. Thank you so much for your help and if you don't mind helping me a little more it is GREATLY appreciated.
     
  7. Sep 25, 2009 #6
    Gladly, but I'm afraid we'll have to pick this up tomorrow since I'm going out in a couple of minutes.

    Now that you've defined the integral properly, all that's left is an exercise in analytic geometry.

    To find the relationship between [tex]x[/tex] and [tex]y[/tex], look at vertical slice through the axis of symmetry. You've got a triangle whose dimensions are known.
    Its height is the height of the pyramid and the length of its base, is the length of the base of the pyramid. I hope you understand what triangle I'm talking about since I don't have time right now to draw it and scan it.

    Now look at a smaller triangle, the height of which is [tex]y[/tex] below the tip of the pyramid instead of the full height of the pyramid and whose base is the side of the square it's resting on, [tex]x[/tex]

    What can you say about these two triangles? Can you see how you can express [tex]x[/tex] in terms of [tex]y[/tex] and known quantities?
     
  8. Sep 25, 2009 #7
    Yes I think I understand it now. I'm pretty sure I can finish it up now, if I get stuck I will come back here tomorrow. Thank you so much for your help, you have no idea how much this means to me!!!!
     
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