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Finding centre of mass of an isosceles triangle

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data

    Explorers in the jungle find an ancient monument in the shape of a large isosceles triangle. The monument is made from tens of thousands of small stone blocks of density 3 800 kg/m3. The monument is 15.7 m high and 64.8 m wide at its base and is everywhere 3.60 m thick from front to back. Before the monument was built many years ago, all the stone blocks lay on the ground. How much work did labourers do on the blocks to put them in position while building the entire monument?

    2. Relevant equations

    the gravitational potential energy of an object-Earth system is given by Ug = MgyCM, where M is the total mass of the object and yCM is the elevation of its centre of mass above the chosen reference level.


    3. The attempt at a solution

    So, I have to find the y-coordinate of the centre of mass, but I'm not sure how to go about doing it... Is integration necessary here? Very confused :(
     
  2. jcsd
  3. Nov 13, 2011 #2

    rude man

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    You're on the right track. Yes, integration is required.
    Hint: do it for a 2-dimensional figure.
     
  4. Nov 13, 2011 #3
    Hm.. I know this sounds stupid but I'm not really sure *what* to even integrate to get the equation to be in terms of y...

    Am I on the right track with this kind of set-up? (I've attached a picture of my 2D diagram)
     

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  5. Nov 13, 2011 #4
    Whoops... maybe this makes more sense?
     

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  6. Nov 13, 2011 #5

    SteamKing

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    Or you could remember what the c.g. of a triangle is from geometry. (Do you work out everything from first principles? That's got to take a lot of time.)
     
  7. Nov 13, 2011 #6
    Integration is certainly 1 way to approach this, but since you're given such exact dimensions, why not just take a geometric approach?

    For any triangle with a constant, continuous mass distribution, the center of mass is the centroid. The centroid is found by drawing 3 lines, each from one vertex of the triangle to the midpoint of the opposite (non-adjacent) side. The intersection of these 3 centroid lines is the centroid.

    So, pretend for now that your triangle is flat (no depth). Using the diagram you drew, assign coordinates to the vertices and use the coordinates to figure out the equations of these 3 centroid lines (you only need to find 2 lines). Then, just find their intersection. That will give you the (x,y) coordinates of the center of mass. To find the z-coordinate, just divide the depth of the triangle in 2.
     
  8. Nov 13, 2011 #7
    Haha that is certainly a lot simpler! But I'd like to also try with integration just as practice, to since I've never done this before...

    I've attached another picture of my progress so far. How is it looking?

    I'm a bit lost on where to go from here, and if I'm even on the right track at this point haha.

    It looks like I have to get that "x" in my 'mass of each strip' equation in terms of y. Should I be using the 'area of a triangle' equation for this? I'm not sure where else to stick it in...

    Thanks for the help so far !
     

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  9. Nov 13, 2011 #8
    Update!

    I think I did it successfully by the integration method :D

    My answer to the original question about energy came out to be 3.21 x 108 J, whereas the textbook answer is 3.57 x 108 J. I think I'm getting decimal place errors since the numbers come out to be so big, or so I'd like to think!

    I've added my updated diagram with a general over-view of my steps, just in case anyone wanted to check. Woot! I'm happy :)
     

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  10. Nov 13, 2011 #9

    rude man

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    Hey, no fair using high school geometry (is that its provenance?) that most of us forgot! :redface:

    Actually - thanks, I realy was unaware of that little gem. Obviously the right way to go here.
     
  11. Nov 13, 2011 #10

    rude man

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    I would set the triangle with the (unknown) c.g. at the origin. The triangle's height is (b+c) located at (0,b) and its base is 2a, located at y = -c.

    Then,

    ∫ from -a to 0 {∫ from -c to (mx + b) of -sqrt(x2 + y2)dy}dx

    + ∫ from 0 to a {∫ from -c to (-mx + b) of sqrt(x2 + y2)dy}dx = 0

    where m = (b+c)/a

    You know a and b, so compute c, then energy = mgc with m = mass of monument = density*volume.
     
    Last edited: Nov 13, 2011
  12. Nov 13, 2011 #11

    rude man

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    EDIT: my integrals are incorrect, sorry.

    May look at this some more later. I still would put the origin at the c.g . though. More gut feeling than anything else.
     
  13. Nov 14, 2011 #12

    rude man

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    Thinking about this some more, maybe looking for the c.g. is not the best way to go.

    Instead: put the base on the x-axis and the opposite vertex at x = 0. Let base = 2a and height = h. Now, the work needed to assemble the monument is the sum of all dm times respective heights, times g. This is an easy integration, and you just have to do half of the triangle. Picking the right part, work lifting an element dx*dy a height y is y*dx*dy (omitting density, g and depth for the moment). So for a sliver dx the work required is ∫ from y = 0 to y = sx+h of ydy where s = -h/a. This computes to (sx + h)^2/2.

    Next we integrate the slivers from x = 0 to x = a:

    ∫ from 0 to a of [(sx + h)^2/2]dx = (s^2)(a^3)/6 + sh(a^2) + a(h^2)

    Substituting,
    s = -h/a = -15.7/32.4 = -0.4846
    a = 32.4
    h = 15.7

    & the integral computes to 1330.7. Multiply by 2 for the left half of the triangele gives
    2661.4
    And work is therefore 9.81*2661.4*3.6*3.8e3 = 357162009 J.
    Which agrees with your textbook!
     
  14. Nov 14, 2011 #13

    rude man

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    I still think finding the c.g is of great interest. I will send you a private message if I work that out, since your post will probably have disappeared from the board by then.

    Sudden dawn: the job is done! Let d be the distance of the c.g. above ground, then d = E/mg where E is the total work (3.57e8 J) and m is the mass of the monument.
     
    Last edited: Nov 14, 2011
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