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Center of Mass of a triangle. Help fast please!

  1. Dec 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the Y component of the Center of Mass

    http://img607.imageshack.us/img607/7122/83370796.png [Broken]


    2. Relevant equations
    ∫(r*dm)/M


    3. The attempt at a solution
    I keep coming up with (2/3)B but i know since that there is more mass near the origin axis, it should be 1/3B. I think it is my definition of area that is wrong. I am using horizontal strips with A * dy but i am unsure how to substitute A since it is a varying as y increases

    ∫(r*dm)/M
    σ=dm/da
    σ*da=dm

    ∫(r*σ*da)/M
    da= area of small strip = dy*Mslope=
    Y=Mslope*x=
    Mslope=B/A

    ∫(y*σ*dy*y(A/B))/M
    y(A/B)=X

    (σA/BM )∫y^2dy=

    σ= M/Area
    Area = 1/2AB
    σAB=2M

    (AσB2/3M)=
    (σAB/3M)*B =
    (2MB/3M)=
    (2/3)B

    Which is wrong. I think the error is somewhere around the da part. I am a little bit unsure of what the area of each horizontal strip would be since the left side of the shape is a function of x, or is it a function of y?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 10, 2012 #2

    haruspex

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    If you write out what you mean by A, B, Mslope etc. it might become apparent to you what's wrong. But if it doesn't, it will give the rest of us a chance to spot it.
     
  4. Dec 10, 2012 #3
    oh wow i forgot the picture. Sorry!

    The lines are the dy's
     
  5. Dec 10, 2012 #4

    gneill

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    It looks like σ is meant to represent the mass density of the material comprising the triangle, and that it is assumed to be constant (triangle material is of uniform density). If that's the case, then you can ignore it (or assume σ = 1 in whatever units you prefer) and just work with the areas.

    If you can find an equation of the line for the hypotenuse of the triangle then you can determine x in terms of y (that is, x(y) ) for the line. With x(y) you can find the length of your area element for a given y (its thickness is dy and its length lies between x(y) and x = A).

    So, begin by finding the equation of the line on which the hypotenuse "lives". It's a linear equation of the form y = m*x. Solve for x in terms of y and carry on...
     
  6. Dec 10, 2012 #5
    Hypotenuse of the triangle would be:
    y=Mslope*x
    Mslope = B/A
    y=(B/A)x
    x=(A/B)y

    is that right? That's what i did on my attempt
     
  7. Dec 10, 2012 #6

    haruspex

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    OK. What do you have for the mass (or area) of the strip width dy at height y?
     
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