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Homework Help: Center of mass in a triangle

  1. Oct 21, 2013 #1
    Hello,, I'm reading about beams in mechanics of matrials. But I'm a bit stuck because of one thing that I either don't understand or just don't comply with!

    So, here's a picture from the book: picture 003

    It says that the center of mass should be at 2/3 of the length..

    I tried to calculate the center of mass on my own using integrals which states that it should be L/√2

    here's my solution using a and b instead of L, am I doing something wrong? : pisture 004

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    Last edited: Oct 21, 2013
  2. jcsd
  3. Oct 21, 2013 #2


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    The site where your pictures are stored requires, "Please sign in or register to access this page.

    You can upload and attach your pictures to a PF post if you use the Advanced editing panel.
  4. Oct 21, 2013 #3


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    When writing integrals, it is ALWAYS important to write the dx or dy, especially so you don't get confused about which variable you are integrating.

    The integral you set up, presumably to calculate the moment of the load about the end of the triangle, is incorrectly formulated, which is why you get an erroneous result.
  5. Oct 22, 2013 #4
    Ok, thanks for help! I found something that were wrong in the calculation,, the integral should equal a*b/4 (area/2), because I'm looking for a x value that gives half of the area, which I call p! but the answer is still b/√2 otherwise, is there something incorrect with the integral? I also tried the result to se if it were half of the area by puting values in a and b, and it were half of the area!

    Is there something that I don't get?
  6. Oct 22, 2013 #5


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    Yes. The center of mass of a triangle is not the on the line which divides the triangle into equal areas.

    If you take a small strip of area under the triangle, you get dA = y*dL. If the load is w0 at x = L,
    then y = w0*x/L for any location between x = 0 and x = L.

    We can rewrite dA = y*dL as dA = (w0*x/L)*dx. The moment dM of this strip of area about the point x = 0 is dM = x*dA = x*(w0*x/L)*dx = (w0*(x^2)/L)*dx

    If you integrate dA and dM between x = 0 and x = L, you should obtain the total load and the moment of that load about x = 0. If you divide the moment by the load, then you should get x-bar = 2*L/3.
  7. Oct 23, 2013 #6
    ok, I don't knowe if i'm making the integration in the right way. But I get W0*(L^2)/3

    dM=(W0*(x^2)/L)dx Integration gives: M=W0*(x^3/3L) -->

    Inserting L and dividing with W0: (L^3/3L) --> L^2/3
  8. Oct 23, 2013 #7


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    Look, I'm not sure why this is so difficult for you, but these are really simple integrations to do:

    dA = (w0*x/L)*dx

    integrating both sides and evaluating from x = 0 to x = L:

    A = (w0*x^2/(2*L) from x = 0 to x = L
    A = w0*L^2/(2*L) - 0 = w0 * L/2

    For the moment:

    dM = (w0*(x^2)/L)*dx


    M = w0 * (x^3)/(3*L)

    evaluating from x = 0 to x = L:

    M = w0 * (L^3)/(3*L) - 0 = w0 * (L^2)/3

    Now, to find the center of mass:

    c.o.m. = M / A

    c.o.m. = w0 * (L^2) / 3 divided by w0 * L / 2

    c.o.m. = (L^2)/3 * 2 / L

    c.o.m. = 2 * L^2 / (3 * L) = 2 * L / 3
  9. Oct 23, 2013 #8
    Neither do I..

    Ok, I think it was the M/A part I forgot,,

    I think I got it now,, thanks!
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