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Homework Help: Integration Center of Mass Problem -Really frustrated

  1. Apr 15, 2008 #1
    Integration Center of Mass Problem --Really frustrated

    1. The problem statement, all variables and given/known data

    An 800 g steel plate has the shape of the isosceles triangle shown in the figure. What are the x and y coordinates of the center of mass?


    2. Relevant equations

    3. The attempt at a solution

    I am getting really frustrated with this one because I can't seem to relate dm to dx. I know that while dx is constant, dm is increasing because the height of the slices you cut the triangle into is increasing. Could someone please give me a push in the right direction? I am having trouble applying integration to center of mass problems.
  2. jcsd
  3. Apr 15, 2008 #2
    The center of mass would be the baricenter of the triangle. If you wanna do it in an analytic way; it's also easy. Just use the superficial density of the plate. density=mass/Area of triangle. Then the center of mass will be: Integral (x*dm)/Total Mass. But dm=density*da. then the center of mass is: Integral (density*x*da)/Total mass
    Where density is constant Total mass is knownk and u just need to integrate in the triangle area.

  4. Apr 15, 2008 #3
    I followed what you were saying, but I still can't integrate it because I don't have dx in my integral, so I'm stilll confused.

    My book says to divide the triangle into vertical strips with width dx, then relate the mass dm of a strip at position x to the values of x and dx.
  5. Apr 15, 2008 #4
    Uhh!! you have da in your integral, you need elevated calculus to make that kind of integrals. Anyway to make it more simple Area=base*height/2 where the base is the x axis and the height can be a function of x, F(x)= straight line1+straight line2. then u can write ir as: Integral(density*x*F(x)*dx)/total mass where dx goes through all the base of the triangle and F(x) is the height function of the trinangle. (Twho straight lines, u have to split the integral in two)
  6. Apr 15, 2008 #5
    I'm not sure I'm getting it yet:

    Ok so dm= (kg/m^2)*F(x)*dx

    kg/m^2= 26.67 kg/m^2
    F(x)= (1/3)x

    Cm= (1/M) * Integral xdm
    Cm= (1/.8kg)* Integral x*(26.67)(1/3)*x dx evaluated from 0 to .3 m
    = I got x= .1 m but the answer for the x coordinate is .2m. What am I doing wrong?

    and then do I go about finding the y coordinate in the same way?
  7. Apr 15, 2008 #6
    dm = (length of the strip) * dx, now you just need to find the length of the vertical strip as a function of x (use similar triangles)
  8. Apr 15, 2008 #7
    How can dm= length of the strip * dx? That doesn't yield units of kg. I found the function for the height to be y=(1/3)x
  9. Apr 16, 2008 #8
    First we assume that the body is uniform, so its mass is proportional to its area in any section. So I am going to use area (A) instead of mass for my calculations.

    dA = height of vertical strip . dx

    height of vertical strip = 2/3 . x

    Then you use the formula for center of mass:

    Integral of x . dm = Total Mass of Triangle . Xo

    Xo is the coordinate of the center of mass

    as I told before you can use area instead of mass (assuming that the body is homogeneus)


    Integral of x . 2/3 . x . dx = Total Area of Triangle . Xo

    You solve this equation and get the answer Xo = 20 cm
  10. Apr 16, 2008 #9
    sorry I thought about integrating over the area.
  11. Apr 16, 2008 #10
    No, it does not. OK, in order to yield units of kg we have to to this:

    Total Mass = density of area . Total Area

    density of area = M / A

    so dm = M / A . dA

    then dm = M / A . lenght of strip . dx ... and you have Kg
  12. Apr 17, 2008 #11
    Ok with the drawing it's easier for me to explain. You did everything right, you just forgot to take into consideration the other half of the triangle. Which means you have to multiplicate your result by two!!! (Remember integrals are really areas so its valid) and the you have your right x coordinate. U can rapeat it in the y coordinate but it's kind of obvious that the y coordiante for the center ofa mass is going to be cero.

    Cheers Hope I helped!!!
  13. Apr 17, 2008 #12
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    Cheers again mate
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