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Centre of Mass of carpenter's L square - please help!

  1. Apr 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A carpenter's square has the shape of an
    "L," as shown at the right. Find the
    coordinates of its center of mass, assuming
    it to be made of uniform, material. (Hint:
    divide the L-shape into two rectangles.)

    2. Relevant equations

    MaXa+MbXb/(Ma + Mb)


    3. The attempt at a solution

    I separated the L shape into two rectangles, with the upper part of the L being a rectangle of 14 cm x 4 cm, and the lower part of the L is a rectangle measuring 4 cm by 12 cm. I found the centre of mass of the upper rectangle to be (2,9) and of the lower rectangle to be (6,2). For the mass, since it's of uniform material, I used the area as its mass (not sure if that's right). The technique I tried was to first find the x-coordinate of the object's centre of mass:

    (2x56)+(6x48) / (56+48) = 3.846 cm

    For the y coordinate I used the y values of each rectangle:

    (9 x 56)+(2x48) / (56+48) = 5.77 cm

    So the centre of mass I got was (3.8 cm, 5.8 cm). Were the steps I used sensible?

    Thanks a lot!
    -Lauren
     

    Attached Files:

  2. jcsd
  3. Apr 8, 2009 #2
    The x is correct, the y is wrong. You are calculating it with respect to the bottom of the L. Is the y-coordinate of the centroid of the upper rectrangle then equal to 9?

    Note: you can check your answer by considering two different rectangles. One large one that covers the entire L, substracted by a smaller one that is enclosed within the legs.
     
    Last edited: Apr 8, 2009
  4. Apr 8, 2009 #3
    Ahh...thank you. Since I separated the two at y=4 cm, then the upper rectangle is 14 cm high. Should I have put

    (7x56)+(2x48) / (56+48) = 4.69 cm for the y coordinate? Thanks for your help!
     
  5. Apr 8, 2009 #4
    The upper is indeed 14 cm high, and if considered alone, its centroid would be at y = 7 cm. However, since there is a horizontal leg element as well, and you place the coordinate system in the lower left corner of the L, the centroid of the upper retrangle moves up by 4 cm, which is the height of the lower leg. See this webpage for some illustrations.
     
  6. Apr 8, 2009 #5
    Hi! Thanks a lot for attaching that link, it really helped. I think I've got it this time, since I'm using the right coordinates -
    (11x56)+(2x48) / (56+48) = 6.85 cm
    I appreciate your time srvs!
     
  7. Apr 8, 2009 #6
    Pleasure. To double-check you could different rectangles as well. Haven't got a picture but if you consider a rectangle that spans from the most bottom-left corner of your L, and across the entire L then its dimensions are 18 by 12. The rectangle between the legs of the L is 14 by 12, so you could also say that with respect to the bottom of the larger rectangle, y = [ (14+4)/2 * (14+4 * 12) - (4 + 14/2) * 14 * (12 - 4) ] / [ (14+4 * 12) - 14 * (12 - 4) ] = 6.85 cm. So you consider instead the surface area and y coordinate of the large rectangle and substract from that the surface area and y coordinate of the smaller one.
     
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