1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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!

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook