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Math Help - Center of Mass (easy)

  1. #1
    Member WhoCares357's Avatar
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    Center of Mass (easy)

    Edit: Nevermind. I figured out that I was using the wrong coords and the wrong mass of the second object. The correct answer was -.25m.

    A uniform square plate 6. m on a side has had a square piece 2. m on a side cut out of it (see the figure). The center of that piece is at x=2 m, y = 0. The center of the square plate is at x = y = 0. Find the coordinates of the center of mass of the remaining piece. x-component ( m)?
    The formula for center of mass is r_x=\frac{x_1*m_1+x_2*m_2+x_3*m_3}{m_1+m_2+m_3}
    m=(density)Ah Where A is area and h is height of the object.

    This is a very easy topic and a very easy question. I'm just having trouble seeing where the coords are. In my attempt I had the following values:
    x_1=x_2=4
    x_3=1
    m_1=m_2=2^2*1*1=4
    m_3=6^2*1*1=36
    My final answer was 1.54m. The online system tells me this is wrong. Does anyone see where I went wrong?

    Thanks.
    Attached Thumbnails Attached Thumbnails Center of Mass (easy)-9-24.jpg  
    Last edited by WhoCares357; November 1st 2009 at 09:21 AM.
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by WhoCares357 View Post
    Edit: Nevermind. I figured out that I was using the wrong coords and the wrong mass of the second object. The correct answer was -.25m.



    The formula for center of mass is r_x=\frac{x_1*m_1+x_2*m_2+x_3*m_3}{m_1+m_2+m_3}
    m=(density)Ah Where A is area and h is height of the object.

    This is a very easy topic and a very easy question. I'm just having trouble seeing where the coords are. In my attempt I had the following values:
    x_1=x_2=4<<<<<< here: x_1 = x_2 = 0
    x_3=1<<<<< x_3 = -1
    m_1=m_2=2^2*1*1=4
    m_3=6^2*1*1=36
    My final answer was 1.54m. The online system tells me this is wrong. Does anyone see where I went wrong?

    Thanks.
    The mass and the area of the partitions of the plate have the same measures. m_1 and m_2 are the masses of the top and bottom rectangles. The centers of mass of these rectangles must be at x = 0.

    m_3 is the mass of the reduced rectangle. The center of mass of this rectangle must be at x = -1

    Now using your formula you'll get:

    r_x=\dfrac{-1 \cdot 8}{12+12+8}=-\dfrac8{32} = -0.25

    I've modified you sketch a little bit to illustrate my considerations.
    Attached Thumbnails Attached Thumbnails Center of Mass (easy)-schwerpkt_ubrett.png  
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