The letter E can be constructed of rectangles one unit wide and unit squares as in the following:
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Find the length of the top and bottom vertical rectangles such that the center of gravity of the letter is exactly at the center of the middle square.
... you are referring to the two horizontal rectangles:
Let the zero x value be 1/2 unit to the right of the left edge of the E, along the vertical centerline of the column of unit squares (and portions of the horizontal rectangles).
If s is the length of each of the nonsquare rectangles, then the length in excess of the 1 unit in the first column is s1, and its distance to the right of the 0 line is (s1)/2 + 1/2 = s/2. There are two of them at the same distance, so the area is 2 per unit length.
You want the CG to be at x=1:
(0*5 + 1*1 + (s/2)*(2*(s1))) / (5 + 1 + 2*(s1)) = 1
1 + s*(s1) = 6 + 2*(s1)
s^2  s  2*s + 2 = 6  1
s^2  3*s  3 = 0
s = (3 +/ sqrt(9+12)) / 2
s = (3+sqrt(21)) / 2 ~= 3.79128784747792
Edited on April 30, 2007, 8:54 pm

Posted by Charlie
on 20070430 10:50:56 