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A nice letter V is crudely shown above.
The goal is to get its center of gravity to be at the exact point where the two inner diagonal segments meet.
Case 1: Find the width of the V if the two horizontal segments are one unit long and the overall height of the V is three units.
Case 2: The width and the height are each one unit. Find the lengths of the horizontal segments.
Since the horizontal centroid is obvious the only question is the vertical.
Cut the figure into top and bottom.
w width of "cut out" triangle
x width of bottom triangle (2 * width of horiz segment)
y height of the "cut out" triangle
z height of bottom triangle
Note that given any angle, the cut out triangle and the bottom triangle are similar (w/x) = (y/z). Also, the top "wings" are equivalent to a rectangle of width x and height y wrt vertical centroid.
The centroid of the bottom is at z/3 down (triangle) and the top is y/2 up (rectangle).
The area of the bottom is xz/2 and the top is xy
so, for the center to be at the cut, (z/3)(xz/2) = (y/2)(xy)
z = y√3
by similarity, x = w√3
Case 1 : width = 2+w = 2+2/√3 = 2(1+1/√3)
(note that the height was not needed)
Case 2 : 1 = x+w = x+x/√3 = x(1+1/√3) = x(3+√3)/3
x = 3/(3+√3)
horiz segment = x/2 = (3/2)/(3+√3)
Once again, height not needed.
Posted by Joel
on 2007-05-07 15:19:44