A letter F is composed of 6 unit squares and two rectangles of unit width as in the figure:

Find the lengths of the two rectangles such that the center of gravity is at the center of the middle square.

(In reply to re: maybe -- numerical confirmation and question by Charlie)

>How does one analytically solve the two simultaneous quadratics?

I don't know: this particular one could be simplified.

Let's L1 and L2 are the lenghts of the upper and lower rectangles and

x=0.5+L1/2
y=0.5+L2/2

Setting the origin (0,0) at the center of the square which is left from the gravity center, we will get the following eqs:

2x^2-x-2y-3=0
2y^2+y-4x-3=0

Subtracting one from the other:

2(x^2-y^2) +3x-3y=0  then

2(x-y)(x+4)=0 then

x=y

Posted by Art M on 2007-05-17 19:56:14