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.
Let H be the length of the long horizontal rectangle on the top and let V be the height of the long vertical rectangle on the side.
Divide the F into three rectangles A, B and C. Rectangle A consists of the long horizontal rectangle on the top plus the two squares which are also on the top of the F. Rectangle B consists of the long vertical rectangle on the side plus three squares also on the side of the F. Rectangle C is just the square containing the center of gravity. The dimensions of A are 1x(H+2), the dimensions of B are (V+3)x1.
Let the center of rectangle C be located at the origin of a coordinate system, call that point G. Then the center of rectangle A (point X) is at ((H-1)/2, 2), and the center of rectangle B (point Y) is at (-1, -V/2).
For G to be the center of gravity of the F, the points G, X, and Y must be colinear and the net torque along that line must be 0. Two equations can be generated as follows:
Colinear: (2-0)/((H-1)/2 - 0) = (0 - -1)/(0 - -V/2)
Torque: 1*(V+3)*sqrt[(-1)^2 + (-V^2)^2] = (H+2)*1*sqrt[((H-1)/2)^2 + (2)^2]
Simplifying each equation a bit yields:
V = 8/(H-1)
(V+3)*sqrt[V^2 + 4] = (H+2)*sqrt[H^2-2H+17]
Substituting the expression for V and more simplifying yields:
2*(3H+5)*sqrt[H^2-2H+17] = (H-1)^2*(H+2)*sqrt[H^2-2H+17]
Since sqrt[H^2-2H+17] is always positive, the equation can be reduced to:
(H+1)(H^2-H-8) = 0
This polynomial has one positive root H = (1+sqrt(33))/2, which makes V = 8/((1+sqrt(33))/2 - 1) = (1+sqrt(33))/2.
Edited on May 17, 2007, 3:38 pm
Solution
