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Well Balanced Letter: E (Posted on 2007-04-30) Difficulty: 2 of 5
The letter E can be constructed of rectangles one unit wide and unit squares as in the following:
+---------+
|         |
+--+------+
|  |
+--+--+
|  |><|
+--+--+
|  |
+--+------+
|         |
+---------+
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.

No Solution Yet Submitted by Jer    
Rating: 4.3333 (3 votes)

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Solution Solution Comment 4 of 4 |

The geometric centroid -- indicated at the crux of the >< -- acts as the fulcrum and "center of gravity" for the lamina shaped as a capital letter E. With an imaginary vertical axis of a unit square in width passing through this pivotal point, we can see that the the "mass" of the three unit squares and a unit square portion of each of the two horizontal rectangles should be in static equilibrium with the "mass" of the two portions of the horizontal rectangles to the right of the "fulcrum".

The "mass" of each lamina is equal to it square area, and with equal "forces" exerted on the left and right, the equation for static equilibrium becomes mLDL = mRDR.

DL is 1 unit in distance from the geometric centroid of the left portion and that of the geometric centroid of the "E".

 DR is (1/2 unit square + 1/2 the unit length of the portion of the rectangle, L - 2) in distance from the geometric centroid of the right portion and that of the geometric centroid of the "E".

As the width of each square and rectangle is 1 unit,
mL is equal to 5 and mR is equal to 2(L - 2).

   5 x 1 = [2(L - 2)] x [1/2 + 1/2(L - 2)]
> 5      = (2L - 4) (L/2 - 1/2)
> 0      = [(2L - 4)(L/2)] - [(2L - 4)(1/2)] - 5
> 0      = L2 - 2L - (L - 2) - 5
> 0      = L2 - 3L - 3

Using the quadratic equation and discarding the negative result, L is found to be (3 + 211/2)/2 ~= 3.79128785


  Posted by Dej Mar on 2007-05-01 14:03:44
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