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Precariously Balanced (Posted on 2009-12-22) Difficulty: 3 of 5
Three identical weights are to be suspended from the ends of a rigid light1 "Y"-shaped frame. Each arm of the frame is to be of a different length.

How is this to be accomplished (ie, how do you shape the "Y") so that the 'system' is in equilibrium within a horizontal plane?

















1 "light" is meant as being weightless, having no concern for mass.
Note too, the colours are the radial ring extremeties of the "Y" arms within the horizontal plane.

See The Solution Submitted by brianjn    
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Another general soution method | Comment 7 of 10 |

If the lengths of the three legs are a,b & c, respectively, arm "a" is on the x-axis, and the angle from a to b is B, and a to c is C (typical counter clockwise notation), then the following system of equations will give all correct unique answers:

a + b*cos(B) + c*cos(C) = 0  &

b*sin(B) +c*sin(C) = 0

This system will give multiple answers, depding on the constraint of B>C or C>B.  Again, maybe the trig can be worked out by someone - it's Christmas Eve, so maybe someone will give us a present and do that.


  Posted by Kenny M on 2009-12-24 11:17:55
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