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 Precariously Balanced (Posted on 2009-12-22)
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 No Rating

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 Rather Simple Solution | Comment 5 of 10 |

Because the same mass is being hung at the end of each arm, each arm can be represented by a vector with a magnitude of the arm's length.  Say the arms' lengths are x, y, and z.  Then the vectors have magnitudes of kx, ky, and kz.  Let A be the angle between arms x and y, B be between x and z, and C be between y and z.

For the Y to balance, the resultant of the vectors must be zero.  This means the vectors form a polygon when arranged tip-to-tail, in this case a triangle.  The angle between vectors kx and ky equals 180-A, similarily the angle between kx and kz equals 180-B and the angle between ky and kz equals 180-C.

By the law of sines and sin(180-t)=sin(t), a simple compound equation results:
x/sin(C) = y/sin(B) = z/sin(A)

 Posted by Brian Smith on 2009-12-23 22:59:57

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