All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Logic > Weights and Scales
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    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts Somthing missing in "simple soultion"? | Comment 6 of 10 |

Since the lengths of arms of the "Y", as defined by the vectors in Brian Smith's notation, are always positive, I would submit that the law of sines equation is not enough.  The sines of all three angles would necessarily end up as positive values for this relation to hold, and threrefore there would be multiple solutions, only some of which would be valid.  I believe you would need to add the additional equation of A+B+C=360 degrees.

For example. the simpler case of equal length arms should result in A=B=C=120 degrees, but sin(120)=sin(60), so you could end up with any combination of 60 or 120 for A, B, & C using only the law of sines.


  Posted by Kenny M on 2009-12-24 10:59:05
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (8)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information