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Advaita
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January 28th, 2008, 4:39 pm

A woman with 2 balls of weight 10 each has to cross a bridge that is 100 feet long. She weighs 100.The bridge is old and breaks at weight > 115. How will she cross the bridge if she can cross it only once?
 
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tetrabit
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January 28th, 2008, 4:50 pm

If she has a rope, she could tie one end of the rope to ball A and cross with ball B in her hands. Then she can pull ball A over.
 
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MCarreira
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January 28th, 2008, 5:00 pm

Juggling ?
 
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Advaita
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January 28th, 2008, 7:39 pm

that's right.
 
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Traden4Alpha
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January 28th, 2008, 8:07 pm

Perhaps my dynamics knowledge is rusty after almost 20 years, but I could not make juggling work.15 weight units of upward force (the max that the bridge can handle) exerted on a 10 unit ball for 2 time units (1 time unit to decelerate the falling ball & 1 time unit of accelerate the ball) only provides 1 time unit of air time which isn't enough to handle the second ball. Unless she is Zeno's juggler (each successive juggling cycle being half the height of the previous one), I don't see how this works.What am I missing?!?!?!
 
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Vassili
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January 28th, 2008, 8:35 pm

QuoteOriginally posted by: AdvaitaA woman with 2 balls of weight 10 each has to cross a bridge that is 100 feet long. She weighs 100.The bridge is old and breaks at weight > 115. How will she cross the bridge if she can cross it only once?The way you ask it, she can just drop one, or both, balls and cross the bridge, she doesn't need to carry both balls to the other side ;-)
 
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MCarreira
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January 28th, 2008, 9:21 pm

QuoteOriginally posted by: Traden4AlphaPerhaps my dynamics knowledge is rusty after almost 20 years, but I could not make juggling work.15 weight units of upward force (the max that the bridge can handle) exerted on a 10 unit ball for 2 time units (1 time unit to decelerate the falling ball & 1 time unit of accelerate the ball) only provides 1 time unit of air time which isn't enough to handle the second ball. Unless she is Zeno's juggler (each successive juggling cycle being half the height of the previous one), I don't see how this works.What am I missing?!?!?!It seems the problem does not ask for physics knowledge, but for "out-of-the-box" thinking, even if it will not work in reality (like finding a specific name in the phone list).
 
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Advaita
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January 28th, 2008, 10:07 pm

Traden, I tried this on the Brooklyn Bridge and it works, OK??! Case closed.
 
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TheVulture
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January 28th, 2008, 10:32 pm

Roll both balls across the bridge - 100 feet isn't that far. Then walk over, carrying no balls.
 
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NoMark
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January 29th, 2008, 2:33 pm

Throw the first ball, then walk half way up to the first ball and throw the second ball over it, then repeat?Assuming that if she can juggle the balls, she can throw one far enough away for this to work. This is a better strategy as it requires much less physical co-ordination and concentration and so she will be less likely to make a mistake and end up falling to her doom
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