# The driver and the chickens

In a sealed container van, there are 5 chicken cages with a chicken each. The driver has to cross a dilapidated bridge by jumping over it. If he can minimize the weight of his truck by even just 3 lbs, he can get over the bridge. Each chicken cage with a chicken on it weighs 5 lbs. He plans to knock on his container as soon as he is about to jump the bridge so that the chickens will fly from their cages and minimize its weight. Will this work?

• This is tagged as a logic puzzle. Are you looking for an unusual out of the box solution? Otherwise this seems like the relatively familiar physics problem... Jun 9, 2016 at 0:57
• Welcome to Puzzling! I've retagged your question: [logic-puzzle] should only be used for puzzles about pure logical deduction.
– Deusovi
Jun 9, 2016 at 1:23

It will

not work

firstly because

chickens are scarcely able to fly, so knocking on the container is unlikely to produce a van full of airborne chickens

and secondly because

the chickens are still subject to gravity, so if they aren't sinking into the earth there must be an upward force equal to that of gravity, and they will in turn be exerting a corresponding downward force; the only difference is that now what they're directly exerting that force on is the air in the van instead of their cages; the overall effect is the same.

There could be

short-term variations in the force on the van as they take off, fly around, and land -- but I would expect these to be on too short a timescale to help with crossing a bridge.

I think this is more a (famous) physics problem than a puzzle, though. See e.g. here for some recent-ish research related to it.

This problem was done by Mythbusters in real life, which concluded that

There will be a negligible difference.

This is because

The force of the wings will be in all directions - not just up, so the force of the weight downwards will still be the same.

One proposed solution

Now if he were to, say, put holes in the cages, then this would be more likely to work.