# This one may be easier than cannibal logic

• naeblis
In summary, the dilema of the bridge is that a person weighing 185lbs cannot cross it due to the reaction force of the balls being thrown or rolled across. There are three possible solutions: 1) leave your mallet behind, 2) remove a couple non-load bearing pieces of the bridge, or 3) tie something to the mallet and cross with it.
naeblis
The dilema of the bridge

You are going into the jungle to play croquette with an old friend. on your way you come to a bridge. there is a sign that says weight capacity 185lbs.

you think to your self i weigh 175lbs, my mallet weighs 5lbs, and my croquette balls weigh 2 lbs a piece [you have 3].

the bridge is too long to throw them or roll them across. you are going to be late so you can't make more than one trip [meaning if you cross there's no going back]. There is no fraction and / or decimals of lbs involved and if you are even 1 lb over the capacity you will fall to your doom. What do you do to make it to the match on time? [note you must make it across with mallet and all 3 balls and your self]

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i can't believe you guys got my cannibal one in 15 minutes and no one knows this one lol.

Juggle the balls as you walk across, that way you only have two balls in your possession at a time.

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Juggling won't work. When you toss a ball up, there is an equal and opposite reaction downward and the reaction force acts like weight on the bridge. Besides you've got that mallot.

jimmysnyder said:
Juggling won't work. When you toss a ball up, there is an equal and opposite reaction downward and the reaction force acts like weight on the bridge. Besides you've got that mallot.

Are you 100% sure?

I'd count on the structural engineer having incorportated a factor of safety in his design, and cross the bridge on the assumption that an extra half kilo isn't going to make a difference...

Take of 1lb worth of clothes, go for a pee first, anything to reduce your weight a little.

I have three possible solutions:

1) Leave your mallet behind. Come on, it's just a stick, and surely an "old friend" would be more than happy to let you borrow one during your turn.

2) Remove a couple non-load bearing pieces of the bridge. After you have removed 1lb, then the maximum allowable load is 1lb higher.

3) Tie some sort of string to your mallet and leave the mallet on one side. When you get to the other side, start pulling it in.

...thats the best i got.

whozum said:
Are you 100% sure?
Yes.
(10char)

juggleing was what i was looking for the mallet was useless info

So then juggling isn't really the correct answer, according to Dave and jimmy?

Juggling?? I feel sorry for the guy who actually tries this...

Personnaly, I'd use my acme hover-pod and fly there...

I don't think it's a perfect answer, for I don't know how to juggle(or we can't presume atleast). Suppose if the guy starts juggling before stepping on the bridge we can avoid the reaction force equal to 6 lbs, if any. One ball is always in the air and second ball produces a 2 lb reaction or weight depending upon whether it is leaving your hand or in your hand.

But juggling 2lb balls is what I fancy to see unless I am kind of Hercules.

Further, even if the balls are made up of mercury, my rough calculation shows the diameter to be more than 2"(56mm).

Juggling does not work. Period.

If it did, just imagine what we could do:

We could make a juggling machine in the cargo hold of a 747, we could all fly our luggage to Italy for free. We could have the Juggletron keep 499 peices of luggage in the air and only be holding one at at any given time. 500 pieces of luggage and we only have to carry 40 pounds!

Note that the juggling machine itself has a weight that is negligible. It only has to weigh enough to hold one piece of luggage (since it's only ever holding one piece at a time). It does not have to contend with inertia or momentum of the falling luggage. No matter how high you throw the luggage, as the force of the luggage coming down does not impart any force (and thus increased weight) to the juggling machine. If it were on a weigh scale, it would be juggling five hundred pieces of luggage and all the while never register more than 40 pounds.

How many contradictions can you spot in the above Juggletron device?

Juggling does not work.

Well, you can't throw or roll them, but you have a mallet. Toss them into the air and whack them across.

This is actually a well-known physics riddle. It is equivalent to the 'birds in the truck' riddle.

Scroll down to 'Juggling physics' and 'Fly away birdie' on http://www.lhup.edu/~dsimanek/scenario/miscon.htm

P.S. If you're going to post a riddle, don't just make it up - you got to at least make sure you know the answer!

I like BicycleTree's answer though... if you're not good enough to do that, then you probably won't have fun playing croquette anyway.

I guess you could find a sharp rock and cut the ropes on that side and make sure you hang on to the bridge so you can swing to the other side and then climb up...

Or take your chance walking across and when it breaks make sure you're hanging onto the side that will end you up on the correct side.

Another answer... this is the jungle right? Just use some of those Tarzan vines that must be lying around everywhere in the jungle... tie all your supplies up in your shirt and lower it to the bottom of the ravine and then carry the vine across (as long as the weight of the vine is less than all 4 other items you'll be fine). Then after you're on the other side you can pull the items back up using the vine.

In reality there are a few things to try... for the riddle, the answer would be to juggle the balls (of course this wouldn't work as explained).

whozum said:
Are you 100% sure?
I am 100% sure that the current model used by physicists to explain the world won't allow the juggling solution here. However, I am not 100% sure that the current model is correct. Is anyone?

jimmysnyder said:
I am 100% sure that the current model used by physicists to explain the world won't allow the juggling solution here. However, I am not 100% sure that the current model is correct. Is anyone?

I'm 99.9999% sure that it won't work. Unless there's a massive magnet floating overhead and the balls are metal...

Since the person has three balls couldn't they begin juggling before they stepped on the bridge as long as they had two balls in the air at any time? That way they are only holding one and have 3lb left over to provide an upward force on that one ball before catching the next.

Hmm, now that I think about it 3lb isn't much force against a 2lb ball. Throwing one ball up and catching it would probably put the guy over the 5 lb. Maybe if he could juggle sideways.

Alkatran said:
I'm 99.9999% sure that it won't work.
Since the question was are you 100% sure, I'll take that as a no.

I got it!: You carry the mallet and one ball. After you've reached the other side, your friend comes across with the other two balls

How long is the bridge? Guys, can you tell me what is the shotput record in the Athens Olympics?

this require a SUPER JUGGLER.

at first you juggle
but you throw the ball extreamely high
you start juggling before walking on the bridge
and you throw high enough that it gives you time for you to walk to the other side
before that ball land back at your palm

a question arise from the fact that juggling normally is impoosible to solve the trick. a helichopter without rotating blade sink in water, but the one with rotating blade fly, does not sink ?

I don't think throwing the ball super high will work. The problem states that the balls cannot be thrown across the bridge. That means that he must catch any balls he throws. When he catches that super high ball its velocity will be much greater and exert a much greater downward force, thus causing him to fall to his doom.

Helicopters fly by the same principles as airplanes or jets. The shape of their wings create a difference in air pressure above and below the wing. This produces lift. A helicopter blade is just another device that functions as a wing. Airplanes are also reffered to as fixed-wing aircraft, and helicopters are called rotary-wing aircraft.

My question is, if the balls were juggled with two in the air at any given time and they were juggled in a lateral direction instead of an up down direction then could the juggler avoid enough of the downward force to remain below the weight limit? He can throw the ball as hard and fast as he likes in a lateral direction without adding to a downward vector of force.

If he had very long arms this would be easier for him. If he changes the vertical velocity of the ball slower he will still create the same total upward force to change the direction of the ball, but it will be stretched over more time.

Maybe he could MacGuyver a giant slingshot using his mallet and a pair of super-elastic waistband underwear to launch the balls across the bridge. The problem does say he is wearing clothes.

Huckleberry said:
My question is, if the balls were juggled with two in the air at any given time and they were juggled in a lateral direction instead of an up down direction then could the juggler avoid enough of the downward force to remain below the weight limit? He can throw the ball as hard and fast as he likes in a lateral direction without adding to a downward vector of force.

Did you take grade 11 physics? When determining what happens on the Y axis, iIt doesn't matter what you do on the x-axis (unless your balls have wings).

So you end up with the same force downwards + a bit to the side.

No need for insults, just answers and questions and polite discussion.
When determining what happens on the Y axis, iIt doesn't matter what you do on the x-axis (unless your balls have wings).
This is what I was saying. The juggler can't juggle the balls with much vertical force because when he throws the ball up a downward force acts on him putting him above the weight limit of the bridge. And when he catches the ball he exerts an upward force on it and it exerts a downward force on him, again putting him over the weight limit.

Juggling laterally he can avoid most of the vertical forces. And this is where I was wondering, If the vertical force of throwing and catching the ball can be removed then with two balls in the air at anyone time can the balls be juggled without going over the weight limit? Is it possible to work around the force of gravity acting on all three balls if two of them are in the air at any given moment? Why or why not?

Huckleberry,

What Alkatran is saying is that you can add horizonal force to your juggling, but it will have no effect on the vertical force necessary to keep the balls in the air. If you were to look at the juggler from the side, you wouldn't able to detect the horizontal motion of the balls but you would see the vertical motion just as if there weren't any horizontal motion. This is true no matter how much or how little horizontal force you exert.

Anyone who thinks they can fool the bridge into not knowing what is going on is going swimming.

This is probably true.

Can you explain at which point the force is greater than what the bridge can hold if the juggler is only holding one ball and has two in the air and is juggling horizontally? This is what I need to know to understand why it is impossible.

Huckleberry,

Remember that what you are calling juggling horizontally still must have a vertical component to it. Also, you should know that your weight and the weight of all you carry is a force acting on the bridge. When you juggle, there will be other forces acting on the bridge that have the same effect as weight.

When one of the balls comes down and you catch it. It not not imparts its weight to the bridge, but you must stop and then reverse its free fall acceleration. To do this you must exert a force upward on the ball and this force equals the weight of the ball. The reaction to the upward force is an equal force downward on the bridge. That downward force acts exactly like weight.

In the scenario I just described, you are accelerating the ball upward with a force equal to its weight. This is enough to stop its downward motion and get it moving upward again. However it takes a relatively long time to do. Therefore, you will only be keeping one ball in the air at any time. So you've got one ball not accelerating and one ball accelerating for a total force equal to the weight of three balls.

Now juggle faster so that you have two balls in the air at any given time. You must accelerate the ball faster too and more acceleration means more force. While you accelerate that one ball, it is effectively three balls of weight.

You will get damp.

"In the scenario I just described, you are accelerating the ball upward with a force equal to its weight. This is enough to stop its downward motion and get it moving upward again. "

This is untrue, if you apply a force upward equal to gravity, the ball will continue in whatever state it was originally in. To accelerate the ball upward, you need to apply a force greater than that of gravity.

Here's another way to look at the problem: when you lower the object you are holding, you lower it's force on you and thus your force on the bridge (your effective "weight"). If you can raise the objects to a height such that you can lower them for enough time to cross the bridge, and at a rate which lowers your effective weight to the needed amount, you can, in theory, safely cross the bridge. However, this is clearly impracitcal because of the height required to do so.

This leads to the interesting consideration that if the bridge is angled downward enough, and you traverse it fast enough, you can safely cross the bridge regardless of your weight. (I had assumed it was horizontal).

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Greg825 said:
This is untrue, if you apply a force upward equal to gravity, the ball will continue in whatever state it was originally in. To accelerate the ball upward, you need to apply a force greater than that of gravity.
You are correct. However, the numerical difference between equal to and greater than is rather small wouldn't you say?

Not if you're catching a falling ball. For example, if you want to accelerate the falling ball to its original vertical velocity in exactly the same amount of time as it spent in the air since last leaving your hand, you would have to exert twice as much force as the ball weighs. And normally when juggling you would accelerate the falling ball to its original vertical velocity in much less time than it spends in the air.

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