This one may be easier than cannibal logic

[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]

 

[naeblis]

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

 

[whozum]

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

 

[Jimmy Snyder]

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.

 

[whozum]

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?

 

[brewnog]

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...

 

[matthyaouw]

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

 

[minger]

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.

 

[DaveC426913]

Are you 100% sure?

Yes.
(10char)

 

[naeblis]

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

 

[whozum]

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

 

[Alkatran]

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

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

 

[quark]

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).

 

[DaveC426913]

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.

 

[BicycleTree]

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

 

[DaveC426913]

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!

 

[Rahmuss]

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).

 

[Jimmy Snyder]

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?

 

[Alkatran]

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...

 

[JamesU]

Is your friend already there?

 

[Huckleberry]

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.

 

[Jimmy Snyder]

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.

 

[JamesU]

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

 

[quark]

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

 

[ArielGenesis]

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

 

[ArielGenesis]

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 ?

 

[Huckleberry]

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.

 

[Alkatran]

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.

 

[Huckleberry]

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?

 

[Jimmy Snyder]

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.

 

[Huckleberry]

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.

 

[Jimmy Snyder]

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.

 

[Greg825]

"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).

 

[Jimmy Snyder]

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?

 

[BicycleTree]

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.

 

[BicycleTree]

For example, if you wanted to throw it back up in a third the time it spends in the air, you would have to exert four times as much force as it weighs.

 

[Jimmy Snyder]

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.

The necessary force is just a smidge more than the weight of the ball and in that case, the time spent accelerating the ball would be roughly equal to the time that it spends in the air. You are correct that this is not the normal way to juggle, but that was not a condition of the puzzle.
By the way, there is no way to juggle the balls without accelerating your own fingers, or some part of your own body. Also, you need to overcome air resistance on the ball. In short, you will actually increase the downward force on the bridge by juggling. You would be better off just carrying the balls and hoping for the best.

 

[Greg825]

The necessary force is just a smidge more than the weight of the ball and in that case, the time spent accelerating the ball would be roughly equal to the time that it spends in the air. You are correct that this is not the normal way to juggle, but that was not a condition of the puzzle.
By the way, there is no way to juggle the balls without accelerating your own fingers, or some part of your own body. Also, you need to overcome air resistance on the ball. In short, you will actually increase the downward force on the bridge by juggling. You would be better off just carrying the balls and hoping for the best.

I'm not sure that average downward force would increase, if you take the person and the three balls as a system, what net external force acts on these? the air resistance acts to decelerate the balls on the way up but also on the way down. Anyway, I'm not refuting your claim that juggling won't work, because it wont, I'm just nitpicking :).

 

[Huckleberry]

The juggler has 5 lbs to play around with while juggling the balls. When he catches a ball he must overcome gravity, 2lbs
and he must overcome the inertia of the ball with the 3 remaining lbs.
With 3lbs remaining it seems enough to me to be able to overcome the inertia that one ball would produce. Remember that he doesn't have to throw these balls up very much at all. He only needs to throw them high enough to counteract the force of gravity. If he only handles one ball at a time then when he throws one ball he will have the full 5lbs to deal with the next. Is 5lbs of force enough to arrest the motion of a 2 lb ball and provide a small lift to counter gravity? At what point in this process does he go over 5 lbs?

Just saying it is impossible according to Newton's laws doesn't explain to me why it is impossible.

This guy doesn't need to juggle the balls at all. If he crawls across the bridge on hands and knees he will be just fine. The problem states that the bridge can support 185lbs but it doesn't say over what area. Every location on a bridge does not support the same load. The weakest points on this bridge support at least 185lbs. If he was crawling then he would only need to disperse 1lb of his weight into an area that could support more. For the most even distribution of weight this bridge is likely a suspension bridge. It could be an arc, but it would have to be steep like a McDonalds arch to have a almost even load bearing capacity.

 

[Jimmy Snyder]

Just saying it is impossible according to Newton's laws doesn't explain to me why it is impossible.

This is an extremely unfair description of the generous efforts of several people to explain those laws to you. Once again, if you juggle with small force, you will need to have two balls in your hands at the same time. If you juggle with enough speed so you can hold one at a time, you will need more force. If you insist that Newton's laws don't apply because you don't understand them, then you are going to be all wet.

 

[Greg825]

If he only handles one ball at a time then when he throws one ball he will have the full 5lbs to deal with the next. Is 5lbs of force enough to arrest the motion of a 2 lb ball and provide a small lift to counter gravity? At what point in this process does he go over 5 lbs?

No, the five pounds is not enough.

In order to minimize the most force exerted on the bridge at anyone time, he needs to apply a contsant

amount of force to any of the balls and to apply it consistently through the entire time. Let's suppose he does

this succesfully. Each ball will then spend 2/3 of a unit of time per cycle in the air and 1/3 of a unit

of time being accelerated by the person.

the ball will have mass M and velocity V. Acceleration due to gravity is 9.81 m/s^2, "gravity" will denote acceleration due to gravity.

V1 = - 9.81 * (2/3)t [this is the initial velocity as it reaches the person's hand, this comes from velocity = acceleration * time]

v2 = - 9.81 * (-2/3)t [it needs the opposite velocity to attain the original height. this comes from velocity = acceleration * time]

V2 = V1 + Net Acceleration* (1/3)t [from resultant velocity = initial velocity + acceleration * time]
V2 = V1 + (applied acceleration - gravity)* (1/3)t

solving for applied acceleration we find that

acceleration = (v2 - v1) * 3/t + gravity

acceleration = (- 9.81 * (-2/3)t - -9.81 * (2/3)t) *3/t

acceleration = 29.43 m/s^2

we now have mass ( a 2 lbs ball has a mass of .90718 kg) and acceleration.

Force = mass * acceleration

Force = 26.6983 Newtons

A Newton to lbs conversion yields that he needs 6 pounds of force, just as he would need if he carried

all three at once and in compliance with Newton's laws.

 

[DaveC426913]

(Why will juggling the balls not work? Because catching the balls and throwing them again will - at least temporarily - increase your weight beyond the tolerance of the bridge.)

How do we know? Let's simplify the experiment.

Forget the juggling balls for a moment. Jump up and down. Every time you land and push off you increase the weight that the bridge experiences.

Only a fool will step onto a bridge that barely holds him and begins jumping up and down. We KNOW this will immediately cause the bridge to break.

 

[Huckleberry]

I don't think they were unfair questions. I think they were relatively simple questions.

Is 5lbs of force enough to arrest the motion of a 2 lb ball and provide a small lift to counter gravity? At what point in this process does he go over 5 lbs?

I never insisted that Newton was wrong or that the laws of gravity don't apply to me. I stated a few times that I just want to understand why this won't work. I was looking for understanding, not trying to debunk Newton. If I just accept Newton's laws because Newton said so then I'm accepting it on faith and that would be dogmatic, and I would be learning nothing.

I apologize for upsetting anyone.

 

[whozum]

The amount of force you need depends on how high you initially threw the ball in the first place. If you apply a force of m_{ball}g[/tex] then you are displacing the force of gravity, the net force is zero, however you have a nonzero momentum. You will need to apply additional force to counter this momentum, which inherently depends only on how high the ball was at its peak. <br /> <br /> Then you have a circle of the amount of force needed to raise the ball to its peak, you end up going in circles. <br /> <br /> If the ball was let&#039;s say 3m high, then when it falls to your hand it will have velocity v= \sqrt{2gh} = 7.7m/s with mass 2lb = 0.91kg, momentum of 7kgm/s. Say you are just &quot;slapping&quot; the balls back up, then we can say impact time can be half a second, you know that F = \Delta mv / \Delta t, 7 / 0.5 = 14N. 14N to just stop the ball. To throw it back up 3m you will need even more force.<br /> <br /> The ball that initially weighed 8.9N requires 14N to stop it from falling.

 

[BicycleTree]

The necessary force is just a smidge more than the weight of the ball and in that case, the time spent accelerating the ball would be roughly equal to the time that it spends in the air. You are correct that this is not the normal way to juggle, but that was not a condition of the puzzle.
By the way, there is no way to juggle the balls without accelerating your own fingers, or some part of your own body. Also, you need to overcome air resistance on the ball. In short, you will actually increase the downward force on the bridge by juggling. You would be better off just carrying the balls and hoping for the best.

Yes--I hadn't considered the inertia of your own arms.

However, "the necessary force is just a smidge more than the weight of the ball" is not correct. Say, for example, that the force used is 10% more than the weight of the ball. Then the ball would spend ten times as long in your hand as it would in the air. It would only leave your hand at the very end of your swing, almost immediately falling back into your hand, and you have to deal with the other balls too.

Juggling very slowly, though not as slowly as that--with the ball in the air half of the time and in your hand the other half--would not require a "smidge more" force than the ball weighs. As I said, you would need to exert force equal to twice the weight of the ball.

 

[BicycleTree]

Huck:
If you use the full 5 lbs of force to accelerate each ball, then this is force equal to the weight of the ball + 1.5 times the weight of the ball. So the ball will spend 1.5 times as much time in the air as it will in your hand. But you're trying to juggle three balls.

Call the time it takes for you to accelerate one ball from your hand 1 time unit (1 time unit can actually be any length of time, depending on just how you are juggling). So you throw ball 1 into the air, and this takes 1 time unit. Ball 1 will be up for 1.5 time units. Now you throw ball 2 into the air, and this takes another time unit. Ball 2 now has 1.5 time units to go before you have to toss it up again, and ball 1 now has 0.5 time units to go. Now you throw ball 3. But, whoops--you only get half done with that (you decelerate the falling ball 3 so that it is stationary but have yet to throw it up) before ball 1 smacks back into your hand, sending you over the weight limit and destroying the bridge.

The thing is that the force exerted by the bridge on you is ultimately responsible for the location of the center of gravity of you and the balls. If that center of gravity undergoes no average motion up or down, then the average force the bridge exerted on you must have been equal to the weight of the system of you and the balls.

 

[Greg825]

The amount of force you need depends on how high you initially threw the ball in the first place.

more nitpicking :) : I don't think this is (directly) so, increasing the height increases the time allowed to accelerate the ball back upward. I showed In my proof that time cancels and height is irrelavent (this was assuming the maximum amount of time was used to accelerate a ball). What really effects the force you need is the impulse you apply, which is dependant on time spent accelerating the balls upward.

 

[BicycleTree]

Yes, it is not true that throwing it higher or lower affects the amount of force you need. It does, however, affect the impulse per throw; therefore the force does not depend on the impulse.

The only thing that the force depends on is the proportion of time the ball is in the air. Given the proportion of time the ball is in the air, I can tell you the force required to throw it, and given the force used to throw it, I can tell you the proportion of time it will be in the air. If the force is f, the proportion of time in the air is r, and the weight of the ball is w, then r / (1 - r) = f / w - 1.

 

[Huckleberry]

I see my error now. Juggling with high vertical distances creates a large difference in maximum and minimum values. Juggling with a minimum vertical height to keep the balls aloft only keeps the force required closer to the average of the weight of the balls. That would be 6lbs and more than the bridge would bear.

Thank you to everyone for putting up with my persistent questions. I think I understand now what jimmy was trying to say. Seeing the formula helped. I feel foolish for not seeing this sooner.

 

[BicycleTree]

No, you could theoretically throw the ball three miles into the air using only 5 pounds of force on each throw. The height does not matter for the force; only the proportion of time in the air or in your hand matters.