This one may be easier than cannibal logic

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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.
  • #36
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.
 
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  • #37
BicycleTree said:
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.
 
  • #38
jimmysnyder said:
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 :).
 
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  • #39
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.
 
  • #40
Huckleberry said:
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.
 
  • #41
Huckleberry said:
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.
 
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  • #42
(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.
 
  • #43
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.
 
  • #44
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 [itex]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.

Then you have a circle of the amount of force needed to raise the ball to its peak, you end up going in circles.

If the ball was let's say 3m high, then when it falls to your hand it will have velocity [itex] v= \sqrt{2gh} = 7.7m/s [/itex] with mass 2lb = 0.91kg, momentum of 7kgm/s. Say you are just "slapping" the balls back up, then we can say impact time can be half a second, you know that [itex] F = \Delta mv / \Delta t, 7 / 0.5 = 14N [/itex]. 14N to just stop the ball. To throw it back up 3m you will need even more force.

The ball that initially weighed 8.9N requires 14N to stop it from falling.
 
  • #45
jimmysnyder said:
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.
 
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  • #46
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.
 
  • #47
whozum said:
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.
 
  • #48
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.
 
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  • #49
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. :blushing:
 
  • #50
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.
 
  • #51
BicycleTree said:
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.

And to minimize the maximum (there's got to be a better way to phrase this ) force needed, all of the time spent must be used accelerating the balls upward. ie carrying them. As someone said, you're better off carrying them than juggling them, unless you can manage to utilize all of the time perfectly.
 
  • #52
Conversely, you could theoretically juggle the balls using 1 ton of force and still only have a throwing height of 1 inch. For the three mile case with 5 lb force you'd just apply the force over a long period of time, and for the 1 inch case with great force you'd just apply it over a very short period of time.
 
  • #53
BicycleTree said:
As I said, you would need to exert force equal to twice the weight of the ball.
If you exert a force exactly equal to the weight of the ball, then there will be no net force on the ball and it will move with constant velocity. A smidge more and it will accelerate upward.
 
  • #54
A smidge more and it will decelerate and eventually accelerate upward very, very slowly, and will spend almost all of its time in your hand. To have it spend even half the time airborne, you must provide much more than a smidge more.
 
  • #55
Huckleberry said:
I don't think they were unfair questions.
Now you have twisted my words.
 
  • #56
jimmysnyder said:
Now you have twisted my words.
Private praise and public apologies.
You are right. Rereading the thread I can see that I have made many many mistakes here. I made incorrect assumptions of Newton's laws, which you and others have been patient enough to answer.
I also made some statements that were misguided. For this I apologize to you. Early on in this thread I was set off balance by a comment directed against me. I've been having a bad time lately and was defensive even before I read that comment. I didn't even realize how poorly I misinterpeted your intentions. Here are two things I never should have said.
Huckleberry said:
Just saying it is impossible according to Newton's laws doesn't explain to me why it is impossible.
Huckleberry said:
I don't think they were unfair questions.
I hope that you will accept my apologies and we can go on without any hard feelings. I am sorry.
 
  • #57
Greg825 said:
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.


Thats pretty much exactly what I said, The force you need to apply must be greater than the impulse of the ball's impact and the impulse to get rid of it again.

I'm looking for your proof now though.

edit:
increasing the height increases the time allowed to accelerate the ball back upward.

I don't tihnk this is so at all. Regardless of how high the person threw it, assuming it is not along the magnitude of his maximum throw ability, that he can maintain the same contact time with the ball regardless of how high it comes down on him. Howver, your and my experiments are under different premises.

My statement says that to reach the original height of the preceding throw one must apply an impulse with twice the magnitude of the ball as it hits his hand. How long it takes him to do this doesn't really matter until you start considering time constraints of the other balls, which I didnt. I merely simplified the juggle so Huckleberry could see that it takes more force to catch and throw something than it does to just hold it.
 
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  • #58
I have some apologies to make too.

To Huckleberry, I wrote the 'twisted' post without having read an earlier post of yours in which you had already said you understood the reason juggling fails. In any event some of my responses to you were born of impatience.

To Greg825 and BicycleTree, In my post #34, I thought I had explained things rather well, but in rereading, I see that I did not make it clear. I meant that you would have to exert a force equal to twice the weight of the ball, just as BicycleTree said. One unit of weight to support the weight of the ball and one unit to reverse the velocity. I think Greg825's post #35 missed that point (quite understandably so) and my posts on the subject after that point are incoherent.
 
  • #59
whozum said:
Thats pretty much exactly what I said, The force you need to apply must be greater than the impulse of the ball's impact and the impulse to get rid of it again.

I think the fault may be mine here, I didn't really understand what you were saying in most of your post when I responded, I was only refuting that first statement.

whozum said:
I don't tihnk this is so at all. Regardless of how high the person threw it, assuming it is not along the magnitude of his maximum throw ability, that he can maintain the same contact time with the ball regardless of how high it comes down on him. Howver, your and my experiments are under different premises.


My statement about more time allowed was based on the two premises that the force applied would be constant and this force is desired to be minimal.



I'm also unsure of what you mean by "I don't tihnk this is so at all. Regardless of how high the person threw it, assuming it is not along the magnitude of his maximum throw ability, that he can maintain the same contact time with the ball regardless of how high it comes down on him.", my saying he is allowed more time to accelerate it upward doesn't conflict with the fact that he can also not use more time if he doesn't want to. Neither of these statements are contradictory as they are simply two proposed possibilities. Can you please clarify?
 
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  • #60
I was just pointing out that they arent necessarily requirements, just possibilities. I was oging for constant time, you were going for constant force.
 
  • #61
if the bridge is vertical, i could just jump with the ball and the mullet. it doesn't matter much wether i DO survive right.
 

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