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Juggler on bridge 
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#1
Dec705, 02:17 PM

P: 176

Please tell me if my reasoning is correct.
a juggler carrying three balls wants to walk across a bridge that can only support the weight of the juggler and two balls. in order to make one trip, the juggler decides to juggle the three balls while walking across the bridge. that way, the juggler can keep one ball in the air at all times. Will the juggler cross saftely? Yes or no? using newton's 3rd law, I found that since the juggler needs to exert a force to propel a ball upward, the ball will exert an equal force down on the juggler. The following is what my book says: If M = juggler's mass, m = one ball's mass, and g = the acceleration of gravity, then according to newton's 2nd law, the juggler's total weight, holding all three balls is: F = (M + 3m)g the juggler's total weight, holding two balls is: F = (M + 2m)g the force needed to accelerate the third ball is mg + ma. Based on this, the total force downward on the bridge is: F = (M + 2m)g + mg + ma = (M + 3m)g + ma therefore, according to this book, juggling a ball will actually be worse than not juggling at all. So the answer is No, the juggler will destroy the bridge. I have only one question about my book's reasoning. Why isn't the force downward on the bridge simply F = (M + 2m)g + ma? Because 1 ball is in the air, 1 of the 2 balls in the juggler's hands has to be accelerated. The third ball is in the air. The first ball will currently stay in the juggler's hands. The second ball will then have the force of ma exerted on it to propel it into the air. Therefore, shouldn't the force downward be: F = (M + 2m)g + ma? 


#2
Dec705, 07:18 PM

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While he's accelerating the ball upward he is still supporting the weight of the ball being accelerated, i.e. he has to exert a force greater than the weight of the ball in order for it to accelerate upward.



#3
Dec705, 07:52 PM

P: 176

I think I understand what is happening except I'm confused about the equations.
My book says the total downward force on the bridge is: F = (M + 2m)g + mg + ma = (M + 3m)g + ma (with "mg + ma" being the force of the third ball while the juggler is exerting a force to accelerate it upward) According to the book, the juggler starts (or is always) holding all three balls, and begins to throw one upward. But I'm thinking that the juggler is already juggling, and therefore has one ball in the air at all times. Then the reason for the juggler breaking the bridge is that the force exerted to accelerate one ball (one of the two balls in the jugglers hand) adds to the force of holding two balls. Algebraically, F = (M + 2m)g + ma or F = [M (juggler mass) + m (1 ball) + m (1 ball)] * g (gravity) + m*a (the force to accelerate one ball upward) Because of Newton's third law, we have m*a, the equal and opposite force of the juggler's propelling a ball upward. So who is right? The book or me? Are we both right? For sure, either way, the answer to the question is that the juggler will not make his way safely across the bridge. 


#4
Dec705, 08:03 PM

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Juggler on bridge
I think endeavor's question is why does the weight of the third ball (the one in the air) enter into the answer at all, since the person is obviously not exerting a force on that ball. Good question! I think the book's reasoning is a bit sloppy. (I'll have to think about the specific forces in more detail.)
But one way of thinking about it makes it easy to answer: What's the average force that the juggler exerts on the balls? Since the balls (considered as a system) are not accelerating on average, the average force the juggler exerts on them must just equal their weight = 3mg. Of course, since whenever he catches or throws a ball the force spikes up a bit, the force does exceed the weight of all three balls at times. 


#5
Dec705, 08:11 PM

P: 176

I have to apologize. I did not state my question explicitly.
yes, that's my question Doc Al. I don't know if my next question is silly or not, but: Assume the 1st & 2nd balls are in his hands, and the 3rd is in the air. What if the juggler, while throwing up ball 1, moves his hand that is holding ball 2 downward? Would this relieve some stress on the bridge, while allowing the juggler to throw up ball 2? 


#6
Dec705, 09:57 PM

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It's a losing proposition since he must also catch the balls on the way down and, on average, the weight threshold of the bridge will be exceeded.



#7
Dec705, 10:34 PM

P: 949

Tide hit the nail on the head  you can figure it out second by second if you want, but if he walks across the bridge with three balls, no matter what, the bridge is "on average" going to experience the weight of one juggler and three balls.
If you want to model the problem in a different way, imagine a person holding one ball, and just throwing it up and catching it. It's easier to imagine this "evening out" to being the same as just holding it. The original question is essentially this, only you could also imagine a threearmed person with three balls, each arm playing catch with itself. Each one will even out. Or, as a third way of thinking about it, you could say this: Neither the juggler nor any of the balls fall down (as in, dropped off the bridge)  so in order for the bridge to not have an average load of at least the juggler+3 balls, there would have to be an outside force involved. As you can see, there is not. 


#8
Dec805, 02:16 AM

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Ok, there IS a way for the jugler to get over the bridge safely, and that is by throwing one ball so high (just before he gets on the bridge) that he catches it only when he has crossed it



#9
Dec805, 02:56 AM

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#10
Dec805, 03:10 AM

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#11
Dec805, 04:09 AM

P: 84

If we are going to nit pick about a butterfly’s wing why not do the same about the jugglers arm? When the ball is accelerated upwards the hand is having the same acc. while the shoulder is at rest. Unless the balls are made of one ton of steel, we are dealing with the normal kind of ball, the mass of arm is about 100x bigger.
This gives an extra downwards force of F=100/2a. He/she needs a good insurer! Don’t you just love physics? 


#12
Dec805, 04:21 AM

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Good point, erick! And it's more than just his hand that is accelerating, it's his whole arm with recoil and rotation of the whole body so you would need some moments of inertia  rotation of forearm about the elbow, whole arm about the shoulder, etc.



#13
Dec805, 09:05 AM

P: n/a

If the guy starts juggling before entering the bridge, the force (M + 3m)g + ma is exerted on firm ground and causes no problem. If, after entering the bridge, he tosses the second ball before the first drops, he will be exerting on the bridge the force (M + 2m)g + ma, as you said and when he catches the first ball, while the second is in the air he will exert again a force of (M + 2m)g + ma.
If a < g, he could cross by maintaining at any instant 2 balls in the air. The force when throwing or catching each ball woul be (M + m)g + ma. 


#14
Dec805, 09:48 AM

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I don't think it matters how many balls he maintains in the air (unless the bridge is really short ). He still must support the weight of the balls on average.



#15
Dec805, 11:51 AM

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Congrats on 6,000 posts. 


#16
Dec805, 12:06 PM

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#17
Dec805, 12:19 PM

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In practice, no juggler could achieve it. 


#18
Dec805, 12:34 PM

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Got a weird question for you here. Since the math stuff totally eludes me, I'm wondering about the speed of crossing.
If the bridge would support him and two balls crossing at a normal walking pace, would creeping along very carefully reduce the impact loading of his steps enough that it might let him across with 3? 


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