Bobo the Clown wants to cross a bridge over dangerous waters while performing a three ring juggle, in which one or two rings are in the air at all times. The bridge can support a maximum weight of $\SI[per-mode=symbol]{1000}{\newton},$ above which it will collapse. Each ring weighs $\SI[per-mode=symbol]{100}{\newton}$ and Bobo himself weighs $\SI[per-mode=symbol]{800}{\newton}.$

Coco the Clown tells Bobo that it will be safe to cross the bridge as long as Bobo is
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never
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holding all three rings at once. Dodo the Clown disagrees, arguing that the juggle will collapse the bridge, even if at least one ring is in the air at all times.

Who is correct, Coco or Dodo?

Coco
Dodo

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If Bobo starts to throw two ring into air before he sets feet onto the bridge and keep two rings in the air all time he is on the bridge, then he should have a chance to make his way to the other end, shouldn't he? Providing he catches the ring gently enough. Any one agrees?

Chuanli Zhao
- 4 years ago

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No, he's saying that there are 2 rings in the air at all times. It doesn't seem implausible. I will try here. For this situation to be plausible, Bobo must throw a ring in the air for exactly half the time that it is the air (since there are 3 rings, they must each be in the air twice as long as in the hand).

How long is a ring going to be in the air? We know that position must equal 0 when it lands. Using kinematics (and g = 10)...

$position = 0 = F_\textrm{toss} * t - 5t^2$

$t_\textrm{in air} = F_\textrm{toss} / 5$

(The 0 root is the start of the throw, and we only care about the second root.)

So, since this time is twice the time to apply the force, we have that

$t_\textrm{to throw} = 2 * F_\textrm{toss} / 5 = F_\textrm{toss} / 2.5$

Now, we have to relate the time to throw to the amount of upwards force. F = ma. Thankfully, the optimal scenario is constant downwards acceleration, so I can replace acceleration with velocity times time, saying

$F_\textrm{toss} = mv_\textrm{final}t$

However, note that the force of the toss is half that of the actual force, since slowing a falling ring will take as long as much force as throwing it back up.

$F_\textrm{toss}/2 = m * v_\textrm{final} * t_\textrm{to throw}$

Okay, now we have the equations, so let's do math. Substituting for t...

$F_\textrm{toss}/2 = mv_\textrm{final} * (F_\textrm{toss} / 2.5)$

$5/4 = mv_\textrm{final}$

With a mass of 100...

$5/400 = 1/80 = v_\textrm{final}$

Yay, equation! But, eh, what does it mean? Well, simply put, if Bobo throws his rings up in the air such that their velocity is equal to 1/80 meters per second or less, then he can, actually, make it across the bridge. Is this realistic? Well, with an upwards velocity of 1/80, we're expecting each ring to be in the air (by the above equation) for about 1/400 seconds. Which means each throw takes place in 1/800 seconds. So, if Bobo can manage 800 carefully balanced throws in a second, while walking across the bridge, he can, actually, get across with 3 rings using your strategy.

Alex Li
- 4 years ago

Yes that makes sense. Thank you for clarifying that. When I said he he was pushing down onto the bridge, he is actually accelerating his body downwards and although his mass does not increase, the force he is exerting on the bridge is increasing. Which produces the extra force needed to throw the ring into the air and to catch the other ring when it falls into his hand.

Kevin Quigg
- 4 years ago

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So basically, this whole time, you were saying that my solution should have clarified a couple of things like that?

If so, sorry, I'm kind of new with this.

Deva Craig
- 4 years ago

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When I joined the discussion I wasn't trying to criticise you personally. I genuinely didn't understand what you were trying to say and I wanted you to explain further. Your solution concentrated on the motion of the juggling ring. You seemed to imply that when the ring was accelerating it produced a force that caused the bridge to collapse, without explaining how that force was transmitted downwards onto the bridge. When I asked you to explain further you kept talking about the speed of the ring, but what I was trying to understand, was the forces on the bridge that would caused it to collapse. I knew that the ring was being accelerated upwards by Bobo throwing it into the air, but what I forgot was that when the ring was being accelerated upwards, Bobo was accelerating downwards, thus increasing his 'weight' or downwards thrust, onto the bridge. In order to throw the ring into the air, Bobo had to push downwards onto the bridge and it was that downward push that collapsed the bridge. Discussing problems can be useful and lead to a more complete understanding of the solution. Although I found our discussion confusing, it did help me to think more clearly. When Josh posted his solution with the equations, describing the forces involved, I was eventually able to understand what was going on.

Kevin Quigg
- 4 years ago

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@Kevin Quigg – Thanks! I was trying to explain that to you, but I kind of had trouble expressing it in a way you'd understand it. I mean, I guess I don't really have to worry about that so much because I am only learning how to write good solutions, but I guess the moral of the story is that the main purpose of the discussions is to communicate our ideas and to learn something new from all of it, not so much to argue about who's right and who's wrong or to show off your intelligence to other people.

And to be perfectly honest, I feel as though most of these "smart" people tend to use their intelligence or just learn stuff they're probably never going to use as a means of showing off rather than learning about the world due to its beauty and magnificence.

Deva Craig
- 4 years ago

Interesting Thought about Elevators: When it says weight limit 1 Ton, does it consider if all occupants jump into the air at the same time?

J C
- 4 years ago

Coco is not correct. Once Dodo is holding two rings, he and the two rings are at 1000 N. And at that same moment, as he launches the third ring upward, he has just increased his effective weight downward, beyond 1000 N. The bridge collapses.

Dennis Rodman
- 2 years, 1 month ago

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Relevant wiki:
Newton's Second Law
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According to Newton's Second Law of Motion, an object's net force is the result of its change of momentum over time. Suppose that Bobo throws one of the three rings 1 meter high. If that were the case, then v $^{2}$ = 0 + 2(10)(1), which means that the final velocity is approximately 4.5 m/s. The momentum of one of the rings would then be the mass of the ring (Newtons divided by g) multiplied by 4.5 m/s, 10 kg times 4.5 m/s, or 45 $\frac{kg * m}{s}$ . If it took 0.2 seconds for Bobo to catch the ring, then the amount of downward force exerted by the ring would then be 225 N.

This same problem can also be solved by considering Newton's Third Law of motion. According to Newton's Thrid Law of Motion, for every action, there is an equal and opposite reaction. In order for Bobo to throw any of the three rings into the air, he must apply a force that is higher than the upward force of one of the rings themselves. This creates a larger downward push on him than if he had just carried the rings across, causing the bridge to collapse.

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What if he threw the first one before he stepped onto the bridge and then always kept one in the air?

Joshua Hall
- 4 years ago

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That doesn't change the conditions of the problem. He is still exerting a higher upward force on the rings when he gets into the bridge, which creates a high downward force.

Deva Craig
- 4 years ago

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Yes, when he catches the ring then the ring will weigh more than 100N due to its retardation.

Rohit Gupta
- 4 years ago

But the bridge needs simultaneously 1000N to break, so could you not distribute the force of supporting the rings over time through juggling?

George Salafatinos
- 4 years ago

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@George Salafatinos – Not necessarily. The reason the downward force increases in this problem is because the downward force of Bobo increases as he throws the rings. Even if he successfully catches the rings, the ring will end up exerting more than 100 N due to its retardation, or its application of negative acceleration.

Deva Craig
- 4 years ago

He would need to always have at least two rings in the air. Any force used to propel any ring upward would be matched by an increased force downwards with Bobo pressing harder on the bridge.

Richard Desper
- 4 years ago

Hmmm, the assumption seems to be that the forces exerted by all three rings on the bridge are simultaneous, maybe even continuous, which seems doubtful. Also, as a juggler I can say that usually two of the three objects are in the air. As Mr. Fleming said, it's kind of a duff question.

Mike Dyer
- 4 years ago

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That wasn't really the assumption, my solution just wasn't clear enough. I've fixed it.

Deva Craig
- 4 years ago

The ring in the air does not exert any force on the bridge.

Rohit Gupta
- 4 years ago

This seems to discuss the difference between static weight and dynamic weight. The motion and effort uncreasing the total weight. A similar argument could me made if Bobo just carried two rings across. The movement and shift of his weight would create in excess of 1000 lbs of dynamic weight. Think about a runner weighing 1000 N approaching this bridge. His first step will exert more than 1000 N of weight onto the bridge.

Thom Disch
- 4 years ago

If Bobo had thrown only 2 rings at the same time, the bridge could be easily collapsed.

And if Bobo threw the ring in such a way that the net acceleration on a single ring was upward then only one ring could have collapsed the bridge.

Kaushik Chandra
- 4 years, 1 month ago

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if you could keep force of each throw, catch to less that 200N (something which criteria however unlikely is not specified) then the bridge will not collapse - Reality bridge will collapse but it is based on assumption - its a bit of a duff question. ie if we were to say this clown had upward motion of 100N in a swinging hand catch realease of a descending ring then we could theorise that no impact on weight other than the 1 ring in his hand

David Fleming
- 4 years ago

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See, the ring will move upward only when the upward acceleration acting on it is greater than the acceleration due to gravity. So, net upward acceleration becomes 10+x ms^-2. By the newton's third law, we get the reaction force acting on the ground to be more than 200N. Thank you!

Kaushik Chandra
- 4 years ago

Has this been demonstrated experimentally? As in using a scale accurate to multiple decimal points.

Veiran X
- 4 years ago

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Go on a weighing scale and jump. You will see that your weight increases significantly as you push yourself into the air.

Alex Li
- 4 years ago

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Unfortunately, the scale I have isn't quick enough to "see" that.

Veiran X
- 4 years ago

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@Veiran X – I've added on Newton's Second Law onto my solution for a more complete explanation.

Deva Craig
- 4 years ago

@Veiran X – Hmm. Well, imagine that you are on a diving board. When you jump (or land), you push the diving board down much more than when you simply stand on it.

If you imagine jumping off verses just stepping off, hopefully you will imagine that jumping off makes the diving board go down quite a bit more.

An alternative way of thinking about things is perhaps if you're on a fragile surface. If you want to break it, you will jump on it. Why? This applies a larger downwards force than simply standing.

Alex Li
- 4 years ago

I agree that the bridge would fall and I thought that when I read the problem. But the problem states "The bridge can support a maximum weight of above which it will collapse." Not a real world problem, but your correct answer includes force, which is "real world".

Ron Tha
- 4 years ago

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What do you mean by "real world"? The weight limit of the bridge is in fact a limit on the normal force that can be applied on it before it breaks.

Pranshu Gaba
- 4 years ago

I do not understand your logic. I do not understand how you calculate a speed of 4.5 m/s or how it is relevant to the problem. If he throws the hoop upwards at 4.5 m/s it will decelerate due to gravity until it reaches a speed of 0 m/s at which point it will start to fall under the influence of gravity, exerting a force of 100N. An object does not gain weight, when it is thrown into the air! If the ring is thrown into the air with an initial speed of 4.5 m/s then it will reach its maximum height at approximately 0.45 s, it will then fall under gravity coming back to its original position after approximately 0.12 s, so it will be in the air for approximately 0.57 s. How did you calculate 0.2 s?

Kevin Quigg
- 4 years ago

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The problem here seems to be that you mistook the velocity I found as the average velocity of the ring. In reality, the velocity I found was the final velocity. I didn't make that clear in my original solution, sorry! The final velocity squared is calculated by adding the initial velocity squared onto the product of twice the acceleration and the displacement of the ring. We need to find the final velocity of the ring in order to find the momentum of the same ring when Bobo catches it, which is directly proportional to the velocity of the ring.

Deva Craig
- 4 years ago

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I am still confused, the only force acting on the bridge is the weight of the clown and the rings due to gravity. When the ring is pushed up into the air the clown can only use his own weight to push the ring upwards. When the ring falls back down it is falling under gravity. If the clown only has two rings in his hand at any one time there is only1000N pushing down on the bridge. There are no extra forces!

Kevin Quigg
- 4 years ago

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@Kevin Quigg – Remember that Newton's Third Law states that for every action, there is an equal but opposite reaction. When Bobo throws a ring, a downward force is also be exerted on him, which exceeds the 1000 N limit.

Deva Craig
- 4 years ago

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@Deva Craig – Yes when Bobo throws the ring up into the air he pushes down onto the bridge with the same force. But he is generating that force using the weight of his own body. If he throws the ring with 225N, as you say in your answer, he also pushes down onto the bridge with 225N, But that 225N is part of the 800N that he has by virtue of his mass and the force of gravity acting on his body. He is only re distributing his own weight. He can NOT generate extra mass in order to throw the ring into the air. He can only use the weight that he has. There is NO extra force.

Kevin Quigg
- 4 years ago

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@Kevin Quigg – Once again, you're confusing the values I've found for average values, the values I've really found (The 225 Newtons and 4.5 m/s) were the final values, the values of the force and velocity when Bobo catches the rings, 225 Newtons is not the amount of force Bobo exerts on the ring as he throws it, and 4.5 m/s is the final velocity, in other words, the velocity of the ring when it stops in its path.

Also, you have to keep in mind that an object does NOT always need to change its mass in order for its net force to change. Mass is not the only factor that determines how much net force an object has while it's in motion. Net force can also change if the acceleration changes, but the mass remains constant as demonstrated in Newton's famous equation F=ma. Another way an object's Net Force is determined is if we find the change in momentum and divide it by how much time it takes for the object to stay in the air, which is in fact where we can derive Newton's famous equation in the first place. In other words, how an object's net force changes depends on the circumstances.

Let's take this problem for example. Before Bobo throws a ring in the air, the only force acting on the ring right now is the force of the pull of the earth due to gravity. When Bobo throws a ring there are two forces acting on the ring. There's (1) The force of the pull of the earth due to gravity and (2) The upward force exerted by Bobo to make the ring get in the air in the first place. The latter force is known as an applied force, a force which is put on an object through the application of a push or pull on it. The value of the latter force will decrease due to the acceleration of the ring as it makes its trip. At one point though, the total velocity of the ring will be zero. That is when the ring reaches its maximum height. After that, the force you've already put on the ring becomes a downward force since it is still decreasing due to acceleration, which therefore increases the total downward force of the object.

Let's also talk about the amount of time I've used to prove the solution to this problem. The reason why the amount of time would be arbitrary would be because the amount of time doesn't really change the fact that downward force would increase. That downward force is really the result of the acceleration of the force of the pull of the earth due to gravity.

Deva Craig
- 4 years ago

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@Deva Craig – No! When the ring stops in its path the velocity is 0 (by definition). The ring reaches its maximum upward velocity when it leaves Bobo's hand. The ring only accelerates upwards when it is in Bobo's hand. Once Bobo releases the ring there is no upward force on the ring and its upward acceleration will cease, it will however accelerate downwards due to gravity. It will still travel upwards, but its velocity will decrease until it reaches zero, then it will travel downwards. The amount of time is not arbitrary. The amount of time it spends in the air is determined by the velocity it has when it leaves Bobo's hand, this determines how high it will travel and the time it takes to reach its maximum height can be calculated. Similarly the amount of time it takes to fall back into Bobo's hand can also be calculated.

Kevin Quigg
- 4 years ago

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@Kevin Quigg – Sorry about that, I've confused the two in the explanation. What I meant to say was that the velocity would be zero at that point. I probably should have never written a solution to this problem because I'm so horrible at it and I'm making your more confused! I really wanted to help you understand what's going on!

Deva Craig
- 4 years ago

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@Deva Craig – I find your solution good. No matter what values you plug in, the bridge will break if Bobo throws or catch a ring on the bridge.

Rohit Gupta
- 4 years ago

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@Rohit Gupta – I'm having a hard time explaining to Kevin why the time would be arbitrary in this case, which I feel is confusing him a bit.

Deva Craig
- 4 years ago

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@Deva Craig – Ahh, it is ok. Sometimes it is difficult to explain yourself. But, we should always try. I am able to make perfect sense of your solution and understand what you are trying to say.

Rohit Gupta
- 4 years ago

@Kevin Quigg – When the clown catches the ring, the ring pushes on him with a force greater than its weight. That is why the ring deaccelerates and stops in the hands of the clown. This extra force is transferred to the bridge through the legs of the clown and the bridge collapses.

Rohit Gupta
- 4 years ago

All the numbers chosen in this solution are arbitrary and could be replaced with any positive number. The point is the total weight will be greater than that of both rings at 1 point.

Alex Li
- 4 years ago

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The numbers cannot be arbitrary, the juggler and the rings can only act in accordance with the laws of physics. The position and speed of the rings along with the forces acting on the bridge can be calculated using the laws of physics, you cannot just pull numbers out of a hat!

Kevin Quigg
- 4 years ago

I think your answer is fine but the problem should say that the ring is thrown to a non zero height relative to his hand otherwise I Can think that he is switching infinitely fast between the two rings without throwing them upward

Antonio Di Bacco
- 4 years ago

If he could toss 2 rings high into the air before stepping onto the bridge, then hurry across the bridge while juggling the third ring in any fashion, then catch the 2 rings after reaching the end of the bridge, then it could be done.

Garth Kroeker
- 4 years ago

In "theory" catching-force could be equal and opposing throwing-force, and both actions could occur simultaneously. In practice, there would be some variance and maximum would be exceeded by at least some small amount.

Jake Abernethy
- 4 years ago

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In catching as well as in throwing in both the cases Bobo has to apply an upward force on the rings.

Rohit Gupta
- 4 years ago

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Oh shoot. Duh. You're right. Should've thought about that a little longer before commenting. (c:

Jake Abernethy
- 4 years ago

How does an 80kg man juggle 3 off 10kg rings?!

Douglas Hamilton
- 4 years ago

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Yes, exactly. To stop the incoming ring, the clown has to apply a greater force than its weight. An equal amount of extra force is exerted by the ring to the clown. This extra force is transferred to the bridge through the legs of the clown and the bridge will experience a force greater than 1000N and will break.

Rohit Gupta
- 4 years ago

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Where does the extra weight come from? The only force exerted by the ring is the force of gravity i.e. 100N. There is no extra force.

Kevin Quigg
- 4 years ago

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While catching the extra force is required to stop the incoming ring.

Rohit Gupta
- 4 years ago

When Dodo came into scene? ;-)

Félix Pérez Haoñie
- 2 years, 8 months ago

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Don't confuse Coco and Bobo! :)

Richard Desper
- 4 years ago

When he throws the rings up in the air, an equal but opposite downward force is being put on him, which essentially means the rings are pushing him down as he throws them up.

Deva Craig
- 4 years ago

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And what do you imagine is supplying the normal force to Bobo?

Daniel Newby
- 4 years ago

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If the rings push down Bobo, then the bridge will try to push Bobo up by supplying the normal force. If the bridge is not able to supply enough normal force, then it will collapse, and Bobo will start falling down.

Pranshu Gaba
- 4 years ago

the balls are getting extra motion which give extra force

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As the problem states, Bobo alternates between moments where he has one or two rings in his hands.

Coco claims the bridge will stay intact as long as Bobo never holds three rings simultaneously, which implies that in the periods when two rings contact Bobo's hands, they can't ever require more than $\SI{200}{\newton}$ to support.

Let's consider one of the moments where Bobo goes from holding two rings to holding one ring, by throwing one ring up into the air. In order to simply hold a ring, Bobo (and therefore the bridge) has to oppose the pull of gravity by providing a support force of $m_\textrm{ring}g = W_\textrm{ring} = \SI{100}{\newton}$ upward. To throw a ring upward, Bobo must accelerate it upward with acceleration $a > g$ which means he (and ultimately the bridge) must provide a force $F = ma > mg = W_\textrm{ring}.$

Therefore, in the moment that Bobo tosses one of the rings upward, the total force the bridge must support is $\begin{aligned} F_\textrm{total throw} &= W_\textrm{Bobo} + W_\textrm{ring} + ma \\ &\gt W_\textrm{Bobo} + W_\textrm{ring} + W_\textrm{ring} \\ &= \SI{1000}{\newton} \end{aligned}$ which shows that $F_\textrm{total throw} > \SI{1000}{\newton}$ and the bridge will collapse due to the juggle.

Now, consider the case where Bobo perfectly times the catch of the incoming ring with the upward throw of the other. In this case, things are even worse!

To decelerate the incoming ring, Bobo has to accelerate it upward with acceleration $d > g$ (or else it would still be accelerating downward!), and therefore provide a force upward given by $F_\textrm{catch} = md > mg.$ Thus, both $F_\textrm{catch}$ and $F_\textrm{throw}$ require upward force in excess of $W_\textrm{ring},$ and the bridge will have to support even more force than in the case we analyzed above.