# Juggler on bridge

#### endeavor

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?

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#### Tide

Homework Helper
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.

#### endeavor

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.

#### Doc Al

Mentor
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.

#### endeavor

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?

#### Tide

Homework Helper
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.

#### KingNothing

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 three-armed 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.

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#### vanesch

Staff Emeritus
Gold Member
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 :tongue2:

#### Tide

Homework Helper
vanesch said:
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 :tongue2:
That won't work either. The exertion causes the guy to double over in pain dropping the other two balls into the river while the launched ball strikes a high flying Lorentz butterfly whose wing flapping would have diverted a tornado to Brazil but, as a result of the butterfly's incapacitation, a low pressure system upstream produces a sudden cloudburst with the resulting deluge making the river swell and sweep away the bridge before the guy recovers ...!

#### vanesch

Staff Emeritus
Gold Member
Tide said:
That won't work either. The exertion causes the guy to double over in pain dropping the other two balls into the river while the launched ball strikes a high flying Lorentz butterfly whose wing flapping would have diverted a tornado to Brazil but, as a result of the butterfly's incapacitation, a low pressure system upstream produces a sudden cloudburst with the resulting deluge making the river swell and sweep away the bridge before the guy recovers ...!
Damn, you're right, didn't think of that :tongue:

#### erickalle

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?

#### Tide

Homework Helper
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.

#### SGT

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.

#### Doc Al

Mentor
I don't think it matters how many balls he maintains in the air (unless the bridge is really short :tongue2:). He still must support the weight of the balls on average.

#### KingNothing

Doc Al said:
I don't think it matters how many balls he maintains in the air (unless the bridge is really short :tongue2:). He still must support the weight of the balls on average.
I think what he meant was that if the juggler threw a ball up just before leaving, on average it could support slightly less than 1J+3B on average. I think that's what he meant, anyway.

Congrats on 6,000 posts.

#### Doc Al

Mentor
KingNothing said:
I think what he meant was that if the juggler threw a ball up just before leaving, on average it could support slightly less than 1J+3B on average.
But unless that tossed ball never comes back down again while he crosses the bridge, he's going to end up supporting its weight--on average. (That's why I said "unless the bridge is really short".)

Congrats on 6,000 posts.
Thanks!

#### DrGreg

Gold Member
Doc Al said:
But unless that tossed ball never comes back down again while he crosses the bridge, he's going to end up supporting its weight--on average. (That's why I said "unless the bridge is really short".)
Hypothetically, an extremely skilful juggler could juggle in such a way that the centre of gravity of the juggler-and-3-balls system has a small downward acceleration throughout the crossing. Then the force on the bridge would be slightly less than the weight of the juggler plus three balls.

In practice, no juggler could achieve it.

SGT said:
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.
For a < g to be true, the juggler would have to do what I described above.

#### Danger

Gold Member
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?

#### erickalle

Danger said:
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?
That sure is possible even more so when he waits till:
a. there’s a full moon over the bridge
b. the earths spin starts to increase again
c. he’s done some exercising
In fact you mentioned the only way it can be done without cheating.

#### Doc Al

Mentor
DrGreg said:
Hypothetically, an extremely skilful juggler could juggle in such a way that the centre of gravity of the juggler-and-3-balls system has a small downward acceleration throughout the crossing. Then the force on the bridge would be slightly less than the weight of the juggler plus three balls.
Good point! Under those conditions, the average force would be slightly less than the weight of the juggler plus three balls. (That probably won't help him cross this bridge, since it can only support the weight of the juggler plus two balls.)

#### endeavor

He still must support the weight of the balls on average.
Why must he support the weight on average?

If the juggler started juggling before he started to cross, isn't what SGT said right?
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.
and therefore, the reason why he does not cross safely is the force of tossing up a ball (ma).

**wow, i never imagined this post would generate so much interest....

#### Doc Al

Mentor
endeavor said:
Why must he support the weight on average?
To keep the balls from falling!

Since the balls have weight, unless the juggler exerts a force balancing that weight, the balls (or at least the center of mass of the set of balls) must accelerate and fall.

It's like trying to reduce your weight by hopping from one foot to the other (or just hopping up and down). It won't work! (In fact, it makes it worse--while the average force that the floor must exert to support you will equal your weight, the peak force generated each time you land will be greater.)

#### Danger

Gold Member
endeavor said:
**wow, i never imagined this post would generate so much interest....
You're mistaking amusement for interest. :tongue:

It is actually a very interesting question in that even if the answer is easily calculated, the reasoning that goes into thinking it out is far from simple. Common sense, which is what I use rather than math, leads to several different conclusions, depending upon how you approach it. It's also the kind of question that is likely to grab the interest of newcomers to PF, and thus get them hooked.

#### krab

This question is like What is the weight of a box of gas? Since all the gas molecules are suspended, how can they exert any force? the answer is of course that they must sometimes hit the bottom of the box and so give impulses. Actually, they hit the top too, but with slightly less force, and the difference in forces is on average exactly the weight of the gas.

#### DarkMind

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.
No you can't reduce the weight of the balls, even with 2 in the air and one in the hand each time, and that's why: let's see, that way above, the force is of (M + m)g +ma, 79 kgforce +a/g kgforce, and he will take "t" seconds maintaining this acceleration, so the ball velocity will be of "at" and it will take "2at/g" seconds untill it goes back to the hand, with the same velocity, so it will take more "t" seconds to catch it with an "a" desaccelaration, so, calling of "x" the time that it takes until one ball land in one hand since he just dropped one another in the air, we have to say that the "2at/g" is higher than "4t + 2x", but even if "x" was zero, istill "2at/g" can't be higher than "4t" until "a" is higher than "2g", and than the bridge will not support ^^ . Obs: the horizontal forces accelarations and velocities don't matter and the only things being considerated are the vertical ones. And sorry for the english I am Brazilian by the way.

Resuming: You need to reestring the bridge so that you can pass, and I think you can't do it juggling. Maybe considerating the buoyancy this task could be solved.

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