Juggling Across a Bridge: Will It Work?

[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?

 

[Tide]

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]

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]

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.

 

[vanesch]

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]

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]

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]

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]

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]

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]

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]

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.

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]

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]

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 Earth's 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]

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]

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]

**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 until 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.

 

[Hakuin]

Average weight of the juggler + three balls

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...

DarkMind has made the whole point, but maybe failed to write the solution clearly enough let me try to help:

Let us consider the cycle of throwing a ball in the air until one catches it again. Repeating that cycle will suffice to describe any attempt at crossing the bridge. The following will happen:
Let M be the juggler's mass, and m a ball's mass
His own weight plus three balls is g(M+3m)
Let us also consider the case of a constant force used by the juggler to throw the ball or catch it. Any deviation from that would be equivalent but would need integrals to solve the problem...so let's keep it simple :yuck:

As the juggler throws the ball up, he has to deliver an upward force m.A
The first remark is that it is impossible that A be smaller than the acceleration of gravity g. Indeed, the ball is at rest in the juggler's hand because it sustains a force equal in amplitude and opposite to that caused by gravity on its mass: m.g (Newton's first law)
In order to "force" the ball up, the juggler has to exert a force opposite and greater in amplitude than the force of gravity: m(-g+A) has to be positive (upwards), so a>g. (NB: this is already enough to have the bridge collapse ) Let us assume the juggle exerts that force on the ball for a time t1, and for ease of notation, let us say that A=g+a, so that the net force on the ball is m(-g+g+a)=m.a, corresponding to a constant acceleration a.
As DarkMind says, after this time t1, the upward speed of the ball will be
v_up = a.t1

The ball now leaves the juggler's hand and goes up, being decelerated by the action of gravity, that is -g.
This takes a time t2=a.t1/g (because v=-g.t+a.t1, so v=0 -> t2=a.t1/g)

The ball starts to fall, it takes it a time t2 to fall back to the juggler's hand, by which time it is falling at speed v_down=-a.t1
The juggler has to catch it and decelerate it to finish the juggling cycle:
He applies a deceleration A during a time t1, and finally the ball is again at rest, ready to restart a cycle.

Let us look at the forces exerted on the bridge (assuming the juggler hold the two other balls all the time, but that doesn't change anything to the problem):
t: 0 => F= -g (M+3m)
t: >0 - t1 => F= -g (M+3m)+a.m
t: t1 - t1+2t2 => F= -g (M+2m)
t: t1+2t2 - 2(t1+t2) => F = -g(M+3m)+a.m

We already see two things:
1. throwing the ball upwards exerts a force LARGER than that caused by the total of juggler+three balls, otherwise the ball wouldn't go up
2. the same happens when CATCHING the ball.

Trained physicists will already know that on average, the weight of the system (juggler + three balls) is on average g(M+3m), regardless of what the juggler does. THERE IS NO ESCAPE ! (except throwing the ball sufficiently strongly up over the bridge from the ground, crossing the bridge within the time 2t2, and catching the ball on the other side)

For those less trained in physics, let us calculate the average weight of the system on the bridge during one juggling cycle:
A cycle = 2(t1+t2)
During 2.t1, the weight is [(M+3m).g+m.a]
During 2.t2, the weight is (M+2m).g
Therefore the average weight is :
<F> = {(2.t1.[(M+3m).g+m.a]+2.t2.(M+2m).g}/{2.(t1+t2)}

Let us substitute t2 using t2=a.t1/g to confirm that conservation of mass has taken care of the little "miracle":
<F> = {(2.t1.[(M+3m).g+m.a]+2.a.t1/g.(M+2m).g}/{2.(t1+a.t1/g)}

Rearranging the above solution gives:
<F> = {(t1.[(M+3m).g+m.a]+t1(M+2m).a}/{t1.(g+a)/g}
<F> = {(t1.(M+3m).g+t1(M+3m).a}/{t1.(g+a)/g}
<F> = t1.{((M+3m).(g+a)}/{t1.(g+a)/g}
<F> = (M+3m)g
QED ! (not quantum electrodynamics, quod erat demonstrandum :tongue2: )

So the average weight of the juggler is exactly his weight plus that of the three balls. He can throw them as subtely as he wants, turn it whichever way you like, THERE IS NO ESCAPE: since on average the center of mass of the whole system remains in place, on average no momentum is added to the system, and since on average the system consists of the juggler plus three balls, its average weight remains constant. This explains why the extra force needed to counteract gravity for a while by throwing the ball up and catching it back exactly balances the fact that the ball is in the air for a while...

Needless to say, we neglected air friction but that doesn't essentially change the problem.

PS: someone said that the arm and hand is moving and therefore creates even greater a force onto the bridge. This is correct, but the juggler can do something against that (and good jugglers intuitively do that): bring his other arm and hand back down while he throws up and so on: he will create a torque, but no net force. The torque will have to be balanced by an opposite -- reaction -- torque from the bridge, which could also contribute to it yielding, but that is another story

 

[Hakuin]

why worry about a ball if you can have it collapse without ?

I forgot to mention an interesting aspect:
Juggling balls are usually very light :rofl:, at any rate much lighter than the juggler. Yet the juggler will also exert extra force on the bridge while walking, since our normal gait cycles our center of mass up and down. Unless (s)he takes extreme care not to have anybody parts moving up, (s)he is in trouble, because that is going to cause a force
M(g+a) on the bridge, and the chance of M.a > 2m.g (the bridge collapsing with the juggler not carrying ANY balls :yuck:) is not small, since that means
a > 2m.g/M. With M >> 2m ... you better slide your feet :tongue:

(as an example 75kg >> 2 x 0.1 kg, so the vertical (upwards) acceleration of your body should be less than about 0.0025 g (g=acceleration of gravity) !) If you want to be convinced, it only needs a fast-response bathroom scale. Stand on your scale and watch your weight (errr ) now lift your leg as if to walk, staying on te scale, and watch the weight change... any change larger than two juggling balls will bring you down with the bridge )

 

[PebblesRox]

Thank you, Hakuin. You explained this problem so clearly that I could understand it even though I haven't had Physics yet.
I think though that when the problem says that the bridge supports the weight of the juggler and two balls, it assumes that the juggler will walk across with them. The bridge is strong enough to support the added force of the juggler's steps. If this is the case, could you get across with three balls if you shuffle across very carefully, using that extra wiggle room to support the weight of another small ball?
I love this website! I just found it today, but I can't wait to explore. :shy: