## A Conversation for Juggling

### the bridge conundrum

dr. dunc Started conversation Nov 20, 2000

>

i seem to remember that a few years ago a question appeared in the back of New Scientist magazine - the situation was thus:

a guy wants to walk across a bridge -

he weighs 100kg,

he wants to carry 3 cannon balls across with him, each of which weighs 10kg

the bridge will collapse with 125kg weight on it.

..can he get across?

..would it work if he juggles the balls (3 ball cascade, i.e. one ball in the air at all time)

the only prob is, i can't remember the answer - anyone fancy a go???

ta-ta- dunc

### the bridge conundrum

jazzman Posted Dec 12, 2000

He could always carry one, come back, carry the next etc... It does kind of bypass the big question though. Thinking way back to High School physics, every action has an equal and opposite reaction, so lobbing 10kg into the air must press down with 10kg (the strength of the throw shouldn't matter, it'll only increase the hieght of the arc and delay the 10kg landing again), so as long as he starts juggling before he gets to the bridge he should be OK.

Maybe we could get the NFL to stick it into the half time show at the Superbowl: they seem to have the most 254lb guys around.

### the bridge conundrum

Hoovooloo Posted May 17, 2001

It's an enormously long time since this message was posted, but I'm going to answer it as I'm (a) a juggler and (b) an engineer.

I haven't done the calculation (I might later...) but there is a basic flaw in the argument given. Specifically, the force exerted by your man would be the 100kg. The force exerted by your man plus one ball would be 110kg. If he now throws that ball into the air, he imparts acceleration to the ball. The force he exerts upwards (and hence also downwards onto the bridge) is equal to the mass of the ball multiplied by the acceleration. I can't wait any more, I simply have to go away and actually do the calculation.

### the bridge conundrum

vogonpoet (AViators at A13264670) Posted May 16, 2002

To continue the rather disjointed flow of this conversation, depending on the accelerations required, it might be better to juggle the three canons in one hand - that way he would only have one cannon ball in his hand at a time, but would be exerting more force downwards per cannon ball per throw, as he would have to throw them higher than he would the simple three ball cascade.

Ever get those calculations done Hoovooloo?

### the bridge conundrum

Researcher 202192 Posted Sep 4, 2002

Thinking laterally on this one, nothing is falling, all are maintaining approximately the same height, so no energy will be lost by the time they reach the other side of the bridge. In order to prevent them losing height the AVERAGE upward force must equal the weight of the juggler plus each of the three balls: 130 Kg. Hence the bridge is gonna break 'cos at some time the total force is gonna be > 130Kg, and certainly > 125 Kg.

AAAAARRRGGGGHHHHH!

### the bridge conundrum

Hoovooloo Posted Apr 30, 2004

OK, I'm dizzy with the speed of this conversation, and yes, I did get those calculations done, so here's the news.

Remember that weight is not the same as mass.

Units of mass are kg.

Units of weight are newtons, which in SI units are kg.m/s^2. 1 kg exerts a force of 10 newtons downwards on earth.

So, yer man has a mass of 100kg, so he WEIGHS 1000 N.

Each ball WEIGHS 100 N. So the force downwards of yer man carrying one ball is 1100 N. The bridge will successfully resist a force of 1250 N, so he's safe.

BUT he's thinking of juggling. So he carries ONE ball out onto the bridge. Force downwards = 1100 N.

He works out that he's going to need to be able to throw the ball upwards about half a metre. So he needs to work out how fast the ball must be travelling UPWARDS when it leaves his hand in order that gravity will bring it to a halt half a metre up.

v2 - u2 = 2as

v (the final velocity) = 0 - the ball stops at the top of its flight.

u = ? how fast must it travel?

a = acceleration due to gravity, 9.8m/s2

s = 0.5m

so u2 = 2*9.8*0.5

so u = sqrt 9.8 = 3.13m/s.

So, yer man now realises he needs to accelerate that ball from a standing start to 3.13m/s. But he needs to do it with a vertical arm movement of no more than about 0.1 m. So what acceleration must he apply to get the ball moving?

again: v2-u2 = 2as

v=3.13 m/s (the ball ends up going the right speed...)

u = 0 (... from a standing start)

a = ? how much acceleration

s = 0.1m (a typical distance for juggling, although the balls are very heavy...)

So a = 9.8/(2*0.1)

a = 49 m/s2

Now,

F = ma

So the FORCE yer man must apply to accelerate the ball to the correct speed is:

m = 10kg

a = 49m/s2

F = 490 N.

So, if yer man walks out onto the bridge carrying just ONE ball, the force downwards on the bridge - capacity 1250 N - will be 1100 N.

But if he then throws that ONE ball up just half a metre, the downward force increases to 1590 N, and the bridge collapses.

Ouch.

I haven't even bothered to take account of the fact that when he catches a ball, decelerating it to rest will also cause an increase in downward force, because frankly he never gets that far.

### the bridge conundrum

Madent Posted May 22, 2004

Won't his knees, arms and back, etc. have a shock absorbing potential?

I mean that while I don't disagree with the math, it doesn't take account of the stiffness of the man, which will go a long way towards reducing the force applied to the actual bridge.

### the bridge conundrum

Mu Beta Posted May 22, 2004

" it doesn't take account of the stiffness of the man"

**snickers childishly**

B

### the bridge conundrum

Madent Posted May 23, 2004

Almost all "engineering" terminology is a double entendre.

Cock = valve, bastard = file or rasp, nipple = hydraulic fitting, etc

In this case "stiffness" refers to the spring like nature of the human body. As in when you stand on the bathroom scales and flex your knees sharply, first you weigh less, then you weigh more. The idea comes from Hooke's Law and is the basic principle behind suspension systems on cars. Springs and dampers.

The question is whether there is sufficient "stiffness" in the man's body system to counter the rapidly changing momentum of his "balls".

Now, snicker away

### the bridge conundrum

Madent Posted May 30, 2004

Knowing you have a degree doesn't help a lot, as so many researchers are pretty well qualified.

However knowing your discipline is useful

### the bridge conundrum

Fathom Posted Jun 3, 2008

In that case you'll know that however much the man absorbs or damps the acceleration it will all ultimately be transmitted to the bridge.

F

### the bridge conundrum

Mu Beta Posted Jun 3, 2008

Ah - another longer-than-a-year gap.

Having realised that nearly the entire contents of my degree are now useless, I have forgotten not only this entire conversation but all three years of lectures. Apart from some smutty jokes about cementite.

Having realised that I haven't yet sensibly answered this question, I would agree that it isn't possible.

B

### the bridge conundrum

Mu Beta Posted Jun 3, 2008

That is, it isn't possible to cross the bridge, not that it isn't possible to answer the question.

**wishes he'd unsubscribed in 2004**

B

### the bridge conundrum

Alfster Posted Jul 13, 2009

Another year long gap.

Of course, thinking laterally, the guy could turn the cannon balls into makeshift hammers for hammer-throwing. A sporting hammer weighs 7.257 kg and can be thrown over 80metres.

Of course, this is defeating the object of the exercise...but I am an engineer...and a lazy one so I can't be bothered with equatoins of motion.

Key: Complain about this post

### the bridge conundrum

- 1: dr. dunc (Nov 20, 2000)
- 2: jazzman (Dec 12, 2000)
- 3: Hoovooloo (May 17, 2001)
- 4: vogonpoet (AViators at A13264670) (May 16, 2002)
- 5: Researcher 202192 (Sep 4, 2002)
- 6: Hoovooloo (Apr 30, 2004)
- 7: Mu Beta (May 11, 2004)
- 8: Fathom (May 20, 2004)
- 9: Madent (May 22, 2004)
- 10: Mu Beta (May 22, 2004)
- 11: Madent (May 23, 2004)
- 12: Mu Beta (May 23, 2004)
- 13: Hoovooloo (May 23, 2004)
- 14: Mu Beta (May 24, 2004)
- 15: Madent (May 30, 2004)
- 16: Mu Beta (Jun 1, 2004)
- 17: Fathom (Jun 3, 2008)
- 18: Mu Beta (Jun 3, 2008)
- 19: Mu Beta (Jun 3, 2008)
- 20: Alfster (Jul 13, 2009)

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