A Juggler's Weight

[Shnoz]

I was studying for the Physics Olympiad (as you do:)) when I got a question wrong in the exam paper, and I cannot figure out why.
The question is as follows:



An expert juggler, carrying five juggling pins, has to cross a swing bridge which has a maximum safe load rating of 50 kg. The juggler weighs 47 kg and each of his five pins weighs 2 kg. He believes he can make it across safely in one trip by juggling the pins, so that he is never holding more than one pin. His skill enables him to juggle smoothly without any jerking. He is
A. incorrect – more information is needed.
B. correct – the total weight will never exceed 49 kg
C. correct – no jerking means no extra weight.
D. correct – the total weight can be made to exceed 49 kg by an arbitrarily small amount.
E. incorrect – the total weight is 47 kg + 5x2 kg = 57 kg.



I chose D, because I though that the juggler's weight, plus the weight of the single pin would equal 49kg, but would be slightly exceeded by the force of acceleration.
But the correct answer is E, according to the examination's solution papers. This completely baffled me. How can the pin's weight affect the juggler's weight and the bridge when they contact neither? :help:

[xelasnave]

What do Myth Busters say?
Could we use E=MC^2?
I have no idea, I wonder if the height the pins are throw enters into it...

alex

[sjastro]

It makes sense.

It's Newton's third law in operation. For every action there is an equal an opposite reaction. The maximum reaction force of the bridge is it's load rating.

You need to consider that the juggler is also applying a force to each pin to overcome the effects of gravity in the act of juggling.

The total reaction force of the bridge is therefore the weight of the juggler plus the weight of the pins, plus mass X acceleration of each pin during juggling.

This will exceed 57 kg and the bridge will come tumbling down.:lol:

Regards

Steven

[Shnoz]

Ah, I'd overlooked Newton's Third Law!:D I understand now, thanks for helping.

[OneOfOne]

I don't know if I agree with their answer, I consider it to be the LEAST correct. The jugglers static weight for most of the time will be 47+2kg. At any instant he will need to provide energy to accelerate a single pin into the air, which will temporarily increase his "weight" to something above the static 49kg, this amount being dependent on the height to which he throws the pin, the higher it goes, the more the force, the greater the increase in his weight. Once the pin has been released, his weight will drop to 47kg as all pins will be in the air and so will have no influence on his weight. At some instant, he will need to catch a single pin and his effective weight will increase again, probably by the same amount it increased when he threw the pin into the air. So his maximum weight will average to something around 49kg plus the force to accelerate a single pin....

If you pushed the experiment to the limit, he could throw 4 pins high enough into the air that he would not need to catch them until he gets to the other side and would only need to throw a single pin into the air whilst on the bridge...

[sjastro]

The question is what is the total load on the bridge. The total load is the sum of all the gravitational and inertial forces applied to the bridge. In this context the gravitational force is the weight of the juggler and the pins.

The inertial force needs to be greater than 2 kg-force for each pin. The inertial force acts on the bridge irrespective if the pin is about to leave the juggler's hand or is airborne.

The total load on the bridge exceeds 57kg.

Regards

Steven

[xelasnave]

What force is required to "throw" each pin.
Each pin is 2 kg so to throw one pin say 10 feet up maybe we need more than 2 kgs force and that inertial force may be many fold the weight of the pin...no doubt it can be worked out... and I wonder do we add the force when the pin accelerates from say a height of ten feet to the hands of the juggler... I expect the force of a pin falling from ten feet will be more than its weight of 2 kilos....

I think the problem is more complex than it seems ... well it is to me and I can only observe my reality not the reality as experienced by others...

AND how long is say force from the falling pin time wise... if force needed to be cumulative time separation would be relevant maybe.

alex

I am going out to find a juggler and a bridge

[Shnoz]

I think perhaps what xelasnave said could have a interesting effect upon the answer. Since in this respect, force is going to be transferred into weight upon the bridge, throwing the pins higher would require more force. Therefore, a larger reaction would occur on the bridge. If the juggler threw even a single pin high enough, he could theoretically exceed the bridge's safe load rating.
Perhaps the answer is neither D nor E, but A; because we require to know how much force the juggler is placing upon the pins.

[xelasnave]

Yes:D if he threw it high enough the pin may smash his hand and the bridge:eyepop::lol::lol::lol: .
alex:):):)

[sjastro]

Anythings possible. The juggler could be superman and throw the pins out of orbit:P.

Remember the Juggler is walking across the bridge as he/she is juggling. Assuming that juggling involves throwing the pins vertically, a pin thrown too high will not be retrieved.

The Juggler doesn't have much scope as to how far a pin can be thrown, and while E is not technically correct, it's still probably closest to being the correct answer.

Steven

[OneOfOne]

Yes, I think the solution is rather involved as you would need to know how much force is required to accelerate a single pin to a suitable height, I expect it could exceed the 2kg of the pin itself. The greater the height, the greater the effective increase in the "apparent" weight of the juggler for that instant. Once the pin has left the hand of the juggler, the weight of that pin and the reactionary force generated by the throw is no longer an influence on the bridge. Any pin that is in the air is of no consequence to the equation at all until it is caught again as each one is in free fall.

Myth Busters did a similar experiment with a remote controlled helicopter in an enclosed room and found that the total mass of the room did not change when the helicopter was flying as the force expended in keeping it airborne created a downward pressure onto the floor of the room equal to the weight of the helicopter. However, if the object were in free fall, as the pins are, they do not present a force onto the ground (or bridge in this case) and so are only part of the equation whilst in contact with the juggler.

Do we have a juggler in our midst who has access to a good pair of scales? This would provide the best way to get some real experimental results.

[sjastro]

That is simply not true. The reaction/inertial force is a function of the juggling action and has nothing to do with the pins. A juggler can generate the same inertial forces with or without pins. It's all in the the acceleration of the arms and upper body. And since juggling is a continuous process the inertial forces are there throughout the entire cycle.

There is also the consideration that walking adds to the inertial forces.

Regards

Steven

[Kal]

This is an old quiz question, a 'Millergram' from Professor Julius Sumner Miller from back in the 60's last century.

Q32: A juggler comes to a foot-bridge of rather flimsy design. He has in hand four balls. The safe load is no more than the juggler himself and one ball. Can he get across the bridge by juggling the balls, always having at most one ball in the hand (and three in the air)?
A: No. A falling ball exerts a force on the hand greater than its own weight.
Rather, a 'thrown' ball exerts greater force than a 'held' one. That is, the additional force equal and opposite to that imparted to a flung ball, in addition to the juggler's mass, would exceed the bridge's tolerance (the bridge can tolerate a juggler and held ball, but not the additional downward force associated with forcing a ball 'up').

The correct answer is E because in order to juggle and have only 1 pin in his hand at a time he will have to impart a minimum accelerating force on that pin equivalent to the gravitational force of 4 pins. Lets assume 1 second in the hand with a constant acceleration for that second, plus 4 seconds in the air for each pin. Therefore to throw the pin and have it come back in 4 seconds you need to accelerate it at 4 x 9.8ms-1. This equals the same force as holding 4 pins in your hand with only the basic 9.8ms-1 acceleration of gravity on them, add the weight of the pin being thrown and you have the 5 pins weight accounted for.

[g__day]

I would have said insufficient information to agree with E) - but its on the right track. You need to know how much time he has each pin in his hand - this will give you the minimum acceleration needed to loft each pin into the air.

If he exceeds the minimum acceleration by a large enough factor he would make it across. For example he throws a pin 500 yards in the air - smoothly - then does the same to three more pins - walks across simply holding the fifth pin - which if he wants he could launch once he's on solid ground - before the first has yet descended. If the bridge is short enough and he is fast and powerful enough - he makes it!

[Terry B]

So who can throw anything 500 yards?
It has to be realistic to work.

[Shnoz]

If we assume our juggler is not Superman and must cross the bridge juggling:D, then if he intends to keep the pins in the air for 4 seconds, then for every pin he accelerates into the air, its inertia causes the pin to experience 5 g's, as Kal said. So each pin weighs as much as five pins when accelerating. So after you apply Newton's third law, the juggler's weight, plus the accelerating pin's weight equals 57kg. This seems like the kind of logic the answer would work on.
Although, I assume the 'weight' of the pins would change depending on how high they were thrown. If the pins only had to stay in the air for 1 second (he must be a VERY skilled juggler:)), would that make it possible for the juggler to cross the bridge?

[Kal]

If teh pins are in the air for 1 second then he must take exactly 0.25 of a second to throw the pin, with guess what - the EXACT same acceleration as used in my previous example. Acceleration and force is the same, the only thing that changes are length of acceleration while in his hand & how high he throws the pins (and hence how long they are in the air)

[Kal]

And if he throws teh pins 500 yards in the air he is breaking the rule descibed in the original question "His skill enables him to juggle smoothly without any jerking."

[Shnoz]

Ah yes, 1:4 seconds and 0.25:1 seconds are the same ratio. I was never great at maths.:lol: So indeed, the acceleration would remain the same, as would the force and reaction, independent of the time taken.
Thanks for helping!:)

[sjastro]

Hmmm. 4Gs coming out of the hand. I don't think so with a juggling action. A quick calculation shows that if the pin is in the hand for one second (perhaps a reasonable time frame), the pin will exit the hand at over 140 km/hr.

Regards

Steven

[xelasnave]

and the moral to the story...

things can be more complex than they first appear.

I used to hate such questions where if you really looked at the problem there was more to it than what you had been told in class.

Some many times I was at odds with the teachers with such issues..they called me disruptive and not interested when in fact I was more interested and concerned with truth than anyone else in class.

alex

[Kal]

You failed to take into account that the pin arrives at a speed and exits at the same speed but in the opposite direction. If the pin arrives in the hand at 70km/h and is accelerated in the opposite direction and leaves the hand at 70km/h, then you have a differential speed of 140km/h, but only a throwing speed of 70km/h, which is reasonable.

-70 +140 = +70

I would imagine the pin would be in the hand for a shorter period than this, and I only used 1 second in the hand as an example because it made the underlying math more easy to understand.

An exact acceleration equivalent to 4G's (upwards) is the only acceleration acceptible and possible for the juggler to juggle 5 pins with a smooth motion.

[Kal]

The moral to the story for me in school was "teachers can be idiots that don't know jack apart from what is written in the book infront of them". I had some fantastic teachers in school, & I had some real dunces. I consistantly scored distinctions in the school science competitions however, and I have a fair amount of faith in my own ability to disseminate a problem and work out an answer.

[xelasnave]

Well Steven (with respect to all those wonderful teachers who I think are great doing what they do) teachers never have left school and therefore somewhat insulated from some of lifes realities:).

But heck ex teachers always made the best sales people which I attribute to management of groups of children...management of buyer lists and listing prospects is very similar because you need to understand each individual and what makes them tick.

I employed ex teachers readily because they mostly were great performers in sales.

AND Steven your ability is very evident you always provide very well considered views:thumbsup:... just dont fall into the teachers trap of believing everything you read...particulary about dark matter;)

alex:):):)

[DJDD]

which realties are those? :shrug:

[avandonk]

Please folks this is simple year 11 or 12 physics. The juggler as a time average weighs 47 + 10 kg. Those pins dont hover by themselves.

Remember F delta t and change in momentum.

See here for the theory! It even goes into Relativity.

http://en.wikipedia.org/wiki/Newton's_laws_of_motion


If I was the teacher I would explain it as gravitational potential energy being equivalent to the kinetic energy of the pin depending on how far it was thrown up! The impulse needed F delta t gives you the change in momentum and hence the energy and then you can calculate a time averaged force. As a time average the juggler weighs his 47kg + 10kg of pins.


Bert

[sjastro]

That makes no sense at all. The juggler catches the pin. The physics of catching an object involves bringing the object to rest while the hand is subjected to a short term impulse force. If the object isn't brought to rest you have a collision or hitting action.

If in fact the pin drops into the juggler's hand at 70 km/hr under 1g acceleration, then the juggler has thrown the pin to a height of over 19 metres. Yet another improbable scenario for juggling.

Steven

[Kal]

What? You aren't even making any sense.

[Kal]

So change the 1 second throw time ffs. It was just used as an example to explain the freaking maths

[sjastro]

You are stating the obvious but the point of the discussion is to select conditions that are within the physical capabilities of the juggler to prove the point.

Regards

Steven

[sjastro]

Alex,

It's the scientist coming out of me wanting to perform a test to verify the assertion.:)

Steven

[Kal]

No, the point of the discussion is to explain why the correct answer is 'E'

The conditions are irrelevant, and were only selected to allow me to explain the mathematical concept behind why the answer is 'E'

[avandonk]

Are you trying to say a good juggler could do it? If you integrate the force over the time no matter how gradual, the momentum change still is the same. This is a waste of my time. The answer is self evident to anyone who knows 12th year physics.

Bert

[xelasnave]

That you are not always right.
alex

[sjastro]

The juggler cannot cross the bridge. Have I made that clear enough?

Should I also make it clear that the solution be realistic that factors in the physical limitations imposed by the juggler?

In mathematical terms it's known as a boundary value condition.

Thanks for wasting my time having to explain this to you.

[Kal]

No problem, glad I could help!

[sjastro]

Your maths is not even right. Go back and do your sums.

[Kal]

Troll

[sjastro]

Reaction force on bridge caused by accelerating pin at 4G.

F= 2 X 4 X 9.8 = 78.4 kgf. Definitely not answer E).

Let the facts speak for themselves.

[Zuts]

You are joking?

(1) He posted in this thread before you.
(2) He has more posts in this thread than you.

Why not just present your maths?

Paul

[Shnoz]

Clearly we have two conflicting sides to this problem. Rather than be less than polite about it, wouldn't it be funner to simply point out what you think is wrong and/or don't get, so it can be explained and move on?
We're all fans of astronomy here; I'd welcome the oppurtunity to explain something to someone, it shouldn't be something to spurn.:)

[Kal]

Paul,

Both of the above facts are irrelevant to the act of trolling. He is trolling because he is baiting me. I presented my maths completely, and yet he refuted it with a blanket statement saying I was incorrect without even having the courtesy to point out exactly where he thought I was incorrect. For his previous attempts at discrediting the information I have provided in this thread, I have come back and pointed out the flaws in his thinking every time.

He is now attempting to troll me, plain and simple, and I will not fall into his game any more.

[xelasnave]

Come on guys lets settle down I know the problem is of Earth shattering importance but the only way to get thru it is by staying cool:)...

AND it does not matter who thinks who started it:).

We are all individuals and all entitled to have our say:).

I am sorry if my remarks have contributed to anyone getting upset:sadeyes:.

It upsets me to see this as you are all such wonderful people and I hate to think that any of you are less than happy:sadeyes:.

alex:):):)

[Kal]

Fine, I'll quote myself with some bits bolded to point out where I know Steven is getting confused.

[rally]

Kal,

Doesn't he also have to accelerate against gravity the mass of his arm and hand to some degree ?
Although the Question said the jugging action was "smooth so from that maybe we are supposed to infer he is able to transfer some circular motion starting fore and aft into up ward motion without generating a downward reaction ?

I dont know the answer there - it could be argued that it ends up balancing to zero net force as both arms are operating 180º with respect to one another.

Interesting problem and lots of good points (ignoring the usual forum temper !!!)

Cheers

Rally

[sjastro]

"impart a minimum accelerating force on that pin equivalent to the gravitational force of 4 pins."

The criteria to get a pin airborne is >1g. To suggest a juggler with an underarm action can obtain 4g is ridiculous.

This is the total force on the bridge at 4g acceleration.

1 pin= 2g, gravitational force of 4 pins= 4 X 2g.
Weight of juggler = 47 kg. (Should have included this in my last post).

Weigth of Juggler + reaction force on bridge = 47 kgf + 4 X 2g = 125.4 kgf.

Total force = 125.4kgf

According to you the answer should be 57 kgf.
As Paul enquired show us how you obtained this result

This is my final contribution in this thread.

Regards

Steven

[Kal]

F=MA (Force = Mass x Acceleration)
Force (aka weight) of a juggler standing still with 5 pins = (47 + 5x2) x 9.8 = 558.6
Force on bridge with a juggler standing still with 1 pin = (47 + 2) x 9.8 = 480.2
Force required by juggler to throw 1 pin with acceleration equivalent to 4G = 2 x 9.8 x 4 = 78.4
Therefore Force excerted downwards by the juggler while juggling = 480.2 + 78.4 = 558.6, which is exactly the same as the force excerted downwards by the juggler standing still with 5 pins.
I can't make it any clearer than this, so if you want to shoot holes in it I can't help you I'm afraid.

[Kal]

To be perdantic weight = mass x acceleration by gravity.

Therefore the mass of the juggler is 4.8 kilograms, and the mass of each pin is 204 grams.

All ratios are the same when solving the problem though, and if you use the fundamentally correct values you will still come to the conclusion that (E) is the correct answer.

[theodog]

Maybe you should look into teaching, sounds like they need your help.:rofl:Not.



I left school for a trade. While building I became interested in astronomy (a hobby for the last 28 years. I gained my HSC at TAFE, payed my way through Uni, started teaching and now run a faculty (18 years). I feel I have some knowledge of life -as do the vast majority I work with.

What I don't do is patronise, criticise or judge collectively.

I teach and I'm proud of my Profession.

[xelasnave]

Point taken Jeff and entirely valid my sincere appologies for making a sweeping statement

alex

[theodog]

Accepted.:D
This thread reminds me of a story form a wise bricklayer.

The drivers of live chicken transporters would hit the side if their trucks (to make the birds fly) and allow them to cross bridges with low weight restrictions.

I do use this statement as a problem for my Physics group.;)

[Kal]

This was on a mythbusters episode! "Birds in a Truck"

http://mythbustersresults.com/episode77

[theodog]

Ok, I have watched them on many occasions. I have not seen this one. Love their entertainment, science is Q'able.

Of all names, the bricklayer was Kal Kelly. He entertained us with this story one afternoon returning from work most probably in 1983.

How's that for memory? Didn't read about it in a text book.

Good luck with your abilities.

My thoughts on the Juggler are;
Weight force=mass of Juggler x g + (mass of two balls x g and 2 bits of g, to get one in the air, and stop the other from falling) down, at any one time. The bridge should not have such a reaction force available so should collapse, only because of "a ball & 2 bits". One assumes that one arm is raised to throw a ball as the other moves down to catch another ball, thus =force (from arms) on the bridge.
Maybe as the balls are in the air there is an extreemly small attraction force between them & the bridge.

With such skill he could throw them across the bridge, even bouncing them, and still get his balls across safely.

[xelasnave]

As someone walks do they exert more force when they move their weight onto one foot ...in other words does forward motion when contacted with the gound in effect increase their weight...

if I had a set of scales it would be interesting to walk across the room then step on the scales as one passed... I suspect that the scales would register more than my stationary weight.

alex

[Kal]

J = Mass of juggler
P = Mass of Pin
G = Gravity

Now a juggler standing on the bridge with 5 pins exerts the following force (also known as weight) on the bridge:
F=MA
F=(J+5P)G

The same juggler with 1 pin standing on the bridge excerts this force on the bridge:
F=(J+P)G

The force required to throw the pin in the air at the correct speed to juggle as per the quiz question is:
F=4PG

Therefore the force excerted on the bridge while juggling equals:
F=4PG+(J+P)G = 4PG+JG+PG = JG+5P = (J+5P)G which is exactly the same force excerted on the bridge when he is standing still with the 5 pins!!!

[iceman]

How did this thread turn into a slanging match with personal attacks and why didn't anyone use the report bad post button?

Criticising other forum members like I see in this thread is not on and won't be tolerated.