A Juggler's Weight

[Shnoz (Sophie)]

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 (Andrew)]

Originally Posted by 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

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.

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.

alex

[Kal (Andrew)]

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

Originally Posted by 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.

[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 (Andrew)]

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 (Andrew)]

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 (Jeff)]

Originally Posted by 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.

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

Originally Posted by 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... just dont fall into the teachers trap of believing everything you read...particulary about dark matter

alex

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]

Originally Posted by theodog

Maybe you should look into teaching, sounds like they need your help.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.

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

alex

[theodog (Jeff)]

Originally Posted by xelasnave

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

alex

Accepted.
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 (Andrew)]

Originally Posted by theodog

Accepted.
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.

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

http://mythbustersresults.com/episode77

[theodog (Jeff)]

Originally Posted by Kal

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

http://mythbustersresults.com/episode77

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 (Andrew)]

Originally Posted by sjastro

So much for staying out of this thread.

You don't multiply the the weight of the juggler and pins by 9.8.
Their values are in kgf weight (kgf= kgX9.8) not mass (kg).

Hence force of juggler and five pins = 47 + 10 = 57 kgf
Force of juggler and one pin = 47 + 2 = 49 kgf.

Force to accelerate one pin at 4g = 2 X 4 X 9.8 = 78.4 kgf.

Total force on the bridge while still holding the pin = 78.4 + 49 = 127.4 kgf

Steven

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 (Mike)]

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.