AssemblyBoard

Let me ask you all something.  You know the rules...if you know the answer, please don't post it.  But I promise you I will post the answer within one week (unless I get run over by a freight train or win a vacation to Antarctica).

You're walking through the woods, and you only have one match.  You come to a log cabin.  Inside is a candle, a wood-burning stove and a lantern.  What do you light first?

mithrandir

Oh, come on, Clarence.  That one is easy.  You first light the golden lampstand.   ;D

Clarence,

     That's IT?!   i need more input:  Is the cabin unlocked?  Is anyone at home?  How long do i have until sunset?  Is this cabin in the northern or southern hemisphere?  Are there bears in the woods?  Is the pope a catholic?

 ;D al

Okay, okay, the riddle wasn't that hard, I guess.  You light the match first. ;D

mithrandir

Mr. and Mrs. Mustard have six daughters and each daughter has one brother. How many people are in the Mustard family?

NINE  ;D

I'm not sure how many are in the Mustard family, but they all relish one another, and they're sure to help each other out if they get in a pickle. Lettuce think about that and do the same. Beets me why I like puns so much.

And let's hope they serve Canadian beef at the table with them there condiments.

Moo!      Get out of here, cow!  What are you, crazy?



drj

How many squares are on a chessboard?   :P

--
lucas

I have a chessboard that has only 32 squares.

They are slightly smaller than the spacing so that their edges don't join to make continuous lines across the board.  Also, since the board is folded up in its box, the outer border doesn't make a square either.

Funny. But what about a regular chess board?

     I keep hoping to get the jump on Joe one of these times, but he's too quick -- I guess I'll never ketchup.  Then again, the opportunity may turnip.  I'd probably be more motivated if I received a celery for my efforts.  Orange you glad you read this thread?

Unless this is a trick question (so not counting pieces, players, or decorations), I get 204.

That's what I get.

How many rectangles? Similar formula but now your adding 1x1;1x2;1x3 . . . ;2x1;2x2; . . . ;8x1;8x2 . . .

The number of rectangles in a rectangle is a(a+1)b(b+1)/4.

The number of squares in a rectangle is a(a+1)(3b-a+1)/6, where a<=b.


How about harder questions:

A. How many 8 by 8 mazes are there?

B. How many ways can a chessboard be cut up into 32 dominos?

C. How many ways can a hexagon be cut up into unit rhombuses?


Answers:

A: 126231322912498539682594816

B: 12988816

C: For a hexagon with sides a, b and c:

the # of tilings = product(n^e_n)

where sum(e_n x^n) = (x^a - 1)(x^b - 1)(x^c - 1) x/(x-1)^2.

Stephen,
     Is that a heat image of Lucas' brain?  ;D ;D ;D

Imagine John, a party magician, is carrying three pieces of gold each piece weighing one kilogram. While taking a walk he comes to a bridge which has a sign posted saying the bridge
could hold only a maximum of 80 kilograms. John weighs 78
kilograms and the gold weighs three kilograms. John reads the sign and still safely crossed the bridge with all the gold. How did he manage this?



A:John is a juggler. When he came to the bridge he juggled the
gold, always keeping one piece in the air.

Which, by the way, is wrong (although it is the commonly stated answer).  Juggling objects doesn't change the persons average "weight".  The weight lost while the object is in the air is made up for by the force needed to throw it up, and then to catch it.  Since there is a brief time when the object is in free fall (Johns force against the bridge is smaller), there will also be a time (during catch and throw) when his force against the bridge is greater than the average force (81 kg * accel_of_gravity).

The only way it would work would be for John to throw and catch the extra piece of gold while standing on the ground on each side of the bridge.  This essentially results in John throwing one of the pieces of gold across the bridge, to the other side.

The proof is left as an exercise to the reader.

BTW, the picture is even worse than I declared in my previous post.  It is unlikely that John would be able to make it across the bridge, even by himself.  Each time John takes a step, he accelerates himself upward, which requires a downward force on the bridge.  He would have to walk very carefully to avoid applying the additional 2kg(*9.8m/s^2) of force to the bridge.

Realistically, a bridge rated with a given maximum weight will hold quite a bit more, because the maximum weight accounts for ordinary acceleration that objects.  The relatively light gold bars that John is carrying are much smaller than the extra force John will apply just to walk.

So, a good answer is that John should just carry the three bars across the bridge, and try to walk softly.

(http://media.theinsiders.com/Media/Other/63032_Eek.GIF)

     If I were John, I would use the gold to buy the bridge, then go on across, set up a toll booth, and quickly regain my investment & then some, but...

     I guess I'll just have to cross that bridge when I come to it.  ;D

al ;)

I guess we should question John's wisdom, since he is trying to carry about $30,000 US worth of gold.  He should have already invested it.

And where exactly is this bridge? I think my band of merry men and I might have to relieve this wealthy gentleman of his burden so he can feel safe crossing his bridge!(http://media.theinsiders.com/Media/Other/63032_guns.GIF)

Um...it was just a riddle...a RIDDLE...no need to get so darn technical-GEESH! ;)

The other version of that kind of mistake has to do with a truckload of birds crossing a bridge.  The truck driver figures that if he scares the birds into flying while in their cages in the truck, the truck will be lighter, but he is mistaken.  The downward motion of the air from the flying birds will still be pushing down on the truck.

Ok, so it is probably isn't all that important.  It is just kind of frustrating to have riddles like that where there is no correct answer, and the normal answer is COMPLETELY wrong.

It is almost as bad as getting a trivial pursuit question where the answer in the game is wrong.  Try convincing people of that.

in full respect towards your understanding of physics, and lmao at my own geekiness, i must disgree. if he cleverly juggled with one hand, only one gold bar would be exerting force downwards at one time. granted, the force downwards would be considerably stronger than if he was just holding one bar in his hand, but it would NOT be greater (at any given moment) than if he was throwing a single gold bar into the air repeatedly. and the specs on the bridge stated 80kg, which is not a measure of acceleration. so it should be able to handle the momentum 80 kg would exert under full gravitational acceleration for at least a few feet (what if an 80kg someone jumped on the bridge?), so it should be able to handle the acceleration of stopping the downward motion of one gold brick at a time.

the interesting question would be what is the force required to stop a gold brick that has fallen for about a meter, versus the force required to hold up a horizontially motionless gold brick. if it requires more than 3 times the amount of force to stop the falling brick's momentum, then carrying is definitely the better option. what would the equation be? momentum is calculated from the velocity and the mass, and the amount of force exerted by a given momentum is some sort of moment of impulse calculus calculation, i think.

i am such a dork :P

brian

*sigh* I feel compelled to be the biggest dork ever....so here i go (I had to do this stupid problem in highschool physics) ::)

The person could juggle each bar as his right foot steps on the ground or bridge, so he would do something like:

Right foot (juggle 1 bar), Right foot (juggle 1 bar), Right foot (juggle 1 bar) just before the bridge (thereby getting all 3 bricks moving).

To minimize the instanteous force, he wants to juggle each bar so that the upward force he exerts on the bar is constant.

So, the impulse-momentum principle approximately reduces to

F delta time = m(v2 - v1)
where F is the force vector

There is a time when the velocity of the bar relative to the bridge and the person is zero. At that point, the person will also have his weight about equally balanced between his feet. So, at this time, the total weight on the bridge would be (78 kg + 1 kg)*g + a, where *g is the gravitational acceleration and a is an additional upward acceleration.

Since the magnitude of F is supposed to be constant, we can take |F| = (1 kg)*g + a

Each bar goes up and down in a parabola given by
x = ut,
where u = the person's speed (assumed constant) across the bridge.
y = vsin(theta)t - gt^2/2,
where v = initial speed of the bar
theta = the angle to the ground at which it leaves the hand
 t = time
g = gravitational acceleration

The velocity vector at the beginning of the parabola, when it leaves the hand,
v2 = (u, vsin(theta))

The velocity vector at the end of the parabola, when it is coming in,
v1 = (u, vsin(theta) - g(th))
where (th) = hang time

Hence,
v2 - v1 = 0, g(th)
and
|v2 - v1| = g(th)

Therefore,
|F|delta t = m|(v2 - v1)|
or
((1kg)*g + a) delta t = (1kg)*g(th)


The ideal solution would assume delta t to be equal to the hangtime, making a = 0. The bridge thus never needs to bear a weight exceeding (79 kg)*g.

 :P


on a more serious solution, the above is way optimistic. Simply holding one bar (no throwing) gives you 79kg*g. So, applying *any*force to toss it up would push you over 79kg*g.

kinda hard to average 80 when the wieght average alone pushes you up to 81 :)

but we can just say this juggler is the best ever and gets all the timing and magnitudes perfect  :o
--
lucas

In other words (Lucas), you're saying that if he juggles and times it perfectly with his steps, he can maintain a constant force.  But, at that point, he might as well just carry the bars across the bridge, and hope for the best.

Another way to put it: the best he can do by juggling is cause the same force on the bridge as the total weight.  Most likely, he will not juggle perfectly, and the juggling will cause more force than just walking across the bridge.

Is it then possible that proper juggling can help with the force imbalance from walking?  It would probably be better to just put one bar in his pocket, and hold the other two bars, moving them up and down in time with his steps.

Dave

The sign says 80kg because people wouldn't understand a sign that said 784N.  The gravitational constant is implied.  The only thing the bridge could be affected by is the force applied to it.  If the bridge were to truly collapse when more than 784N were applied to it John had best tread very lightly, since his steps will exhibit considerably more force.

That is why this problem isn't terribly realistic.  When they rate a bridge for a given weight, they include a significant amount of extra padding.  The assumption is that a rating of 80kg means that an 80kg person could walk across the bridge with low risk.

Since bridge weight ratings are usually much larger, and large trucks aren't much into jumping, this problem doesn't come up much.

The horizontal component of motion can be eliminated (we'll assume a flat bridge, since a sloped bridge would affect the problem).  Any horizontal component of force has no effect on the downward force on the bridge (this isn't completely true in real life).

Once a bar of gold is in free-fall (in the air), it's weight has no bearing on the force John applies to the bridge.  Let's assume that the bar is freefall for time 't'.  During this time, gravity has applied a constant downward force of 1g.  In order for the bar to have the same ending state as when it started, John had to apply the same acceleration to the bar (partly during the throw, partly during the catch, both equal if he catches it at the same height he released it on the throw).  The integral of this force over time across the bridge will be equal to the force of the static system (81*g).  Since we know that when the bar is in freefall, and therefore the force on the bridge will be less, it must be greater when John is throwing or catching.

John can carefully move himself up and down in opposition and balance this force out, but I doubt this is something a human could coordinate.  In any case, the best he can do is limit the force to 81*g.  In all likelihood, the juggling will only make matters worse.

But it would look oh so cool 8)

--
lucas

     No, Brian, you're certainly no dork.  A heretic, possibly, but not a dork.  You see, this isn't a question of religion, so it was just wrong of you to try to involve "the mass."
     Besides, the question of calculus is best addressed by a dental hygienist.

     You guys are all leaving out all kinds of important variables.  Is it light outside or dark?  Is there a breeze, and if so, what is its direction?  In fact, in what directions does the bridge run?  And in which of those directions is the dude with the gold traveling?
     Does the bridge cross over a dry gulch or over water, and if over water, is it flowing or still?  Of what substances is the bridge composed and what is its architectural style?  How near the bottom of the bridge is the water or land beneath it?
     What is the season of the year?   Is the weather wet or dry?  In what phase is the moon, and what is the alignment of the planets?  How much wood could a woodchuck chuck?  Where's the peck of pickled peppers that Peter Piper picked?

     I know, at this point all you genii suppose that I'm just making fun of you because I can't keep up with your ideas...  Hah!  ;D
     Just consider:  If the bridge is of a simple truss design, made of wood, and crosses a stream which has been backed up by a clogged beaver dam until the water's surface is touching the bottom of the bridge, and these conditions have remained constant well into a cold winter which has frozen the water's surface to a depth of, say, a foot, while the dark of night and a stiff breeze have seen to solidifying what slight melting the day's sunlight may have produced, well...  this guy can juggle his gold or his checkbook while he crosses the bridge in an Abrams tank & be safe.  Nothing short of a direct bomb strike is going to break that bridge.
     On the other hand, if an extremely wet spring has eroded away the supporting ground at both end of the bridge, the thaw to the north has caused the creek beneath to become a rushing torrent which is now slamming broadside into the bridge's lowest two feet, and the winds are approaching hurricane velocity at the same cross-direction to the bridge as the water, a mouse wouldn't want to set foot on it, 'cuz it's gonna go at any second...

     Eulaha, please find another thread to post on-- this is giving me a splitting headache, and I haven't slept since Saturday!!! :P  Some of us find this stuff addictive...

al ;)