Another Clever Riddle

[al Hartman]

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

[Eulaha L. Long]

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.

[d3z]

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.

[d3z]

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.

[Scott McCumber]

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.

[al Hartman]

     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.  

al

[d3z]

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.

[Scott McCumber]

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!

[Eulaha L. Long]

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

[sfortescue]

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.

[d3z]

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

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.

[brian]

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).

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

brian

[jesusfreak]

*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.

 


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  
--
lucas

[d3z]

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

[d3z]

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 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.