Can You Cross A Bridge With Three Gold Bars?

[darty]

could some kind person please clarify a long standing argument within my circle of friends.It goes like this .There is a bridge spanning a gorge that can take a maximum of 80 kilograms(sorry for those not used to kilos ), a person wieghing 78 kilos must transport three gold bars weighing one kilo each to the other side in one go.Can it be done? Some argue that it can be achieved if the person juggles the bars, never holding more than one in any hand at anyone time, ie there are always two in the air.Others argue that it cannot be done. None of us have any great knowledge of physics and would be most grateful for the answer
Cheers

 

[Mr. Robin Parsons]

2 (or 3) trips...cause you forgot to 'bar' that option, juggling should work...just don't drop anything...

 

[turin]

Originally posted by Mr. Robin Parsons
2 (or 3) trips...cause you forgot to 'bar' that option, ...

Actually, darty said:

Originally posted by darty
... in one go.

I don't see any reason why juggling would not work (in principle). You have to make sure that you don't throw the one bar up to hard, though, because too much impulse will look like too much weight. This is precisely why you can only hold one at a time, not two, while you juggle. But I personally wouldn't juggle my own bars across a gorge. If I was doing it for the govt I would try it.

 

[ZapperZ]

Originally posted by darty
could some kind person please clarify a long standing argument within my circle of friends.It goes like this .There is a bridge spanning a gorge that can take a maximum of 80 kilograms(sorry for those not used to kilos ), a person wieghing 78 kilos must transport three gold bars weighing one kilo each to the other side in one go.Can it be done? Some argue that it can be achieved if the person juggles the bars, never holding more than one in any hand at anyone time, ie there are always two in the air.Others argue that it cannot be done. None of us have any great knowledge of physics and would be most grateful for the answer
Cheers

It was alluded to in one of the responses of an "impulse". When you throw something, Newton's 3rd Law will dictate that the force you exert on throwing this object will exert a reaction force AGAINST you. So even if the person who is juggling this is holding only 2 bars at any given time, the moment he tries to throw one up into the air, his total weight will be the sum of all the weight he is holding PLUS the reaction force from the bar onto him.

You can easily verify this by standing on a weighing scale and throw something up in the air. You can see that the scale shows short fluctuation in weight each time you toss something up.

Final answer: it cannot be done this way.

Zz.

 

[HallsofIvy]

There's a long bridge near me and I'm a terrific juggler!

You send me the gold bars and I will do the experiment for you!

edit:
Fixed tags.
Integral

 

[LURCH]

Originally posted by ZapperZ
It was alluded to in one of the responses of an "impulse". When you throw something, Newton's 3rd Law will dictate that the force you exert on throwing this object will exert a reaction force AGAINST you. So even if the person who is juggling this is holding only 2 bars at any given time, the moment he tries to throw one up into the air, his total weight will be the sum of all the weight he is holding PLUS the reaction force from the bar onto him.

You can easily verify this by standing on a weighing scale and throw something up in the air. You can see that the scale shows short fluctuation in weight each time you toss something up.

Final answer: it cannot be done this way.

Zz.

And in case you think you can get around this by throwing the first bar into the air before stepping onto the bridge, the same forces apply to catching a falling bar. Your total force against the bridge is going to be the same.

 

[turin]

Ah. I was not considering the proper factor. Applied impulse per time is the important factor, not just impulse alone. Duh, oh well. To sum up, I will solidify my reversal and say: "It cannot be done by juggling."

 

[Mr. Robin Parsons]

Oooops...errrrr...uhmmm, hows about, Juggle higher...?

(I'll shut up, now)

 

[dfz111]

i don't know but being a physics forum i guess you were seeking a logical answer, but the bridge I believe is a misdirection, there's nothing saying that the man could simply scale gorge and pop up on the other side? I am assuming this is a lateral thinking problem, in which case that seems the most obvious answer.

 

[Mr. Robin Parsons]

Fast for three days, walk across carrieing all three bars

 

[turin]

Originally posted by Mr. Robin Parsons
... hows about, Juggle higher...?

The higher you throw the bars, the more impulse you must give them per throw. The lower you juggle the bars, the sooner they will return to your hand and so the sooner you will have to provide the impulse again. I still haven't calculated, but I imagine that if you did caculate, then the impulse per time would always average to ... 3 kg worth for the 3 bars. Oh hell, I'm due for some wasted time; I'll go ahead and calculate right now.

EDIT:
Calculation complete.

Not only will juggling not work, but it doesn't even seem close to a solution.

For instance, consider only 1 bar. In order to keep the bar from falling, you must give it 10 N (like the normal force, or whatever you want to call it). In order to throw it up, you have to give it extra force. Let's say you push the limit by giving it 10 N extra (which puts your applied weight to the bridge at exactly 780 N + 10 N + 10 N = 800 N, the maximum weight limit) and that you do this for 1 s. This will give it an impulse of 10 Ns and therefore an initial upwards velocity of 10 m/s.

How long will it take to come back? Well, it takes gravity 10 m/s², thereofore 1 s, to stop it, then another s to bring it back to you, for a total hang time (relief time) of 2 s.

At this point it looks deceivingly hopeful, because you only had to endure the extra force for half of the time that you are relieved. However, the ideal situation that you started with (applying impulse to a bar initially at rest) is by no means repeatable because now you must supply an impulse to stop the bar from falling (at 10 m/s). To do this, let's say you again push the limit by applying 10 N. The bar takes 1 s to stop and an additional 1 s to achieve the previous initial velocity condition, for a total of 2 s.

Now this can be repeated indefinitely for steady state. But look what has happened. For the same amount of time that we have been relieved of the weight we must endure double the weight, which shows that juggling does not change the average impulse per time at all.

Juggling with only two bars would be just as dangerous across the bridge as holding them the whole time. Three is not possible unless the bridge is short enough that you could make it across before you had to pay the impulse debt; that is, if you could throw the bar all the way across and then catch it on the other side. One thing that could possibly save you is air resistance; I have not calculated with that.




How about a He balloon? Any rule against that?

 

[Mr. Robin Parsons]

...and the fasting idea?

 

[outandbeyond2004]

The person could juggle each bar as his right foot steps on the ground or bridge. He starts juggling three times before the bridge while walking like so:

Right foot Right foot Right foot bridge .... bridge
juggle 1 bar juggle 1 bar juggle 1 bar just before the bridge

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) EQUATION 1

where F is the (vector) force on the bar, m is the mass of each bar, and v1 is the initial (vector) velocity (just before the bar returns to his hand) and v2 is the final (vector) velocity (just after the bar leaves his hand).

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

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, shortly after it leaves the hand, theta = the angle to the ground at which it leaves the hand, t = time, and g = gravitational acceleration.

The velocity at the beginning of the parabola, when it leaves the hand, is (vector, with 2 components!)
v2 = u, vsin(theta)

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

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

From equation 1,
|F|delta t = m|(v2 - v1)|
or
((1kg)*g + a) delta t = (1kg)*g(th) EQUATION 2

I don't see why we can't make delta t = (th), meaning that the person juggles the bars constantly, that is, as soon as one bar leaves his hand, another one is already coming into place to be handed. Hence, a = 0. The bridge thus never needs to bear a weight exceeding (79 kg)*g.

It would require coordination, timing, constancy of performance, stamina that are perhaps beyond the ability of any juggler, but I don't see why it can't be done in priciple.

 

[LURCH]

Catch the bar while your foot is in the air, and the force of the impact will be added to the impact of your foot when it comes down.

 

[jdavel]

outandbeyond said: "The bridge thus never needs to bear a weight exceeding (79 kg)*g."

This number is way too optimistic! Just holding one bar (no throwing) gives you 79kg*g. So, applying ANY force to toss it up would cause you to exceed 79kg*g.

I think Turin nailed this one in post #11. No matter how you juggle the bars, the average force you apply to the bridge over any complete juggling cycle will be your weight plus the weight of the bars you're juggling. With three bars that's 81kg*g. And you can't average 81 without ever getting over 80!

 

[outandbeyond2004]

Perhaps jdavel is assuming that there is no way to catch a bar so that the bridge-load of bar + person does not ever exceed (80 kg)*g. Besides, a relevant average is really a little less than one bar handled per right-foot-to-next-right-foot stride, not 3.

He is right in asserting that 79 is too optimistic, though. One reason is that the arm doing the juggling must take some time to go from tossing a bar to catching the next one. Still, I think the arm can be fast enough.

 

[Mr. Robin Parsons]

...And the Fasting Idea? [/size]

 

[jdavel]

outandbeyond2004:

"Perhaps jdavel is assuming that there is no way to catch a bar so that the bridge-load of bar + person does not ever exceed (80 kg)*g."

No I'm not. It's possible, depending on how high it was thrown.

"Besides, a relevant average is really a little less than one bar handled per right-foot-to-next-right-foot stride, not 3."

I don't understand why you've introduced this "right-foot-to-next-right-foot" issue. I also don't understand why you introduced parabolic motion or the sine of the angle that the bars leave and return to your hand. If you just stood on the bridge and juggled 3 bars do you think you could avoid ever exceeding 80kg*g? I don't.

 

[outandbeyond2004]

Edit == I need to think more about this. Ignore the rest. Wait for a later post

I didn't work out the procedure in detail. Forget what I said about moving the arm fast enough to catch the incoming bar. I revised the catch-tossing procedure:

Step: (each odd-numbered step is with left foot forward; each even-numbered with right)
1. toss first bar up and forward. (This must be done just before the bridge.)
2. second bar (step on the bridge)
3. third bar (on the bridge)
4. catch first bar and toss it up and forward
5. second bar
6. third bar
7. repeat step 4 and so on until you are safely off the bridge.

Think of this way: As long as you can catch and toss ANY bar so that the bridge load does not exceed 80 kg *g, you can do it with 3 bars as in the above procedure.

 

[jdavel]

outandbeyond2004 said: "As long as you can catch and toss ANY bar so that the bridge load does not exceed 80 kg *g..."

You can do that with one bar. You can just barely do it with two bars. But you can't do it with three bars.

Read post #11. With 3 re bars you have to meet 3 conditions to avoid exceeding the limit:

#1 No bar can stay in the air for MORE than 2 seconds.

#2 No bar can stay in your hand for LESS than 2 seconds.

#3 You can never be holding more than one bar.

So if you keep one in the air and one in your hand, you have to swap them every 2 seconds. But now there's no place for the 3rd one. If it's in your hand your dead. If it's in the air you're going to be dead when you swap.

 

[outandbeyond2004]

Okay, here's the procedure.

At the start, toss up a bar with right hand, wait a while then toss second bar with right hand, again. Again wait a while and finally toss up third with right hand, again.
Meanwhile the left catches the first bar. The procedure goes like this:

1. left slows down to zero while right gets ready to receive the bar.
2. left transfers bar to right.
3. right tosses the bar up while left moves up to catch the next bar
4. left catches the bar, and go back to step 1 and so on.

When the bars are moving around like they should, walk across the bridge, juggling that way all the while.

The bars have to be evenly spaced out in time, with two bars in the air.

The following equation from a previous post is still relevant:

((1kg)*g + a) delta t = (1kg)*g(th)

with delta t = one-half the hang time (th). Therefore, a = 1 kg *g.

Just barely doable. In reality, professionally engineered bridges are designed with large safety margins or safety factors, so maybe a few jugglers can really do the job safely enough.

 

[jdavel]

outandbeyond2004,

I'll try this once more. But it seems to me as if you're not reading any of the other posts.

You say: "At the start, toss up a bar with right hand"

Question: How high does it go?

 

[turin]

In reality, professionally engineered bridges are designed with large safety margins or safety factors, so maybe a few jugglers can really do the job safely enough.

Apparently you're not very familiar with this safety margin concept. If we wanted to consider the realistic safety margin, then this question would have died before it was finished being asked. The rule of thumb for most civil engineering applications (those that involve a significant human life factor like bridges do) is to overdesign in favor of safety by three times the nominal requirement.

If we assume that the nominal requirement is 800 N, then we have, by proper civil engineering standards, a hell of a lot more than a few gold bars worth of leeway.

If we assume that 800 N already accounts for this extra design margin, then there's no way in hell that I would even consider stepping onto that bridge even while holding onto 100 He balloons and fasting for three days.

But, bottom line concerning realism: this is just not one of those questions that you can be realistic about in a pub. There are far too many factors to consider, like pressure/stress is actually a more appropriate parameter than force.

 

[outandbeyond2004]

Well, folks, so far, based on my work (which is not yet totally accepted apparently) and the comments of others, the answer is theoretically (better yet, "fantasically") yes and practically no.

I believe I understand Turin's comments, except the remark about pressure/stress. My thought is that we have one person going across the bridge, so we should be concerned with one-point force (more or less anyway) rather than force spread over a wide area.

Unless someone wants to offer a different solution or a critique of my solution, I've written all I want to write. I am going to leave jdavel's last question in post #22 as an exercise for the student.

 

[turin]

... based on my work ... and the comments of others, the answer is theoretically ... yes ...

Who else says this besides you? I admit that I hastily replied in the affirmative at the beginning, but I have since recanted.

I believe I understand Turin's comments, except the remark about pressure/stress. My thought is that we have one person going across the bridge, so we should be concerned with one-point force (more or less anyway) rather than force spread over a wide area.

Do you understand why bridges have weight limits?

 

[outandbeyond2004]

Turin, I do not understand why you asked that question "why bridges have weight limits?" Your reply would be offtopic, so I am going to start a new topic and there you can hold forth, I hope it would be all right with you.

 

[turin]

Turin, I do not understand why you asked that question "why bridges have weight limits?" Your reply would be offtopic, so I am going to start a new topic and there you can hold forth, I hope it would be all right with you.

No, I don't really want to get into it, primarily because I don't have a deep knowledge of it. But it does most certainly have much more to do with shear stress, torsion, and stuff, rather than just plain vanilla force. We can drop the issue here and now; I don't really want to go to another thread about it. I am not trying to raz you or anything, I just wanted to demonstrate that civil engineering was irrelevant to the original question by virtue of the extra complication and obviation of direct force considerations.

 

[Mr. Robin Parsons]

You fast for two days, no water, no food, and then you carry all three bars across with you in one shot...

Problem solved! case closed?

 

[outandbeyond2004]

The problem specifies that the person is 78 kg. This thread would not exist if the person was just 77 kg.

 

[jdavel]

outandbeyond2004 said: "This thread would not exist if the person was just 77 kg."

Actually, it might have been more interesting that way. The question would have been, is it possible to set 80kg on a bridge carefully enough to avoid exceding applying a force of 80kg*g to the bridge. In principle, maybe. In reality, probably not.

 

[benzun_1999]

How about kneeling down and moving instead of walking upright the force will be produced over a large area so there are chances, and if at all the bridgeis going to break it will do so when the person is in the middle.

how about Creating more support for the bridge?

look if this is any foolish please forgive me. I am just trying to find a solution.

 

[outandbeyond2004]

You may keep changing the assumptions all you want. I think I will unsubscribe this thread. However, anyone with a solution you REALLY think is better than mine may pm it to me.