Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pub trivia

  1. Mar 23, 2004 #1
    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 any one 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
     
  2. jcsd
  3. Mar 23, 2004 #2
    2 (or 3) trips....cause you forgot to 'bar' that option, juggling should work...just don't drop anything....
     
  4. Mar 23, 2004 #3

    turin

    User Avatar
    Homework Helper

    Actually, darty said:

    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.
     
    Last edited: Mar 23, 2004
  5. Mar 23, 2004 #4

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    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.
     
  6. Mar 23, 2004 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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
     
    Last edited by a moderator: Mar 23, 2004
  7. Mar 23, 2004 #6

    LURCH

    User Avatar
    Science Advisor

    Re: Re: pub trivia

    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.
     
  8. Mar 23, 2004 #7

    turin

    User Avatar
    Homework Helper

    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."
     
  9. Mar 23, 2004 #8
    Oooops....errrrr....uhmmm, hows about, Juggle higher...?

    (I'll shut up, now)
     
  10. Mar 23, 2004 #9
    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.:smile:
     
  11. Mar 24, 2004 #10
    Fast for three days, walk across carrieing all three bars
     
  12. Mar 24, 2004 #11

    turin

    User Avatar
    Homework Helper

    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 acheive 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?
     
    Last edited: Mar 24, 2004
  13. Mar 24, 2004 #12
    ...and the fasting idea?
     
  14. Mar 25, 2004 #13
    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 perfomance, stamina that are perhaps beyond the ability of any juggler, but I don't see why it can't be done in priciple.
     
  15. Mar 25, 2004 #14

    LURCH

    User Avatar
    Science Advisor

    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.
     
  16. Mar 26, 2004 #15
    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!
     
  17. Mar 26, 2004 #16
    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.
     
    Last edited: Mar 26, 2004
  18. Mar 26, 2004 #17
    ...And the Fasting Idea???
     
  19. Mar 26, 2004 #18
    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.
     
  20. Mar 26, 2004 #19
    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.
     
    Last edited: Mar 26, 2004
  21. Mar 26, 2004 #20
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Pub trivia
Loading...