Is it not possible for the magician to simply take 2 of the coins (leaving the 3rd) dance across the bridge and return to pick up the 3rd and final coin, thus completing his task without damaging the bridge or himself?
Or was the question specifically about juggling?
The question is just a variation of the “birds in a truck” thought experiment, where one has to figure out if a flying object exerts force on the surrounding environment. (Answer in both cases: Yes.) So the practical solutions, such as not carrying all three pieces at once, is moot for the purpose of the discussion.
In this thought experiment, while the gold coins are flying they exert no downward force on the bridge. However, as the juggler catches each, and as her throws eachj back into the air, there will be a downward force on his hand much larger than the one pound weight of the coin, and this force will be transmitted by his body to the bridge.
If he could throw all the coins in the air, walk across the bridge, and catch them all on the other side, he could satisfactorily solve the problem, but not if he catches and throws them while on the bridge.
Isn’t there a subtle difference between an enclosed truck and the open environment surrounding a bridge? Why is it necessary for the juggled coin to exert force on the bridge? Suppose he were to keep two coins in the air? Would that change anything?
A related question : If the magician were to walk across the bridge with only two coins, and a bird flew by overhead, would the bridge collapse?
I guess what I’m really asking is what constitutes a “system?” Thanks.
Because in order to juggle the coins, he must throw them upwards. Throwing the coin up exerts an equal and opposite force on the person pushing down, which in turn pushes on the bridge, increasing the person’s weight. There’s also the matter of catching the coins as they fall, as Giles points out.
Probably not, since the bird is acting on the air and only a small amount of the surrounding air will affect the bridge directly. In the truck scenario, the air inside the truck is trapped in there along with the birds.
That’s what I was thinking.
Suppose he begins jugging before he steps on the bridge, and keeps two coins in the air at all times; thus, whatever force is exerted is related to only one coin. Can he cross safely?
I suppose there is a subtle difference between the two systems, but it doesn’t matter, because it is necessary the juggled coin to exert force on the bridge (ignoring the “clever” answers, like throwing the coins across the bridge, or entering low-Earth orbit, or whatever).
Consider a simplification: the 68kg magician stands on a scale, tossing a single 1kg coin up and down. What does the scale read? While the coin is in the air, the scale will read 68kg (ignoring the magician’s motion). But when the magician catches the coin, he’s got to decelerate it, and then accelerate it back upwards. And, as a basic principle, F=ma. So the scale sees not only the weight of the magician and the coin, but also the upward thrust required to accelerate the coin – the scale sees more than 68+1kg (equivalent force), in other words.
The magician could, of course, accelerate the coin very lightly, to minimize the additional force. But, in that case, the coin is in his hand for a long time, and it’s not in the air very long at all, so the scale shows a little more than 68+1kg for a long time.
Or, the magician could accelerate the coin very firmly, to minimize the time it stays in his hand. But, in that case, the required force is very large, so the scale shows a lot more than 68+1kg for a little time.
Either way, if you go through a time average, you will (I think–I did this once but I’m too lazy to do it again) come up with a total time-averaged scale reading of 68+1kg.
If you figure in losses due to friction I bet you’d bump that number to just a hair over 68+1, no?
You guys are missing the point his a Magician. He used magic!
Could the magician put the gold in a bag, tie a string to the bag, and twirl the bag around over his head (at a velocity great enough to make the bag’s arch parallel to the bridge) and then walk across? Just a thought…
In order to keep that bag twirling you have to move your arm back and forth with sufficient force to overcome that gold’s weight. And that causes you to push really hard with your legs in order to maintain your balance, adding to your weight. This may not seem immediately obvious based on your experience with twirling things above your head. But try it with something heavy. Your legs will get quite a work-out.
So all twirling does is confuse matters by moving the gold sideways instead of up and down, but it doesn’t solve the problem.
The question used weight as a shorthand way of expressing the force the magician and the coins applied to the bridge. But force is really mass times acceleration. It’s the difference between dropping a hammer on your foot and bending over and placing a hammer on your foot - same weight (mass), different speed (acceleration).
So, as others have pointed out, when you throw a coin up in the air you are applying a force that is great enough to accelerate that coin’s mass upwards against gravity. This force is greater than the coin’s resting weight. The force you are applying upwards is counterbalanced by an equal force you apply downwards - you are in effect pushing your body away from the coin. And this downward force is transferred to the bridge you are standing on.
Yeah. I get it. The poster I was responding to stated that the coin exerted force on the bridge while it was in the air. Hence the “juggled coin.” I understand that juggling the coin exerts a downward force. I just don’t see why it continues to exert the force after it has been juggled.
When the coin is in the air, it isn’t exerting any force. If it suddenly got “magically” suspended from outside the system, everything would be fine. But you can think of the closed system as if the coin is exerting force to help understand that you aren’t getting any free lunches by trying to juggle the coins…
As I said, I get it.
Those are impressive coins - with a thickness of 50mm they’d have a diameter of 115mm - about 4.5". The next step beyond this would be juggling manhole covers.
I dunno - that’s pretty small beer. 
What if you used an airplane-wing-shaped bag, so that you got one kg worth of lift? 
Sorry, I guess I didn’t understand that part. 
Correction to my post above: thickness should be 5mm.