There is a popular riddle that you may have heard about traveling in a boat with 3 heavy balls such that the total weight is more than the boat can handle. The “solution” is to juggle the balls so that one is always in the air, which will make the total weight low enough for the boat to handle.
When I heard this riddle as a kid I though it was very clever. Once I had learned some physics I began to wonder if it was really accurate. Once I had a decent intuition, it struck me that this idea could never work. But it wasn’t until this morning that I actually calculated anything — I think I had a dream about this last night that inspired this activity.
Heuristic #1: When you’re juggling, you have to apply an upward force to the ball to throw it and during this time the ball is perceived as heavier. The same goes for stopping the ball when you catch it. The fact that the ball in the air contributes nothing is counteracted by these other two forces.
Heuristic #2: The boat with the traveler and the balls is a closed system. It can’t get lighter unless it’s acted on by an outside force. This trick is like picking yourself up off the ground.
I’m pretty sure someone taking a freshman physics class would have the tools to bust this riddle. Imagine a piston that periodically launches and catches a ball. It’s straightforward to calculate the perceived weight of the ball during launching and catching and, including the zero weight during the flight of the ball, find that the time averaged weight of the ball is just the gravitational weight.
Lunchtime! 
The limiting factor here is the normal force, ie the contact force of you on the boat. It’s true that you do have to apply an upward force on the ball to get it to launch into the air; however, while the ball is still in contact with your hand, it’s still contained within the system of your body. (The normal force–hand on ball–counteracts the applied force.) So normal force (you on boat) would not change in the throwing of the ball upwards. However, when you catch the ball, you’re accelerating it (slowing it down). The ball hitting your hand creates a force that would indeed increase the normal force, by even more than the normal weight of the ball. So for that split second, the force would exceed the weight limit of the boat. What you’re doing is trading in a constant force greater than the maximum for short “bursts” of force (impulse, right?) periodically. So if the boat can handle shorts bursts of large forces–and I think they can, what breaks the boat is a constant strain–you’ll be fine. So your reasoning was mostly good, but I think the answer holds. It would work.
Are you saying that while accelerating the ball upward, while it’s still in contact with your hand, that the boat feels no additional force where you are standing on it? As you put it, the ball is is still contained within the system of your body, so the force that accelerates the ball upward is coming from where? It better be coming from your contact with the boat.
You need a model for what goes wrong with the boat. The impulse given to the ball (and through your body, the boat) will be double the ball’s downward momentum. If the boat can temporarily withstand that impulse, you’re fine. But, if the boat will actually break from the weight of the three balls, then you’re in trouble and juggling won’t work. If the boat will only capsize, then you pick up a time constant based on how quickly the boat can sink when it’s weight exceeds the weight of the displaced water. My guess is that the boat will sink slower than the timescale of the impulse, sho you’ll be find.
I believe the boat is not breaking, but drowning. With the balls in it, the perceived density of the boat system is larger than water.
If you’ve done much juggling, you know that the balls are not caught and thrown with burst impulses. I would imagine that at least one ball, is always being caught or thrown. The time average perceived weight of a single ball is just it’s gravitational weight. With one ball always contributing zero, and one or two balls always contributing more than their gravitational weight, I would guess that the total perceived weight of the three balls during juggling doesn’t change all that much from the total gravitational weight.
I think at this point someone should make an experiment. 🙂
How about someone with a bathroom scale try juggling three balls while standing on it and see how far the needle moves? Assume the three-balls-plue-you system is just heavy enough to sink the boat. If the scale crosses above the resting mark then juggling won’t work.
ps – I’d do this but we don’t have a scale at home.
You need to have one of those cheap dial scales — this will run you about $6 at Walmart. Most people these days get the digital scales and I don’t believe those report a changing value.
You’ll probably want some pretty heavy balls. This is not a euphemism for courage, you actually want to juggle something substantial so that you have good signal to noise.
I wonder if there is a force guage (sounds more professional than ‘scale’ — which is the professional sounding name for ‘ruler’, who’s confused??) handy around the department that interfaces with a computer that will measure the weight of a person? How come when I search online for these things they cost $500-1000 but a bathroom scale is $20?
Thinking about it more, you don’t even need a scale, you don’t want an actual number, you just want an answer to “Did my weight go up?”. Along these lines I stood on a 1×6 spanning two paint cans (good adjustable fulcra) and watched the sag (displacement) on a vertical ruler behind the center of the 1×6. Juggling on a 1×6 is a new experience but with three water bottles (1/2 L) the 1×6 clearly crossed the mark corresponding to the static situation. I’ll post a video when I get to my office next (too slow to upload from home).
I think this is a good example of why I find experimentalists frightening… Nice job, Andy!
Good job Andy!
Since force is change in momentum with respect to time, the juggling force is the vertical momentum of the water bottle doubled near instantly by catching and throwing it back upwards, while the static force is just good old mass times gravity.
We need to see if:
mg < 2mv/dt
g/2 < v/dt
Assuming that Andy catch and release time, dt, is faster than 1s
g/2 < v/1second < v/dt
4.9 m/s < v * 1second/dt
So, if the bottle reaches a velocity around 5 m/s then we don’t need to know the time it takes Andy to throw the bottle back up in the air. However, I get that
mv^2 < 2 mgh
25/20 < h
1.25 m = h
So Andy would have to be throwing bottles over his head as he juggles (unlikely) and we need to know how fast his catch and release is to calculate this.
That wasn’t terribly illuminating. You’re welcome.
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Ok, so I can’t spell, that last comment should link to my post here:
Final ruling on Juggling riddle (for now) << clevertitlehere
Sorry about that.