Juggling and weight.

F = MA

- as the balls are accelerated upwards the scale will read in excess of 194

- as the balls are decelerated downwards the scale will read in excess of 194

- if both actions are precisely equal and opposite in time and in all other aspects, then the net effect on the scale will be the added weight of the balls, this however is very unlikely

Effectively you'll be supporting you're weight (mass) and the weight (mass) of the ball. It is analogous to the question of birds flying in a cage on a scale--if the birds are airborne does the scale include their weight (mass). Now with juggling, the weight measurement will not be constant because you're going to be accelerating and de-accelerating the mass of the balls as you're catching and tossing, and with three ball juggling that perturbation will be affected by juggling style---are you doing a cascade or a shower.

Rap

rap must have three balls to be able to pronounce on a matter like that with such authority and wit.

At least I hope it was wit and not just a happy accident.
And yet.

But it does sound like rap knows what he's talking about.

Won't the scales, I presume they are scientifically tuned, vary, infinitessimally, as the equal and opposite reaction comes into effect.

You can't juggle in space without whirling around in some pattern determined by your catch and pitch techniques.

I thought you were scientists.

Alright, thanks everyone for confirming my thoughts. It seems logical, to me, that the reading of the scale would fluctuate as you applied force to the balls, but the average reading would be the same as if you were simply holding them statically.

I was prompted into thinking about this after hearing a riddle about a magician with three pieces of gold who needs to cross a bridge that can only support himself and two of the pieces of gold. The answer was that he juggles them as he walks accross the bridge, always keeping one in the air. I suppose, however, that we have debunked that.

If the magician even juggled one piece of gold (due to F = MA) he could possibly exceed the strength of the bridge.

In fact I would recommend that the magician either not cross the bridge or drop the gold and strip down to his undies and cross it. Although his average weight does not exceed the bridge's tolerance, the acceleration from his legs will, temporarily - and that is all it takes.

/Kayyam

And then there's the one about the guy that smuggled birds in from Mexico. He's be driving along, and suddenly stop to whang away at the camper with a bat. When asked why, he would say that with a 1/2 ton pickup and a ton of birds, he had to keep at least 1/2 the birds in the air at once.

If the person weighed 200 lbs and the balls weighed 10 lbs each (three balls) then statically the scale would read 230 lbs.

As the juggler tossed the balls in the air, the scale would read the reaction as increased weight. As the balls fall back to the hands, they release the energy you gave them against the scale.

Its equivalent to standing stationary on a scale and pressing your hands against the ceiling. You're holding 30 pounds and simply using your body's energy to move that 30 lbs around.

Assuming the release of any one ball happens at the exact moment the previously thrown ball is caught, so that the loss of weight from the thrown ball is balanced by the added weight of the returning ball, the net result would be 198# (weight of juggler plus essentially two balls) plus the kinetic energy contributed by the falling ball, which is dependent on the height it's thrown. This would register as a momentary gain in weight, which would disappear (the scale returning to 198#) until the next thrown ball is caught.

If thrown 10 feet into the air each two pound ball would contribute about 20 additional pounds when caught. Theoretically (were it not for the tempering effect of the act of catching, which distributes the energy over time) the scale would momentarily register 118 pounds.