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a juggler carrying three balls wants to walk across a bridge that can only support the weight of the juggler and two balls. in order to make one trip, the juggler decides to juggle the three balls while walking across the bridge. that way, the juggler can keep one ball in the air at all times. Will the juggler cross saftely? Yes or no?

using newton's 3rd law, I found that since the juggler needs to exert a force to propel a ball upward, the ball will exert an equal force down on the juggler.

The following is what my book says:

If M = juggler's mass, m = one ball's mass, and g = the acceleration of gravity, then according to newton's 2nd law,

the juggler's total weight, holding all three balls is: F = (M + 3m)g

the juggler's total weight, holding two balls is: F = (M + 2m)g

the force needed to accelerate the third ball is mg + ma. Based on this, the total force downward on the bridge is:

F = (M + 2m)g + mg + ma = (M + 3m)g + ma

therefore, according to this book, juggling a ball will actually be worse than not juggling at all.

So the answer is No, the juggler will destroy the bridge.

I have only one question about my book's reasoning. Why isn't the force downward on the bridge simply F = (M + 2m)g + ma? Because 1 ball is in the air, 1 of the 2 balls in the juggler's hands has to be accelerated.

The third ball is in the air. The first ball will currently stay in the juggler's hands. The second ball will then have the force of ma exerted on it to propel it into the air. Therefore, shouldn't the force downward be:

F = (M + 2m)g + ma?