Solution for "Where does the weight go?"

WE HAD a classical riddle of a man crossing a bridge juggling all the way thereby reducing the effective weight acting on the bridge. We had to find a way to mathematically disprove this solution.

Note that when the man applies some force on a ball (to throw it up), the ball applies an equal and opposite force on him (downwards). This force gets transferred to the bridge as there is no acceleration of the man in the vertical direction. So, the total force acting on the bridge is the sum of the weight force of the man and the force applied by the ball.

( Let < signify "less than or equal to" and > signify "more than or equal to")

Let T be the time for which a ball remains in air. And let t be the time for which the ball remains in the hands of the man. Since the man has to handle 9 more balls before the first ball returns in his hands,
T > 9t

Let v be the velocity with which the ball leaves the man's hands. This is also the maximum velocity of the ball. Time required for the ball to reach the highest point is T/2. Hence,

v=gT/2

Let the man applies a constant force F on the ball in his hand for a period t while throwing it up. The ball comes down with a velocity v downwards and is thrown back up with a velocity v upwards. Hence the change in momentum of the ball is m(2v), where m is the mass of the ball. Hence,

F = 2.m.v/t + mg

=> 2mv = (F - mg).t

by the inequality of t,

(F - mg).T/9 > 2.m.v

Substituting the value of v,

(F - mg).T/9 > 2.m.(gT/2)

(F - mg)/9 > mg

F - mg > 9 mg

F > 10 mg

It implies that the force applied by each ball on the man would be greater than or equal to the weight of all ten balls. If one juggles, he is effectively putting more force on the bridge than he would otherwise have. Hence, juggling does no good. The strength of the bridge has to be at least 70.9 kg for that man to pass through with those balls; and even with that strength, it's not prudent to juggle.