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If a man enclosed inside a chamber and the chamber is on a weighing scale and if he throws a ball straight up in air will the weight of the whole system changes or not? If yes, then what is the weight change at 1) just moments after throw; 2) before reaching the highest point of flight; 3) at highest point of flight; 4) on descending downwards; and 5) if there is any change in above observations in vacuum.

Supposing that scale is fast enough to register the changes before ball comes back to the man.

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In order to throw the ball up in the air, you must apply a force to give the ball a greater acceleration then the downwards force from gravity. By Newtons third law, the ball will exert a force of equal magnitude and opposite direction on the person. So since you are giving the ball a force in the upwards direction, you will feel a force downwards and the reading on the scale will deflect since the net force increases (not just gravity acting). After the brief deflection where your apparent weight increases, the scale will go to the equilibrium for the weight of the person from gravity.

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    $\begingroup$ the thing is, the weight of the ball will be removed temporarily, because if the man throws it, he must have had it in hand on the scales, so his weight is not recorded. It is a badly stated problem. $\endgroup$ – anna v May 2 '18 at 14:10
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    $\begingroup$ Before the mass of the ball is removed, there will be an instantaneous downward force and therefore increased apparent weight from the force. The scale will then come to a new equilibrium for the mass of the person without the ball. $\endgroup$ – fhorrobin May 2 '18 at 14:16
  • $\begingroup$ I'd add that the presence of the chamber is superfluous. $\endgroup$ – V.F. May 2 '18 at 14:53
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The force on the entire system (measured by the scale), must equal the weight of the entire system, plus any required to accelerate the center of mass upward.

If the center of mass is not accelerating, the scale will measure the weight of all the contents. If the center of mass is accelerating upward (such as when the person is starting to push the ball upward), then the scale will read higher than the weight.

If the center of mass is accelerating downward (such as when the ball is in the air), then the scale will read lower than the weight.

So assuming the man has had time to stop moving, then 1, 2, 3, and 4 are all identical. The center of mass of the system is accelerating downward, so the weight indicated by the scale is less than it was when the ball was held still.

A vacuum would only change the answer to the extent that we want to model air resistance. It's a small correction in this case.

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  • $\begingroup$ How can the ball affect the scale when it is up in the air? $\endgroup$ – V.F. May 2 '18 at 16:34
  • $\begingroup$ Is there a part of the answer that suggests that it should? That is reflected in the fact that the scale indicates a lower weight while the ball is in the air (and accelerating downward). $\endgroup$ – BowlOfRed May 2 '18 at 16:40
  • $\begingroup$ This is the sentence: "If the center of mass is accelerating downward (such as when the ball is in the air), then the scale will read lower than the weight." When the ball is in the air, CM could accelerate downward or upward, depending on whether the ball moves up or down, while the scale will just reflect the weight of the man. If I misinterpreted your statement, feel free to disregard my comment. $\endgroup$ – V.F. May 2 '18 at 17:42
  • $\begingroup$ Do not confuse acceleration with velocity. When the ball is in the air, it is constantly being accelerated downward (and that acceleration is $g$). $\endgroup$ – BowlOfRed May 2 '18 at 17:45
  • $\begingroup$ Good point. Now I see that your statement is entirely correct. Thanks for the clarification. $\endgroup$ – V.F. May 2 '18 at 20:21

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