if a beaker was filled with 200 cc of water and we tethered a ping pong ball to it's floor and that ping pong ball size is 10 cc .

Q1: what will be the weight of the beaker at that time ? (ignore the empty weight, ball and thread)

A1 : 200 grams

Q2: what will be the weight of the beaker at the first few milliseconds if we cut the thread ?

A2: 210 grams

(for the next analysis , suppose that the ball is like balloon and we just popped it with a needle so the air inside it is free)

for those who agree, they might explain it by the buoyancy force of the ball acting against the water. but the ball is almost weightless (presumed with no weight in this experiment , think of it as air bubble) , so how come a weightless object (say 0.0122 grams) can cause a reaction of 10 grams ?

the way i see it is that it all depends on the total water pressure acting on the beaker floor and that force is 210 before and after the thread being cut or the balloon exploded but in the first case there was upward tension of the thread acting on the beaker while in second case, that tension was gone.

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    $\begingroup$ The air is neither weightless, massless, nor without volume. The water must accelerate around the ball but cannot move past the bottom. The water above the ball must be stopped, resulting in a force against the bottom of the beaker which consequently results in a force against the weighing device. You have moved from a static equilibrium to a dynamic situation. So your premise of a "weightless object" is flawed. And it's not clear what your specific question is regarding a physics concept. $\endgroup$
    – Bill N
    May 20 '17 at 2:12
  • $\begingroup$ Possible duplicate of changing force without mass or acceleration alteration $\endgroup$ May 22 '17 at 13:33
  • $\begingroup$ I never said the air doesn't have mass, I actually gave it mass of 0.0122 gram in that example, which is the weight of 10 cc of air $\endgroup$ May 23 '17 at 21:18

Your answer to question $2$ is incorrect for the following reason.

When the ball is tethered, the forces on the ball and the forces on the water are shown in the left-hand diagram.

The reading on the spring balance (weighing scale) is the force which the spring balance exerts on the water (and the beaker) which is the weight of the water $W$ and the weight of the ball and the air inside it $w$.
The net force on the ball is zero due to the tether applying a downward force on the ball equal to $W-w$.
In turn (using Newton's third law) the tether applies an upward force on the pan of the spring balance of the same magnitude $W-w$.

enter image description here

Your mistake was to say that when the tether was cut the reading on the spring balance changed.
I think that you did this because you wanted the forces to balance and removing the tether did reduce an upward force on the pan of the spring balance.
However there must be a net force on the water because as the ball accelerates upwards the water must be accelerating downwards because the centre of gravity of the water moves downwards.

In the right-hand diagram there is a net force upwards on the ball $W-w$ which causes it to accelerate upwards.

There is a net downward force on the water of $W+X-(X+w) = W-w$ which causes the centre of mass of the water to accelerate downwards.

  • $\begingroup$ I studied the downward acceleration of water and the change of the water mass center. It's like the sand downward flow in a sand watch, it doesn't weigh less during that process because as the sand is released to respond to gravity, it hits the bottom or previous sand layers and causes a down force. I totally understand and agree for the diagram on the left, although I see both side with different approach, where X instead of being mass, it's the water pressure multiplied by the beaker floor area, which is more than just the mass because of the extra water displacement. $\endgroup$ May 23 '17 at 21:39
  • $\begingroup$ Technically the inverse square law of. Gravity also plays a part here. When the ball is submerged more of the water molecules are higher up and farther from the center of the earth. Total acceleration due to gravity should therefore be smaller. If you release the ball more of the water will fall closer to the center of the earth increasing its total acceleration due to gravity. $\endgroup$ May 24 '17 at 21:04
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    $\begingroup$ We can't include distance to "center of the earth" in this experiment. It may count at heights like low earth orbit and above. But even 1 meter of change for water level can be neglected as it represents 0.00000015 of the distance to the center of the earth $\endgroup$ May 24 '17 at 22:01

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