A weight and a glass of water stand on a scale. Compare the reading of the scale $P$ with the reading of the weights in the following cases:
1) the kettlebell is suspended by a thread from an external tripod and is submerged in water ($P_{1}$);
2) the suspension is broken and the kettlebell begins to sink ($P_{2}$).
Decision:
The suspension will act on the system with some upward force. Carrying out reasoning similar to the solution of Problem 51, it is easy to obtain that the reading of the weights in this case decreases: $P_{1} < P$.
After the suspension breaks, the action of the external force ceases, but the center of mass of the system moves accelerated downward, hence $P_{2} < P$. To compare the values of $P_{1}$ and $P_{2}$, consider the forces acting on a glass of water in two cases. In the first case, the beaker of water is acted on by the downward force of gravity mg, the upward force $P_{1}$ and the downward force of pressure from the kettlebell, which is equal numerically to the hydrostatic force pressure on the weight $F_{A}$. The center of mass of the beaker of water is stationary, hence,
$P_{1}=mg+F_{A}$.
In the second case, the same gravity force, support reaction force $P_{2}$ and some pressure force from the kettlebell $F_{A}^{\prime}$ act. It is easy to show that $F_{A}^{\prime}>F_{A}$.Indeed, the kettlebell on the suspension was pressing on the surrounding water, and the water was in a stationary state.At the moment of collapse, the kettlebell begins to move downward at an accelerated rate and displace the nearby water layers in the same direction, hence the force of the kettlebell's pressure on the water must increase.Under the action of all the forces the center of mass of the water in the beaker moves upwards, hence $P_{2}>mg+ F_{A}^{\prime}>mg+F_{A}=P_{1}$.
The final result is $P>P_{2}>P_{1}$.
1) the kettlebell is suspended by a thread from an external tripod and is submerged in water ($P_{1}$);
2) the suspension is broken and the kettlebell begins to sink ($P_{2}$).
Decision:
The suspension will act on the system with some upward force. Carrying out reasoning similar to the solution of Problem 51, it is easy to obtain that the reading of the weights in this case decreases: $P_{1} < P$.
After the suspension breaks, the action of the external force ceases, but the center of mass of the system moves accelerated downward, hence $P_{2} < P$. To compare the values of $P_{1}$ and $P_{2}$, consider the forces acting on a glass of water in two cases. In the first case, the beaker of water is acted on by the downward force of gravity mg, the upward force $P_{1}$ and the downward force of pressure from the kettlebell, which is equal numerically to the hydrostatic force pressure on the weight $F_{A}$. The center of mass of the beaker of water is stationary, hence,
$P_{1}=mg+F_{A}$.
In the second case, the same gravity force, support reaction force $P_{2}$ and some pressure force from the kettlebell $F_{A}^{\prime}$ act. It is easy to show that $F_{A}^{\prime}>F_{A}$.Indeed, the kettlebell on the suspension was pressing on the surrounding water, and the water was in a stationary state.At the moment of collapse, the kettlebell begins to move downward at an accelerated rate and displace the nearby water layers in the same direction, hence the force of the kettlebell's pressure on the water must increase.Under the action of all the forces the center of mass of the water in the beaker moves upwards, hence $P_{2}>mg+ F_{A}^{\prime}>mg+F_{A}=P_{1}$.
The final result is $P>P_{2}>P_{1}$.