The weight $P$ of a system consisting of a beaker of water and a cork ball is measured in five cases:
1) the ball floats freely in the beaker (scale reading $P_{1}$);
2) the ball is lying on the scale next to the beaker ($P_{2}$);
3) the ball is held fully submerged by a thin weightless string attached to the bottom of the beaker ($P_{z}$);
4) the ball is held in a fully submerged state by a thin weightless spoke attached above the beaker ($P_{4}$);
5) the ball held in a drowned state is released and begins to float freely ($P_{5}$$).
Arrange the readings of the scales in ascending order.
Decision:
In the first three cases, two external forces act on the system consisting of a ball and a glass of water: the force of gravity $Mg$ ($M$ is the total mass of the glass, water and ball) directed downward, and the reaction force ($P_{1}, P_{2}$ or $P_{z}$) from the side of the scale directed upward. Since the center of gravity of the system is at rest, both forces compensate for each other. Следовательно, $P_{1}=P_{2}=P_{з}=Mg$.
In the fourth case, in addition to the forces $P_{4}$ and $Mg$, another external force $F$ acts on the ball from the spoke side, preventing the ball from floating and directed downward. Since the system is in equilibrium, it is clear that $P_{4}=F+Mg$, i.e. $P_{4}>Mg$.
In the fifth case, only two external forces act on the system: $Mg$ and the reaction force $P_{5}$ on the side of the stationary scale.But now, unlike in the first three cases, the ball floats. As a consequence, the center of gravity of the system falls, although the cup is stationary. The character of motion of the center of gravity is determined by the character of motion of the ball.At the initial moment the ball had no velocity, and in the following moments its velocity is directed upwards, hence, at the initial moment it moves with some acceleration directed upwards. At the same time, the center of mass of the system falls downward with some acceleration. Let this acceleration be equal to $a$. By Newton's law $Ma =Mg - P_{5}$. Hence $P_{5}=M(g-a) < Mg$.
We finally obtain: $P_{5} < P_{1}=P_{2}=P_{3} < P_{4}$.
1) the ball floats freely in the beaker (scale reading $P_{1}$);
2) the ball is lying on the scale next to the beaker ($P_{2}$);
3) the ball is held fully submerged by a thin weightless string attached to the bottom of the beaker ($P_{z}$);
4) the ball is held in a fully submerged state by a thin weightless spoke attached above the beaker ($P_{4}$);
5) the ball held in a drowned state is released and begins to float freely ($P_{5}$$).
Arrange the readings of the scales in ascending order.
Decision:
In the first three cases, two external forces act on the system consisting of a ball and a glass of water: the force of gravity $Mg$ ($M$ is the total mass of the glass, water and ball) directed downward, and the reaction force ($P_{1}, P_{2}$ or $P_{z}$) from the side of the scale directed upward. Since the center of gravity of the system is at rest, both forces compensate for each other. Следовательно, $P_{1}=P_{2}=P_{з}=Mg$.
In the fourth case, in addition to the forces $P_{4}$ and $Mg$, another external force $F$ acts on the ball from the spoke side, preventing the ball from floating and directed downward. Since the system is in equilibrium, it is clear that $P_{4}=F+Mg$, i.e. $P_{4}>Mg$.
In the fifth case, only two external forces act on the system: $Mg$ and the reaction force $P_{5}$ on the side of the stationary scale.But now, unlike in the first three cases, the ball floats. As a consequence, the center of gravity of the system falls, although the cup is stationary. The character of motion of the center of gravity is determined by the character of motion of the ball.At the initial moment the ball had no velocity, and in the following moments its velocity is directed upwards, hence, at the initial moment it moves with some acceleration directed upwards. At the same time, the center of mass of the system falls downward with some acceleration. Let this acceleration be equal to $a$. By Newton's law $Ma =Mg - P_{5}$. Hence $P_{5}=M(g-a) < Mg$.
We finally obtain: $P_{5} < P_{1}=P_{2}=P_{3} < P_{4}$.