A beaker of mass $m = 100 g$ and volume $V = 200 cm^{3}$ is suspended upside down from a dynamometer, with the edges of the beaker touching the surface of water and the inside volume of the beaker filled with water. Why does the water not pour out of the beaker? What does the dynamometer show?
Decision:
Three forces act vertically on the water in the beaker: gravity $\rho gV$, pressure force $F_{1}$ from the bottom of the beaker, directed downward, and pressure force $F$ of the remaining water at the surface level, directed upward. The water in the beaker is in equilibrium if
$F = F_{1} + \rho gV$. (1)
Dividing (1) by the area of the bottom of the beaker $S$, we get
$p_{atm} = p + \rho gh$, (2)
where $p_{atm}$ is the water pressure at the surface level, equal to the atmospheric pressure; $p$ is the pressure of the bottom of the beaker on the water (according to Newton's third law, it is equal to the water pressure at the bottom of the beaker); $h$ is the height of the beaker. This formula gives the known dependence of pressure on depth inside an incompressible liquid. Water does not pour out of the beaker, because the difference of pressure forces at the surface and at the bottom of the beaker
exactly balances the weight of the water in the beaker.
Consider the forces acting vertically on the beaker: the force of gravity $mg$ and the force of atmospheric pressure $F_{atm}$ directed downward, the force of tension of the dynamometer spring $T$ and the force of water pressure on the bottom of the beaker $F_{1}$ directed upward. The beaker is in equilibrium, i.e..
$T+F_{1}=mg + F_{atm}$. (3)
From relation (3), taking into account equality (2), we have
$T = F_{atm} - F_{1} + mg = (p_{atm} - p)S + mg = (m + \rho V)g$. (4)
Hence $T = 3 N$. From formula (4) we see that the dynamometer readings are equal to the total weight of the beaker and the water filling it.
Decision:
Three forces act vertically on the water in the beaker: gravity $\rho gV$, pressure force $F_{1}$ from the bottom of the beaker, directed downward, and pressure force $F$ of the remaining water at the surface level, directed upward. The water in the beaker is in equilibrium if
$F = F_{1} + \rho gV$. (1)
Dividing (1) by the area of the bottom of the beaker $S$, we get
$p_{atm} = p + \rho gh$, (2)
where $p_{atm}$ is the water pressure at the surface level, equal to the atmospheric pressure; $p$ is the pressure of the bottom of the beaker on the water (according to Newton's third law, it is equal to the water pressure at the bottom of the beaker); $h$ is the height of the beaker. This formula gives the known dependence of pressure on depth inside an incompressible liquid. Water does not pour out of the beaker, because the difference of pressure forces at the surface and at the bottom of the beaker
exactly balances the weight of the water in the beaker.
Consider the forces acting vertically on the beaker: the force of gravity $mg$ and the force of atmospheric pressure $F_{atm}$ directed downward, the force of tension of the dynamometer spring $T$ and the force of water pressure on the bottom of the beaker $F_{1}$ directed upward. The beaker is in equilibrium, i.e..
$T+F_{1}=mg + F_{atm}$. (3)
From relation (3), taking into account equality (2), we have
$T = F_{atm} - F_{1} + mg = (p_{atm} - p)S + mg = (m + \rho V)g$. (4)
Hence $T = 3 N$. From formula (4) we see that the dynamometer readings are equal to the total weight of the beaker and the water filling it.