A tube of diameter $d$ is lowered into a vessel with water. A ball of the same diameter is placed in the tube. The center of the ball is at a depth $h$. Find the density of the substance of the ball. There is no gap between the pipe and the ball. The frictional force between them is zero.
Decision:
Let $m$ and $\rho$ be the mass and density of the ball, $V =\frac{4}{3}\pi \left ( \frac{d}{2} \right )$ is its volume. Consider the forces acting on the ball. These are the downward directed force of gravity $mg$ and the upward directed force of water pressure $Q$ acting on the downward facing surface of the balloon. Since the ball is at rest, by Newton's law II we have:
$mg-Q = 0$. (1)
Let's find the force $Q$. To do this, mentally remove the balloon and pour water into the pipe so that the water level in the pipe is the same as outside. Obviously, the system will remain in equilibrium. Hence,
$Mg-Q = 0$, (2)
where $M$ is the mass of the refilled water. Comparing (1) and (2), we find that
$m = M, \rho_{1}V_{1}=\rho V$, (3)
where $\rho_{1}$ is the density of water and
$V_{1}=\pi \left( \frac{d}{2} \right)^{2}h+\frac{2\pi}{3} \left( \frac{d}{2} \right)^{3}$
- is the volume of water (it is the volume of a cylinder of height h with base area $\pi \left ( \frac{d}{2} \right )^{2}$ and the volume of half of the balloon).
From (3) and (4) we find:
$\rho = \rho_{1} \frac{1+ \frac{3h}{d}}{2}$
Decision:
Let $m$ and $\rho$ be the mass and density of the ball, $V =\frac{4}{3}\pi \left ( \frac{d}{2} \right )$ is its volume. Consider the forces acting on the ball. These are the downward directed force of gravity $mg$ and the upward directed force of water pressure $Q$ acting on the downward facing surface of the balloon. Since the ball is at rest, by Newton's law II we have:
$mg-Q = 0$. (1)
Let's find the force $Q$. To do this, mentally remove the balloon and pour water into the pipe so that the water level in the pipe is the same as outside. Obviously, the system will remain in equilibrium. Hence,
$Mg-Q = 0$, (2)
where $M$ is the mass of the refilled water. Comparing (1) and (2), we find that
$m = M, \rho_{1}V_{1}=\rho V$, (3)
where $\rho_{1}$ is the density of water and
$V_{1}=\pi \left( \frac{d}{2} \right)^{2}h+\frac{2\pi}{3} \left( \frac{d}{2} \right)^{3}$
- is the volume of water (it is the volume of a cylinder of height h with base area $\pi \left ( \frac{d}{2} \right )^{2}$ and the volume of half of the balloon).
From (3) and (4) we find:
$\rho = \rho_{1} \frac{1+ \frac{3h}{d}}{2}$