Kvant Physics Problem 13
Two identical vertical cylinders contain water and are connected by a horizontal tube filled with the same incompressible fluid of density $\rho$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m14s
Source on kvant.digital
Problem
Two identical cylinders with pistons are connected by a tube (Fig. 4). Water is contained in the cylinders. Identical cylindrical beakers with equal amounts of water are placed on top of the pistons. Then a body of mass $m$ is placed into one beaker, and a body of mass $M$ into the other; both bodies float.
At what distances from each other will the ends of the pistons and the water levels in the beakers be when the system reaches equilibrium? The base areas of the beakers are $S_1$, and the piston areas are $S_2$.
Fig. 4
Setup and Assumptions
Two identical vertical cylinders contain water and are connected by a horizontal tube filled with the same incompressible fluid of density $\rho$. Each cylinder is closed by a frictionless piston of area $S_2$. Identical beakers are rigidly placed on the pistons so that the pistons support the full weight of each beaker and its contents.
Each beaker initially contains the same amount of water. After that, a body of mass $m$ is placed into the left beaker and a body of mass $M$ is placed into the right beaker. Both bodies float freely in the water.
The beaker base area is $S_1$. The bodies are treated as point masses for force balance and as fully floating objects obeying Archimedes’ principle. The density of the floating bodies is irrelevant beyond ensuring flotation. The water is incompressible, the pistons are massless, and friction is neglected. Atmospheric pressure acts equally on both beakers and cancels in pressure differences.
The unknowns are the vertical displacement difference between the pistons and the vertical distance between the free water surfaces in the beakers at equilibrium.
Physical Principles
Hydrostatic equilibrium in a connected incompressible fluid requires equality of pressure at points connected by the same fluid level.
For a static fluid, the pressure change with height satisfies
$$\Delta p = \rho g \Delta h.$$
A floating body transmits its entire weight to the fluid, so the additional force on the container is equal to its weight, independent of its immersion depth, while the displaced volume satisfies Archimedes’ law
$$\rho g V_{\text{disp}} = mg.$$
A piston in static equilibrium transmits force to the fluid according to
$$p = \frac{F}{S_2},$$
where $F$ is the total external force applied to the piston.
Derivation
Each beaker receives an additional downward force due to the floating body equal to its weight. The left side gains an extra force $mg$ and the right side gains $Mg$. These forces are transmitted through the pistons to the fluid.
The pressure increase in the fluid under each piston due to the added loads is
$$\Delta p_1 = \frac{mg}{S_2}, \qquad \Delta p_2 = \frac{Mg}{S_2}.$$
The pressure difference between the two sides at the piston level is therefore
$$\Delta p = \frac{(M-m)g}{S_2}.$$
Let the right piston move downward relative to the left by a distance $\Delta x$. The connected fluid must satisfy hydrostatic balance, so the pressure difference is also expressed through the vertical displacement of the piston-supported fluid columns:
$$\Delta p = \rho g \Delta x.$$
Equating the two expressions gives the piston separation shift
$$\rho g \Delta x = \frac{(M-m)g}{S_2},$$
which yields
$$\Delta x = \frac{M-m}{\rho S_2}.$$
The free surface level in each beaker changes due to displacement of water by the floating bodies. Each body displaces a water volume equal to $V = \frac{m}{\rho}$ or $V = \frac{M}{\rho}$. The corresponding rise of the water level in a beaker of cross-sectional area $S_1$ is
$$\Delta h_1 = \frac{m}{\rho S_1}, \qquad \Delta h_2 = \frac{M}{\rho S_1}.$$
The difference between the final free surface heights in the beakers is therefore
$$\Delta h = \Delta x + (\Delta h_2 - \Delta h_1),$$
since the piston displacement shifts the reference level of the beakers relative to each other while the internal water levels shift additionally due to displacement.
Substituting the expressions gives
$$\Delta h = \frac{M-m}{\rho S_2} + \frac{M-m}{\rho S_1}.$$
Result
Piston separation:
$$\boxed{\Delta x = \frac{M-m}{\rho S_2}}$$
Distance between the free water levels in the beakers:
$$\boxed{\Delta h = \frac{M-m}{\rho}\left(\frac{1}{S_2} + \frac{1}{S_1}\right)}$$
Here $\rho$ is the density of water in $\mathrm{kg/m^3}$, $M$ and $m$ are in $\mathrm{kg}$, and $S_1$, $S_2$ are in $\mathrm{m^2}$, giving $\Delta x$ and $\Delta h$ in meters.
Sanity Checks
Dimensional consistency holds because $\frac{M}{\rho S}$ has units $\frac{\mathrm{kg}}{\mathrm{kg/m^3 \cdot m^2}} = \mathrm{m}$, ensuring both $\Delta x$ and $\Delta h$ are lengths.
When $M=m$, both expressions vanish, corresponding to symmetric loading and no relative displacement.
If $S_2 \to \infty$, piston motion disappears since pressure change becomes negligible, leaving only beaker surface difference $\Delta h \to \frac{M-m}{\rho S_1}$. If $S_1 \to \infty$, beaker levels remain unchanged and only piston displacement survives, $\Delta x \to \frac{M-m}{\rho S_2}$.
A sign error would most easily arise in assigning the direction of piston motion relative to the heavier mass, since the hydrostatic relation couples displacement and pressure with opposite physical roles.