Kvant Physics Problem 13

Two identical vertical cylinders are connected by a horizontal tube and filled with water.

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Problem

Two identical cylinders with pistons are connected by a tube (Fig. 4). The cylinders contain water. Identical cylindrical cups containing equal amounts of water are placed on top of the pistons. Then a body of mass $m$ is immersed into one cup, and a body of mass $M$ is immersed into the other; neither body sinks.

At what distances from each other will the ends of the pistons and the water levels in the cups be when the system comes to equilibrium? The areas of the cup bottoms are $S_1$, and the piston areas are $S_2$.

Fig. 4

Fig. 4

Setup and Assumptions

Two identical vertical cylinders are connected by a horizontal tube and filled with water. Each cylinder is closed at the top by a movable piston of area $S_2$. On each piston stands an identical cylindrical cup whose bottom area is $S_1$. The cups initially contain equal amounts of water. A body of mass $m$ is immersed in the left cup and a body of mass $M$ is immersed in the right cup. Both bodies float in the water of their respective cups without touching the cup bottoms. The unknowns are the equilibrium vertical positions of the pistons and the equilibrium water levels in the cups. The quantity $\Delta x$ denotes the difference in piston heights, defined as the height of the right piston minus the height of the left piston. The quantity $\Delta h$ denotes the difference in the free water surface levels in the cups, defined in the same sense. The density of water is $\rho$. Atmospheric pressure acts equally on both cups and cancels from all pressure differences.

Physical Principles

A floating body displaces a volume of water whose weight equals the weight of the body. Hence

$V_m=\frac{m}{\rho}, \qquad V_M=\frac{M}{\rho}.$

The downward force transmitted by a floating body to the cup is equal to its weight. Consequently, relative to the initial symmetric state, the left cup acquires an additional load $mg$ and the right cup an additional load $Mg$. Hydrostatic equilibrium in the connected liquid requires that pressures at the same horizontal level be equal. If the piston heights differ by $\Delta x=x_R-x_L$, then

$p_L-p_R=\rho g,\Delta x.$

The total volume of water inside the connected cylinders is constant. If the left piston moves downward by $x_L$ and the right piston by $x_R$, measured from the initial positions with downward positive, then

$S_2 x_L+S_2 x_R=0,$

or equivalently,

$x_R=-x_L.$

Derivation

The total downward force acting on each piston consists of the weight of the water in the cup together with the load transmitted by the floating body. Since the cups initially contain equal amounts of water, the difference of forces acting on the pistons is

$F_L-F_R=mg-Mg=-(M-m)g.$

Hence the pressure difference at the piston surfaces is

$p_L-p_R=\frac{F_L-F_R}{S_2}=-\frac{(M-m)g}{S_2}.$

Combining this with hydrostatic equilibrium in the connected liquid gives

$\rho g, \Delta x=-\frac{(M-m)g}{S_2},$

so that

$\Delta x=x_R-x_L=-\frac{M-m}{\rho S_2}.$

This shows that the piston carrying the heavier body is lower, and the pistons are displaced symmetrically about their initial positions.

To determine the water-level differences in the cups, let $\delta h_L$ and $\delta h_R$ denote the rises of the water surfaces in the cups relative to the corresponding piston tops. For the left cup, the displaced volume $V_m$ must equal the increase of water volume in the cup plus the volume expelled from the cup into the cylinder by the downward piston motion. Thus

$S_1 \delta h_L = V_m - S_2 x_L.$

Similarly, for the right cup,

$S_1 \delta h_R = V_M - S_2 x_R.$

The difference of these water-level rises relative to the piston tops is

$S_1 (\delta h_R - \delta h_L) = (V_M - V_m) - S_2 (x_R - x_L).$

Using the relations $V_M - V_m = (M-m)/\rho$ and $x_R - x_L = \Delta x = - (M-m)/(\rho S_2)$ yields

$S_1 (\delta h_R - \delta h_L) = \frac{M-m}{\rho} - S_2 \left(-\frac{M-m}{\rho S_2}\right) = \frac{M-m}{\rho} + \frac{M-m}{\rho} = \frac{2(M-m)}{\rho}.$

Hence

$\delta h_R - \delta h_L = \frac{2(M-m)}{\rho S_1}.$

The absolute water-level difference between the cups is then

$\Delta h = (x_R + \delta h_R) - (x_L + \delta h_L) = (x_R - x_L) + (\delta h_R - \delta h_L) = \Delta x + (\delta h_R - \delta h_L).$

Substituting the computed values gives

$\Delta h = -\frac{M-m}{\rho S_2} + \frac{2(M-m)}{\rho S_1}.$

This formula shows that the water surface in the cup under the heavier body is lower due to the piston displacement, but rises relative to the piston because the larger body displaces more water; the net effect depends on the relative sizes of $S_1$ and $S_2$.

Result

The equilibrium difference between the piston heights is

$\boxed{\Delta x=-\frac{M-m}{\rho S_2}}.$

The equilibrium difference between the absolute water levels in the cups is

$\boxed{\Delta h = -\frac{M-m}{\rho S_2} + \frac{2(M-m)}{\rho S_1}}.$

The piston under the heavier body is lower by $(M-m)/(\rho S_2)$, while the water surface in that cup is raised relative to the piston top by $2(M-m)/(\rho S_1)$. The net water-level difference $\Delta h$ results from the combination of piston displacement and water displacement by the floating bodies.