Kvant Physics Problem 37

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Problem

In two vertically arranged cylinders, whose cross-sectional areas are $S_1$ and $S_2$, there are two weightless pistons connected by a thin weightless string of length $l$ (Fig. 1). The space between the pistons is filled with water. Find the tension in the string if the ends of the vessels are open to the atmosphere. The density of water is $\rho$.

Fig. 1

Setup and Assumptions

Two vertical cylinders of cross-sectional areas $S_1$ and $S_2$ are filled with water between two weightless, frictionless pistons. The pistons are connected by a massless, inextensible string of length $l$, so the vertical separation between the pistons is fixed at $l$. The upper end of the upper cylinder and the lower end of the lower cylinder are open to the atmosphere, so the external pressure at both free ends is $p_0$.

The fluid is incompressible with density $\rho$, and the system is in static equilibrium in a uniform gravitational field of acceleration $g$. The pistons are assumed to seal perfectly, and the string carries only tension $T$, identical along its length.

The unknown quantity is the tension $T$ in the connecting string.

Physical Principles

Hydrostatic equilibrium in a stationary incompressible fluid implies that pressure increases with depth according to $p = p_0 + \rho g h$, where $h$ is the vertical depth below a reference free surface.

For each piston in static equilibrium, the sum of vertical forces must vanish. The force exerted by the fluid on a piston equals the pressure at that interface multiplied by the piston area. Atmospheric pressure acts on the opposite side of each piston.

Derivation

Let the upper piston be at depth $y_1$ below the upper free surface. The pressure of water immediately below it is $p_1 = p_0 + \rho g y_1$. Above the piston the pressure is $p_0$. The upward force from water on the upper piston is $p_1 S_1$, while downward forces are atmospheric pressure $p_0 S_1$ and the string tension $T$, which pulls the upper piston downward.

Static equilibrium of the upper piston gives

$$p_1 S_1 = p_0 S_1 + T.$$

Substituting $p_1$ yields

$$(p_0 + \rho g y_1) S_1 = p_0 S_1 + T,$$

hence

$$T = \rho g y_1 S_1.$$

Let the lower piston be at depth $y_2$ below the upper free surface. The string constraint gives

$$y_2 = y_1 + l.$$

For the lower piston, the pressure above it is $p_2 = p_0 + \rho g y_2$, while below it the pressure is $p_0$. The upward forces acting on it are atmospheric pressure from below, $p_0 S_2$, and the string tension $T$, which pulls the lower piston upward. The downward force is the fluid pressure above, $p_2 S_2$.

Force balance on the lower piston gives

$$p_0 S_2 + T = p_2 S_2.$$

Substituting $p_2$ yields

$$p_0 S_2 + T = (p_0 + \rho g y_2) S_2,$$

and therefore

$$T = \rho g y_2 S_2.$$

Equating the two expressions for the same tension,

$$\rho g y_1 S_1 = \rho g y_2 S_2.$$

Canceling $\rho g$ gives

$$y_1 S_1 = y_2 S_2.$$

Substituting $y_2 = y_1 + l$ produces

$$y_1 S_1 = (y_1 + l) S_2.$$

Expanding,

$$y_1 S_1 = y_1 S_2 + l S_2,$$

$$y_1 (S_1 - S_2) = l S_2.$$

Solving for $y_1$,

$$y_1 = \frac{l S_2}{S_1 - S_2}.$$

Substituting into $T = \rho g y_1 S_1$ gives

$$T = \rho g S_1 \cdot \frac{l S_2}{S_1 - S_2}.$$

Result

The tension in the string is

$$T = \frac{\rho g l S_1 S_2}{S_1 - S_2}.$$

For numerical evaluation, the expression is already in SI units when $S_1$ and $S_2$ are in $\text{m}^2$, $l$ in $\text{m}$, $\rho$ in $\text{kg},\text{m}^{-3}$, and $g$ in $\text{m},\text{s}^{-2}$. The tension is then in newtons:

$$\boxed{T = \frac{\rho g l S_1 S_2}{S_1 - S_2}\ \text{N}}.$$

Sanity Checks

Dimensional analysis gives $[\rho g l S_1 S_2 / S] = (\text{kg},\text{m}^{-3})(\text{m},\text{s}^{-2})(\text{m})(\text{m}^4)/(\text{m}^2) = \text{kg},\text{m},\text{s}^{-2}$, which matches force units.

If $S_1 \to S_2$, the denominator tends to zero while the constraint forces equal hydrostatic levels that cannot differ by a finite $l$, indicating the model requires a divergent pressure gradient, consistent with the physical impossibility of maintaining a finite separation with equal areas under this constraint.

If $l \to 0$, the tension tends to zero since the pistons coincide and no hydrostatic imbalance is sustained.

The sign of the expression depends on the ordering of $S_1$ and $S_2$; a physically meaningful tension corresponds to $T = \rho g l S_1 S_2 / |S_1 - S_2|$, ensuring positivity regardless of which cylinder has larger area.