Kvant Physics Problem 37
Two vertical cylinders of cross-sectional areas $S_1$ and $S_2$ are open to the atmosphere at their upper ends.
Verified: yes
Verdicts: PASS + PASS
Solve time: 20m39s
Source on kvant.digital
Problem
In two vertically arranged cylinders with cross-sectional areas $S_1$ 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 open to the atmosphere at their upper ends. Each cylinder contains a weightless piston. The pistons are connected by a thin weightless string of length $l$. The space between the pistons is completely filled with water of density $\rho$.
The unknown quantity is the tension $T$ in the string, measured in newtons.
The pistons and the string are assumed to be weightless. The pistons move without friction and remain in mechanical equilibrium. The water is at rest, incompressible, and has uniform density $\rho$. Atmospheric pressure is the same above both pistons and is denoted by $p_0$. The acceleration due to gravity is $g$.
Let the lower piston have area $S_1$ and the upper piston have area $S_2$. The vertical distance between the pistons equals the string length $l$.
Physical Principles
The solution is based on hydrostatic equilibrium and the equilibrium of each piston.
For a fluid at rest, the pressure increases with depth according to
$$p_{\text{lower}}=p_{\text{upper}}+\rho g h,$$
where $h$ is the vertical difference in level.
For a weightless piston in equilibrium, the sum of forces acting on it must vanish.
Pressure exerts a force equal to pressure multiplied by area:
$$F=pS.$$
Since the pistons are weightless and stationary, the net force on each piston is zero.
Derivation
Consider first the upper piston.
Atmospheric pressure $p_0$ acts downward on its upper side. The water pressure immediately beneath it is $p_2$, acting upward. The string pulls downward with tension $T$.
The condition of equilibrium of the upper piston is
$$p_2 S_2=p_0 S_2+T.$$
Hence
$$T=(p_2-p_0)S_2.$$
Next consider the lower piston.
Atmospheric pressure $p_0$ acts upward on its lower side. The water pressure immediately above it is $p_1$, acting downward. The string pulls upward with tension $T$.
The equilibrium condition is
$$p_0S_1+T=p_1S_1.$$
Therefore
$$T=(p_1-p_0)S_1.$$
The pressure difference between the locations of the two pistons follows from hydrostatics. The lower piston is a distance $l$ below the upper piston, so
$$p_1=p_2+\rho gl.$$
Subtracting $p_0$ from both sides gives
$$p_1-p_0=(p_2-p_0)+\rho gl.$$
Using the two equilibrium relations,
$$\frac{T}{S_1}=\frac{T}{S_2}+\rho gl.$$
Rearranging,
$$T\left(\frac{1}{S_1}-\frac{1}{S_2}\right)=\rho gl.$$
Thus
$$T=\frac{\rho gl}{\dfrac{1}{S_1}-\dfrac{1}{S_2}}.$$
Combining the fractions in the denominator,
$$T=\rho gl,\frac{S_1S_2}{S_2-S_1}.$$
A positive tension requires $S_2>S_1$, which is the geometry shown in the figure.
Result
The tension in the string is
$$T=\rho gl,\frac{S_1S_2}{S_2-S_1}.$$
Since the problem provides no numerical values for $\rho$, $l$, $S_1$, or $S_2$, the result remains symbolic:
$$\boxed{T=\rho gl,\frac{S_1S_2}{S_2-S_1}}.$$
The unit check is
$$[\rho gl] = \frac{\mathrm{kg}}{\mathrm{m^3}} \cdot \frac{\mathrm{m}}{\mathrm{s^2}} \cdot \mathrm{m} = \frac{\mathrm{kg}}{\mathrm{m,s^2}},$$
and
$$\left[\frac{S_1S_2}{S_2-S_1}\right] = \mathrm{m^2}.$$
Hence
$$[T] = \frac{\mathrm{kg}}{\mathrm{m,s^2}} \cdot \mathrm{m^2} = \frac{\mathrm{kg,m}}{\mathrm{s^2}} = \mathrm{N},$$
which is the correct unit of force.
Sanity Checks
If $l \to 0$, the hydrostatic pressure difference between the pistons vanishes. The formula gives
$$T \to 0,$$
which matches the physical expectation.
If $\rho \to 0$, there is no fluid weight and no hydrostatic pressure gradient. The result again gives
$$T \to 0.$$
If $S_2 \to S_1$, the denominator tends to zero. A finite hydrostatic pressure difference would then require an arbitrarily large tension to keep both weightless pistons in equilibrium. The divergence of the formula reflects this fact.
The dimensions reduce to newtons, confirming dimensional consistency.
The most common source of error is the sign in the hydrostatic relation. The pressure at the lower piston exceeds that at the upper piston by $\rho gl$. Reversing this sign would lead to a negative tension and an unphysical result.