Equilibrium of Fluid with Lever System
We are asked to find the height $h_P$ of the water column under piston P. We will analyze the forces and torques acting on the system. Let us assume the cross-sectional area of the column under P is $S$. Note that the area under Q (let’s call it $S’$) does not need to be equal to $S$ for this derivation.
1. Mechanical Equilibrium (Torque)
The lever arm is in equilibrium under the tension forces from the two pistons. Taking moments about the fulcrum $F$:
$$ T_Q l_1 = T_P l_2 \quad \dots(1) $$2. Fluid Equilibrium (Forces)
Let’s analyze the forces on the water columns. The piston R is at the bottom and is free to slide. For the system to be in equilibrium, the pressure at the bottom must be equal to atmospheric pressure $P_0$ (balancing the outside air).
Consider the water column under piston P. The forces acting on this water column are:
- Downward force due to atmospheric pressure: $P_0 S$
- Upward force due to tension lifting the piston: $T_P$
- Upward force from bottom pressure: $P_0 S$
However, a simpler way to view this is that the tension $T_P$ supports the weight of the water column above the equilibrium level. If the column has height $h_P$ and area $S$:
$$ T_P = \text{Weight of water under P} = (\rho S h_P) g $$Similarly for piston Q (even if its area $S’$ is different):
$$ T_Q = \text{Weight of water under Q} = (\rho S’ h_Q) g $$3. The “Volume Moment” Relation
Now, substitute these weight expressions back into the torque equation (1):
$$ (\rho S’ h_Q g) l_1 = (\rho S h_P g) l_2 $$Notice that $\rho$ and $g$ cancel out. Also, observe that $S’ h_Q$ is the volume of water in the left column ($V_Q$), and $S h_P$ is the volume of water in the right column ($V_P$).
$$ V_Q l_1 = V_P l_2 $$This is a powerful result: The moments of the water volumes about the fulcrum balance each other. This holds true regardless of the cross-sectional areas.
4. Solving for Height using Total Volume
We are given the total volume of trapped water $V$.
$$ V = V_Q + V_P $$From our volume moment relation, we can express $V_Q$ in terms of $V_P$:
$$ V_Q = V_P \left( \frac{l_2}{l_1} \right) $$Substitute this into the total volume equation:
$$ V = V_P \left( \frac{l_2}{l_1} \right) + V_P $$ $$ V = V_P \left( \frac{l_2 + l_1}{l_1} \right) $$Solving for the volume under P ($V_P$):
$$ V_P = \frac{V l_1}{l_1 + l_2} $$5. Final Calculation of Height
Since we defined $V_P$ as the volume of the right column with cross-section $S$ and height $h_P$, we have $V_P = S h_P$.
$$ S h_P = \frac{V l_1}{l_1 + l_2} $$Note: $S$ in this formula refers specifically to the cross-sectional area of the pipe section containing piston P.