Solution to Question 7
We have a horizontal cylinder supported by a fulcrum at its geometric center.
Total length $L = 33.0$ cm. Radius $r = 10.0$ cm.
A piston separates the water (left side) from the air (right side). The right side is open to the atmosphere via an orifice, so pressure there is $P_{atm}$.
Equilibrium Condition 1 (Mechanical): “When mass of water equals mass of piston, the cylinder stays horizontal.” This implies the center of mass of the system remains at the fulcrum.
Let $x$ be the position of the piston face measured from the left end of the cylinder.
The water occupies the length $x$. The piston is at $x$.
Mass of water $M_w = \rho A x$.
Mass of piston $M_p = M_w = \rho A x$.
Torque Balance about the Fulcrum ($L/2$):
1. Water: Treat as a point mass at $x/2$. Distance to fulcrum = $\frac{L}{2} – \frac{x}{2}$.
2. Piston: Point mass at $x$. Distance to fulcrum = $x – \frac{L}{2}$.
(The cylinder mass is uniform and centered at the fulcrum, so it creates no torque).
$$ \tau_{water} = \tau_{piston} $$ $$ M_w g \left( \frac{L-x}{2} \right) = M_p g \left( x – \frac{L}{2} \right) $$ Since $M_w = M_p$: $$ \frac{L-x}{2} = \frac{2x – L}{2} $$ $$ L – x = 2x – L \implies 3x = 2L $$ $$ x = \frac{2}{3} L = \frac{2}{3}(33.0) = 22.0 \text{ cm} $$
The piston is in equilibrium. The forces acting on it are:
- Fluid Force (Right): $F_{fluid} = P_{center} A$.
The pressure at the center of the cylinder (axis) is determined by the height of the water column relative to the axis.
Total height of water column from axis = $r$ (cylinder radius) + $h$ (tube height).
$P_{gauge} = \rho g (h + r)$.
$F_{fluid} = \rho g (h+r) A$. - Spring Force (left): The spring is compressed.
Relaxed length $l_0 = 15.0$ cm.
Current space for spring = Total Length ($L$) – Piston Position ($x$).
$L_{spring} = 33.0 – 22.0 = 11.0$ cm.
Compression $\Delta x = l_0 – L_{spring} = 15.0 – 11.0 = 4.0$ cm $= 0.04$ m.
Spring Force $F_s = k \Delta x = 400\pi (0.04) = 16\pi$ N.