Problem 13: Floating Stack of Beads
Physical Analysis:
Let the mass of the beads be $m$ and the mass of water be $M$. The initial height of both columns is $h$. Since the density of plastic $\rho$ ($800 \, \text{kg/m}^3$) is less than that of water $\rho_0$ ($1000 \, \text{kg/m}^3$), the beads will float. When water is added, it fills the voids between the beads. The total volume of the mixture is determined by the volume of water plus the volume of the solid plastic material of the beads.
Let $A$ be the cross-sectional area of the vessel. The volume of water is $V_w = \frac{M}{\rho_0}$. The volume of the solid plastic material is $V_p = \frac{m}{\rho}$.
The buoyant force acts on the submerged volume of the plastic. For the beads to float in equilibrium, the weight of the entire stack of beads must equal the buoyant force provided by the displaced water.
$$ mg = F_B = \rho_0 V_{sub} g $$where $V_{sub}$ is the volume of the plastic submerged. Thus:
$$ V_{sub} = \frac{m}{\rho_0} $$The total volume below the water surface consists of the volume of the water added plus the volume of the submerged plastic:
$$ V_{below} = V_w + V_{sub} = \frac{M}{\rho_0} + \frac{m}{\rho_0} $$The height of the water level $h_w$ is:
$$ h_w = \frac{V_{below}}{A} = \frac{M + m}{\rho_0 A} $$However, there is a portion of the beads that remains “dry” above the water surface. The volume of this dry plastic is:
$$ V_{dry} = V_p – V_{sub} = \frac{m}{\rho} – \frac{m}{\rho_0} = m \left( \frac{1}{\rho} – \frac{1}{\rho_0} \right) $$To find the height of this dry layer, we must consider the packing fraction. Originally, the beads of mass $m$ occupied height $h$. Thus, the effective density of the stack was $\rho_{stack} = m / (Ah)$. The fraction of volume occupied by plastic is $\eta = V_p / (Ah) = m / (\rho Ah)$. Assuming the packing remains constant, the height of the dry portion $h_{dry}$ corresponds to the volume of dry plastic $V_{dry}$:
$$ h_{dry} = \frac{V_{dry}}{A \eta} = \frac{m (\frac{1}{\rho} – \frac{1}{\rho_0})}{A \cdot \frac{m}{\rho A h}} = h \left( 1 – \frac{\rho}{\rho_0} \right) $$We also know from the initial conditions that $M = \rho_0 A h$, so $A = \frac{M}{\rho_0 h}$. Substituting this into the expression for $h_w$:
$$ h_w = \frac{M+m}{\rho_0 (\frac{M}{\rho_0 h})} = h \left( 1 + \frac{m}{M} \right) $$The total height $H$ is the sum of the water level and the dry portion:
$$ H = h_w + h_{dry} = h \left( 1 + \frac{m}{M} \right) + h \left( 1 – \frac{\rho}{\rho_0} \right) $$Numerical Substitution:
Given: $h = 1.0\,\text{m}$, $m = 500\,\text{kg}$, $M = 1000\,\text{kg}$, $\rho = 800\,\text{kg/m}^3$, $\rho_0 = 1000\,\text{kg/m}^3$.
Answer: The height attained by the top layer of the beads is 1.7 m.