Problem Solution: Glue Strength in Floating Composite Body
Problem Overview: Two objects of different densities are glued together and float fully submerged in a liquid. This implies neutral equilibrium where the average density of the composite body equals the density of the liquid. We need to find the depth $h$ at which the hydrostatic pressure aids the glue in holding the blocks together against the maximum allowable force $F$.
Step 1: Determine Liquid Density
Since the composite body floats fully submerged, the density of the liquid ($\rho_L$) must equal the average density of the two bodies (since they have equal volume $V$).
$$ \rho_L = \frac{m_{total}}{V_{total}} = \frac{\rho_1 V + \rho_2 V}{2V} = \frac{\rho_1 + \rho_2}{2} $$ $$ \rho_L = \frac{400 + 600}{2} = 500 \, \text{kg/m}^3 $$Step 2: Force Analysis on the Upper Block
Upward force on Body 1 = $\text{Buoyancy}_1 – \text{Pressure Force at } S$.
$F_{\text{up}} = \rho_L V g – (\rho_L g h) S$.
Equation of Equilibrium for Body 1:
$$ F_{\text{up}} = W_1 + F_{\text{glue}} $$ $$ \rho_L V g – \rho_L g h S = \rho_1 V g + F_{\text{glue}} $$We need the glue to hold, so $F_{\text{glue}} \le F_{\text{max}} = 500 \, \text{N}$. To find the minimum depth, we set tension to the maximum allowable force.
$$ F_{\text{max}} = (\rho_L – \rho_1)Vg – \rho_L g S h $$Step 3: Calculation
Substitute $\rho_L = \frac{\rho_1 + \rho_2}{2}$:
$$ F = \left( \frac{\rho_1 + \rho_2}{2} – \rho_1 \right) V g – \left( \frac{\rho_1 + \rho_2}{2} \right) g S h $$ $$ F = \left( \frac{\rho_2 – \rho_1}{2} \right) V g – \left( \frac{\rho_1 + \rho_2}{2} \right) g S h $$Rearranging for $h$:
$$ \left( \frac{\rho_1 + \rho_2}{2} \right) g S h = \left( \frac{\rho_2 – \rho_1}{2} \right) V g – F $$ $$ (\rho_1 + \rho_2) g S h = (\rho_2 – \rho_1) V g – 2F $$ $$ h = \frac{(\rho_2 – \rho_1) V g – 2F}{(\rho_1 + \rho_2) g S} $$Substituting Values:
- $V = 1.0 \, \text{m}^3$
- $\rho_1 = 400, \rho_2 = 600 \, \text{kg/m}^3$
- $g = 10 \, \text{m/s}^2$
- $S = 100 \, \text{cm}^2 = 10^{-2} \, \text{m}^2$
- $F = 500 \, \text{N} \implies 2F = 1000 \, \text{N}$