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Tension in a thread holding a ball at the bottom of a vessel
Problem statement
A wooden ball of volume $V = 100$ cm³ and density $\rho = 600$ kg/m³ is held by a thread tied to the bottom of a vessel of water. The ball is completely submerged (see the figure). The density of water is $\rho_w = 1000$ kg/m³. Find the tension in the thread. Take $g = 10$ m/s².
Given
- $V = 100$ cm³ $= 10^{-4}$ m³
- $\rho = 600$ kg/m³
- $\rho_w = 1000$ kg/m³
- $g = 10$ m/s²
- $T = ?$
Hints
Hint
1 / 2
The ball is less dense than water, so the buoyant force exceeds gravity and the thread pulls the ball down. Write the force balance on the ball.
Hint
2 / 2
The buoyant force is $F_A = \rho_w g V$, gravity is $mg = \rho V g$. The tension equals their difference.
Solution
The ball is lighter than water, so the water pushes it up and the thread keeps it from floating. The forces on the ball are the buoyant force $F_A$ (upward), gravity $mg$ (downward) and the thread tension $T$ (downward, toward the bottom). The ball is at rest.
1. Equilibrium condition:
$$F_A = mg + T \;\Rightarrow\; T = F_A - mg$$2. The buoyant force (the ball is fully submerged, so the displaced volume is $V$):
$$F_A = \rho_w g V = 1000 \cdot 10 \cdot 10^{-4} = 1 \text{ N}$$3. Gravity:
$$mg = \rho V g = 600 \cdot 10^{-4} \cdot 10 = 0.6 \text{ N}$$4. The tension:
$$T = 1 - 0.6 = 0.4 \text{ N}$$General formula: $T = (\rho_w - \rho)\,V g$.
Answer
$T = 0.4$ N