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- Homework Statement
- A small empty wooden drum of volume ##0.05\;\text{m}^3## and mass 5 kg floats on water contained in a beaker to a height of 1 m. The beaker itself has a base surface area of ##0.25\;\text{m}^2##. The drum has a heavy solid ball of volume ##5\times 10^{-4}\; \text{m}^3## and mass 4 kg resting on top of it. The drum-ball combination floats to a depth of ##d## m inside the water. With the drum-water-vessel system in equilibrium, the ball is dropped off the drum and allowed to settle at the bottom of the water column, with equilibrium attained once again. Calculate : (a) the depth to which the drum-ball combination floats, (b) the new height of the water column, (c) the new depth (##d'##) to which the drum floats in the water and (d) explain why is this new depth more (or less) than the original one.

- Relevant Equations
- 1. ##\mathbf{Archimedes principle :}## The loss in weight (due to upthrust or buoyancy) of a (fully) immersed body in water = Weight of water displaced. Thus ##\Delta w_B ( = U) = \rho_W V_B g = \Delta w_L ##. (Of course any other liquid would do just as well). The new weight of the body ##w'_B = w_B - U##

2. ##\mathbf{Law of floatation : }## A floating body has no weight OR The weight of a floating body = The weight of liquid displaced. Hence ##w'_B = 0## OR ##w_B = \Delta w_L##.

(The picture below is my drawing. I followed the instructions of the problems and drew for reasons of clarity.)

Let me start by writing down the given details : Volume of drum ##V_D = 0.05 m^3##, mass of drum ##m_D = 5 kg##, height of water column (initially) ##h_W = 1 m##, base area of water column ##A_W = 0.25 m^2##, volume of solid ball ##V_B = 4\times 10^{-4} m^3## and the mass of the solid ball ##m_B = 4 kg##.

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(a) The total mass of the drum-ball combination : ##m_C = m_D + m_B = 5+4 = 9 kg## but this is also the mass of water displaced. Hence ##\Delta m_W = 9 kg##. Thus the volume of water displaced by the combination ##\Delta V_W = \frac{\Delta m_W}{\rho_W} = \frac{9}{10^3} = 9 \times 10^{-3} m^3##. Hence the depth to which the combination sinks into water initially : ##d = \frac{\Delta V_W}{A_W} = \frac{9\times 10^{-3}}{0.25} = 0.036 m = 3.6 cm##, hence ##\boxed{d = 3.6 cm}##.

(b) The volume of water displaced by the ball (at rest on the bottom) : ##\left( \Delta V_W \right)_B = V_B = 5 \times 10^{-4} m^3##. Hence the water will move "up" by a height ##\left( \Delta h_W \right)_B = \frac{\left( \Delta V_W \right)_B}{A_W} = \frac{5\times 10^{-4}}{0.25} = 2\times 10^{-3} m## or the new height of the water column is ##\boxed{h'_W = 1.002 m}##.

(c) The volume of water displaced by the drum only in the second case : ##\left( \Delta V_W \right)_D = \frac{\left( \Delta m_W \right)_D}{\rho_W} \overset{\text{floatation law}}{=} \frac{m_D}{\rho_W} = \frac{5}{10^3} = 5\times 10^{-3} m^3##. Hence the depth to which the drum extends below water ##d' = \frac{\left( \Delta V_W \right)_D}{A_W} = \frac{5\times 10^{-3}}{0.25} = 0.02 m## or ##\boxed{d' = 2 cm}##.

(d) Hence, we find that from the drums point of view, it has

(I ask in particular for case (d). Is it ever possible for the water level to actually fall when the ball is thrown overboard?)

Let me start by writing down the given details : Volume of drum ##V_D = 0.05 m^3##, mass of drum ##m_D = 5 kg##, height of water column (initially) ##h_W = 1 m##, base area of water column ##A_W = 0.25 m^2##, volume of solid ball ##V_B = 4\times 10^{-4} m^3## and the mass of the solid ball ##m_B = 4 kg##.

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(a) The total mass of the drum-ball combination : ##m_C = m_D + m_B = 5+4 = 9 kg## but this is also the mass of water displaced. Hence ##\Delta m_W = 9 kg##. Thus the volume of water displaced by the combination ##\Delta V_W = \frac{\Delta m_W}{\rho_W} = \frac{9}{10^3} = 9 \times 10^{-3} m^3##. Hence the depth to which the combination sinks into water initially : ##d = \frac{\Delta V_W}{A_W} = \frac{9\times 10^{-3}}{0.25} = 0.036 m = 3.6 cm##, hence ##\boxed{d = 3.6 cm}##.

(b) The volume of water displaced by the ball (at rest on the bottom) : ##\left( \Delta V_W \right)_B = V_B = 5 \times 10^{-4} m^3##. Hence the water will move "up" by a height ##\left( \Delta h_W \right)_B = \frac{\left( \Delta V_W \right)_B}{A_W} = \frac{5\times 10^{-4}}{0.25} = 2\times 10^{-3} m## or the new height of the water column is ##\boxed{h'_W = 1.002 m}##.

(c) The volume of water displaced by the drum only in the second case : ##\left( \Delta V_W \right)_D = \frac{\left( \Delta m_W \right)_D}{\rho_W} \overset{\text{floatation law}}{=} \frac{m_D}{\rho_W} = \frac{5}{10^3} = 5\times 10^{-3} m^3##. Hence the depth to which the drum extends below water ##d' = \frac{\left( \Delta V_W \right)_D}{A_W} = \frac{5\times 10^{-3}}{0.25} = 0.02 m## or ##\boxed{d' = 2 cm}##.

(d) Hence, we find that from the drums point of view, it has

**moved up**relative to the water surface. This will**always**be the case, as the loss of the weight of the ball will imply that the drum will have less water to displace.**Is my solution above correct?**(I ask in particular for case (d). Is it ever possible for the water level to actually fall when the ball is thrown overboard?)