Will the water level rise?

[brotherbobby]

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 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?)

 

[jbriggs444]

For part b you are asked to calculate "the new height of the water column". You have calculated the change due to one effect but have ignored a different effect.

 

[brotherbobby]

For part b you are asked to calculate "the new height of the water column". You have calculated the change due to one effect but have ignored a different effect.

I am trying to think of the second effect you referred to. I can't seem to find any. A hint?

 

[jbriggs444]

I am trying to think of the second effect you referred to. I can't seem to find any. A hint?

You take the weight off of the drum. What happens to the drum? What happens to the water previously displaced by the drum?

 

[brotherbobby]

You take the weight off of the drum. What happens to the drum? What happens to the water previously displaced by the drum?

The weight of the drum ($w_D = m_D g$) remains the same obviously, unless you are referring to the fact that weight of the ball ($w_B = m_B g$), which is missing. The drum-ball combination displaced a mass of water ($\Delta m_W$) equal to their own combined mass ($\Delta m_W = m_D + m_B$). Now, the mass of water displaced would be the mass of the drum alone ($\Delta m'_W = m_D$). The volume of this mass of displaced water : $\Delta V'_W = \Delta m_W / \rho_W = m_D/\rho_W$.

Earlier the volume of water displaced (refer above, in this response) : $\Delta V_W = (m_D + m_B) / \rho_W$. Clearly, there is less water displaced now; the difference $= m_B / \rho_W$.

Let me put the numbers in now. The two effects, as you said.

(1) There is less volume of water displaced due to the reduced mass of the drum, as the ball is missing. This reduced amount : $\Delta V_{\text{less}} = m_B/\rho_W = 4/10^3 = 4\times 10^{-3} m^3$. Hence the reduced height : $\left( \Delta h_W \right )_D = \Delta V_{\text{less}} / A_W = 4\times 10^{-3} / 0.25 =- 0.016 m = - 1.6 cm$ (negative because the water comes down).

(2) There is more water displaced due to the ball (settled at the bottom of the vessel and in equilibrium). (see above, in my solution). The extra height with which the water is displaced up $= \left( \Delta h_W \right )_B = 0.2 cm$ (positive because the water goes up).

Hence the net water level comes down : $\Delta h_W = -1.6 + 0.2 =\boxed{ -1.4 cm}$.

Is my solution now correct? If I am right, then the level of water actually goes down, after the ball is thrown over!

 

[jbriggs444]

I agree with your computation and that the water level is reduced by the action of throwing the ball overboard.

 

[brotherbobby]

I agree with your computation and that the water level is reduced by the action of throwing the ball overboard.

Thank you, another non-intuitive effect in physics! That the water level of a lake would actually drop if the boatman threw off a heavy ball he was carrying.

However, even though the level of water in the lake would go down, relative to the level of water, the boat would float up. (See our problem : the drum-ball combination was floating by 3.6 cm under water while the drum itself floats with 2 cm under water. Hence, relative to the water, the drum is floating higher up by 1.6 cm).

Thus, in a different sense, intuition continues to hold. Indeed we would be better off by throwing heavy belongings off the boat into the lake. Of course the water level in the lake would fall, but as passengers and not physicists, we don't care. What we care is how high does our boat stand relative to water.

Please comment when you have the time.

 

[jbriggs444]

Thus, in a different sense, intuition continues to hold. Indeed we would be better off by throwing heavy belongings off the boat into the lake. Of course the water level in the lake would fall, but as passengers and not physicists, we don't care. What we care is how high does our boat stand relative to water.

Agreed. The drum (or boat) rises relative to the water. If we are trying to avoid being swamped, that is a win.