Weight of water with objects floating

[brotherbobby]

Homework Statement

The figure I put below shows three identical open-top containers filled to the brim with water - toy ducks float in two of them. If each of the containers were put on a weighing scale, rank the containers (a), (b) and (c) with the heaviest of them first.

Relevant Equations

1. $\text{Law of floatation : Weight (or mass) of a body = Weight (or mass) of liquid displaced}$. 2. Weight of a body $w = mg = \rho V g$

We understand that the crucial thing about the problem is that the volume of water present in the three containers are not the same. Also, we note that in each case the weight of the container is the total weight of its contents. (A student might be confused as to why should be so - after all, isn't a floating body weightless?! Yes it is - but in order for it to be weightless, the liquid must apply a buoyant force equal to the body's weight on it. The floating body in turn applies the same force back on to the liquid by the third law and this increases the weight of the liquid by an amount equal to that of the floating body).

If we denote the volume of the container itself as $V$, then the volume of water in each container can be ranked : $(V_W)_a [= V] > (V_W)_b > (V_W)_c$. Far as the densities are concerned, the density of water is more than that of the duck ($\rho_W > \rho_D$) and for the ducks in (b) and (c), we have $(V_D)_b < (V_D)_c$.

(a) The weight of the container in case (a) is simple : $\boxed{w_a = \rho_W Vg}$

(b) The weight of the container in case (b) : $w_b = (w_W)_b + (w_D)b$. The weight of the duck is $w_D = \rho_D (V_D)_b g$. As for the volume of water in (b), $(V_W)_b = V- (V_D)_b$. Hence the weight of the water : $(w_W)_b = \rho_W [V - (V_D)_b]g$. Opening brackets and adding the two weights, we find that the weight of the container in (b) $w_b = \rho_D (V_D)_b g + \rho_W Vg - \rho_W (V_D)_b g$ or we have $\boxed{w_b = \rho_W Vg - (V_D)_b g\underbrace{(\rho_W - \rho_D)}_{>\;0}}$. Given that the density of the duck is less than that of water, we have $\boxed{\color {green} {w_a > w_b}}$.

(c) This is identical to (b) above, except that the volume of the duck is more : $(V_D)_c > (V_D)_b$. Hence the weight of the container for c is $\boxed{w_c = \rho_W Vg - (V_D)_c g(\rho_W - \rho_D)}$. Subtracting a larger amount from the weight in (a) here than in (b) implies that $\boxed{\color {green} {w_b > w_c}}$.

Hence, finally : $\boxed{\color {green} {\mathbf{w_a > w_b>w_c}}}$.

Is my solution correct?

 

[Doc Al]

Hint: Make use of Archimedes' Principle.

 

[phinds]

Hint: Make use of Archimedes' Principle.

You are way overthinking it.

 

[PeroK]

Hint: Make use of Archimedes' Principle.

... and don't start thinking about surface tension!

 

[kuruman]

A neat application of the (Archimedes) principle involved here is the Fallkirk wheel. Does the weight load on each of the two arms depend on whether it's just water, a rubber duckie or a boatful of tourists floating on water?

 

[haruspex]

(b) The weight of the container in case (b) : $w_b = (w_W)_b + (w_D)b$. The weight of the duck is $w_D = \rho_D (V_D)_b g$. As for the volume of water in (b), $(V_W)_b = V- (V_D)_b$. Hence the weight of the water : $(w_W)_b = \rho_W [V - (V_D)_b]g$.

You are confusing the volume of the submerged part of the duck with its whole volume. If you were to compress the duck into its submerged volume its density would necessarily be the same as that of water.

 

[brotherbobby]

Hint: Make use of Archimedes' Principle.

According to the Archimedes' Principle - the loss in weight of a body (or upthrust) is equal to the weight of the liquid displaced ($U = \Delta w_L\; \text{where}\; U = w_B - w_B'$).

For both the floating ducks, the loss in weight is equal to their weights and so their floating weights are zero. The duck in (c) displaces more water than the duck in (b) but also has more weight. The amount of water in all the three containers is different.

I don't know how to proceed from here.

 

[haruspex]

the loss in weight of a body (or upthrust) is equal to the weight of the liquid displaced

Imagine removing a duck (somehow keeping the water in place) and putting the water it had displaced back into the hole created. What, according to Archimedes, would that water weigh? What would be the change in the scale reading?

 

[brotherbobby]

Imagine removing a duck (somehow keeping the water in place) and putting the water it had displaced back into the hole created. What, according to Archimedes, would that water weigh? What would be the change in the scale reading?

The (added) water would weigh the same as the duck and there would be no change in the reading of the scale.

 

[haruspex]

The (added) water would weigh the same as the duck and there would be no change in the reading of the scale.

... and it would look like picture a), right?

 

[brotherbobby]

... and it would look like picture a), right?

Yes. Doing what you said above would make both the containers (b) and (c) look identical to the situation in (a).

 

[PeroK]

Yes. Doing what you said above would make both the containers (b) and (c) look identical to the situation in (a).

Here's another way to look at it. Start with the three containers all full (with the same amount of water). Put the light duck in the second and some water spills out (how much?). Put the heavy duck in the third and some water spills out (how much)?

What is the resulting weight of water plus duck in each container?

 

[brotherbobby]

Here's another way to look at it. Start with the three containers all full (with the same amount of water). Put the light duck in the second and some water spills out (how much?). Put the heavy duck in the third and some water spills out (how much)?

What is the resulting weight of water plus duck in each container?

Excellent. In case (b), water equal to the weight of the light duck spills out - let's call it $\Delta (w_W)_b = (w_D)_b$. ($\Delta$ is used to show the weight of the water that spills out and not the water present). In case (c), water equal to the weight of the heavier duck spills out : $\Delta (w_W)_c = (w_D)_c$. Clearly, $\Delta (w_W)_b < \Delta (w_W)_c$.

However, since the heavy duck also weighs more than the lighter duck by the same amount that the water present weighs less, the resulting weight of water in both containers (b) and (c) are the same as that of (a).

 

[DEvens]

I'm sorry. All I can think about while reading this question is the episode from _Monty Python and the Holy Grail_ where they are trying to decide if the woman is a witch. And they decide you burn witches because they are made of wood. And because ducks also float on water, as does wood, then if she weighs the same as a duck then she's made of wood and therefore a witch.

So the water displaced by the duck is made of wood.