Hollow and solid spheres floating in liquid

[brotherbobby]

Homework Statement

Two spheres of the same mass and radius, but one hollow and other solid, float in a vessel containing a certain amount of liquid as shown in the diagram below. Would the height of liquid be the same in the two occasions?

Relevant Equations

Law of flotation : Mass of the body = Mass of liquid displaced ##\left(m_B = \Delta m_L \right)##

The two situations are shown in the figures alongside. The hollow sphere has a thick heavy rim that compensates for the air inside it - both spheres have the same mass ##m_B## and radius ##r_B##.

Since the bodies have the same mass ##m_B##, the mass of liquid displaced is the same : ##\left(\Delta m_L \right)_1 = \left(\Delta m_L \right)_2 = m_B##.
Hence the volume of liquid displaced is the same : ##\left(\Delta V_L \right)_1 = \left(\Delta V_L \right)_2##.
The vessel is the same, with the same cross sectional area A.
Hence the heights of liquid displaced are the same : ##\left(\Delta h_L \right)_1 = \left(\Delta h_L \right)_2##.
The liquids had the same heights to begin with : ##\left(h_L \right)_1 = \left( h_L \right)_2##.
Hence the final height of the liquid is the same for the two cases :##\boxed{\left(h'_L \right)_1 = \left(h'_L \right)_2}##, which is my answer.This is my problem - given what I read from a textbook : (Knight - Physics for Scientists and Engineers).

The equation (14.14) above (in the text) written in my notation would read : ##\frac{\Delta V_L}{V_B} = \frac{\rho_B}{\rho_L}##. Note ##V_B## is the volume of the whole body.
The boxed warning in red clearly says that the equation above cannot be used if bodies have varying densities within them.
The hollow sphere has a non-uniform density - a very heavy metallic exterior and almost zero density hollow interior.
Yet, I used the same equation from the law of flotation for it.

Is that correct?

 

[etotheipi]

What you did is correct, two floating objects of identical mass will displace the same volume of water. If the surfaces of those objects have the same shape, then they will sink to the same depth.

What I assume Knight is saying is that for equation (14.14) to be valid for an object of non-uniform density, the ##\rho_o## must actually be the average density of the object, and not the density of, for instance, just the shell material. To put it another way, you can start with the condition for static equilibrium for an object of non-uniform density,$$B - mg = 0 \implies \rho_f V_f g - m_o g = 0$$and use that, if ##\rho_o## is the average density of the object (i.e. ##\rho_o = m_o / V_o##), then you'll get$$\rho_f V_f = \rho_o V_o$$but although you could in theory calculate the average density of something like a boat, you'd definitely be better off working with its mass.

 

[brotherbobby]

Thank you ethothepi for your quick response.

To bring the problem above into sharper focus, I have created a slightly different version of it.

A piece of iron is glued to the top of a block of wood. The combined object floats with the iron on the top. The block is now turned over so that it is the iron that is submerged beneath the wood. (1) Does the amount of "solid" immersed in liquid (##\Delta h_L##) change? (2) Is the mass of solid submerged into the liquid change (##\Delta m_S##)?

(Let us agree to call the solid as S and liquid as L)

Solution : Of course the combined solid mass floats on both occasions (same ##m_S##). This might not be so trivial to the student, who might feel that when the heavier portion of the combined mass is lower, it will sink.

(1) The mass of liquid displaced ##\Delta m_L = m_S##, so the height of liquid displaced (or height of solid under liquid) ##\Delta h_L = \frac{\Delta V_L}{A} = \frac{\Delta m_L}{\rho_L A} = \frac{m_S}{\rho_L A}##, where A is the area of cross section of the body and ##\rho_L## is the density of the liquid. Since all the three values remain the same on both occasions, the height of liquid displaced is the same : ##\boxed {\left( \Delta h_L \right)_1 = \left( \Delta h_L \right)_2}##.

(2) The average density of the solid object ##\bar{\rho}_S = \frac{m_S}{V_S} = \frac{\rho_W V_W + \rho_I V_I}{V_W+V_I}##. The same volume of the solid is submerged into the liquid on both occasions : ##\Delta V_S = \Delta V_L = \frac{m_S}{\rho_L}##. However, since the average density of the solid under liquid is greater in the second situation (##\bar{\rho}_1 < \bar{\rho}_2##) (under liquid), there is greater mass of solid under water in the second situation : ##\boxed {\left(\Delta m_S\right) _1 < \left( \Delta m_S \right)_2}##.

Please let me know if am correct. I have not come across the question as to whether there is "greater mass of solid under water". However, I am certain it makes sense if the solid is a combination of bodies of different densities.

 

[etotheipi]

Both parts of your analysis are correct, yes. The key thing to notice is that Archimedes' principle derives from the fact that the buoyant force acting on a body, which arises due to pressure forces of the liquid on the body that scale linearly with depth, is just proportional to volume submerged below the water level (with proportionality constant ##\rho_l##).

So if two bodies of the same total mass float, then they have equal weights and require the same buoyant force for static equilibrium, and thus displace the same volume of water. That in itself doesn't place any constraints on where the highest density regions in the body are. Indeed, you can achieve equilibrium with the iron and wood in any order, and they will sink to the same depth.

What you will notice is that the body is in stable equilibrium when the iron is below, and in unstable equilibrium when the iron is above. Imagine what will happen if you give either of them a small nudge!

 

[Lnewqban]

It is always a balance of weights, independently of how densities combine with volumes to create those weights.
The same variation of densities with height could apply to the fluid (a mix of oil and water, for example).

The pressure gradient exerting the upthrust would be exactly the same for the fluid (liquid or gas) or for a solid occupying the same volume under the surface.
Always, the weight of the displaced liquid must be compensated with something else for balance of forces to exist.

 

[kuruman]

@brotherbobby, here is a multiple choice question that should clarify in your mind Archimedes's principle.

The figure shows a scale on which are about to be placed two identical glasses. The glass on the left contains water in which floats a piece of wood. The glass on the right contains water in which floats a chunk of ice. The level of the water is the same in both glasses. When the glasses are placed on the scale, which arm will go down?

A. The left arm will go down.
B. The right arm will go down.
C. Neither arm will go down.
D. More information is needed to answer the question.

 

[brotherbobby]

When the glasses are placed on the scale, which arm will go down?

I would choose D : More information is needed to answer the question. (That information being the masses of the floating bodies, wood and ice.)

A likely pitfall (error) inherent in the question is the notion of weightlessness of a floating body. The careless but knowledgeable student will argue that both the sides of the balance would weigh the same since the floating bodies, being weightless, have no forces of their own to contribute to the mass of water and vessel already present.

However, why is a floating body weightless? Because the buoyant force of the displaced liquid matches its own weight. By Newton's 3rd law, this bouyant force is also the force the body exerts on the liquid. Hence if the block of wood weighs more, it would also "push" down on the water and the vessel more, increasing the reading on its side of the weighing scale.

 

[haruspex]

D : More information is needed to answer the question.

Sorry, but you'd be wrong.
Archimedes' principle comes from thinking about removing the floating object and replacing it with more of the the fluid, just filling the cavity that results.
Clearly, that would would be a stable arrangement, so the buoyant force on the replacement fluid must exactly equal its weight. There is no reason why the buoyant force would be any different from that exerted on the object that had occupied exactly the same space below the fluid surface, and that force balanced the object's weight. Therefore the weight of the object equals the weight of fluid displaced.

Apply this to the ice and block of wood: mentally replace each with water just filling the cavity in the surrounding water.

 

[kuruman]

I would choose D : More information is needed to answer the question. (That information being the masses of the floating bodies, wood and ice.)

A likely pitfall (error) inherent in the question is the notion of weightlessness of a floating body. The careless but knowledgeable student will argue that both the sides of the balance would weigh the same since the floating bodies, being weightless, have no forces of their own to contribute to the mass of water and vessel already present.

However, why is a floating body weightless? Because the buoyant force of the displaced liquid matches its own weight. By Newton's 3rd law, this bouyant force is also the force the body exerts on the liquid. Hence if the block of wood weighs more, it would also "push" down on the water and the vessel more, increasing the reading on its side of the weighing scale.

Thank you for taking the time to answer my question. A floating body is no more weightless than a book resting on a table top. The upward normal force, normal or buoyant, cancels the downward attraction (weight) of the object in each case. A body will have weight as long as the Earth is there to attract it. I know this is what you meant to say, but you have to be careful about how you say what is in your head.

As for the answer to my question, @haruspex responded almost exactly the way I would have responded, so I will leave it at that.

 

[brotherbobby]

The level of the water is the same in both glasses.

I suppose this was the crucial statement in your problem. Was the level of water in the two vessels the same to begin with? Or was the level of water the same after the objects were put into them?

 

[etotheipi]

I suppose this was the crucial statement in your problem. Was the level of water in the two vessels the same to begin with? Or was the level of water the same after the objects were put into them?

The water level is the same whilst the two objects are floating, in equilibrium. The best way to think about is what @haruspex said, i.e. to replace the object with an equivalent amount of water that exactly fills the cavity. Then you just have two cups with the same water level on both pans.

If you would prefer to put maths to it, then suppose the first object (of mass ##m_1##) displaces a volume ##V_1 = m_1/\rho_w## of water, and the second (of mass ##m_2##) displaces a volume ##V_2 = m_2 / \rho_w## of water. We'll also call the volumes of water in the first and second cups ##V_1^*## and ##V_2^*## respectively. You're given that the water levels are the same, so $$V_1 + V_1^* = V_2 + V_2^* \implies m_1 + m_1^* = m_2 + m_2^*$$where we just multiplied through by ##\rho_l##; i.e. there is equal total mass on both sides.

 

[kuruman]

The level of the water is the same after objects are placed in the glasses and that's all we need to know. Consider this: Suppose both glasses were filled to the brim before the objects are placed in them. The scale will be balanced. Now place the objects in the glasses. Some water will be spilled (displaced). The weight of spilled water from each glass is equal to the weight of the object that displaced it, quoth Archimedes. So now we have two glasses, with water up to their brim that have the same weight because we took out as much weight of water as weight of object that we put in. To get the situation as shown in the picture, with the level below the brim, all we have to do is empty the contents of each glass into larger identical glasses.

The crucial words are "identical glasses" and "same level".

 

[brotherbobby]

Thank you for your replies - in particular https://www.physicsforums.com/members/etotheipi.664237/, kuruman and haruxpex. I must admit that my point is not yet settled for me but I want to get there with your help, if you don't mind. The domain is floating bodies and upthrust, as applications of Archimedes' principle. The main thing to remember is that a floating body is weightless - the weight of the body is equal to the weight of liquid its submerged part displaces.

I want to address Kuruman's problem in post #6 above but want to get there in a few stages. I begin with the following problem.

Shown above is a small wooden box (mass ##m_1##= 200 g = 0.2 kg) and a large hollow marble drum of mass ##m_2## = 800 g = 0.8 kg. There are two vessels both containing the same amount of water to begin with sitting on weighing scales. Let the reading on each scale be 1 kg each.

The objects are put into the vessels as shown below.

What will be the scales read?The right scale would read more than the left scale.
The right scale would read ##w_2 = 1.8 ## kg weight while the left ##w_1 = 1.2## kg wt. Hence ##\boxed{w_1<w_2}##.

Is this correct?

 

[kuruman]

The numbers are correct, if you add 0.2 kg to the scale on the left, you get a total of 1.2 kg; if you add 0.8 kg to the scale on the right, you get a total of 1.8 kg. Your drawing is incorrect. If you start with equal levels in the two glasses, the level on the right must be higher, not lower, than the one on the left. That's because the fluid on the right has to support more weight therefore more, not less, water is displaced.

Now to understand my question, do the same but with the level in each glass filled to the brim so that when you put the objects in the glasses, you spill water that does not get weighed. How do the readings compare under these circumstances? I gave the explanation in post #12.

 

[haruspex]

floating body is weightless - the weight of the body is equal

Can't have it both ways; either it has no weight or its weight equals that of the water displaced.
In my view, it is incorrect to say that a floating body is weightless. Weight, by definition, is the force exerted on a mass by a gravitational field. This does not depend on any other forces acting.

 

[kuruman]

Can't have it both ways; either it has no weight or its weight equals that of the water displaced.
In my view, it is incorrect to say that a floating body is weightless. Weight, by definition, is the force exerted on a mass by a gravitational field. This does not depend on any other forces acting.

To this I add that if the floating objects are weightless, then adding them to glasses, as you describe in post #17, should have no effect and the scales would still read 1 kg. You yourself do not believe that floating objects are weightless, so please stop saying it.

 

[etotheipi]

Can't have it both ways; either it has no weight or its weight equals that of the water displaced.
In my view, it is incorrect to say that a floating body is weightless. Weight, by definition, is the force exerted on a mass by a gravitational field. This does not depend on any other forces acting.

I've always disliked that term also. A similar notion is that of weightlessness on the international space station; sure, in the rotating frame where the space station is at rest we can talk about a gravitational force, or weight, of ##-mg \hat{r}## and a centrifugal force of ##mr\omega^2 \hat{r}##. Some authors define an effective weight, like ##\tilde{g} = (-mg + mr\omega^2)\hat{r}##, and when this is zero say the person is weightless. Much clearer IMO to say that their effective weight is zero.

Same for buoyancy. Sure we can define an effective weight ##(B-mg)\hat{y}## which is zero in equilibrium, but this should not be confused with the actual weight ##-mg\hat{y}##, which does not vanish.