Problem related to Archimedes' Principle and Buoyancy

[Bl4nk]

Homework Statement

"A wooden raft, with dimensions 6.10m x 3.05m x 0.305m, mass 2837.5 kg and density 500 kg/m3, is used to cross a river of density 1000kg/m3. It is used to carry 40 people, each of average mass 70 kg. Will the raft be able to carry all the people across the river? Explain mathematically."

Relevant Equations

Density, rho = mass(m)/volume(v)
F(buoyant) = V(submerged) x rho(fluid) x g

 

[haruspex]

Your method is unusual, and goes a bit wrong at the end.
You calculate a "density" as (total mass to be supported)/(volume of raft), though there isn't anything here which has that density. Then you compute the volume that needs to be submerged in order to derive enough buoyancy and find, mirabile dictu, that it exactly matches the total load.
If you think about that you should realise it was always going to match the load because of the way you calculated it, so you have not actually proved anything. The important part is that this "density" you calculated is less than that of the water. That's what proves it will serve.

A more straightforward way is to calculate the total load and the total weight of water displaced when the raft is fully submerged. Since the latter is the greater, it will float.

More simply, you could note that the density of the raft is half that of the water, so the extra buoyancy it can produce, beyond what is needed to support its own weight, is equal to its own weight. Then you just check 2837.5 kg>40x70 kg.
g doesn’t need to enter into it. The result would be the same on any planet that permits liquid water.

 

[Bl4nk]

Your method is unusual, and goes a bit wrong at the end.
You calculate a "density" as (total mass to be supported)/(volume of raft), though there isn't anything here which has that density. Then you compute the volume that needs to be submerged in order to derive enough buoyancy and find, mirabile dictu, that it exactly matches the total load.
If you think about that you should realise it was always going to match the load because of the way you calculated it, so you have not actually proved anything. The important part is that this "density" you calculated is less than that of the water. That's what proves it will serve.

A more straightforward way is to calculate the total load and the total weight of water displaced when the raft is fully submerged. Since the latter is the greater, it will float.

More simply, you could note that the density of the raft is half that of the water, so the extra buoyancy it can produce, beyond what is needed to support its own weight, is equal to its own weight. Then you just check 2837.5 kg>40x70 kg.
g doesn’t need to enter into it. The result would be the same on any planet that permits liquid water.

Oh so I have to calculate buoyant force with total volume, and show that its greater than the total weight?

 

[haruspex]

Oh so I have to calculate buoyant force with total volume, and show that its greater than the total weight?

That's the obvious way. Your way was ok except for two things:
1. it's hard to define what that density you calculated is
2. you should have stopped when you found it was less than the density of water.

 

[Bl4nk]

That's the obvious way. Your way was ok except for two things:
1. it's hard to define what that density you calculated is
2. you should have stopped when you found it was less than the density of water.

Ok tysm

 

[kuruman]

That's the obvious way. Your way was ok except for two things:
1. it's hard to define what that density you calculated is
2. you should have stopped when you found it was less than the density of water.

I agree with 2 but not with 1. For a floating composite body of non-uniform density, one can define the average density as the total mass divided by the total external volume. Then the composite object will float or sink depending on whether its average density is less or greater than the density of the fluid.

 

[haruspex]

For a floating composite body of non-uniform density, one can define the average density as the total mass divided by the total external volume.

Yes, but it is not the total external volume; it does not include the people. I didn't say it was impossible, just not straightforward. The method can be made valid with a suitably clear definition.

 

[kuruman]

Yes, I should have specified immersed external volume.

 

[pbuk]

Unfortunately the correct answer to this question is that whilst it is possible to calculate a state where the raft with its load of 40 people is floating, the raft will not be able to carry all the people across the river because it has insufficient freeboard/reserve buoyancy to maintain a state of stable equilibrium when perturbed by the dynamic changes to the system involved in this process. This is probably not the answer that whoever set the question is looking for.

Explanation: the centre of mass of a loaded raft (of uniform density) is above its centre of buoyancy, which appears to imply that any equilibrium will be unstable (like a pencil balancing on its point). However providing no part of the raft becomes completely submerged, as its angle changes from the horizontal the centre of buoyancy will move below the centre of mass to maintain the equilibrium (like a pencil balancing on its blunt end). Once any part of the raft becomes completely submerged, the centre of buoyancy can no longer move in the direction of the centre of mass and the raft will capsize.

I find misinformation about loading capacity of small craft upsetting at the moment, as do others particularly in Northern European, Mediterranean and North African countries. Again in the last 36 hours at least four people tragically lost their lives in the English Channel due to the actions of criminals who profiteer from exploiting their victims' desparation to escape suffering.

https://www.bbc.co.uk/news/uk-63982143

May their souls rest in peace.

 

[haruspex]

Once any part of the raft becomes completely submerged, the centre of buoyancy can no longer move in the direction of the centre of mass and the raft will capsize.

Not sure that is sufficient for it to capsize. Yes it will tip a bit, but assuming not quite the whole volume of the raft is submerged, that means one side will rise slightly above the water. That will shift the centre of buoyancy towards the low side.

 

[pbuk]

That will shift the centre of buoyancy towards the low side.

Do the math.

 

[haruspex]

Do the math.

If one side rises a bit out of the water it is providing less buoyancy, so the centre of buoyancy will shift away from that side.

 

[pbuk]

If one side rises a bit out of the water it is providing less buoyancy.

So the other side needs to provide more buoyancy. How is it going to do that if it is already submerged?

If you don't want to do the math draw a FBD.

 

[haruspex]

So the other side needs to provide more buoyancy. How is it going to do that if it is already submerged?

I presume we are not discussing the case where the raft is exactly completely submerged, so there is at least some small margin.
The lower side only has to provide more buoyancy than the side that has risen slightly. That would happen as soon as a tilt begins. But even if the lower edge does go slightly underwater, the total submerged volume would still be enough in total. The mass centre would just be a little lower in the water.

In my experience, a wide shallow object, even when barely floating, is fairly stable.

 

[pbuk]

I presume we are not discussing the case where the raft is exactly completely submerged, so there is at least some small margin.

Correct.

The lower side only has to provide more buoyancy than the side that has risen slightly.

No, it also has to move the centre of buoyancy so that it stays underneath the centre of mass. For the system in the OP the COB is ~15cm below the surface, the COM is ~50cm above the surface so think about what happens when it tilts. Or draw a FBD.

The mass centre would just be a little lower in the water.

I don't think so, there is a problem with potential energy here.

In my experience, a wide shallow object, even when barely floating, is fairly stable.

Yes a wide shallow object of uniform density is stable, I made that point. But the loaded raft in the OP has a mass distribution that is very non-uniform. I am repeating myself, but you need to draw a FBD.

 

[haruspex]

the COM is ~50cm above the surface

Ah, that is the assumption you are making.
What if the people are lying down? Wouldn't it be about on the water surface?

 

[haruspex]

I believe the condition for a rectangular block height z width x, relative density s to float level is $\frac{x^2}{6z^2}>s(1-s)$.
If we model the loaded raft as having width 3.05m (the narrower horizontal) with s=.5 then z can be up to nearly 2.5m. So it should be ok even if they are all standing.

 

[pbuk]

If the people stay lying down then the system is stable: as the raft tilts they become partially submerged and provide additional buoyancy, restoring equilibrium.

But the question did not say that they were lying down, and even if it did when 40 people are lying down on a raft and it tilts so some of them get wet, that is not what happens. What happens is that they panic, stand up and try to move to another part of the raft which becomes more unstable and capsizes.

 

[haruspex]

as the raft tilts they become partially submerged and provide additional buoyancy, restoring equilibrium.

I am not depending on that. And having done the math (post #17) neither am I depending on the passengers' attitude.
Of course I agree that the real world includes waves and winds, requiring a considerable safety margin. But in the academic context of the question, I do not see an issue here.

 

[berkeman]

If the people stay lying down then the system is stable: as the raft tilts they become partially submerged and provide additional buoyancy, restoring equilibrium.

Not necessarily. Many people are fairly neutrally buoyant at the surface (some folks, not so much ), so I don't think that will affect the buoyancy much. Especially in this schoolwork question -- It's mainly looking for a balance between buoyancy of the raft and the overall weight of the raft+passengers and not sinking below the height of the raft, IMO.

 

[pbuk]

Not necessarily. Many people are fairly neutrally buoyant at the surface (some folks, not so much ), so I don't think that will affect the buoyancy much.

No, a person that is being supported by the raft requires the raft to displace his weight in water. A person that is, say, half in the water only requires half of their weight in displacement.

Especially in this schoolwork question -- It's mainly looking for a balance between buoyancy and the overall weight of the raft and not sinking below the height of the raft, IMO.

Which is precisely my problem with this type of question: let's do the math:
the mass of the raft plus load is 5,637.5 kg requiring a displacement of 5.6375 m³.
the surface area of the raft is 6.1 x 3.05 = 18.65 m² so the depth below the waterline is $\frac {5.6375}{18.605} = 0.303 \text{ m}$. With a thickness of 0.305 m the raft has 2 mm of freeboard - can it really carry 40 people across a river?

Would you set a problem saying "there are 40 people on the roof of a burning building each weighing 70 kg, can we rescue them all clinging to a rope dangling from a helicopter which is known to fail at a load of 2,850 kg"?

 

[pbuk]

I believe the condition for a rectangular block height z width x, relative density s to float level is $\frac{x^2}{6z^2}>s(1-s)$.
If we model the loaded raft as having width 3.05m (the narrower horizontal) with s=.5 then z can be up to nearly 2.5m. So it should be ok even if they are all standing.

No that doesn't work - that condition assumes a uniform block.

 

[berkeman]

Thread closed for Moderation. Hopefully the OP has what they need to solve their schoolwork problem, because this Moderation may take a while...