Unstable Equilibrium: Finding Critical Density of Low Density Beam in Water

[sazerterus]

Homework Statement

If a beam with square cross-section and very low density is placed in water, it will turn one pair of its long opposite faces horizontal. This orientation,however, becomes unstable as we increase its density. Find the critical density when
this transition occurs. The density of water is ρ v = 1000 kg/m3 .

Homework Equations

Lacking any idea,do not know how to start.

The Attempt at a Solution

I did not understand the question.Could someone clarify it and tell me the relevant equations.

 

[mfb]

If you do not know how to start, draw a sketch.

You should know equations for swimming objects.

 

[sazerterus]

If you do not know how to start, draw a sketch.

You should know equations for swimming objects.

I am sorry.This wasn't quite helpful.Could you elaborate?

 

[haruspex]

You understand the physical arrangement in the first sentence, yes?
If it becomes unstable, in what way will the orientation be likely to change?
How would you go about determining whether that change will happen?

 

[D H]

Hint, in the form of a couple of questions:

• What distinguishes a stable equilibrium from an unstable one?
• What happens when the system is very close to but not at the equilibrium point?

 

[sazerterus]

Hint, in the form of a couple of questions:

• What distinguishes a stable equilibrium from an unstable one?
• What happens when the system is very close to but not at the equilibrium point?

Well,if the second differential of the potential is negative,then the system will be unstable.But,how to find the equilibrium point.also,how to find the potential energy.

 

[D H]

Think a bit more primitively then: Forces and torques. You can either solve the problem directly with a force/torque approach, or you can use that to lead you toward the potential function.

 

[haruspex]

The first step is this question I asked before: If it becomes unstable, in what way will the orientation be likely to change? Once you have that, you can suppose it changes by small amount in that direction and analyse the forces that will then apply. (The differential you referred to is wrt that change.)

 

[sazerterus]

The first step is this question I asked before: If it becomes unstable, in what way will the orientation be likely to change? Once you have that, you can suppose it changes by small amount in that direction and analyse the forces that will then apply. (The differential you referred to is wrt that change.)

Okay,so then the initial orientation,as i understand,is at first that the beam is parallel to the surface of water.as its density is increased,slowly its centre of gravity will come down,once it is just below the surface of water,the orientation will become unstable.So the answer is 500kg/m^3.is it?could someone verify the answer.

 

[mfb]

once it is just below the surface of water,the orientation will become unstable

Is that a guess, or did you calculate it?

 

[D H]

as its density is increased,slowly its centre of gravity will come down

Correct.

once it is just below the surface of water,the orientation will become unstable.

Incorrect.

So the answer is 500kg/m^3.is it?could someone verify the answer.

That is not the correct answer. Since you just stated a result rather than showing your work I don't know where you went wrong.

 

[sazerterus]

Frankly speaking,it was a guess.Since I still cannot relate how the configuration will become unstable.Since it has square orientation,if the density continues to increase,it will just sink,won't it?
Let me try to tell the forcess.First, of course, a downward force due to weight.
Second,A downward pressure due to Atmosphere.
Third,An upward pressure due to the displaced water,the buoyant force.
Are there any other forces?

 

[D H]

Yes, it will sink a bit. That's not what the question is asking. It's asking about the stable orientation of the beam. If the density is very low the beam will float with two faces parallel to the surface of the water. One of those two faces will be entirely submerged, the other entirely above the surface. At some point as density increases this configuration ceases to be stable. Some other configuration becomes the stable one. There are in fact four different stable orientations, the aforementioned flat top orientation plus three others. Which one is the stable one depends on density. What's perhaps most surprising as that as density increases even further, the flat top orientation once again becomes the stable one. A beam that is just slightly less dense than water will float in that flat top orientation.

The question is asking about that first transition point, where the flat top orientation stops being stable. You don't even need to find the which orientation is stable. You just have find the transition point.

 

[sazerterus]

What about the forces I mentioned?Are there any other forces?

 

[D H]

You can safely ignore atmospheric pressure. It too is a buoyancy force, and it's very small compared to those other two for something with a density that is any sizable fraction of that of water.

 

[sazerterus]

Okay,from what i Understand,due to changes in the centre of buoyancy there will be an unbalanced moment,as a result the stable config will change.Is my assumption correct?

 

[D H]

Correct. So where are those forces applied? What distinguishes an unbalanced torque from a restoring torque?

 

[sazerterus]

Correct. So where are those forces applied? What distinguishes an unbalanced torque from a restoring torque?

Well,the forces are applied at the centre of gravity and centre of buoyancy.If its stable,for small changes it will come back to its original state.But if its unstable,Then due to the unbalanced torque it will not come back.But how to determine the new centre of buoyancy,After i have tilted it a little?And how many axes should i consider when taking calculations?

 

[D H]

Small angle approximation is always a good way to start. As far as which axes you need to be concerned with, there's only one, and it's the obvious one.

 

[sazerterus]

Small angle approximation is always a good way to start. As far as which axes you need to be concerned with, there's only one, and it's the obvious one.

Oh,yeah,the axes chosen as the centre of gravity as origin.Could u suggest me some books that would strengthen my concepts regarding the problem.

 

[sazerterus]

Small angle approximation is always a good way to start. As far as which axes you need to be concerned with, there's only one, and it's the obvious one.

Well,I finally have an idea how to solve it.At first,we tilkt the beam through a small angle θ,The origin being at the centre of gravity.How,two triangles will be formed,one inside the water,and the other outsside.By calculation,it can be shown that the volume of the triangle is (x/8)*(l^2)θ
the volume of the other triangle is also same.Now,we find the new centre of buoyancy By taking this two triangles and the previous centre of buoyancy into consideration.We find the x and y co-ordinate of the centroid of the two triangles,And by taking moments about the origin,we will get the new centre of buoyancy.There we will get two new variables,which are interrelated.The critical density,and the height of the beam below the water.Of course,they are related to each other,by the initial stability condition,with the density of water.Now,all that we have to prove is that the resulting moment is opposite to the direction of the stable condition.

Are my deductions correct?And is my area of the triangles correct?if You tell me these 2 things,I could advance.

 

[D H]

First off, I suggest you look at this as a two dimensional rather than three dimensional problem. The long axis of the beam will be horizontal given a sufficiently long beam. To reduce this to a 2D problem imagine a plane normal to that long axis that intersects the long axis at the center of the axis. This yields a square cross-section of the beam that by construction contains both the center of gravity and the center of buoyancy.

The issue at hand is whether an orientation with two edges horizontal (one below and one above the surface) and the other two edges vertical is stable. It's obviously at least metastable as there are no torques in this orientation. In any orientation, weight and the buoyant force must be equal in magnitude to have balanced forces. In the flat orientation, a rectangular section of the square will be submerged. The area of this submerged section is a function of density and the length of a side of a square. What is this submerged area, and where is the center of buoyancy in this flat orientation?

The answer to the problem lies in determining whether the torque that arises from rotating the beam a bit from this flat orientation tends to restore the beam to the flat orientation or rotate it even further from the flat orientation. Two of the corners of the square are submerged in the flat orientation. This will still be the case for sufficiently small rotation angles. (Note: at some angle the beam will shift from two corners submerged to either one or three corners submerged, one if the beam has a low density but three if the density is high. You can ignore these cases.)

What this means is that the submerged part of the square will form a right trapezoid (American English) / right trapezium (British English). The area of this right trapezoid must be the same as the area of that rectangle described above to maintain balanced forces. From this you should be able to compute the new location of the new center of buoyancy. It's just the geometric center of that right trapezoid.

Imagine a vertical line passing through this new center of buoyancy. This will intersect at some point the formerly vertical line that passed through both the center of gravity and the unperturbed center of buoyancy. This point has a name, the metacenter. You should be able to verify that this metacenter is fixed (independent of rotation angle) for small rotation angles. This is the point about which the beam will roll. Using this point as the center of rotation means that the only force that provides torque is gravitation. (By construction, the buoyant force passes directly through the metacenter, so buoyancy results in zero torque.)

Where does the metacenter need to be with respect to the center of mass for the flat orientation to be stable?

The last thing to do is to develop and solve a mathematical expression that distinguishes between stable and unstable equilibrium. Hint: You should get a quadratic equation in density.