Actually, Steve, the configuration shown is not the only 'minimum' energy solution in terms of the forces acting. Inverting the arrangement to form an arch has the same forces and dispositions.

No, you are missing my point. The approach I recommended is designed to minimize the potential energy of the system. If you invert the arrangement to create an arch, it has higher potential energy. Likewise, any sawtooth arrangement will have higher potential energy. (by potential energy, I mean gravitation potential energy U=mgh)

There is also the issue of stability. An inverted arch like you describe would not be mechanically stable. It would collapse with even the slightest distrurbance since there is no friction on the hinges.

This issue of other possible stable solutions (if they even exist, which would need to be proved first) is not really relevant to the problem because the question makes it clear that the lowest energy state is of interest. The statement that "it is like a suspension bridge" and the diagram make the structure clear.

EDIT: Hopefully my point is clear. A mechanically stable solution will need to have a local minimum (a saddle point or maximum point will be unstable to a disturbance) in the potential energy function. If there is more than one mechanically stable solution possible in this problem, then the lowest energy state is the only correct answer based on the constraints in the problem.

 

studiot is bringing up some good information which leads me to realize that there is one step in our solution that can be criticized. That is the assumption of symmetry in the final stable configuration. I believe that the solution we provided is correct, but it is not a good idea to assume symmetry in this type of problem. In a symmetric situation like this, it is possible to have two non-symmetric solutions. Each solution would be a mirror image of the other. This will happen if a non-symmetric configuration has lower potential energy than the lowest energy symmetric configuration.

I don't believe this is the case here, but if it had been the case, we would have missed the answer by assuming symmetry from the on-set.

I suggest redoing the problem by the minimum energy method, but do not assume symmetry in the final configuration. This means that all seven angles need to be considered. I don't have time right now, but hopefully someone will try it. I think in the end the answer will be verified.

 

No, you are missing my point. The approach I recommended is designed to minimize the potential energy of the system. If you invert the arrangement to create an arch, it has higher potential energy. Likewise, any sawtooth arrangement will have higher potential energy. (by potential energy, I mean gravitation potential energy U=mgh)

There is also the issue of stability. An inverted arch like you describe would not be mechanically stable. It would collapse with even the slightest distrurbance since there is no friction on the hinges.

I haven't missed any points.
a)
Following that line of reasoning it would be true to say that gravitational potential energy would be even more minimised of the seven bars ended up in a heap at the bottom of the gorge. But then if they were in a heap they would be 'unstable' and spontaneously spread out all to the same gound level.

Minimum energy in structural engineering is to do with work against the support reactions, not work against gravity alone. Have you proved that your configuration minimises the horizontal work?

b) Roman arches have been standing for thousands of years. All the forces are reversed so the bars would be in compression and you lack a collapse mechanism as for any bar to move downwards increases the compression in the bar by increasing the horizontal component. Try it and see. It was Hooke who first observed that the arch mirrors the suspended chain.

TNK has indicated he devised the question to study wider issues. I am am simply doing the same.

 

I haven't missed any points.
.

I think you have, but there is no point arguing over it.

 

Just to settle this quickly. I did a quick experiment. I made 7 pieces of plastic and used metal pins as the hinges. Using the span AB equal to 5.6 times the length of the segments allows the problem to be simulated easily. Anyone else can do this in about 5 minutes. I confirmed that the stated solution is correct since I measured 50 deg, 40, degrees and 22 degrees which is well within experimental error. Also, there is no other stable solution for this problem. Any arangement with a sawtooth or an inverted (arch) structure just collapses instantly. It's not even marginally stable requiring a push. It just collapses under its own weight.

Hence, single-mode solution!

 

The extra equation you need is the length constraint:

As I said in my earlier post, I solved it numerically (before you posted your energy solution), these three equations gave much better convergence than just using the length constraint and the energy one. I'm not knocking the energy approach though - I like the simplicity.

Yes, very good. This works well too.

 

Just to settle this quickly. I did a quick experiment. I made 7 pieces of plastic and used metal pins as the hinges. Using the span AB equal to 5.6 times the length of the segments allows the problem to be simulated easily. Anyone else can do this in about 5 minutes. I confirmed that the stated solution is correct since I measured 50 deg, 40, degrees and 22 degrees which is well within experimental error. Also, there is no other stable solution for this problem. Any arangement with a sawtooth or an inverted (arch) structure just collapses instantly. It's not even marginally stable requiring a push. It just collapses under its own weight.

Hence, single-mode solution!

I think studiot is right. The hanging position is stable because it is a minimum of energy. The force trying to move it to a neighbouring position depends on the first derivative of the energy, but as this is a stationary point this vanishes, hence it is stable. As it is a local minimum, forces conspire to move neighbouring configurations to this minimum.

The exact same analysis can be applied to the inverted position, this represents a maximum of energy, also a stationary point, so no forces trying to move it. As it is a maximum, forces conspire to move neighbouring configurations away from the maximum, and it collapses. So unless you managed to get your configuration exactly balanced, it will collapse.

You have to be careful about quick experiments. You could do another one to see which way the bathwater goes down in the various hemispheres.

 

I think studiot is right. .

No he is not right. He didn't say it was a maximum energy solution, he said it was a minimum energy solution. Only local minimum solutions are stable. At best the inverted solution is marginally stable, like trying to balance a pencil vertically on its point. Actually it's even harder than that because there are more degrees of freedom.

I agree the quick experiment does not prove or disprove the presence of unstable stationary points (maxima and saddle points). However, why would we care about unstable solutions or even stable sawtooth solutions in this problem? The given problem is a nice little problem and is very clear about the structure of the solution. It's not inverted and it's not a sawtooth.

Actually, Steve, the configuration shown is not the only 'minimum' energy solution in terms of the forces acting. Inverting the arrangement to form an arch has the same forces and dispositions.

Consider this statement he made. He said that the solution you gave is not the only minimum energy solution. This is false. Perhaps there is one more maximum energy solution - that seems reasonable. However, no one has shown there are any more stationary points (other than these two) and my quick experiment proves there are no other stable stationary points.

There is also the issue of stability. An inverted arch like you describe would not be mechanically stable. It would collapse with even the slightest distrurbance since there is no friction on the hinges.

Consider this statement I made previously. I already acknowledged that the arch is marginally stable by saying that you need a slight disturbance to make it collapse.

 

You have to be careful about quick experiments. You could do another one to see which way the bathwater goes down in the various hemispheres.

By the way, the reason why I did this simple experiment was to show that all of this stuff studiot is talking about is irrelevant to solving the problem. If he's trying to educate people about other ideas, that's fine. However, he seems to be implying that the solution we've shown is somehow not right or not the one we will see in reality, or maybe that we could see another solution. Well, the experiment shows his ideas are not relevant here. The structure settles to the shape we described. It doesn't matter if it's a "mechanism" that can be contorted into different shapes. All these other shapes have higher potential energy and are unstable and require additional forces to create. The solution that is relevant is the natural shape that is created by gravity when there are no external forces trying to change the shape. Actually, the problem asks something even more specific. "What is the ratio of the droop to the span length." There is only one answer to this question and we've provided it, along with two methods to find it.

 

Just to settle this quickly. I did a quick experiment. I made 7 pieces of plastic and used metal pins as the hinges. Using the span AB equal to 5.6 times the length of the segments allows the problem to be simulated easily. Anyone else can do this in about 5 minutes. I confirmed that the stated solution is correct since I measured 50 deg, 40, degrees and 22 degrees which is well within experimental error. Also, there is no other stable solution for this problem. Any arangement with a sawtooth or an inverted (arch) structure just collapses instantly. It's not even marginally stable requiring a push. It just collapses under its own weight.

Hence, single-mode solution!

Just to clarify my comments - I agree that the unqualified remark "Studiot is right" is remarkably broad and I withdraw it!

What I meant to say was that studiot was right about arches, and that you were wrong when you said

Also, there is no other stable solution for this problem. Any arangement with a sawtooth or an inverted (arch) structure just collapses instantly. It's not even marginally stable requiring a push. It just collapses under its own weight.

 

By the way, the reason why I did this simple experiment was to show that all of this stuff studiot is talking about is irrelevant to solving the problem. If he's trying to educate people about other ideas, that's fine. However, he seems to be implying that the solution we've shown is somehow not right or not the one we will see in reality, or maybe that we could see another solution. Well, the experiment shows his ideas are not relevant here. The structure settles to the shape we described. It doesn't matter if it's a "mechanism" that can be contorted into different shapes. All these other shapes have higher potential energy and are unstable and require additional forces to create. The solution that is relevant is the natural shape that is created by gravity when there are no external forces trying to change the shape. Actually, the problem asks something even more specific. "What is the ratio of the droop to the span length." There is only one answer to this question and we've provided it, along with two methods to find it.

I like experiments, and had I the right 'bits' on hand and the time I may have done one. I agree that studiot seems to be reading a bit more into the problem than is needed - it is clear to me that the links simply hang in a symmetric way. It's probably not too hard to show that the minimum energy configuration is symmetric.

I was just commenting that you went a bit far when you used your experiment to conclude that there were no other stable states. This is just like experimentally confirming that a pencil won't stand on its point - when theroetically we know it will.

 

Just to clarify my comments - I agree that the unqualified remark "Studiot is right" is remarkably broad and I withdraw it!

What I meant to say was that studiot was right about arches, and that you were wrong when you said

"Also, there is no other stable solution for this problem. Any arangement with a sawtooth or an inverted (arch) structure just collapses instantly. It's not even marginally stable requiring a push. It just collapses under its own weight."

Yes, I agree that was poorly worded. What I mean by that is "experimentally speaking" is is not possible to balance the arch for even a moment. Sometimes we can balance a ball a the top of a hill, or balance a stick on end for a few seconds. However, the mechanism I made had so little friction and so many degrees of freedom, that it had no hope of balancing in that marginal state for even a millisecond. I'd have more luck trying to balance a sharpened pencil vertically on top of another sharpened pencil.

 

I expect Steve's experiment showed what many builders of failed arches down the centuries have discovered.

You need to make the arch ring heavy enough to contain the curving thrust line within the material of the ring. Then the arch is an incredibly strong structure, which will not collapse even with one hinge. If two or more hinges form and move the thrust line outside the ring, the arch will collapse.

I think the plastic members were not substantial enough.

Since everyone has been going with the inclination of the bars to the horizontal I have been reworking my solution to conform before posting. I do agree with Tesla's numbers however although my equations are in terms of tangent rather than cotangents.

If you are considering energy methods I repeat you also have to consider the horizontal reaction. Don't forget the horizontal tension in the frame as described can be artificially set to any value without altering the geometry. This has often not been appreciated properly by designers/builders of failed suspension bridges.

The fact that the (horizontal) tension in slung (catenary) wires can be independently set is of importance to electrical engineers.

 

If you are considering energy methods I repeat you also have to consider the horizontal reaction. Don't forget the horizontal tension in the frame as described can be artificially set to any value without altering the geometry. This has often not been appreciated properly by designers/builders of failed suspension bridges.

I'm not designing a bridge. I was just solving the given problem. The energy method works for solving this problem without considering horizontal reaction. I get the same answer you did.

You aren't even being clear about what you mean by horizontal tension in the frame anyway. As far as I can tell, you can't arbitrarily set the horizontal forces and tensions. They will set themselves to meet the minimum energy state once the mechanism is freely hanging in equilibrium.

Unless you clearly define what you are saying, or show an analysis with equations and answers, I'm not going to try to figure out what cryptic lesson you are trying to teach. You have a way of creating a great deal of confusion out of a very simple situation. Please be more clear.

 

If you are considering energy methods I repeat you also have to consider the horizontal reaction. Don't forget the horizontal tension in the frame as described can be artificially set to any value without altering the geometry. This has often not been appreciated properly by designers/builders of failed suspension bridges.

I assume that the horizontal reaction is the tension in the central bar (and the horizontal force at every hinge including the end ones). I can't see how this can be set independently without changing the lengths of the bars or the separation AB. Once given the angles it is determined.

As far as I can see, in a catenary this force is set by the length of the cable. To increase it, and reduce the droop, you need to shorten the cable (altering the geometry), which is what you do in power transmission lines.

 

I assume that the horizontal reaction is the tension in the central bar (and the horizontal force at every hinge including the end ones). I can't see how this can be set independently without changing the lengths of the bars or the separation AB. Once given the angles it is determined.

You alter the tension by fixing one end of the cable to an anchor block and jacking the other end. In compression structures this is called prestressing.

Of course most thrust lines are not completely horizontal or vertical so there will be components in both directions. You are able to pull a cable up like this, but if you applied an arbitrary horizontal force to the seven link frame the geometry would not alter.

The analysis shows that the horizontal force applied at the supports is merely transferred from bar to bar through the frame to the other end. All bars carry the same horizontal force, unlike the vertical force which varies from bar to bar.

 

You alter the tension by fixing one end of the cable to an anchor block and jacking the other end. In compression structures this is called prestressing.

Of course most thrust lines are not completely horizontal or vertical so there will be components in both directions. You are able to pull a cable up like this, but if you applied an arbitrary horizontal force to the seven link frame the geometry would not alter.

What you are talking about has nothing to do with the problem under consideration. It's just nonsense to talk about this in this context.

I acutally built a mechanism that models this problem. Are you suggesting that my model is sitting there with an indetermined horizontal force? No, there is one solution here!

If you want to prove your point, post an analysis which shows the same geometry and the same proven solution for the angles, but with at least two different horizontal forces applied against the end supports.

Claims are meaningless without an analysis to prove it. I provided an explaination of a detailed mathematical approach with the final equations, as did Tesla23. You should do the same.