Static stability(Academic problem)

I suppose the person who wrote it was not an English major. Either that or he/she's as inclined to make typos as I am.

 

Prove you know a lot more English than him and know a lot more naval architecture than him. Here's your chance.

 

There is no excuse for a college level exam question to be so poorly written. The question is not particular to Naval Architecture, but basic statics. For it to be significant, there has to be a moment. That is, the weights' CG should be placed not in vertical alignment with the CG of the unloaded vessel. Otherwise, the vessel will increase its draft, but will remain level.

 

Strictly speaking, in an ideal world with no perturbations, what you say is true. But in a real world, the CoG of the loaded vessel has to be below the CoB in order to have a stable equilibrium. Otherwise, the equilibrium is unstable. When CoG=CoB the equilibrium is neutral. And that is true both for an upright and a listing vessel.

Both the neutral and the unstable equilibrium situations are unacceptable. So the objective of this exercise is to find the max. number of blue boxes (containers) which can be stacked up vertically on the deck while maintaining the stable equilibrium (CoG<CoB or KG<KB).

 

Or a scanner has made a mistake. Not uncommon.

I'm surprised this thread has generated posts beyond Ad Hoc's answer. It is correct, and the question (besides the typo, which I personally didn't even notice - I read it as "stacked") is quite clear and understandable - especially when you remember that the context in which it will be seen is that of an NA student who's participating in a static stability course.

 

The answer is six.

 

I see questions written like that in many engineering courses. They teach students to stop challenging absurd questions. Stacked and attacked are not the same. You may decide to use different parameters because what is written makes no sense. However, that is not good engineering. The response is that there is no proper answer and the person posing the question should correct and clarify it.

 

I thought it is 42

 

It's always amusing to see NA's trying to be mathematicians.
The Union say that it will 3 before lunch, and no more than 2 after that.
After the inspector leaves, the top container will be quite a bit lighter that the others, IYKWIM.

 

Ah, but Leo, two plus two is five, if you use sufficiently large values of two and small values of five.

Chris

 

2 + 2 = 5 if you use the ITTC57 line and a form factor

 

LOL

 

Don't feel bad. I was buffaloed too.

The central joke to the is whole thing is the density of the blue boxes.

They are six times as dense as the boat itself, and on sixth as wide.

Assuming the blue boxes are as long as the boat, adding them is going to increase the draft of the boat by 0.25 m, each.

Now the boat itself, having a density of one third of that of water, and being one meter deep, is going to give it a draft of one third of a meter, itself.

This being so, the answer cannot be more than two blue boxes.

This is because: 0.333 m + 0.25 m + 0.25 m = 0.83 m.

Add one more blue box and the boat will be under water!

This really is a stupid problem.

Maybe the intended answer is for the student to state that is unsolvable (no length given for either the barge ship or the blue boxes)

If third dimensions had been given, it would certainly be a useful problem.

I've used it myself.

How deep do I need to make my ballast keel, until it is guaranteed to right my capsized sailboat?

 

sharpii2, I do not know if everything you say is a joke or if you're serious. Prefer assume you're joking.
I would like to point out that a ship can be stable, even very stable, although its deck is below the water.
(Stable ship: one who refuses to leave its equilibrium position, whatever it is that position)

 

Good point. That is why submarines stay upright.