Brain Teasers at Xmas

The partial quote above comes from one of my earlier posts. The bit about "lateral thinking" is derived from the difference in volume between a shpere and the cylinder that just encloses it (divide by two because we have a hemisphere). This gives just under 131kgf. looks like we all thought along similar lines.

My earlier reply to winky, was based on the same thinking, but I simplified it by putting Pi = 3 and calling the answer "about 125 kgf"

Mind is now unblocked. Many thanks.

Cheers

Don

This one (No 12) and also No 11 remain untackled, so I thought a bit of encouragement might help.

The answer to No 12 "Balls in a box" is 288.

And Descartes had something to say about No 11.

Now we just need to know "how"

Cheers

Don

Quote using the icon on the bottom right. Seems to work for me. 

Interesting comment. I think there are several approximations/assumptions we are making in the way we have been tackling this which include:

1) the hemisphere  is massless, and
2) the Hemisphere & surface are perfectly smooth.

WIth 2) the assumption is that so long as the hemisphere  sits on the  surface with no net force on it the water will no leak out. Of course with real materials that may no be sufficient. But as you say in this case one cannot work out an answer so the problem is not interesting anymore.

Your thought experiment of putting a block under the shell is a good one & raises  a question independent of the validity of the assumptions above and deserves some consideration. 

 What you need  to calculate is simply the weight of water needed to fill up the imaginary cylinder encasing  the hemisphere. The use of the 'difference' is simply a convenient way of calculating that, but you could do it simply enough with calculus. 

I'd contend that the force required is independent of the block you insert, but I can't right now find a way of demonstrating that clearly  so I'll give it some thought when I'm fresher. 

Ian

A couple of points.

Its bloody dificult to write a brain teaser that is simple to read (and understand) and at the same time (pedantically) accurate. Secondly, most brain teasers (and A level maths or physics) make unrealistic assumptions such as "a perfect sphere", "ignoring secondary stresses", "assume no loss of heat", "ignore wind resistance". Adam's, (and most of the others in this thread, require such assumptions, even if they are not explicitly stated.

Earlier on I suggested the pressure required was "just" enough to ensure a watertight seal. Because it is material-dependent it isn't possible to even estimate that. I therefore made Ian's assumption that the "massless" bowl and the horizontal surface made "perfect" contact and hence a "perfect" seal without any downward load.

The pressure of water inside the bowl (there isn't any on the outside btw) is only dependent on the depth of the water at each level. So, at 100mm depth the pressure will be 100kgf/sqm. At 250mm depth it will be 250kgf/sqm and at 500mm depth (ie the flat surface) it will be 500kgf/sqm etc. These pressures will be independent of whether there is a block inside the bowl or not - providing the block isn't too close to the inside of the bowl to cause surface tension effects etc.

The pressure at any depth acts equally in all directions. However, because of the boundary condition inposed by the bowl, there will be a net pressure at this boundary which acts nornal (radially) to the surface of the bowl. It is the vertical component of this net pressure that causes the bowl to rise. To prevent this rise, a reaction is required in the form of a downward force.Assuming the bowl is strong enough, this downward force can be applied at the top, around the hole at the pole. This is the force that Adam wanted to find.

By "slicing" the bowl into 100 x 5mm thick horizontal slices, I calculated the vertical component of pressure on each slice and added them up. This is a crude form of integration. I used an Excel spreadsheet to do the arithmetic. The answer came to approx 130.9 kgf.

I could have done an integration - but didn't. I realised, like Ian and Adam, that by surrounding the bowl with a cylindrical column of water that "just" covered the bowl, this external mass of water would "just" provide the necessary downforce. This mass is half the mass enclosed by the hemisphere.

Hope this helps

Don

Just to make sure my grey matter was still functioning I did the integrations of the vertical component of the force as Don outlined and got exactly the same answer F=pi *rho*g *r^3/ 3. 

This approach ONLY involves the surface normal force so the presence or absence of a submerged block makes no difference.

cheers

Ian

I agree. I think Adam was really asking "what force IN ADDITION to that needed to create a seal". At least, that was the basis of my solution.

Of course, if its a rough glass bowl sitting on a rough wooden plane.......well, its just going to leak, isn't it !

Cheers

Don

Don

I think you've misunderstood my comment regarding the block. I know the block will have no effect on the pressure exerted by the water on the internal surface of the hemisphere. I was pointing out the ratio between the mass of water inside and outside the hemishere was irrelivant. Equalibrium would be maintained even though the ratio of mass changed.

Instead of surrounding the hemisphere with a cylinder, it could be surrounded by another hemisphere, say 0.53m radius. Filling the void between the hemispheres with water would equalise the pressure using far less water.

 

That is true but doesn't help you calculate the answer.

 Ian

fatcat, you are absolutely right.

An inverted bowl 530mm radius, placed over Adam's 500 mm bowl with water filling the bowl, would suffice to prevent Adam's bowl from lifting and the weight of the water filling this void would be 50kg.

Indeed, a bowl of 520mm radius would do the trick with only 32.7 kg of water.

If we were to ignore the effects of surface tension, capillary action etc etc, an inverted bowl of 501mm would suffice with only 0.5 kg of water.

However................your bowl (530mm) would need a downward force to hold it in place. This force would need to be .......155.9 kgf. This is the load to hold Adam's bowl in place (130.9 kgf) plus the load to hold the void-water in place (25 kgf).

cheers

Don

I came across a couple of neat little teasers yesterday. Not intended to be topical, but given events in Libya....well, who knows. I'll set them out in separate posts.

The local malitia of 600 men was assembled in the city hall for the final pre-battle breifing. 5% had one weapon each. Of the remainder, exactly half had two weopns each, whilst the others had none.

How many weapons did they have altogether?

Cheers

Don

On their return from battle, 70% had lost an eye, 75% had lost an ear, 80% an arm and 85% a leg.

What was the minimum % of militia that could have lost one of each ie an eye, an ear, an arm and a leg.

Cheers

Don

Feels horribly like 600 as 2 x 1/2 = 1.

I recall an attempted sneaky question in a (national?) IQ or science test.

A glass containing an ice cube is filled to the brim with water. The ice cube melts - what happens?

I suspect the intention was to produce an image of the ice cube floating above the rim of the glass and encourage the common sense answer that the water would overflow as it melted.

I got rather too excited over the effect of cooling on the volume of water and wrestled with answering that the level would drop, slightly, below the rim. I even toyed with whether I should consider the final mix temperature and if this might require consideration of the anomalous expansion of water.

'Wrestled' as I suspected the correct answer sought was that the level remained the same.

Of course (also) how large a glass, how big an ice cube, initial temperature of water and ambient temperature? Could the volume of the glass be altered by sufficient cooling and so on ........?


Luckily I veered off into philosophy -

"The ice cube melts - what happens?"

A butterfly flaps its wings in a hurricane.

Wake, butterfly -
It's late, we've miles
To go together.

Feels right. But its surprising how many of us lack confidence that we have got the right answer, even when the problem is relatively simple.

Cheers

Don

10%

Well done that man !

Cheers

Don

I'm not afraid of answering a simple queation. It's the only chance I get.

Dear Fatcat,

I would not have guessed that, or have been sure how to go about working it out, but would you mind showing your working. I always think the method is more interesting than the answer, and strangely it is what gets many of the marks in Maths Exams.

All the best from George

George

I'm not very good at explaining things but here goes.

30% didn't lose an eye
75% lost an ear
Therefore at least 45% lost an eye and ear. (75 - 30).

55% didn't lose an eye and ear (100 - 45)
80% lost an arm
Therefore at least 25% lost an eye, ear and arm. (80 - 55).

75% didn't lose an eye, ear and arm (100 - 25)
85% lost a leg
Therefore at least 10% lost an eye and ear and arm and leg. (85 - 75).


Also, I've noticed if you add the percentages that didn't lose anything, the total is 90%. 100% less 90% is 10%. I doubt this is a coincedence

Dear FC,

You make it look so easy!

Thanks from George

George,

There are still two teasers from Xmas outstanding.............somebody could make the solution to them look easy !

Cheers

Don

Dear Don,

I doubt I could make them look easy! But I do think in three dimensions and even if you have not followed it my new speakers - to be made in June/July - will be fantastic, and yet be made entirely senza a blue print as everything is precise but in my head - every detail.

That will be for a future thread if Naim will allow me to present the making of speakers in pictures and words on their Forum! After all the nSat will have some seriously heavy-weight competition from my efforts! Haha!

In the meantime you might waste ten minutes and look at my two newest threads in the Hifi place: "Full Frequency Replay and Flat Response," and "Single Driver Speaker?"

ATB from George

A Nait 5i would secure the deal! I love that simple little amp.

No complications or vices. Just does what it says on the box. And 50 Watts RMS is already 20 Watts over-engineered for my speakers.

ATB from George