Brain Teasers ? or 50 Years On........... ?

Just to tidy up ken's sloppy work............

If the load moves from left to right

T1 is the cable AB

T2 is the cable AC

The formulae (which are spot on as we would expect from ken) could beslightly re-arranged so that

T1 = m/2[g cscθ - a secθ]

T2 = m/2[g cscθ + a secθ]

This might make it a little bit more clear that when a = 0 (ie the system is at rest) T1 = T2

and that when the system is in motion from left to right, T2 will be greater than T1

To save you doing a "Google" search, or running for the old A-Level books..........

cscθ = 1/sinθ

secθ = 1/cosθ

ken's terminology makes it look elegant and for the past decade ken has had a phobia about elegant solutions

nice one ken

apologies for my sloppy work -- will (try to) do better next time 

enjoy...

ken

The O-Level results were out today, so a more appropriate O-Level question seems to be in order ?

VABCD is a pyramid with vertex V vertically above the centre H of the horizontal rectangular base ABCD. M is the mid-point of BC.

If AB = 800mm, AD = 600mm and VH = 900mm, what is

• The angle VA makes with the base
• The angle VM makes with the base

tan(angle(AH,AV))=VH/(sqrt(AB^2+AD^2)/2)=1.8

  -> angle(AH,AV)=60.9 degree

tan(angle(MH,MV))=VH/(AB/2)=2.25

  -> angle(MH,MV)=66 degree 

Staying with O-Level for the moment

Football, Cricket or Rugby ?

In a survey of 36 boys, the numbers playing Football, Cricket, Rugby and none of these is given in the Venn diagram.

If a boy is picked at random from this group what is the probability that he plays

1. Football

2. Cricket and Football but not Rugby

3. Only Rugby

4. If a boy who plays cricket is chosen at random, what is the probability that he also plays football ?

Neat ! Today would have been a good day !

Sports answers

1. 19/36

2. 1/36

3. 5/36

4. 3/8

Great sport there sj. All four answers are spot-on !!

Stonehenge (or the pyramids)

There is uncertainty about how the Ancient Brits and the Egyptians moved huge stones great distances.

Some theorists have it that the stones were moved using sledges, rollers and ropes. The sledges and rollers being made from tree trunks and the ropes from leather. As a rough estimate, 500 men might be needed to poll a 50 tonne stone with a further 100 men needed to place rollers in front (and recover rollers from behind).

As a gross simplification, take a stone that is a convenient 2m long that has to be pulled 1km. How many rollers have to placed in the path of the stone ? Assume we want to minimise the the number of rollers.

Assume also that the stone sits firmly on a sledge and there is no slippage between sledge and roller and no slippage between roller and ground. Also assume the ground, roller and sledge are smooth and the rollers are perfectly round.

Two Circles

Two circles radius L and 2l intersect as illustrated. What is the area of the shaded overlapping portion ?

It can be done with 2 rollers, placed under the stone 1001 times.

1 roller under the stone initially and 1 roller per meter travelled.

I think its less than 1 roller per metre. but of course you will need more than two rollers to stop the stone tilting off

Does the size of the roller matter ?

Probably worth a "physics" experiment with 4 round pencils and a fag packet (or similar)

Then try to "explain" the results.

You only need 2 pencils.

Balance cig packet on pencil No.1, place pencil No.2 at one end of the cig packet, push cig packet onto pencil No.2. When the centre of cig packet is above pencil No.2 cig packet will once again be balanced, move pencil No.1 to front of cig packet. Repeat.

Spot-on fatcat. Top marks.

Well done to Fatcat for his explanation of the initial problem, I have now adjusted the text (see above) to formulate the next phase of the problem.

Clearly, its a bit impractical for the old stonemasons to balance a 50 tonne stone on a single roller (even with a few "pit-props") whilst their mates move the second roller. But for the sake of consistency let's stick with two rollers (yes, I know, I'd also prefer a Bentley with a Naim.........) but two rollers it is !

Hmmm...........That reduces to 0.182(L^2)...............I think ?

Not the answer I get...............?

Just realised I'd got it wrong so deleted my first attempt before seeing your response. Apologies.

Tried again and this time got L^2(π/3 - 0.683) i.e. 0.364L^2

Probably just as wrong.

No more brain teasers for me.

Look on the bright side of life sj ! At least you got that bit right 

and more to the point, you had a crack at it. Good on you !

BTW my answer is ≈ 0.108L^2

I started with the drawing below.

The area required is the sum of sectors ACFDA and BCEDB less the area of the quadrilateral ACBDA

The use of the cosine rule also came in handy

Hi Don,

I tried to solve it in a completely different way:-

1. by finding the equations of the 2 circles,
2. then identifying the coordinates of the 2 points of intersection (C and D on your diagram),
3. then using integration to find the difference in areas under the 2 curves between these 2 points of intersection and the x axis.

Succeeded in 1 and 2, but got hopelessly lost trying to integrate a complicated equation! Your way looks much neater (even though I've forgotten the cosine rule).

Regards, Steve

Hi Steve,

Yes, I also thought about tackling this one as the area between two curves, but then decided to try the geometry/trigonometry route instead. It isn't elegant, but it is straight forward - even if you have to be careful.

The sine rule is a/SinA = b/SinB = c/SinC       (it also works the other way up)

The cosine rule is a² = b² + c² - 2bcCosA

The area of any triangle is A = ½abSinC

Hope these bring back happy memories of days gone by...............................

i too went the circle equation route -- but i wasnt planning on intergrating -- but to find the various angles that i could then use for areas of sectors and segments. but the equations didnt simply to something elegant -- all i got was the line CD was at 45° to the base of the square -- but i couldnt really make use of this. 

you new diagram exposes new insights Don, because as soon as you get angles α and β then the problem is solved :-) but the expressions are not 'nice'

enjoy

ken

There is a fairly simple formula to calculate the area of a segment which is cut off by a triangle i.e.

A = R^2/2 [θπ/180 - sinΘ] where R is the radius and Θ is the angle at the centre of the circle in degrees.

Using pythag to calculate the length of AB as 2√2L you can use the cos rule to find angles α and β which when doubled produce the central angles (I get 55.77 degrees for the smaller circle segment and 27.03 degrees for the larger).

Substituting in the above formula produces a figure of 0,0734L^2 for the smaller circle segment and 0.035 L^2 for the larger circle segment. Added together these come to 0.1084L^2 which is fairly close to the answer Don has.

yes, the key is angles alpha and beta:

i get, and this agrees: cos(alpha) = 5/4√2 and cos(beta)=11/8√2

...annoyingly, the symbols for alpha and beta turn into smileys when combined with parens 

these angles are consistent with correct area that you state above

enjoy

ken