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

Don Atkinson posted:

Nice one Mike. we'll have to start calling you "Mike the Mechanic !"..............

............on second thoughts, that might well be an insult, so we'll stick with a "well done sir" 

It took all of 2 seconds.   Belt pulley (& gear) ratios was all part of my daily toil in a previous life.   It gets more interesting with V belts & results that do not follow either outer or inner pulley groove diameter(s).    

Mike-B posted:

Don Atkinson posted:

Nice one Mike. we'll have to start calling you "Mike the Mechanic !"..............

............on second thoughts, that might well be an insult, so we'll stick with a "well done sir" 

It took all of 2 seconds.   Belt pulley (& gear) ratios was all part of my daily toil in a previous life.   It gets more interesting with V belts & results that do not follow either outer or inner pulley groove diameter(s).    

Ah Mike, remember, these are "A" Level Maths and Physics from 50 years ago and Brain Teasers.

Not your real-world problems .

Nobody else interested in the three symbols teaser? I feel a little bad for taking these teasers away from everybody else, but I do like them and you had 17 hours anyway . These are my results:

Helicopter=78

Happy Doctor=35

Ambulance=29

The missing value of the row is 226

Mulberry posted:

Nobody else interested in the three symbols teaser? I feel a little bad for taking these teasers away from everybody else, but I do like them and you had 17 hours anyway . These are my results:

Helicopter=78

Happy Doctor=35

Ambulance=29

The missing value of the row is 226

 

No need to apologise Mulberry, I'm glad you enjoy some of these problems.

Oh, yes. Spot-on with the answers as well. Nice !

I'm surprised nobody has tried the 2kg weight. I know it's easy, but not always that easy for  everybody to see. In this case it's not much more than a classic bit of Pythag and a bit of "g". Even the "g" isn't that important.

However, the follow-up that I have planned is a bit more of a moving beast..............

Three symbols

Each of the three different symbols has a different value associated with it. Adding up the value of the symbols in each row and column gives you the value for that row or column. What is the value of each symbol and what is the missing value of the column ?

Another one, especially for Mulberry......................well,.....................unless somebody can beat him to it

If the 2kg weight problem is easy, might be worth me having a guess.

Is it 750g

Green Cross = 54

Black Tick = 23

Red Cross = 11

Missing column = 111

Thanks SJ, I feel better now . And I do agree with your results.

fatcat posted:

If the 2kg weight problem is easy, might be worth me having a guess.

Is it 750g

A bit of resolution (or should that be resolving ?) to outline your "guess" for the benefit of others ?

But if it was a pure guess Frank, then well done, and don't worry, I'll post an outline later.

BTW, simply multiplying 0.75kg by "g" (ie 9.8m/s²) gives the answer in Newtons --> 7.35 N

sj & Mulberry, Hi !!

You don't need me to let you know you've cracked it again, but good work.

Enjoy the sunshine while it lasts..............................things are about to turn nasty, and i'm not referring to Bojo, Gove & Farage (Estate Agents & Financial Advisers)

On a happier note for all those “Remainers” (like me) who are depressed or disgusted with the way things seem to be going at present…………

Winter is approaching at a small village in Siberia. The soil in the top two metres above the perma-frost will soon refreeze so they need to dig enough graves in the village cemetery in anticipation of the  number of deaths ( I told you this was on a happier note than Brexit, didn’t I ?)

The village population is 1000 and it’s known that each person has a one percent chance of dying during the winter period.

What is the least number of graves they should dig so the probability of having enough is at least ninety percent ?

On reflection, the post above is a bit grim ! But that is the format in which I first encountered it !

I suppose I could have modified it something along the lines of.....

....Naim produce 1,000 NAP250 power amps each year. It is known that each one has a one percent chance of failing the final quality check. With a 90% confidence interval, how many "good" NAP250's should the marketing team assume will be available each year ?

Hope this brightens things up a bit ?

Don Atkinson posted:

Not my best diagram, I know. And it seemed to get left behind on the previous page.

Usual conditions. Frictionless pinned joints and a well greased sliding roller, again no friction !

g = 9.8 m/sec²

The system is at rest. what force F is required to keep it static. The connecting rods are "light" ie you can ignore their mass.

You could give the force in Kg, but just multiplying by 9.8 gives it in Newtons and makes it look so much more difficult to a non-engineer or non-scientist

Yes, I know this has been seen before, and fatcat got the answer with a "guess" ..........

This time, the sliding pin or roller is moving left to right at a steady speed of 2m/s. In other words it's  a machine, in which a piston pushes the sliding roller L --> R then pulls it back at a steady speed. The 2kg weight moves up and down.

When the moving machine is in the configuration shown, what is the force F ?

Don Atkinson posted:

On a happier note for all those “Remainers” (like me) who are depressed or disgusted with the way things seem to be going at present…………

Winter is approaching at a small village in Siberia. The soil in the top two metres above the perma-frost will soon refreeze so they need to dig enough graves in the village cemetery in anticipation of the  number of deaths ( I told you this was on a happier note than Brexit, didn’t I ?)

The village population is 1000 and it’s known that each person has a one percent chance of dying during the winter period.

What is the least number of graves they should dig so the probability of having enough is at least ninety percent ?

I'm informed that the answer is 14.

Hungryhalibut posted:

Don Atkinson posted:

On a happier note for all those “Remainers” (like me) who are depressed or disgusted with the way things seem to be going at present…………

Winter is approaching at a small village in Siberia. The soil in the top two metres above the perma-frost will soon refreeze so they need to dig enough graves in the village cemetery in anticipation of the  number of deaths ( I told you this was on a happier note than Brexit, didn’t I ?)

The village population is 1000 and it’s known that each person has a one percent chance of dying during the winter period.

What is the least number of graves they should dig so the probability of having enough is at least ninety percent ?

I'm informed that the answer is 14.

Hi HH.

You are well informed. Tell Henry he is well up to the mark.

I hope his A Level results get him into his University Course of choice.

Cheers, Don

The STATIC machine - solution. One step at a time.

This one is a reminder of Tan; Sin; Cos in a 3/4/5 Pythagoras triangle

Note that in the next slide the basic triangle will have rotated 90 deg anti-clockwise. But I have retained the two angles α and β

Well, you can't expect to have it all delivered on a plate ! (can you ?)

Just confirming the Pythagoras part of the geometry giving a 3/4/5 triangle - well two triangles in fact, given the symmetry of the system.

This is the simplified diagram showing the 3/4/5 triangles with α and β (rotated 90 deg anti-clockwise).

The loads in the connecting rods are noted as "R" and we have in effect, a vector diagram of the loads.

Because all joints are pinned, all loads are either axial tension or axial compression.

Symmetry means the loads in the two rods are equal when the system is at rest.

The loads R in the two Rods are compressive and illustrated vectorially by the black arrows.

Resolving vertically at A

2 (R.Cosα) = 2 kg

R = 1/Cosα

R = ⁵/₄ kg..............(1)

Resolving horizontally at B

F = R.Cosβ

F = R. ³/_{5...............Sub for R from (1) above}

F = ⁵/₄ x ³/₅

F = ¾ kg (or 750g as “Guessed” by fatcat)

F = ¾ x 9.8 N

F = 7.35 N

Hope the above explanations help.

Most engineers would simply "look" at the diagram and be able to "see" straight away what the loads in the rods would be - (and some of them would get it wrong !!)

I hope I haven't dropped any clangers in providing the above details, but I do hope you will point them out if I have.

Who's going to start the ball rolling, so to speak, with the "dynamic" version of the 2kg machine ?

(well, 2m/s is hardly dynamic I know )

I know I said thing were going to turn nasty. That was only meant in jest ! But I appreciate "many a rue word was said in jest!"

Perhaps if I re-write the question more succinctly, somebody will make a start ? Even if you just outline howothers  might tackle the problem ? The relevant diagram is above and on previous pages.

There are two pivoted rods, each 0.5m long, which carry a weight of 2kg. At the left is a roller that is driven back and forth at a steady speed of 2m/s by some sort of machinery.

What is the force required to do this when the height of the weight is 0.4m and the roller is moving from left to right ie the weight is going upwards.

The rods are light and the pivots and rollers have no friction.

As a guide, this is the sort of problem that was outlined at the start of First Degree Physics courses about 50 years ago..................(it must now be part of elementary GCSE....... )

Hopefully this one isn't too difficult and will "lubricate" a few 70 year old brains

(it bears no relationship to the 2kg moving mass one above !!)

I’ll start off with the easy bit.

The wires and pulleys have the usual conditions, so the wire from the 1kg load to Mass No1 has uniform tension throughout.

The wire from Mass No2 to Mass No 1 likewise has uniform tension M₂kg

The diagram below should help. (edit: Oh dear, the diagram is on the next page !!)

Next is to resolve forces horizontally at X to Find Mass No2

Remember, the system is at rest and is statically stable. So the horizontal forces at X must balance.

I’m pretty sure I remember doing this sort of thing in Applied Maths for A-Levels but i'll give it a day or so before setting out my version of events............