On Sun, 28 May 2017 11:07:10 +0100, dennis@home wrote:

On 28/05/2017 03:38, Johnny B Good wrote:


The only fly in the ointment with this last option is WTF didn't the
daft butcher slide all the hooks to the right hand end of the bar
beforehand? That way, he could have reduced the strain and effort on
his musculoskeletal system even further by arranging for hook V to be
nearer again, allowing him to slide the 'heaviest weight' to the far
end of the bar with even less strain and effort, leaving the remaining
hooks close to hand and available for more '(but slightly less) heavy
weights'.

He doesn't need to he has a hook in the meat already and would just hook
it on the rail nearest him.

Nicely spotted! :-)

If this was a question of observational skills and 'common sense' (as it
seems since it's the only way to make any sense of it), then I'm afraid
I've only got half marks (and the question setter zero marks for failing
to provide any means for the student to demonstrate the 'best answer').



I may be wrong in interpreting this question as one of 'ergonomics'
but
**** it all, that's the only way to make any sense of this one.

Moving onto the cups question which seems to be a question of which
of
the four cups encloses a presumed identical volume of liquid with the
least amount of surface area, I'm rather drawn to B despite answers C
and D looking like they could be equally as good a choice (the 'All
equal' option is rather spoilt by A being quite obviously the one
destined to cool the fastest).

All of them (cogs question)

N

Fall

V (looks closest to the optimal 45 degree angle ignoring air
resistance)

A

H

R

V

C

All equal (assuming we ignore friction effects as Galileo was able to)

N

Fall


truss answer missing here.

Oops! 'My Bad'. :-( I'm afraid I was so hung up on trying to work out
an answer to this one, I decided to 'deal with it later' and moved onto
the rest of the questions, forgetting to return to it before posting my
follow up.


If the truss has been dimensioned correctly they will all have the same
strain but they may well have different loads causing that strain.

I got as far as seeing this as a "vectors" calculation, depending on
(yet more) assumptions that the strains due to the mass of the bridge
components themselves would be insignificant enough compared to the "1
ton load" and largely balance themselves out of the equation for the
purpose of this question anyway as well assuming the structure is made up
entirely from right angled isosceles triangles.

With all those assumptions in place (all pigs prepped up and ready to
fly, so to speak), I can see that members V and X are in tension to the
tune of 0.707 tons with W and Y each carrying a 1 ton force in
compression.

It is impossible to correctly answer this question when complying with
the instruction to select "The one and only correct option" from the list
supplied since I'd want to select the *two* correct options, V and X. If
I ignore my understanding of the examiner's definition of the word
'strain' to decide the most likely singularly correct option, I'd be
forced by such logic to select 'All equal' and hope I'd correctly 'second
guessed' the examiner's definition of a 'correct answer'.

OTOH, it may simply show my ignorance of the mechanics of bridge
construction and the definition of 'strain'. :-)


Rise and then fall

doesn't that depend on the taps being identical?

No, it depends on the taps *not* being identical; in this case the LHS
tap of tank X being a much larger bore, matching the fatter pipework
allowing a faster fill rate than the smaller tap and pipework linking to
the tank on the RHS of tank X will allow it to drain away.

Rise and then fall describes exactly what will happen to the water level
in tank X during the early part of the process. Eventually, the water
levels in all three tanks will level off. The level in the LHS tank will
only fall whilst that in the RHS tank will only rise. Tank X is the only
one of the three that will exhibit this 'interesting behaviour' in the
scenario depicted.

if the flow rate is slow the fall will be impossible to see even if you
know it is there.

The question is about what happens to the water level in tank X
regardless of whether or not it can be observed. The sketch shows three,
apparently transparent tanks, along with quite obviously different sized
'taps' (valves) and plumbing to save the student from thinking up ways to
impose difficulties in arriving at a correct answer. :-)



H

L

R

W

D (as the previous student indicated, assuming a sweeping bend rather
than a tight hairpin bend where the right answer could easily be "All
equal"). Again, yet another question where I can't decide whether I'm
facing a cunningly disguised question concerned with the dangers of
making unwarranted assumptions or just very shoddy question setting.

Its the inside one assuming they depict someone going around the same
bend as you have to lean more the faster you go around the bends which
is why motorcylists scrape their knees and then fall off.

No, the ambiguity lies in the fact that the amount of lean to balance
centripetal force depends not on the speed alone but that of the velocity
change (in this case a change of velocity due to a change in direction
rather than speed).

This sketch could be a snapshot of a group of riders negotiating a
hairpin bend on a wide road where the innermost rider, D, is in fact
moving at the slowest speed but requiring the most lean to balance the
higher change of velocity due to the much tighter turn being made on the
inside of the bend. Indeed, it's just as possible to have this set up so
that all riders are travelling at the same scalar speed and show the same
succession of increasing lean angles.

This yet another badly set question wherein the only way the examiner
could have saved himself from total and utter disgrace would be by
replacing the "All equal" option with "Totally impossible to discern from
the given sketch".



Move in a circle

N assuming disks with holes punched in them (in which case, WTF is
causing M to remain poised in its depicted position?)

The examiners hand.

That lacked a smiley imho. :-)


S

It depends how you define work.
S would have to push hardest but travel less distance.
In reality he wouldn't be able to shift the thing as you wouldn't have
four operating positions if you only needed one man to do the job.


X (assuming equal effort on the part of the 'pushers')

Wow! Yet another imponderable question (about skiddiest car). Yet
again, we are left to make several assumptions from the very poor
quality 'evidence of our eyes' but I'll give it a go.

I'm led to assume we are looking at a **** poor sketch of a snapshot
overhead view of a sharp bend or corner on a race track and further
obliged to assume a dry equally grippy road surface with no adverse
camber or rubber crum to penalise any of the cars which I'm further
obliged to assume all have equally grippy tyres and are all travelling
at the same speed in some sort of race event.

Having been forced to make all these assumptions just to drill down
to
what I *think* is the core of the problem, I can only conclude that car
C is most likely to skid due to its higher rate of change of velocity
needed to negotiate the bend on a tighter radius than the other three
cars which results in higher side forces being applied to the tyres
from the resultant centripetal force.

In real life, there are many reasons why answer C will be most
emphatically wrong but, what the hey, this is just a question on a
1950's mechanics exam paper. :-)


Looking at it C can't be turning yet or the rear wheels will hit the
curb.

B is going to have to turn sharpest or he will hit A.

I would say B because he is going to have to hit the brakes to avoid A.

H

One

All equal

The mechanism will jam (I'm only 99% sure but if I'm wrong then
opposite direction unevenly becomes the only viable alternative)

doesn't the sliding pivot stop it from jamming?

I think the sliding pivot is most likely the reason it *will* jam up imo
(varying ratio of the linking bar as a lever). Now that I've had a break
from pondering this question, it seems to me to be a question of can such
a linkage without the extreme and unusual wear on the centre pin bearing
of the linkage bar even work?

Consider this; shrink the slot in the linkage bar down to a round hole
for the pivot pin bearing and you'll see straight away that such a
linkage cannot allow movement (you land up with two triangles which
cannot be contorted without bending or altering at least one of the
connecting lines. You might think turning the bearing hole in the middle
of the linkage bar into an elongated slot will help but the problem there
is that the varying lever ratios will still result in a jammed machine.

I may not have been entirely sure of my initial answer last night but,
having taken another look at the problem in the bright light of day, I'm
now convinced that the mechanism *will* jam. :-)


I would hope that such shoddy exam question setting as exhibited by
JR
Morrisby's efforts would be rejected today. However, I believe (rightly
or wrongly) that such shoddiness in examination question standards
still abounds to this day.


Well done.
have a gold star.

Thank you very much! You're so kind. :-)

--
Johnny B Good