Only in about half the cases. I looked at every "Question" with a view
to answering each one and noticed that some had been marked incorrectly,
others correctly and some, seemingly not at all.
Starting with the boxes one (which is hardest to turn over?) my answers
are as follows:
M assuming diameters of tops of the LH ones = widths of the RH square
and equilateral triangular cross sectioned tanks
Move to and fro
It wasn't clear what this butcher's weight question was about. I can see
three possible 'correct' answers to this one, each depending on how you
interpret what the real question is.
The previous student's rubbed out answer (V) suggests it was interpreted
as a requirement to minimise bending stress on the hanging beam. Another
interpretation of the problem is to minimise pull out forces on the
beam's anchor rods, suggesting Y as the answer. However, mention of the
*heaviest* weight strongly suggests both those answers are wrong due to
the use of "Trickery" involving common sense and observational skills in
the real world.
We know that such butchers' meathook rails are amply over-engineered for
the butcher's normal every day usage so we can exclude 'structural
integrity' issues from our deliberations (plus, any fatigue induced
failure here can be remedied without expensive and painful medical
procedures being invoked).
This just leaves us with the question of, "If I were that butcher,
carrying the heaviest lump of meat from the direction implied by that
sketch, where would I want to place it to minimise musculoskeletal wear
and tear?" The answer, quite obviously, becomes "The nearest to hand,
stupid!", in this case, Z. :-)
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'.
I may be wrong in interpreting this question as one of 'ergonomics' but
fuck 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)
V (looks closest to the optimal 45 degree angle ignoring air resistance)
All equal (assuming we ignore friction effects as Galileo was able to)
Rise and then fall
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.
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?)
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 piss 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. :-)
The mechanism will jam (I'm only 99% sure but if I'm wrong then opposite
direction unevenly becomes the only viable alternative)
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.
He doesn't need to he has a hook in the meat already and would just hook
it on the rail nearest him.
truss answer missing here.
If the truss has been dimensioned correctly they will all have the same
strain but they may well have different loads causing that strain.
doesn't that depend on the taps being identical?
if the flow rate is slow the fall will be impossible to see even if you
know it is there.
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.
The examiners hand.
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.
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.
On Sun, 28 May 2017 11:07:10 +0100, dennis@home wrote:
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').
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
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
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'. :-)
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
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. :-)
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".
That lacked a smiley imho. :-)
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. :-)
Here are my answers with reasoning:
2. G goes deepest into ground while being angled towards the pulling force
(at 45 degrees)
3. M area is s^2 rather than pi (s/2)^2 (where s is the length of the side
4. Q closest to the pivot: need greatest ratio of length of handle side to
length of twig side (lever - maginfication of force inversely proportional
to distance from pivot)
5. All equal (not entirely confident on this one)
6. A smallest radius
7. H cross-brace, even if it *is* in tension rather than compression!
8. O greatest number of pulleys
9. move to and fro
10. V if the objective is to prevent the hanging rail bending/breaking
11. B smallest surface area of liquid and smallest area of cup
12. F and G (H turns in opposite direction)
13. N at pivot, with greatest distance and therefore greatest turning moment
from children on either end)
14. Fall radius of "fall" side of rope is greater than radius of "rise" side
15. X 45 degrees
16. D longest length of strut from its pivot
17. H load is carried closest to the pivot/wheel, with longest distance from
load to handle
18. L narrowest arch relative to height
19. S goodness knows - hard to estimate how the curved line continues
20. V longest anchor line so greatest component of force sideways compared
21. C if same amount of energy (mgh1-mgh2) = (mgh2-mgh3) is lost on each
22. J smallest surface area so smallest air resistance
23. N roughly in same direction as line through centres of white and black
24. fall balls fly outwards, causing collar to rise and hence pointer to
fall - a governor
25. W guessing - not sure
26. rise and then fall: pipe that fills X is bigger than one that drains it
so inflow is greater than outflow
27. H furthest away from either end (eg F) where it will stop and reverse
28. O greatest turning moment furthest from pivot
29. R others will roll freely only until off-centre bar is at bottom and
will experience retarding force after that
30. W greatest mass concentrated furthest from centre
31. D because he's having to bank over furthest to compensate for (mv^2)/r
32. move in circle (both small cogs are same radius and will turn in same
direction, so no jam)
33. S closest to centre so he's exerting least turning moment to overcome
whatever the capstan is turning
34. X resultant force is midway between directions of pushing forces
35. smallest radius of curvature so greatest (mv^2)/r centripetal force
(assuming all cars doing same linear speed)
37. one ball hits the group of four so one ball leaves group (as for
38. all equal
39. opposite, unevenly (won't jam because ends of rod are 180 deg out of
phase so no vertical component to motion)
I'm open to argument/correction on some of them!
I had in mind the nautical capstan with 4 men and "Which man would have
to work hardest to turn the capstan alone?" ISTM "hardest" is ambiguous
between (a) the force exerted (highest for shortest lever), (b) the
energy required per unit time (ditto if they walk at the same speed -
linear not angular - when walking alone), and (c) the energy required
per revolution (same for all).
reply-to address is (intended to be) valid
If you assume the capstan has to be pushed with a constant rotational
speed then c) is right. If the man has a constant force available he
is doing more work the faster he walks so the furthest out one is the
answer. But unless the capstan has to be pushed at a constant speed it
is a bit hard to see any sensible answer for work done.
HomeOwnersHub.com is a website for homeowners and building and maintenance pros. It is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.