6
$\begingroup$

John, a graduate student in US, walks into his apartment kitchen with his laptop. On the counter his roommate has left a note next to what looked like a door lock. The lock was a rectangular prism with a normal attached hook on top as a latch. Like a Masters lock. The note said,

Hey Mr. Scientist, confirm for me that the weight of this lock is very close to 5.28 pounds. You can only use things in the kitchen. Good Luck. O and just to tease you there is a small ruler for linear measurements for you

John rummaged through kitchen drawers and was shocked to find a spring balance, a linear analog weight scale with a hook at the end. It can’t be that easy he thought. Sure enough. The weight scale was from zero to 5 pounds! Now what?

He sat at the table and started working. He measured the lock first by the ruler. Was about 4x3x1.7 inches. But he could not measure the hook or latch part. He looked around the kitchen to see if there was any other standard measurement tool. NONE.

The usual cutlery, small and large pots and pans, cups, food, drinks like beer, soda cans, snacks. He weighed the lock on the weight scale just in case. Sure enough it went beyond the very last mark of 5 pounds. Looking at the lock he couldn’t tell what material it was made from. But certainly was some hard alloy. Not possible to cut.

Then he sat at the table drinking a beer bottle and thinking. All of a sudden he checked something on his laptop. Then he made ONE weight measurement (of something) on the spring balance. That came out to be 4.5 pounds. He followed that up with something else and jumped with joy. “I got it” he said. What exactly did John do?

$\endgroup$
  • $\begingroup$ I strongly suspect logical-deduction is not fitting here; please read its description. (Specifically, it does not mean "analytical deduction", which would fit most puzzles; it means using formal logic rules and operations, none of which seem to be in evidence here.) $\endgroup$ – Rubio Mar 8 '17 at 20:21
  • 3
    $\begingroup$ Yuck! Imperial units! $\endgroup$ – Victor Stafusa Mar 8 '17 at 20:32
  • 2
    $\begingroup$ Victor If this was in Metric system it would have been quite straightforward and no fun. May be that is a clue?? $\endgroup$ – DEEM Mar 8 '17 at 20:45
  • 1
    $\begingroup$ Seems like a good candidate for science tag $\endgroup$ – humn Mar 9 '17 at 0:11
  • $\begingroup$ On his laptop, he checked the box that says 'Yes, I want to buy a bigger scale' $\endgroup$ – Strawberry Mar 9 '17 at 9:32
5
$\begingroup$

"Hmm, I was planning to take a bath, but I suppose I can tackle this first" he said.

Having measured the dimensions of the bulk of the lock he calculated its volume as being a little over 20 cubic inches. He looked up the weight of that volume of water and saw that it would be nearly three-quarters of a pound.

"That should be enough!", he thought.

He located a container large enough to fit the entire lock and stood that container upright inside a larger container. He then carefully filled the first container with water until it was full to the brim. Next he lowered the lock hanging from the balance into the water until the lock was just fully submerged, displacing an amount of water that was captured in the outer container. He noted that the balance indicated 4.5 pounds.

He then spied an empty 12 oz soda can and found that its ring-pull conveniently allowed it to be suspended from the hook on the balance.

The empty can barely registered on the balance so he took its weight as being negligible and then carefully poured all the displaced water from the second container into the can and noted that it was nearly full. "What a lucky coincidence!" he mused, "that's about 12 fluid ounces right there!". He noted that the balance indicated the weight of the displaced water was about 0.78 lbs.

Being of soundly curious scientific mind, he verified his result by

taking the density of water that he'd looked up as 1.04 lbs per 16 fl oz, so 8 fl oz is 0.52 lbs and 4 fl oz is 0.26 lbs, so the weight of the 12 fl oz of water that had been displaced by the lock should be about 0.78 lbs.

"Eureka!", he said, "that makes it about 5.28 lbs whichever way you look at it."

"Now, what's for lunch? Suddenly I've got a hankering for something Greek!"

$\endgroup$
  • $\begingroup$ Eureka Indeed! Good Thinking $\endgroup$ – DEEM Mar 9 '17 at 14:49
  • $\begingroup$ Archemedes : Wt of displaced water by a submerged body = Wt of body in air - wt of the body in water $\endgroup$ – DEEM Mar 9 '17 at 14:53
9
$\begingroup$

Message back from John:

Hey Mr. Noter, sorry about ruining your lock but at least I can confirm that it weighed 5.28 pounds.   You see, I began by...

...grabbing a cylindrically walled pot and measuring its cross section trigonometrically, ...

... which came to A = 21.6 inch2.   Then I...

...poured 5 inches of water into the pot, as measured with the ruler.

You can guess what happened next to the lock.

Into the water it went, and got jiggled until all air bubbled out from it.   I measured the rise in water, caused by the solid mass of the lock, ...

...and that came to H = 1 inch.   Using a strand of hair to hang the lock from the scale, ...

... so that the scale’s hook wouldn't be in the water whereas hair can be treated as essentially weightless and neutrally buoyant, I weighed the immersed lock...

...and the scale read W = 4.5 pounds.   I had already looked up...

...the density of water, at D = .0361 pound/inch3. Combining all these, because the weight of the volume of the water displaced by the lock is how much buoyant force would reduce the scale’s reading of the underwater lock, ...

...I calculated:

     lock’s weight   =   A × H × D + W   =   21.6 × 1 × .0361 + 4.5   =   5.28 pounds!

With loving apologies (and more explanation/corrections as needed), John


Incidentally, this is a carefully calibrated puzzle as...

...the lock’s volume agrees exactly with the measured volume of water, if the lock’s dimensions are at limits allowed by .1-inch precision:

Approximate lock:         4 × 3 × 1.7   =   20.4 inch3

True lock?:         4.05 × 3.05 × 1.75   =   21.6 inch3

Measured water:      A × H   =   21.6 × 1   =   21.6 inch3

$\endgroup$
  • 1
    $\begingroup$ Did you actually do this? Wow Impressive $\endgroup$ – DEEM Mar 9 '17 at 14:54
  • $\begingroup$ Naw, @Deepak Mahulikar, in an imagined real life I weighed the displaced water too, like the accepted answer, but then realized that the puzzle said "he made ONE weight measurement" and started over. I also realized that the weighing could be very exact by using some hair, which even the worst-equipped kitchens usually have lying around, especially after I've drunk a beer there. $\endgroup$ – humn Mar 9 '17 at 16:37
6
$\begingroup$

I think that the lock:

weighed 5.2734375 troy pounds

The kitchen scale:

measures in multiple units, the original measurement being in troy pounds but the last measurement being in merchant pounds

This works because:

4.5 merchant pounds = 5.2734375 troy pounds

$\endgroup$
  • 2
    $\begingroup$ He weighed the lock on the weight scale just in case. Sure enough it went beyond the very last mark of 5 pounds. Am I missing something here? $\endgroup$ – Mordechai Mar 9 '17 at 4:39
2
$\begingroup$

This is how I would do it.

He used the ruler to measure the length of the spring when he weighed objects
at different lengths. For instance, if he found one object that weighed 4.5
lbs and stretched the spring to a length of 4 inches and another object at
5 lbs that stretched the spring to 5 inches he would be able to calculate
how much weight is required to stretch the spring a given length. Applied
to the doorknob he is able to accurately determine the weight of pull.

$\endgroup$
  • 1
    $\begingroup$ Use <br> to break lines inside spoiler blocks. $\endgroup$ – Victor Stafusa Mar 8 '17 at 20:35
  • 3
    $\begingroup$ This assumes the spring stretches at a constant rate. Maybe the first 1 lb. makes it stretch by 1 inch, but the second 1 lb. only makes it stretch by 1/2 inch, and the scale is calibrated to account for that... $\endgroup$ – GentlePurpleRain Mar 8 '17 at 22:15
  • $\begingroup$ @GentlePurpleRain Good point. $\endgroup$ – MHeath Mar 8 '17 at 22:26
1
$\begingroup$

How about this ...

Balance the ruler on something at its 6-inch mark. Then place the 4.5 lb something at 2 inches, and the lock at 10 inches, so that the center of mass of each is the same distance from the balance point (with the lock's hook facing conveniently upward). Of course, the lock, being heavier than the something, will pull its side of the ruler down.

Now use the spring balance to pull the lock UP until the ruler balances. Read the value on the spring balance, add 4.5 lbs, and that's the weight.

$\endgroup$
1
$\begingroup$

Place the ruler on the edge of a knife supported on either side, so that the blade stacks out past the end of the table, edge upwards. Tie two strings to the ruler at the $6-5.28=0.72$ inch mark (assuming a 1 foot ruler), the other at the $6+4.5=10.5$ inch mark (or vice versa).

.

Tie the lock to the string 4.5" from the centre, and the 4.5 oz. weight to the string 5.28" from the centre.

.

If the ruler remains balanced, then,

.

$5.28" \times 4.5$oz. $=4.5"\times 5.28$ oz

thus, the lock weighs 5.28 oz. Otherwise,

adjust one of the strings along the ruler till it returns to equilibrium, them repeat the calculation to get the weight of the lock.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.