Brain Teasers
Suspended Meter Stick
A meter stick is suspended exactly at its 45 cm mark from a piece of string, and an 8kg is fixed to the stick and hangs from the 0cm mark. You have six 1kg weights you can hang anywhere on the stick (you can put more than one on the same point if you wish). How can you get the system to balance?
Answer
The turning force of the 8kg is 360The max force on the other side would be if you put all 6 on the 100cm mark - 6*55 turning force = 330 Which isn't enough to balance, so the only way you can to this is by moving the point of suspension of the metre stick!
If you change the point of suspension to the 40cm mark, the 8kg has a turning force of 320; so you could put 5kg at the 100cm mark giving a turning force of 5*60, and put 1kg at the 60cm mark, giving a turn of 1*20 =320 in total
(There are other ways to do this if you chose a different point of suspension, but this is the key to the puzzle.)
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Comments
a far less graceful answer: 8x=6(100-x), where x is the distance b/w the fulcrum and the 8kg mass. x ends up being approximately equal to 42.85714 cm. like i said, far less graceful. =)
I've just noticed that some of my solution has been cut off, I was going to say the key is realising you can't balance those weights, and must move the fulcrum
how can u be so sure that the meter stick is of negligible mass? If the meter stick does have some unkown mass couldn't you just move the string and the 8kg mass to the center (50cm mark) and hang 3 1kg weights on either end of the stick? Just a thought--i got the answer the other way(s) too.
how can u be so sure that the meter stick is of negligible mass? If the meter stick does have some unkown mass couldn't you just move the string and the 8kg mass to the center (50cm mark) and hang 3 1kg weights on either end of the stick? Just a thought--i got the answer the other way(s) too.
Sep 01, 2002
"Posted by cathalmccabe May 23, 2002
I've just noticed that some of my solution has been cut off, I was going to say the key is realising you can't balance those weights, and must move the fulcrum "
always an excuse, perhaps it is missing because it wasn't submitted in the first place? Another cathal corker, don't worry the odds are with you, one day you are bound to get one of you teasers right.
I've just noticed that some of my solution has been cut off, I was going to say the key is realising you can't balance those weights, and must move the fulcrum "
always an excuse, perhaps it is missing because it wasn't submitted in the first place? Another cathal corker, don't worry the odds are with you, one day you are bound to get one of you teasers right.
Whoa, there Alphakennwun! Why were you ripping on somebodies teasers like that? Unnecessary to say the least. Be lavish in your praise, or constructive in your criticism, or just plainn quiet!
Well put, Rowsdower.
I was guessing that 6kg. wasn't enough weight, so I did a quick check: 8*(45/55) = 6 6/11, which is the minimum weight to balance it. I was wrongly thinking the fulcrum point was fixed, so hung the six weights on the end of the stick and set the weight of the stick to 10 * 6/11 * 50/45 = 6 6/99 kg. and called it balanced.
Actually, the weight of the stick needs to be 10 * 6/11 * 55/50 = 6 kg., which is a far more pleasing answer.
Personally I don't like these types of "situation" teasers where there is some information missing. The puzzle said the stick was suspended at the 45 cm mark. Now if the answer is that you can just change the point of suspension then you could also suppose that you got an extra weight from somewhere. Maybe the stick was actually longer than 1 metre - some of them have spare timber protruding beyond the scale which goes from 0 to 100.
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