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On December 17 2008 09:35 GearitUP wrote:After much deliberation and reading over einstein's e=mc2 theory, I have come to the conclusion that m(w3) (over elephants ) is the correct way to figure her luggage weight Image 1 will explain... _____ elephants
This makes noooooooooo sense at all. But it so fucking funny that u just HAVE to win it.
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It is easy, just follow these steps:
Mass = m.
m=x
since its unknown, now add 5x to both sides:
m+5x=6x
Then divide both sides by 6x:
m/6x+5/6=1
Take the square of both sides:
(m/6x)^2+10m/36x+25/36=1
Subtract 1 from both sides:
(m/6x)^2+10m/36x-11/36=0
Substitute m/x=u
u^2/36+10u/36-11/36=0
Multiply both sides by 36:
u^2+10u-11=0
Factorize:
(u-1)(u+11)=0
Put in substitution:
(m/x-1)(m/x+11)=0
Multiply by x:
(m-x)(m+11x)=0
Divide both sides by m-x:
m+11x=0
Substitute x=m
12m=0
There, I solved it, all you need to do is bring 12 suitcases and the weight will be zero!
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United States24651 Posts
In about 5 seconds I found the divide by 0 :p
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Divide both sides by m-x:
It's easy to spot buddy 
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Despite the warning in the original post i think a balance is still the easiest and most accurate answer. She can use herself as the counterweight and mark distances based on pretty much anything. If she weighs around 110 lbs then she just has to be roughly 3 times closer to the pivot point than her luggage. All she needs to do this is a long 2x4.
If what you're really looking for is zany solutions then she should find a peice of scrap lumber and hold it in place overlooking the side of a desk like the plank on a pirate ship. Then place the luggage on the end as if it were about to be fed to the sharks. Now you can measure the distance from the ship and the vertical displacement of the board to calculate the weight of the luggage. Displacement = (weight*length from ship^3)/(3EI). The elasticity, E, depends on the type of wood and can be found online or from the shipwright who made your vessel and the I value can be calculated from the cross-area of your plank. I = base*height^3/12
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On December 18 2008 07:25 micronesia wrote:
In about 5 seconds I found the divide by 0 :p
Of course, you always do it except in this one:
The mass "m" of the suitcase.
m=x, since its unknown.
u^u^u^u...=2=z
u^z=2
u=sqrt(2)
Multiply by z to both sides:
m*z=x*z
Now consider:
y^y^y^y...=4
y^4=4
y=4^1/4=sqrt2=u
Therefore
y^y^y^y...=z
and
4m=2x
and
2m=0
, so then again the luggage weight nothing 
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On December 17 2008 14:03 micronesia wrote:
Hang the suitcase from a string, and run it through a straw. Tie to the other end of the string an object of known mass. Spin the object around fast enough to make the suitcase just barely life off the floor. m*v^2/r = W
You will need to measure the radius by holding the string steady and stopping the revolution. You can time the period to find speed.
i wanted to write this 
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guys, the op asks for weight, not for mass hehehe whatever.
Take a sensitive enough torsion pendulum and measure the average density of the luggage. After this, multiply it by the volume of the luggage and the gravitational acceleration
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Try lifting it with one of your hairstrings, if it breaks, try two etc. Weight = Carry capacity of one string times the number of strings it took to lift it. You can measure hairstring carrying capacity with normal groceries and a bag.
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1. find an elevator (or bridge but elevator works better) with a listed weight limit (ie 4000 kilo)
2. use a lot of a dense substance to pile up in the elevator with a few hundred pounds short of the weight limit. for a really dense substance, try osmium, but if you don't have any of that on you i guess lead works fine too
3. put all the osmium (or other substance) in the elevator along with the luggage. if the elevator breaks and it disconnects and falls down the shaft, you know you have too much luggage.
4. find a new elevator and new luggage because anything you used in the first test is probably smashed.
5 repeat steps 1-4 until you run out of elevators
tips:
security guards = bad
be sure to remove all surveillance cameras from the vicinity
honestly i have no idea, but wouldn't they have a luggage weigh in at the airport? or if she brought a friend to take a bag or 2 of things that aren't necessary if its over the limit.
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ask a neighbor for a scale?
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On December 17 2008 09:27 Wysp wrote:
so easy. grab something that weighs 15 kilos then grab the bag. if you can't figure it out I gotta ask you where your man sense is
this is just a practical solution, someone else can enjoy the shirt
how would you know that something is 15 kilos?
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friend of mine said:
I would find one of those pneumatic tube systems, which I'm guessing many airports have. After knocking out the guards, I would stuff a cylinder attached to a string into the tube, and attach the other end to the suitcase. I would feed the string up until it cannot be pulled higher, and holds the full weight of the suitcase. Based on the atmospheric pressure (or an approximation), the top of the tube being a vacuum, the length of the string, and the diameter of the tube, I could figure out how much force the suitcase is pulling.
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Simple hold a gun up to everyone until someone tells you the weight of the object
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Easy, weight = mass x gravity, just have her find the mass of the bags using a scale and standardized masses, then multiply by the gravity of the earth, 9.8 meters per second squared.
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i'm getting near a solution.
Should work if my understanding of TS is correct.
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Netherlands6142 Posts
Ants can carry up til 20 times their body mass. The average ant weighs 0.003 grammes. An average ant, therefore, can carry 0.060 grammes of freight. My proposal therefore would consist of rounding up 250000 average ants (15000 / 0.06) and put the suitcase on them. If they are squashed and die the suitcase is too heavy. You're welcome.
Edit: Alternatively one could use eggs or matchsticks but these methods can't outweigh the merits of animal brutality.
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On December 18 2008 07:29 patrick321 wrote:
Despite the warning in the original post i think a balance is still the easiest and most accurate answer. She can use herself as the counterweight and mark distances based on pretty much anything. If she weighs around 110 lbs then she just has to be roughly 3 times closer to the pivot point than her luggage. All she needs to do this is a long 2x4.
If what you're really looking for is zany solutions then she should find a peice of scrap lumber and hold it in place overlooking the side of a desk like the plank on a pirate ship. Then place the luggage on the end as if it were about to be fed to the sharks. Now you can measure the distance from the ship and the vertical displacement of the board to calculate the weight of the luggage. Displacement = (weight*length from ship^3)/(3EI). The elasticity, E, depends on the type of wood and can be found online or from the shipwright who made your vessel and the I value can be calculated from the cross-area of your plank. I = base*height^3/12
Engineer? ^^ Engineering Pirate, perhaps?
I'd say a reasonable Young's Modulus for pine would be 11-12 GPa. When using it in the formula be sure to convert to MPa, so 11,500 or so to take an average.
Be sure to make it a sturdy cantilever. Try to get a plank with no knots or imperfections which would drastically change the cross-sectional area. Sit another plank beside it to have a reference point to measure the deflection from. Stick bent nails or something in the end to hold the suitcase on the very edge. Measure length from where the cantilever makes 90° with the face it is on, not the whole plank.
You'd also put some values in the formula to work out what kind of a deflection you'd get from the 15kg for certain plank dimensions to make sure it is sensitive enough (like, if it would only deflect a max of 1cm for 15kg then less than that is going to be hard to measure accurately), so like a 5cm+ deflection for 15kg.
So far I vote the above idea (I was going to post similar but was beaten to it) for serious one, and Pholon's for most amusing ^^
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On December 18 2008 08:25 RoC)Ninjah wrote:
1. find an elevator (or bridge but elevator works better) with a listed weight limit (ie 4000 kilo)
2. use a lot of a dense substance to pile up in the elevator with a few hundred pounds short of the weight limit. for a really dense substance, try osmium, but if you don't have any of that on you i guess lead works fine too
3. put all the osmium (or other substance) in the elevator along with the luggage. if the elevator breaks and it disconnects and falls down the shaft, you know you have too much luggage.
4. find a new elevator and new luggage because anything you used in the first test is probably smashed.
5 repeat steps 1-4 until you run out of elevators
tips:
security guards = bad
be sure to remove all surveillance cameras from the vicinity
honestly i have no idea, but wouldn't they have a luggage weigh in at the airport? or if she brought a friend to take a bag or 2 of things that aren't necessary if its over the limit.
Your elevator idea fails, because it will be able to take more weight that it is specified for. The weight limit is just a safety limit, although it's not unthinkable that elevators are fitted with something that causes them to not work when exceeding the safety limit.
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EPT![[image loading]](http://i66.photobucket.com/albums/h254/luciferbw88/hahatheanswer.jpg)
