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Long story short, a friend is planning to take a plane rather soon and must pack her shit. She, being of the female variety, obviously doesn't make a border-crossing move without a sizable amount of hotpants, toiletries and British souvenirs (*last item not confirmed). She has less than 24 hours to figure out how much shit she can actually pack into that suitcase of hers and, for the life of her, doesn't want to be caught dead with overweight baggage at check-in, which is rated at 15 kilos (approximately 33 lbs) or above.
What is the quickest, most imaginative or most accurate method for her to determine the weight of her freight without using a scale in the traditional sense? Please note that using a makeshift scale (such as one consisting of a pivoting plank and roughly 15 kilos worth of, say, leftover toiletries) is out of the question as said toiletries have been used up, as is any other such contraption. Damaging the suitcase and its contents is permitted as long as the weight-determining method is far-out enough, man! When the determining process is finished, at least half the contents by weight must be left over, as must half the suitcase. (Of course she would want to have more than half her suitcase and its contents when boarding the plane; ideally, no damage whatsoever should occur.)
Do note that the question in this topic is not purely of utilitarian value, but is to a greater extent meant to actuate your creative minds, contribute to general scientific knowledge or simply churn out the most incredible and otherworldly suggestions that, to a reasonable degree of accuracy, actually work. The best answer, chosen after an adequate period for answer submissions, by a tribunal consisting of me, myself and I, will be rewarded with a free TL shirt.
Disclaimer: No conditions apply (except in cases where the shipping costs of the T-shirt would equal the price of the T-shirt alone or exceed it). The grant period for answer submissions will be until the thread disappears from page 1 or, at most, 2 (two) weeks from the date of thread submittal. I will choose 2 (two) best answers, one from the category of answers that include damaging the suitcase and its contents in any reasonably unacceptable way, and one from the answers that will leave the suitcase and its contents intact after the weight-determining procedure is finished. Among those two, the winner of the TL shirt will be drawn either by a poll (moderators willing) or a coinflip. This is not a joke thread. This is not a homework thread.
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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
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liter = kilogram.
Place the suitcase in a plastic container, which has been allowed to come to rest in a bathtub filled to the brim with normal water. catch water which runs off in a cut in half dryer tube, and use it to fill milk cartons or what not to count the amount of liters of runoff. Once you've hit 7 2L cartons, switch to a 1L carton for the next addition and you should come up slightly below 15kg due to a tiny bit of extra water being siphoned off due to surface tension.
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 is dota's trojan action done?
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On December 17 2008 09:27 L wrote:liter = kilogram. Place the suitcase in a plastic container, which has been allows to come to rest in a bathtub filled to the brim with normal water. catch water which runs off in a cut in half dryer tube, and use it to fill milk cartons or what not to count the amount of liters of runoff. Once you've hit 7 2L cartons, switch to a 1L carton for the next addition and you should come up slightly below 15kg due to a tiny bit of extra water being siphoned off due to surface tension. Show nested quote +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 is dota's trojan action done?
archimedes you dog! how i s'pose to defraud gold buyer now?
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go to the nearest university or supermarket and let them weight the package..
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Get a big spring, and hang it from the ceiling. Find something that you know the weight of, like a can of soup or whatever, and hang it from the ceiling. Measure how far the spring stretches. If your known weight is 1 kilo, a 33 kilo suitcase stretches the spring 33 times as far.
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how i s'pose to defraud gold buyer now? melt more dense materials in too. Little lead never hurt anyone.
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On December 17 2008 09:34 L wrote: melt more dense materials in too. Little lead never hurt anyone.
then when they bite it they die of lead poisoning, brilliant!
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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
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On December 17 2008 09:30 Rev0lution wrote:
go to the nearest university or supermarket and let them weight the package..
I think you missed the point 
But i cant come up with anything other than displacement in water, which was already said. This is gunna have me thinking for a while though, funny question
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again on the water theme. you need:
1) plastic bags, or other water containers.
2) a few meters of rope or equivalent. (worst case you could even tie clothes together)
Fill bags with 15 liters of water. If yuo dont have a litre container in the kitchen, then use a milk/soda bottle that says how much it contains, and take as many bottles you need.
Tie one end of the rope through the handles of the bags (or make holes, or tie the bags to rope directly...).
Find something to hang the rope over. For example a branch on a tree, or some part of a building. If needed, use a plastic bag or something to make the rope slide better.
Tie the other end to the backpack.
You can now fill the backpack until it can lift the water on the other end.
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You don't even really need a liter container. Just read the flow through rate on the tap to determine how long you need to fill the bucket or whatever other container.
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On December 17 2008 09:27 L wrote:
liter = kilogram.
Place the suitcase in a plastic container, which has been allows to come to rest in a bathtub filled to the brim with normal water. catch water which runs off in a cut in half dryer tube, and use it to fill milk cartons or what not to count the amount of liters of runoff. Once you've hit 7 2L cartons, switch to a 1L carton for the next addition and you should come up slightly below 15kg due to a tiny bit of extra water being siphoned off due to surface tension.
if you dont want to set up the mess with catching all the overflowing water, you could instead add the backpack until it floats (let the overflowing water go down the drain) and then see how much you need to add afterwards to refill the tub.
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but doesn't water displacement measure volume? wouldn't you just be finding the weight of the suitcase if it was full of water? unless you know the density of its contents. edit: o wait, i totally missed the "put it in a plastic container" part. my bad.
though if you have a long surface, such as a ironing board or support boards for your matress, you could try to make your own scale using 15L of water on one side and the luggage on the other.
water - - suitcase
(_)(_)..........[__]
```````````^``````````` <- macgyvered board
too bad i can't lend you my awesome portable xscale
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On December 17 2008 09:58 Not_Computer wrote:but doesn't water displacement measure volume? wouldn't you just be finding the weight of the suitcase if it was full of water? unless you know the density of its contents. though if you have a long surface, such as a ironing board or support boards for your matress, you could try to make your own scale using 15L of water on one side and the luggage on the other. water - - suitcase (_)(_)..........[__] ```````````^``````````` <- macgyvered board too bad i can't lend you my awesome portable xscale
u want some gold?
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but doesn't water displacement measure volume? Not if you do it in the method I prescribed.
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There is a trick using paper on your car to check the weight when u put it under your tires and measure the dirt. I forget how it works, maybe can be applied to other things as well.
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Mark a line on the floor. Hold the suitcase and jump staight up in the air. In mid air you push both the suitcase and yourself backwards. As we all know, the momentum is always preserved. So the equation is p1=p2=m1*v1x=m2*v2x. Where m1 is the mass of the person doing the experiment, m2 is the weight of the luggage, v1x is the velocity in x-direction of the person and v2x is the velocity of the luggage in the x-direction. To obtain the velocities, if you don't have a speed-sensor, you have to derivate the measurable position, dx1(t)/dt=x1-0/(t1-0)=x1/t1 where t1 is the time it took to travel the distance x1. Easiest way to make it time-independent is if you can make both object land at the same time. This yields the equation: m2=m1*x1/x2. This all of course assumes zero air resistance, or at least that the air resistance is the same for both the objects. Which it will not be as it has been proven that air resistance is a viscous non-linear effect (e.g. proportional to the square velocity, v^2).
A more serious suggestion is to just use a board that rests on a cylinder (glass bottle) and put on a known weight, m1 (suggestible a person) on one side at distance x1 from the middle and the luggage with weight m2 at distance x2 from the middle so that the system is in equilibrium and the board is horizontal. This will give the weight of the luggage m2=m1*x1/x2. Much easier to measure and does not involve air resistance.
Edit: Sorry my second suggestion was already posted...
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If she can carry it with one hand, for 1 min, that's < 15 kg
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intrigue
Washington, D.C9933 Posts
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
someone please explain why this made me laugh
lol it's so stupid but great WTF
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On December 17 2008 10:32 xiashan wrote:
Mark a line on the floor. Hold the suitcase and jump staight up in the air. In mid air you push both the suitcase and yourself backwards. As we all know, the momentum is always preserved. So the equation is p1=p2=m1*v1x=m2*v2x. Where m1 is the mass of the person doing the experiment, m2 is the weight of the luggage, v1x is the velocity in x-direction of the person and v2x is the velocity of the luggage in the x-direction. To obtain the velocities, if you don't have a speed-sensor, you have to derivate the measurable position, dx1(t)/dt=x1-0/(t1-0)=x1/t1 where t1 is the time it took to travel the distance x1. Easiest way to make it time-independent is if you can make both object land at the same time. This yields the equation: m2=m1*x1/x2. This all of course assumes zero air resistance, or at least that the air resistance is the same for both the objects. Which it will not be as it has been proven that air resistance is a viscous non-linear effect (e.g. proportional to the square velocity, v^2).
A more serious suggestion is to just use a board that rests on a cylinder (glass bottle) and put on a known weight, m1 (suggestible a person) on one side at distance x1 from the middle and the luggage with weight m2 at distance x2 from the middle so that the system is in equilibrium and the board is horizontal. This will give the weight of the luggage m2=m1*x1/x2. Much easier to measure and does not involve air resistance.
Edit: Sorry my second suggestion was already posted...
fuckin win lol I dont understand any of this but it beat my paint picture I think.. im still gona pushin for the elephant theory
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First, you need a 10kg weight. It can be any kg, as long as its measured, but for simplicity's sake, lets make it 10.Take a board that is made of the same material throughout and set it at an angle. Measure the angle. Now, put the 10kg weight on the board and give it a small nudge. Keep increasing the angle and repeating the step until you get to the point where the small nudge makes the 10kg weight move very slowly at a constant velocity. Measure that angle and take the inverse tangent of it to get MUkenetic.
Now, you need a string. Attach the string to the ceiling or something. Attatch the bag to the string so that the bag just barely hangs above the floor. Mark the spot with a big X. Move the bag sideways now until it is exactly 1M above the ground. Now put the objects as shown in this picture. Let the bag go, and record how far the 10kg weight went.
[img=http://img126.imageshack.us/img126/2310/physicsyg3.th.png]
If everything went right, these equations should apply.
M1=Bag. V1=velocity of bag.
M2=Weight. V2=velocity of weight
(m1)(g)(h)=(.5)(m1)(v1^2).
Masses cancel out and you are left with
(19.6)(1)=v1^2. v1=4.42 m/s.
Momentum never changes, so pi=pf .
(m1)(v1initial)+(m2)(v2initial)=(m1)(v1final)+(m2)(v2final)
(15)(4.42)+0=0+(10)(V2final)
V2final= 6.63.
Now, we need the total initial energy, which we can measure with the .5(m2)(v2^2) equation.
(.5)(10)(6.63^2)=220
This is where your coefficient of friction you measured earlier comes in. (Mu)(N)=force of friction. (Force of friction)(distance)=work done by friction. Ef-Ei=work done by friction. Since EF is zero, work done by friction is equal to the negative of the initial energy. For simplicities sake, we can say they are equal.
(Mu)(98)(distance traveled of 10kg weight)=220. Repeat this experiment until you get the desired distance.Distance is measured in meters.
Lets say the Mu turned out to be .9. I would solve the earlier equation
(.9)(98)(distance)=220. Distance=2.5m. I would then repeat the experiment while adding or subtracting things until I get the 10kg weight to travel 2.5 meters.
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is awesome32274 Posts
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
HAHAHAHAHAHAHH
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This may be cheating but here goes...
This is stuff to do before your trip.
Get something that weighs 15kg. (If you want to play it safe make it 13 or 14 kgs)
Put it on the floor.
Sit on the floor cross-legged.
Put the object arms length away from you. Grab the object with an underhand grib: when you bend your elbow it should go towards the ground. Try to lift the object straight up without bending your arm. If you can, this won't work. But I don't think you will be able to unless you are really quite strong. So, assuming you can, move the object towards you tiny bit by tiny bit bending your elbow bit by bit as it gets closer. It will get easier to lift the object but stop once you can lift it a couple inches off the ground.
Now put the object down and measure the distance from you to the base of the object in terms of your hand. Remember this number: ie. 3 and a half hands or w/e.
Now whenever you want to know if something weighs over 15lbs just see if you can lift it in that position at that distance. This can be done anywhere you have your luggage and yourself. Notably, you can do it in an airport before you head through security.
This isn't particularly exact so you might want to play it a little bit safe by using an object weighing less than 15kgs by a bit. But in terms of practical and easy its pretty good.
The main problem other than inaccuracy is that if you can't lift the object at all or can lift it at any distance then you would need to find some other lifting method so that you could use this method.
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On December 17 2008 10:32 xiashan wrote:
Mark a line on the floor. Hold the suitcase and jump staight up in the air. In mid air you push both the suitcase and yourself backwards. As we all know, the momentum is always preserved. So the equation is p1=p2=m1*v1x=m2*v2x. Where m1 is the mass of the person doing the experiment, m2 is the weight of the luggage, v1x is the velocity in x-direction of the person and v2x is the velocity of the luggage in the x-direction. To obtain the velocities, if you don't have a speed-sensor, you have to derivate the measurable position, dx1(t)/dt=x1-0/(t1-0)=x1/t1 where t1 is the time it took to travel the distance x1. Easiest way to make it time-independent is if you can make both object land at the same time. This yields the equation: m2=m1*x1/x2. This all of course assumes zero air resistance, or at least that the air resistance is the same for both the objects. Which it will not be as it has been proven that air resistance is a viscous non-linear effect (e.g. proportional to the square velocity, v^2).
A more serious suggestion is to just use a board that rests on a cylinder (glass bottle) and put on a known weight, m1 (suggestible a person) on one side at distance x1 from the middle and the luggage with weight m2 at distance x2 from the middle so that the system is in equilibrium and the board is horizontal. This will give the weight of the luggage m2=m1*x1/x2. Much easier to measure and does not involve air resistance.
Edit: Sorry my second suggestion was already posted...
May i say...nice first post.
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jeez, just go the the mall or a pharmacy and try out their scales, which you wont buy. much simpler.
edit: use milk cartons for measuring as best as you can 15L of water, which you will put into a water-tight trash bag. then you will just have to compare them by hand.
for accuracy you will have to attach a makeshift "handle" to both the suitcase and the trashbag with 15L of water, and adopt the same lifting stance to carry both, so that your perception of weight will not be swayed by other factors. maybe have friends help you do it blind folded if possible.
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displacement in water will just measure the volume of hte suitcase. AKA, the suitcase may displace 1L, but that doesnt mean the suitcase weighs 1Kg. For example, a baloon might displace a liter but it is certainly lighter than a kilo
Unless I missed something..
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United States24649 Posts
Well I guess this one is right up my alley...
Bring the suitcase to the large hadron collider. When a black hole forms, launch the suitcase at the black hole. If the black hole absorbs the suitcase instead of the other way around, then the suitcase's weight is acceptable.
Ok here's a serious one expanding on the spring suggestion:
Determine spring constant of spring by using F=kx and a mass of known weight. Hang suitcase from spring and start it oscillating. Measure the time it takes to complete ten oscillations, and then divide the result by ten. Use the formula T = 2pi*sqrt(m/k) to determine m and then convert to weight.
Another: Place small charged objects inside the suitcase of known mass. Place the suitcase inside a steady magnetic field. Give the suitcase a kick so that it can slide freely without friction. Measure the radius of the circle formed by the Lorentz Force: (M+m)*v^2/r = Q*v*B and solve for m. Convert to weight. (You will also need to time one complete circular trip in order to find the velocity)
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United States24649 Posts
Place the suitcase on a cart. Take another cart and attach a large box to it. Place objects of known mass into the box. Launch the cart with the suitcase into the cart with the miscellaneous objects. If the suitcase cart comes to a dead stop, and the other cart takes off at the exact speed that the first cart started with, then the mass of the suitcase is equal to the sum of the masses of the objects of known mass. If not, grab different objects and try again! (You can tell whether to add or subtract mass by if the first cart continues forward after the collision or reflects)
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what is the suitcase filled with?
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United States24649 Posts
Create a perfectly vertical (variable) electric field. Place a charged object inside the suitcase. Vary the electric field until the suitcase approximately hovers. (M+m)*g = E*Q
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On December 17 2008 13:41 himurakenshin wrote:
what is the suitcase filled with?
A bunch of 1 lb diving weights. Why?
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United States24649 Posts
Ballistic Pendulum: Tie suitcase to string and hang it from ceiling to form a pendulum. Tie large box of known mass to a string. Pull back suitcase to a measured height, and let go. At the bottom it collides inelastically with the box, and both objects swing upwards together. Measure the height of the subsequent swing... and calculate the mass of the original suitcase... convert to weight.
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United States24649 Posts
On December 17 2008 13:42 micronesia wrote:
Create a perfectly vertical (variable) magnetic field. Place a magnetized object inside the suitcase. Vary the magnetic field until the suitcase approximately hovers.
This works also!
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United States24649 Posts
Place a crushable material on the floor. Using objects of known mass, drop them on to the material, and measure the distance the surface deforms. Create a best fit approximation and extrapolate to the crush achieved by the suitcase to find weight.
Another: Throw small bits of anti-matter at the suitcase until you've completely annihilated half of the suitcase. Make sure you keep track of how much anti-matter you throw at it in order to predict the mass of the suitcase. Beware: this will destroy half of the suitcase and its contents.... and probably you.
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This is a great thread.
On December 17 2008 13:29 Steelflight-Rx wrote:
displacement in water will just measure the volume of hte suitcase. AKA, the suitcase may displace 1L, but that doesnt mean the suitcase weighs 1Kg. For example, a baloon might displace a liter but it is certainly lighter than a kilo
Unless I missed something..
What they're talking about is they're putting it into a boat-like object and measuring displacement, which will measure weight. But they have to measure the water runoff.
My proposal:
Fill a bathtub, put a container than can hold 15 kg/suitcase in it (basically to build the boat-like setup that L proposed).
Put the container in, put 15 kg of something (could be water, could be weights hijacked from the weight room that your friend's hotel room). Balance the boat, note the line that water comes up to on the boat (with a pen, if you must).
Now take your suitcase, put it in the boat. If the water line is below the original water line (with the 15 kg in it) that means your suitcase is under 15 kg; if the water line is above the original water line, you're above 15 kg.
The main problem is to find the boat; it's not nearly as hard as it sounds. Use a large plastic trash can or use a large flat surface, put your suitcase in a trash bag (for water proofing) and attach the suitcase to the trash bag and use where the waterline comes up to accordingly.
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United States24649 Posts
Bring the suitcase to a wind tunnel and determine the drag coefficient. Then measure the terminal velocity of the suitcase after dropping it from a helicopter (stand at the target, and point a radar gun straight up... if this doesn't work drop it alongside a large building or tower, and use markers alongside a stopwatch to find the average speed across that distance). Weight = constant times velocity squared
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Haha, micronesia is having a blast.
Use a mattress, put 14 kg of weights (subtracting for the weight of the suitcase) and put it in the center of the bed. Measure the distance the bed compresses.
Now do exactly the same, for the items in the suitcase. The method should be sound as long as you use something with the same base both times (ie the suit case).
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United States24649 Posts
Oh what is the limit for the maximum number of submissions?
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He said no limit. Keep going Micro.
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United States24649 Posts
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.
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United States24649 Posts
Stand on a frictionless cart near the edge of a cliff. Install a pulley on the corner so that a string tied to the cart can run through the pulley and hang off the side of the cliff. Hang a mass of known weight from the string. While standing on the cart, hold an accelerometer (you can use a calibrated system of a marble inside a circular or parabolically bent straw) and measure the acceleration of the system. Fnet=ma so W_known = (M+m_known)*a_measured
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16974 Posts
Hahaha micronesia's having a blast here. Too bad I bombed my physics final so at the moment I hate physics...though I the plus side I still ended up with a B+ (though I could have easily gotten an A...fuck my life).
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Dude micronesia, you are a physics hero man. Seriously, if micro doesn't win this thread, I don't know what can. gogo micro!
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On December 17 2008 14:21 Empyrean wrote:
Hahaha micronesia's having a blast here. Too bad I bombed my physics final so at the moment I hate physics...though I the plus side I still ended up with a B+ (though I could have easily gotten an A...fuck my life).
haha same exact thing
i had such a high grade in the class i blew off the final, failed, and still pulled off B+
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On December 17 2008 14:06 micronesia wrote:
Stand on a frictionless cart near the edge of a cliff.
You mathematicians and your silly assumptions 
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Bring luggage to nearest Fedex / UPS store and ask them how much it would cost to ship. After they weigh it and tell you the answer, tell them its to expensive and your going to just bring it with you.
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You could also go to the nearest Walmart and buy a scale. They have a 30 day return policy.
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Here is what you do. Fill the bag as much as you want with whatever crap you want, but make sure to take an empty handbag with you to the airport (Don't fill more than you know is too much). Once you get there and they measure your bag, they will tell you how much your bag weighs and by how much it is over. Proceed to empty the suitcase stuff into your empty handbag until the suitcase is of proper weight.
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On December 17 2008 13:53 micronesia wrote:
Bring the suitcase to a wind tunnel and determine the drag coefficient. Then measure the terminal velocity of the suitcase after dropping it from a helicopter (stand at the target, and point a radar gun straight up... if this doesn't work drop it alongside a large building or tower, and use markers alongside a stopwatch to find the average speed across that distance). Weight = constant times velocity squared
This.
A scale might be hard to find, but I'm sure there's a helicopter lying around somewhere for your experiment!
Brb gonna try this now
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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
hahahah this just HAS TO WIN IT!
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On December 17 2008 17:44 strongwind wrote:Show nested quote +On December 17 2008 13:53 micronesia wrote:
Bring the suitcase to a wind tunnel and determine the drag coefficient. Then measure the terminal velocity of the suitcase after dropping it from a helicopter (stand at the target, and point a radar gun straight up... if this doesn't work drop it alongside a large building or tower, and use markers alongside a stopwatch to find the average speed across that distance). Weight = constant times velocity squared This. A scale might be hard to find, but I'm sure there's a helicopter lying around somewhere for your experiment! Brb gonna try this now
LOL
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Your friend has to do the following:
1. Fill the suitcase with whatever stuff she needs for the journey.
2. Go to the airport and let the luggage guy weigh the suitcase.
3a. <15kg: All good.
3b. >15kg: Hurl shoes at the luggage guy screaming "It is the farewell kiss, you dog!" until he lets her pass.
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On December 17 2008 09:35 GearitUP wrote:+ Show Spoiler +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
ahahah you win; they look so happy in this pic!
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Haha i believe i deserve a t-shirt for that drawing
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wet everything a little bit and put it in the microwave for 1 minute. Compare the temperature difference that occured with the amount of Watt your microwave pumped into it.
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On December 17 2008 10:36 Murlox wrote:
If she can carry it with one hand, for 1 min, that's < 15 kg
On December 17 2008 13:43 fight_or_flight wrote:Show nested quote +On December 17 2008 13:41 himurakenshin wrote:
what is the suitcase filled with?
A bunch of 1 lb diving weights. Why?
nice fellas
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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 States24649 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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United States24649 Posts
Er this still winding down?
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MURICA15980 Posts
So it's 33 lbs, eh? So if she weighs around 120, the bag should be at most 4 times her weight. So pick up the bag to test it out a bit. Then when you pick her up, immediately comment that she's at LEAST 5 times the bag and then start taking shit out of the bag.
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god this thread reminded me how incredibly useless I am in math and physics haha
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United States17042 Posts
On December 28 2008 05:14 Klogon wrote:
So it's 33 lbs, eh? So if she weighs around 120, the bag should be at most 4 times her weight. So pick up the bag to test it out a bit. Then when you pick her up, immediately comment that she's at LEAST 5 times the bag and then start taking shit out of the bag.
ROFL- this.
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Water displacement does not measure mass at all.... it's all about volume. a 10kg box that's 5L in volume will displace just as much water as a 5L box that weighs 1000kg.... the only way would be to MacGuyver a scale. Plus, why the hell would you dump your suitcase in the bath just to find out how much it weighs? lol
User was warned for this post
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EPT![[image loading]](http://i66.photobucket.com/albums/h254/luciferbw88/hahatheanswer.jpg)
