EPTHow to measure weight without a scale? - Page 2
| Forum Index > Closed |
|
Murlox
France1699 Posts
| ||
intrigue
Washington, D.C9934 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 ![[image loading]](http://i66.photobucket.com/albums/h254/luciferbw88/hahatheanswer.jpg) someone please explain why this made me laugh lol it's so stupid but great WTF | ||
|
GearitUP
United States337 Posts
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 | ||
|
hacpee
United States752 Posts
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. | ||
IntoTheWow
is awesome32277 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 | ||
|
Zortch
Canada635 Posts
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. | ||
|
sith
United States2474 Posts
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. | ||
|
gusbear
333 Posts
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. | ||
|
Steelflight-Rx
United States1389 Posts
Unless I missed something.. | ||
micronesia
United States24747 Posts
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) | ||
|
micronesia
United States24747 Posts
| ||
|
himurakenshin
Canada1845 Posts
| ||
|
micronesia
United States24747 Posts
| ||
|
fight_or_flight
United States3988 Posts
On December 17 2008 13:41 himurakenshin wrote: what is the suitcase filled with? A bunch of 1 lb diving weights. Why? | ||
|
micronesia
United States24747 Posts
| ||
|
micronesia
United States24747 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! | ||
|
micronesia
United States24747 Posts
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. | ||
|
thunk
United States6233 Posts
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. | ||
|
micronesia
United States24747 Posts
| ||
|
thunk
United States6233 Posts
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). | ||
| ||
