// So I was given this physics project to do as a final ISU project. Engineers only!! :) //

[Martin]

You know what, if you you feel like doing tons of work..build your own digital scale!


take a spring made of conducting material and run electricity through it. The force of the object on the spring will change is resistance and you can use that to calibrate.

 

[CTho9305]

Putting the obejct on a boat lets you weigh it even if it is denser than water.

 

[Martin]

Originally posted by: brxndxn

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Originally posted by: brxndxn

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these water displacement suggestions are no good, that only gives you volume
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hahaha... what happens if you measure the mass of the water displaced? ... or, if you find the volume and asume that the mass of the water is 1gm/cm^3
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then you would know what the object would weigh if it were made of water

If the object weighs less than water, it will float. If it floats, you measure the volume of the water displaced. Mass of displaced water will equal mass of object.

what if it weighs more? A centrimeter cubed of uranium will displace the same amount of water as a centimeter cubed of magnesium.

 

[CTho9305]

Originally posted by: MartyTheManiak

Originally posted by: brxndxn

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Originally posted by: brxndxn

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these water displacement suggestions are no good, that only gives you volume
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hahaha... what happens if you measure the mass of the water displaced? ... or, if you find the volume and asume that the mass of the water is 1gm/cm^3
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then you would know what the object would weigh if it were made of water

If the object weighs less than water, it will float. If it floats, you measure the volume of the water displaced. Mass of displaced water will equal mass of object.

what if it weighs more? A centrimeter cubed of uranium will displace the same amount of water as a centimeter cubed of magnesium.

Not if you put it on a boat... the boat needs to be big enough that it doesn't sink. Then you look at how low it floats, get the water displaced, multiply by density, and there is your mass.

 

[Sukhoi]

So just get a boat that's more than 2000 cm^3 in size, right?

 

[Martin]

ah, thank you enlightening me!


that certainly seems like the most consistent method.

 

[Rhombuss]

Since I'm a civil engineer, I'll just talk about the two topics that really apply to what I'm doing .

The cantilever beam deflection is a neat idea. If you go indepth, you might come across statically indeterminate methods such as the Force Method, Displacement Method, or Moment Distribution procedures to calculate vertical deflection of cantilever supports. Typically you can relate the force application with overall mass, and I guess that could give you an estimation of what mass quantity you're dealing with. For fully determinate systems (and I guess this would apply since you know what the support values are), there are equations in almost any structural analysis textbook for cantilever deflection.

For the Young's Modulus (or what we like to call Modulus of Elasticity), that might be hard for anyone equipped with highschool tools. You're going to need some type of electronic gauge of shear stress and strain within the beam that you're testing on. Generally:

stress [MPa] = strain[mm/mm] X Elastic Modulus [MPa]

Each material has it's own elastic modulus, like steel has an E value of 200 GPa. So once you have a strain gauge, you can determine the stress at the bending point. Once you have the stress:

stress [MPa] = Force [N] / Area [mm²]

So you can derive the specific equivalent force to generate that stress, and ultimately getting the mass from that value.

Hope this helps Simon!
-Phillip

 

[Kelvrick]

Run a current through the pendulum. When the pendulum reaches either end, the current can flow from it into an electrical circuit, thereby giving you an accurate measurement of time.

 

[raptor13]

I'm a sophomore in mechanical engineering and I want to know why you wouldn't just use the compression or extension of a spring method? It doesn't even matter which you do, they're exactly the same.


Get a spring with a relatively low k value so hanging your mass from it will result in a fairly large extension or compression. Hang a known mass from the spring, measure the deflection as accurately as you can. Take the mass off. Put it back on. Measure again. Do that a few times. Use a different known mass. Do it a few more times. Do it as many times as you want to get a whole crapload of measured k values. They should all be extremely close to the same value if you do it carefully. Average all your k values together to get your final k value which you'll assume to be the real value. Hang your unknown mass, measure deflection, solve. Done. It doesn't get any simpler than that and because distance is so easy to measure, you should end up with remarkably accurate results.

Originally posted by: Sukhoi
Oh, IMHO (I chould be way wrong here) I would avoid springs and other deformable objects. Who knows how much your spring is going to change properties from when you're calibrating it to when you do the test.

Don't listen to this kid! He's only a freshman at my school and hasn't taken his materials class yet. So, like he said he might be, he's wrong. The reason for that is so long as your spring stretches elastically and isn't stretched beyond it's normal limits, a spring retains its properties almost indefinately. Think about the springs in a car... they compress and extend millions and millions of times throughout their life and they're still good. If you compress and/or extend your spring 50 times, you really think it'll make a difference?

The answer to that last question is no.

 

[WarCon]

I like the displacement experiment, but am wondering if the teacher wants you to use a known constant like the weight of water for example. I mean a simple hanging balance scale is easy enough to construct if you say know the weight of a known volume of water. Construct a scale that is in perfect balance with a container on one side that will hold up to 2000 grams of water. Set the mass on one side and slowly fill the container with accurately measured amounts of water until it balances. Then just figure the weight of the water = weight of unknown object.

 

[WarCon]

Originally posted by: raptor13
I'm a sophomore in mechanical engineering and I want to know why you wouldn't just use the compression or extension of a spring method? It doesn't even matter which you do, they're exactly the same.


Get a spring with a relatively low k value so hanging your mass from it will result in a fairly large extension or compression. Hang a known mass from the spring, measure the deflection as accurately as you can. Take the mass off. Put it back on. Measure again. Do that a few times. Use a different known mass. Do it a few more times. Do it as many times as you want to get a whole crapload of measured k values. They should all be extremely close to the same value if you do it carefully. Average all your k values together to get your final k value which you'll assume to be the real value. Hang your unknown mass, measure deflection, solve. Done. It doesn't get any simpler than that and because distance is so easy to measure, you should end up with remarkably accurate results.

Originally posted by: Sukhoi
Oh, IMHO (I chould be way wrong here) I would avoid springs and other deformable objects. Who knows how much your spring is going to change properties from when you're calibrating it to when you do the test.

Don't listen to this kid! He's only a freshman at my school and hasn't taken his materials class yet. So, like he said he might be, he's wrong. The reason for that is so long as your spring stretches elastically and isn't stretched beyond it's normal limits, a spring retains its properties almost indefinately. Think about the springs in a car... they compress and extend millions and millions of times throughout their life and they're still good. If you compress and/or extend your spring 50 times, you really think it'll make a difference?

The answer to that last question is no.

The only issue I see with using a single spring would be the great variance in weight he is measuring 100g-2000g. If he wants an accuracy of 1 gram then he will need the spring to be able to stretch a measureable distance with the 100g weight and then only stretch 190 more measuring units with the large weight, since he could be measuring anywhere in-between. If you use a nominal value of 12 inches of stretch you have to measure finer than 1/10 of an inch to even have 1 gram accuracy.

Not sure if you teacher would allow this, but if you could applie a series of springs (with progressive k values) to the same weight and they all have an extension of 12 inches. You would calculate your mass from the springs (springs) with the larger extensions. A fine spring will more accurately measure a smaller weight, but a heavier weight would distort the spring. A heavier spring for heavier weights. If you can only use one spring, then it will have to handle the 2kg mass (since you don't know which one you will get)

If the teacher will only allow one spring to be used then you could use a moveable pivot so that you can adjust the moment (ratio) of the balance arm. If you put a stop so the spring can never be over extended (keep from skewing results from deformed spring) and adjust the pivot with the weight in its dangling basket until the spring is extended to a set distance. This will increase your accuracy. You would just throw the ratio of the pivot arm in with the distance and K to get your weight.

Edit : Now with basic pic........Experiment.jpg

 

[zephyrprime]

Use the compression of a spring. The compression of a spring is the same method used by real scales that people use when they weigh themselves.

The ballistic pendulum is accurate enough that it's used in grandfather clocks but it's kinda tough to measure the periods if you're stuck using stop watch. You'd need some sort of automatic method to timing to attain good accuracy.
If the teacher gives you some sort of object to weigh that's really un-aerodynamic, that'd screw it up too.

Don't use hooke's law. Ever use a scale that uses hooke's law? I have. They suck. (actually though, compression of a sping is really hooke's law too.)

Don't use young's modulus. The compressions/expansions involved would be too small for you to accurately measure.

Intertial system: hard to measure accurately. Hard to delivery precision amounts of impulse.

As far as the compound spring goes, I'm not really sure I understand it. It seems to me like it'd be essentially the same as a single spring.

It'd be interesting if your teacher gave you something really light to weight like a piece of paper. Hmmm...I wonder what the best way to deal with that would be. Probably the pendulum if you were allowed to crumple up the piece of paper.

 

[jamesave]

use hooke's law

find a spring that has k which can determine the x with weigh 100-200g
(eg. if you need to be exact within 1 g- make sure you have 1g/10mm)

use known masses to "marks" it between 100-200g.

you prolly can get the spring in scientific catalog, which usually pre-calibrated by them also (with estimated k value)

 

[simms]

I had already thought of using Hooke's law as my first method of doing so, and I already know how to calibrate, etc etc. What I need to account for is the mass of the SPRING ITSELF - say I use a SHM (Simple Harmonic Motion) with Hooke's Law, eg: The spring is held by a retort stand on a table, with the spring dangling below.

The camera seems like a good idea and is feasible, but wouldn't the measure of how high I adjust my tripod still have parallax? If I use a camera I still can't expect to get it dead on.

By Hooke's Law, do you mean COMPRESSION of the spring or EXPANSION of the spring? I only did expansion because I figured compression would be too hard to find extension for.

-Simon

 

[simms]

BTW there are SO many replies I will have to come home tonight and read them all, I gotta go for school now, but up for the morning engineering crew

 

[Rhombuss]

Simon,

About the use of the Elastic Modulus - since your highschool PROBABLY doesn't have the equipment inorder to properly conduct the experiment. I could hook you up with a TA or Prof at the Civil department down at UT to book some lab time for the project. We have some structural testing labs that deal with cantilever deflections and/or strain gauge measurement from applied loads.

Let me know if you're interested and I'll give you some contacts.

 

[raptor13]

Originally posted by: simms
I had already thought of using Hooke's law as my first method of doing so, and I already know how to calibrate, etc etc. What I need to account for is the mass of the SPRING ITSELF

No you don't. Hang the spring up, let it completely stabalize, then make all your measurements from where the end of the spring is when hanging by itself. It's only change in displacement that matters with a spring. Total displacement is irrelevant here.

 

[simms]

WarCon has a pretty good idea, but I'm not sure on how it would work (especially the ratio bit). Anyone can clarify?

 

[jamesave]

Originally posted by: simms
I had already thought of using Hooke's law as my first method of doing so, and I already know how to calibrate, etc etc. What I need to account for is the mass of the SPRING ITSELF - say I use a SHM (Simple Harmonic Motion) with Hooke's Law, eg: The spring is held by a retort stand on a table, with the spring dangling below.

The camera seems like a good idea and is feasible, but wouldn't the measure of how high I adjust my tripod still have parallax? If I use a camera I still can't expect to get it dead on.

By Hooke's Law, do you mean COMPRESSION of the spring or EXPANSION of the spring? I only did expansion because I figured compression would be too hard to find extension for.

-Simon

I'd go with expansion.
Since the mass is 100-200g, and the calibration should be done many times (like at least 10 times, the more the better) you should have a grip on what the resonances when you put the mass on. OTOH, your hand should be crafty already to do it. To avoid paralax, you can put a horizontal (straight line) mark at the end of the spring where you hook the mass. This mark should be used during your calibration, and make sure it doesn't move, especially during the test time!

I think there is also cathetometer that can help with the marking/paralax-
see http://www.titantoolsupply.com/tc.html or google it.

 

[simms]

To avoid paralax, you can put a horizontal (straight line) mark at the end of the spring where you hook the mass. This mark should be used during your calibration, and make sure it doesn't move, especially during the test time!

Already thought of that one! Thanks again for the reminder though.

I think there is also cathetometer that can help with the marking/paralax

How much is that? I doubt our local high school would have something like that.

 

[Sukhoi]

Originally posted by: raptor13
I'm a sophomore in mechanical engineering and I want to know why you wouldn't just use the compression or extension of a spring method? It doesn't even matter which you do, they're exactly the same.


Get a spring with a relatively low k value so hanging your mass from it will result in a fairly large extension or compression. Hang a known mass from the spring, measure the deflection as accurately as you can. Take the mass off. Put it back on. Measure again. Do that a few times. Use a different known mass. Do it a few more times. Do it as many times as you want to get a whole crapload of measured k values. They should all be extremely close to the same value if you do it carefully. Average all your k values together to get your final k value which you'll assume to be the real value. Hang your unknown mass, measure deflection, solve. Done. It doesn't get any simpler than that and because distance is so easy to measure, you should end up with remarkably accurate results.

Originally posted by: Sukhoi
Oh, IMHO (I chould be way wrong here) I would avoid springs and other deformable objects. Who knows how much your spring is going to change properties from when you're calibrating it to when you do the test.

Don't listen to this kid! He's only a freshman at my school and hasn't taken his materials class yet. So, like he said he might be, he's wrong. The reason for that is so long as your spring stretches elastically and isn't stretched beyond it's normal limits, a spring retains its properties almost indefinately. Think about the springs in a car... they compress and extend millions and millions of times throughout their life and they're still good. If you compress and/or extend your spring 50 times, you really think it'll make a difference?

The answer to that last question is no.

Hey just because I haven't taken ME 231 yet doesn't mean that I can't pretend that I have!