finding mass of a single screw lab! No calculations XD

jayjay713

I have been given the masses of 5 cups, each with a different amount of screws inside. I am asked to find the mass of a single screw.

The cups themselves are all equal in mass and the screws are all equal in mass as well.

I subtracted all the cups from eachother to get a bunch of numbers. I know the common factor is supposed to be the mass of the screw but is there not a more accurate way!?

thanks

spacelike

Do you know how many screws are in the cups?

are the "masses of the 5 cups" including the mass of the screws contained within those cups, or just the mass of the empty cups?
If it is the former, can we assume the cups have equal masses?

jayjay713

I do not know the number of screws.

the masses of the 5 cups are all including the masses of the screws and yes all the cups are of equal mass.

cup1: 28.7g
cup2: 32.20g
cup3: 37.6g
cup4: 45.80g
cup5: 60.50g

jayjay713

think this works:
(c=mass of cup, v=bottle2-1=mass of screws more in bottle 2 then 1 and w=bottle3-1=mass of screws more in bottle 3 then 1, x is number of screws)

1) cup2 = c + vx
2) cup3 = c + wx
so
1) 32.20 = c + 3.5x
2) 37.60 = c + 8.9x

1) c = 32.20 - 3.5x
add into 2
2) 37.60 = (32.20 - 3.5x) + 8.9x

5.4 = 5.4x
x = 1g

This is all relative to bottle number 1! Think this is okay?

spacelike

Sorry for deleting my previous post, I had to rethink it a bit, but it was right.
I'll quote it here for the record:

Then I think the way you described was the best method, I don't think given that information that there is any way to deduce with 100% certainty what the mass of a single screw is.

consider the total masses (where cup1, cup2, etc.. are the values you provided, "c" is the unknown mass of an empty cup, and "s" is the unknown mass of a screw)
cup1=c+vs
cup2=c+ws
cup3=c+xs
cup4=c+ys
cup5=c+zs
v, w, x, y, and z are the number of screws in each cup.
You want to find solutions to those equations such that all v, w, x, y, and z are integer values. (unless you can have a fraction of a screw, but I'll just assume not :) )

Considering that, there are technically an infinite amount of solutions.

say you find solutions of v, w, x, y, and z, and "s" turns out to be some value S.

You can start there and find another solution quite easily, try "s"=S/2 and this will simply mean that all of the v, w, x, y, and z values are twice as much as they were before.. but if they were all integers before, then multiplying by 2 would mean they are still integers

Therefore that is also a valid solution.

If it can be solved exactly with a single solution then it will probably have to involve some other kind of measurements, such as moving half the screws from one cup to another and re-measuring the mass, and repeating this in such a way that you can deduce the mass of a single screw.

But, you cannot have a method that will give you a unique solution mathematically, because one does not exist with the information provided.

As for your last question, the 'v' and 'w' you used was not the same as the ones I used, just to be clear.

I don't think that works because the values you actually plugged in to find that answer were:
cup1=c+(cup2-cup1)x
cup3=c+(cup3-cup1)x
which doesn't make sense.

You would have to do:
cup2-cup1=(c+ws)-(c+vs)=(w-v)s=3.5
cup3-cup1=(c+xs)-(c+ws)=(x-w)s=8.9

So you have 4 unknowns and 2 equations. You need to involve more equations than this.

spacelike

Sorry if I am being a bit confusing.

I guess a better way to word it is, you have 5 equations, and 7 unknowns. It is not possible to find a unique solution to that system. You can actually find many solutions.

So unless your lab allows you to perform some other kind of measurements that will give you 7 different equations with those same 7 unknowns, then you can't find a unique solution.

However, if you are simply stuck with the 5 equations and 7 unknowns that you currently have then I am sure your lab instructor would take the largest value of "s" that gives a valid solution as the "correct answer"

jayjay713

no teacher said it was a simple calculation! I think i'm overlooking it :(

technician

I have not had a go at it yet but......would calculating differences between the sets of numbers be an approach?
It reminds me of millikans experiment to find the charge on an electron in the oil drop experiment. It was not known how many electron charges were on each drop but the smallest difference was taken to be the charge.

gneill

Algorithmically, you could:
1. Take the differences between each number in your current list
2. Sort these differences (ascending) to create a new current list. Throw away duplicates.
3. If minimum difference is smaller than last pass, goto 1.

Eventually the new minimum difference will be the same as the previous one and the loop ends. The minimum difference is then the best guess at the unit screw size obtainable from your given information.

technician

I agree with gneill... I am going to try it!

jayjay713

wow i have no idea what you guys are trying to tell me! Example please

technician

Each weighing = cup+(number of screws x mass of 1 screw)
mass1= cup +(n1 x m)
mass2=cup +(n2 x m) and so on.
If you take all the difference in your table of masses then you have only the masses of the screws. The trouble is you do not know the n1, n2 n3....etc
If you keep taking differences then you should end up with a smallest mass and that is the best estimate of the mass of 1 screw.... it is a tedious exercise!!!!

gneill

Quote by jayjay713

wow i have no idea what you guys are trying to tell me! Example please

The idea is, if you're dealing with numbers that are integer multiples of some value then any sums or differences of those numbers must also be integer multiples of that value.

When you first took differences between the given set of numbers you effectively tossed away the common weight of the cups and were left with "pure screw" weights. By taking differences again you will toss away another common chunk, leaving a new set of numbers which are smaller and represent smaller differences, yet they are all still multiples of the screw weight.

Every time you toss out common chunks of screws you're left with piles that are smaller in total number of screws, and the least of the differences between them should approach the smallest indivisible group of screws discernible from the data.

jayjay713

Okay I am going to try that now. I will have to do every possible combination of the 5 cups correct? in ascending order? :O

gneill

Quote by jayjay713

Okay I am going to try that now. I will have to do every possible combination of the 5 cups correct? in ascending order? :O

Yup. It would help to program the algorithm to save time. Bonus if the language has built-in sort routines you can call.

Probably do-able in Excel, too.

jayjay713

i got -198.8! in the 10th difference haha what... i did it on excel as well, same answer! what am i doing wrong?

jayjay713

YIIIIIIIIIIP i did it without the negatives and i got 0.9, 0.3, 0.9 .. which equals 0.6, 0.6. BUT if i use negatives for the last step it would be 0.6 - (-0.6) = 1.2 g! is that a good guess? or should i stick with 0.6?

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spacelike	Do you know how many screws are in the cups? are the "masses of the 5 cups" including the mass of the screws contained within those cups, or just the mass of the empty cups? If it is the former, can we assume the cups have equal masses?

jayjay713	I do not know the number of screws. the masses of the 5 cups are all including the masses of the screws and yes all the cups are of equal mass. cup1: 28.7g cup2: 32.20g cup3: 37.6g cup4: 45.80g cup5: 60.50g

technician	I have not had a go at it yet but......would calculating differences between the sets of numbers be an approach? It reminds me of millikans experiment to find the charge on an electron in the oil drop experiment. It was not known how many electron charges were on each drop but the smallest difference was taken to be the charge.