Finding mass of a single screw lab! No calculations XD

  • Thread starter jayjay713
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  • #1
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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
 

Answers and Replies

  • #2
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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?
 
  • #3
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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
 
  • #4
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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?
 
  • #5
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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.
 
  • #6
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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"
 
  • #7
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no teacher said it was a simple calculation! I think i'm overlooking it :(
 
  • #8
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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.
 
  • #9
gneill
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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.
 
  • #10
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I agree with gneill... I am going to try it!
 
  • #11
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wow i have no idea what you guys are trying to tell me! Example please :blushing:
 
  • #12
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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!!!!
 
  • #13
gneill
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wow i have no idea what you guys are trying to tell me! Example please :blushing:
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.
 
  • #14
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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
 
  • #15
gneill
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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.
 
  • #16
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i got -198.8! in the 10th difference haha what... i did it on excel as well, same answer! what am i doing wrong?
 
Last edited:
  • #17
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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?
 
  • #18
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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.
Interesting that you say it reminds you of Millikan's oil drip experiment. How so?
 
  • #19
gneill
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There should be no negatives if you sort the list each time and just take differences between items forward in the list and the current item. Of course you could always take the absolute value of the differences, but I think you're generating a lot more difference pairs than required. No matter, if you're discarding duplicates.

With the given data I think the algorithm should converge on 0.1. You can check the result by dividing each of the original screw pile weights by your proposed value to see whether or not it dives each evenly.
 
  • #20
Delphi51
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Yes, all the differences will converge on 0.1 . . . because that is the last significant digit in all the data! Some idea of the accuracy of measurement is necessary here, isn't it?

It is like the Millikan experiment where he got a lot of values for charges on oil drops, then deduced the charge on the electron from them by this method. In the high school version of that experiment, you measure the voltage necessary to suspend the oil drop and the drop size, from which you get mass. So Fe = Fg, mg = qV/d. You then graph mg vs V/d for many oil drops. Good data forms a series of lines on the graph with slopes e, 2e, 3e, etc. If most of the points fit one of the lines within the error bar, you can reach a conclusion.

Another approach possibly worth trying is to "fit" the first differences to a formula M = n*Mo, with a computer trying many values of Mo and calculating the sum of the differences between the data and the nearest n*Mo value. If you get a sum of difference that is less than the number of points values * estimated experimental error, then you have a solution.
 
  • #21
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3.5 1.9 0.9
5.4 2.8 -2.1
8.2 0.7
8.9

see what I mean? absolute value algorithm ends with 0.6
0.6

how do i finish it and get the mass? is 0.6 grams the mass?:S
 
  • #22
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Well the point I was making still stands I think.. Correct me if I am wrong.

But you can do all that and no matter what answer you get, you can always ask "but how do I know the mass of the screw isn't half of what I just got and each cup has twice as many screws as I am predicting now?"

I showed it mathematically that you still get a solution. and you can keep cutting them in halves, probably thirds or whatever you want, and there are any number of solutions.

Although I am betting that the largest value of the screw mass which is a solution is probably the answer they are looking for.
 
  • #23
gneill
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3.5 1.9 0.9
5.4 2.8 -2.1
8.2 0.7
8.9

see what I mean? absolute value algorithm ends with 0.6
0.6

how do i finish it and get the mass? is 0.6 grams the mass?:S
0.6 does not divide the values evenly. For example, 3.5/0.6 = 5.833. So 0.6 is not the unit mass. The algorithm I described zeroes in on the only value that works, 0.1, which, as Delphi51 pointed out, also happens to be the limit of accuracy of the data values in this case (it's the "last resort" value when nothing larger can be found). Perhaps there is a bug remaining in your implementation if it doesn't perform likewise.

There is no value greater than 0.1 that divides them all evenly, and no information is to be had for values less than 0.1 because that represents the decimal accuracy of our values (only one decimal place). All you can say is that the mass is either 0.1 or some unit fraction of that (0.1/n for n = 1,2,3,4,...).

Delphi51 also suggested a best fit strategy where the sum of the differences between the values and a closest integer multiple of some test value is minimized. In other words, try out a range of test values and for each of the screw piles, calculate and sum the difference between the screw pile mass and the nearest integer multiple of the test value. I wrote something similar that would calculate the square root of the sum of the squares of the differences, and the result was the same: the only values that "hit pay dirt" were 0.1 and its fractions (0.1/n for n = 1,2,3...).

I don't know what more to suggest. I'm not seeing any evidence of the "simple calculation" that your teacher suggested!
 
  • #24
Delphi51
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I did a little trial and error on a spreadsheet. My formulas are shown in the upper image, and the results in the lower image. I modeled the Mass of a cup of screws as m*n+b, where b is the mass of the cup, m the mass of a screw and n the number of screws in the cup. I played with the m and b numbers and watched the average deviation of the model from the given masses.
screws.jpg

Curiously, I found almost immediately a "sweet spot" of sorts at the values shown above. If you make m or b slightly different, you get a larger average deviation.
 
  • #25
PeterO
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i got -198.8! in the 10th difference haha what... i did it on excel as well, same answer! what am i doing wrong?
Clearly a mistaken formula in there somewhere???

I get an extremely simple, perhaps obvious answer - though I cannot determine the mass of a cup.
 

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