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Finding mass of a single screw lab! No calculations XD

  1. Nov 21, 2011 #1
    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
     
  2. jcsd
  3. Nov 21, 2011 #2
    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?
     
  4. Nov 21, 2011 #3
    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
     
  5. Nov 21, 2011 #4
    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?
     
  6. Nov 21, 2011 #5
    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:

    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.
     
  7. Nov 21, 2011 #6
    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"
     
  8. Nov 21, 2011 #7
    no teacher said it was a simple calculation! I think i'm overlooking it :(
     
  9. Nov 21, 2011 #8
    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.
     
  10. Nov 21, 2011 #9

    gneill

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    Staff: Mentor

    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.
     
  11. Nov 21, 2011 #10
    I agree with gneill... I am going to try it!
     
  12. Nov 21, 2011 #11
    wow i have no idea what you guys are trying to tell me! Example please :blushing:
     
  13. Nov 21, 2011 #12
    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!!!!
     
  14. Nov 21, 2011 #13

    gneill

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    Staff: Mentor

    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.
     
  15. Nov 21, 2011 #14
    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
     
  16. Nov 21, 2011 #15

    gneill

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    Staff: Mentor

    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.
     
  17. Nov 21, 2011 #16
    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: Nov 21, 2011
  18. Nov 21, 2011 #17
    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?
     
  19. Nov 21, 2011 #18
    Interesting that you say it reminds you of Millikan's oil drip experiment. How so?
     
  20. Nov 21, 2011 #19

    gneill

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    Staff: Mentor

    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.
     
  21. Nov 21, 2011 #20

    Delphi51

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    Homework Helper

    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.
     
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