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Exam question that I have no idea how to solve (regarding forces and mass)

  1. Oct 14, 2014 #1
    1. The problem statement
    Peter threw 10 (same) coins into an empty container (a tube/drum), which was floating in A LIQUID. The diameter of the container was 2,257 cm, and so because the container is a tube the surface of the tube was equal to π times the diameter squared. He measured, after 4 to up to 10 coins, how much the tube sank deeper into the water, and displayed them on this chart (n = number of coins, Δh = change of height in centimeters).

    The issue I had here was mostly because it wasn't stated that the container was floating in water, but I assumed it did.

    n | 4 | 5 | 6 | 7 | 8 | 9 | 10
    Δh | 9,4 | 10,3 | 12 |13,49 | 15 | 16,9 | 17,6

    Q1: What is the mass of one coin?
    Q2: What is the mass of the container?

    2. The attempt at solution
    I know that the volume of water moved = the volume of coins we put in, but to calculate the mass wouldn't we have to know the density of the coins?

    ρ = m/V

    Now I am guessing that the denisty of the water moved = to the density of the coins moved so;

    ρ(H2O) = ρ(coins)

    ρ(H2O) = 1000 kg/m^3

    Now the second issue:
    I have no idea how to proceed with this, since from the change of height it seems like we are using different coins, since the height doesn't increase linearly, but I used the first 2 coins measured.

    I used the change of height between the 5th and 4th coin;

    h(coin) = 10,3 cm - 9,4 cm = 0,9 cm (not the same with every coin)

    The volume of the coin equals the volume of the pushed water so it equals to the height times the surface;

    V(coin) = π*r^2 * h
    = π*2,257cm^2 * 0,9cm
    = 14,4 cm^3 = 1,44 * 10^-5 m^3

    And so we use the density from above to calculate the mass of one coin:

    ρ(coins) = m(coin) / V(coin)

    m(coin) = ρ(coins) * V(coin)
    = 1000 kg / m^3 * 1,44 * 10^-5 m^3
    = 0,0144 kg = 14,4g

    This is where I finished, I do not know how to calculate the mass of the container. Feel free to also correct me on my way to this point, as I am sure I have went wrong somewhere.
     
  2. jcsd
  3. Oct 14, 2014 #2

    mfb

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    This is indeed bad, but the problem statement mentiones water later on.
    I guess this should be radius instead of diameter.
    The volume of the coins is not relevant. How do you calculate the volume of water moved for floating objects? Not with the volume of coins.
    You can draw a graph and estimate the change per coin. Taking a specific difference is quite imprecise. This graph will also allow to estimate the height at zero coins (just the container).
     
  4. Oct 30, 2014 #3

    Mark Harder

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    The formula the problem gives for the surface area would be the area of the sides of the tubular (i.e. cylindrical) container if the height of the cylinder = the circumference of the ends of it. The surface area = Pi D^2 because the cylinder is isotropic ( Can you derive this formula from the geometry of the cylinder?). I guess you are supposed to infer the form of the cylinder from the formula given for the surface area (of the sides). Pi r^2 is the surface area of one end of the cylinder, all of which is immersed. Does this affect the solution? Gotta go. Whoever wrote this problem is very sly. Good luck.
     
  5. Oct 30, 2014 #4

    NascentOxygen

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    Archimedes had something interesting to say about floating coins in a rubber duck, and contrasted it with dropping the coins directly into the bath. Well, it was something like that, I think. ;)
     
  6. Oct 30, 2014 #5

    haruspex

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    I don't believe there was any intent to imply a length of the tube; it's just very sloppily worded. It should say something like "the cross-sectional area of a tube is ##\pi r^2##", or maybe "the volume of a tube is ##length* \pi r^2##".
    With regard to the data set, there is a possible wrinkle associated with the 10 coin data point, but again, I don't think any such trap is intended.
     
  7. Oct 31, 2014 #6

    Mark Harder

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    Yeah, subject to interpretation. I didn't bother to work the whole thing out; maybe that would clarify which assumption needs to be made.
     
  8. Oct 31, 2014 #7

    mfb

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    We know the container is a cylinder and we know its diameter. The confusing surface value is irrelevant.
     
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