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Archimedes' Principle- Popping lid off of a barrel

  1. Dec 3, 2008 #1
    1. The problem statement, all variables and given/known data
    A barrel with diameter 80cm is completely full of water and sealed except for a piece of tube extending vertically through the lid with a diameter of 0.8cm. It's asking what height would you have to fill the small tube with in order to make the lid of the barrel pop off. The upward force required to do this is 450N.

    2. Relevant equations
    Boyant F= density (fluid) x A x h x gravity=density (fluid) x gravity x height

    3. The attempt at a solution
    I tried finding the areas of both the barrel and the tube, but I may be getting confused with the units. I got: A(barrel)=10053.09; A(tube)=4.021 by using 2Pi x r^2, and from the P=F/A formula, i got the pressures to be: P(barrel)=.01119 and P(tube)=111.9. From here I'm kind of lost. I tried to plug these numbers into the density x g x h equation, but they came out wrong. Any help as to where I should go from here (or how to start it if it's that bad) would be much appreciated.
  2. jcsd
  3. Dec 3, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Forget Archimedes' principle for this one. Instead, just think in terms of pressure. What does the pressure have to be inside the barrel (at the top) to pop off the top?

    Once you know the water pressure that you need, then figure out how high the column of water in the tube must be to create such a pressure.
  4. Dec 4, 2008 #3
    Ok, I plugged the numbers into the Pressure=Force/Area [450/(.8m^2)(Pi)] for the barrel and got the necessary pressure to be 223.81 pa.
    Then to find what force would be required in the tube to cause this, I used the same P=F/A formula, to solve for F [223.81=F/((.008m)^2 x Pi), and got it to be 0.04512N (required to cause enough pressure in the barrel, which sounds reasonable).
    I plugged this into the classic F=ma equation to find the mass of the water needed to cause this force [0.045124901=9.8m] and got m to be .004604582kg.
    Since the density of water is given as 1000 kg/m^3, I used density to find the V this water would take up [p=m/V] and got .000004605 M^3.
    Using V=Area x height [.000004605 M^3= (.000201062 M^2) x h] and got h to be 0.022901M, or 22.901cm (what answer is supposed to be in).
    I was wondering if this looked like a reasonable method for solving this problem (and if I made any simple errors) and if there is a simple way that I could check my answer?
  5. Dec 4, 2008 #4

    Doc Al

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    Staff: Mentor

    OK, except that 80 cm is the diameter of the barrel, not the radius. Correct that.
    That's the hard way. (Too many opportunities for error.)

    All you need to do is find the height of the column of water needed to produce that pressure. How does pressure in a fluid depend on height?
  6. Dec 4, 2008 #5
    Thanks, I used the P=density x height x gravity and got the same answer. Thank you very much!
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