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An Open Cylindrical Tank of Acid...

  1. Apr 3, 2016 #1
    1. The problem statement, all variables and given/known data
    An open cylindrical tank of acid rests at the edge of a table [itex]2.20\cdot 10^0\ m[/itex] above the floor of the chemistry lab. If this tank springs a small hole in the side at its base, how far from the foot of the table will the acid hit the floor if the acid in the tank is [itex]7.00\cdot10^{-1}\ m[/itex] deep?
    Let [itex]\rho[/itex] be the mass density of acid, [itex]g[/itex] be the acceleration of gravity, [itex]h_{tank}[/itex] be the depth of the tank, and [itex]h_0[/itex] the height of the base of the tank.

    2. Relevant equations
    [itex]p_1+\rho g y_1+\frac{1}{2}\rho {v_1}^2=p_2+\rho g y_2+\frac{1}{2}\rho {v_2}^2[/itex] (Bernoulli's Equation)

    3. The attempt at a solution
    [itex]\Delta x = vt[/itex]

    [itex]\frac{1}{2}gt^2=h_0[/itex]
    [itex]t^2=\frac{2h_0}{g}[/itex]
    [itex]t=\sqrt{\frac{2h_0}{g}}[/itex]

    [itex]\rho g h_{tank}=\frac{1}{2}\rho v^2[/itex]
    [itex]gh_{tank}=\frac{1}{2}v^2[/itex]
    [itex]v^2=2gh_{tank}[/itex]
    [itex]v=\sqrt{2gh_{tank}}[/itex]

    [itex]\Delta x=\sqrt{gh_{tank}} \sqrt{\frac{2h_0}{g}}[/itex]
    [itex]\Delta x=\sqrt{\frac{2gh_{tank}h_0}{g}}[/itex]
    [itex]\Delta x=\sqrt{2h_{tank}h_0}[/itex]
    [itex]\Delta x=\sqrt{2\cdot 7.00\cdot 10^{-1}\ m\cdot 2.20\cdot 10^0\ m}[/itex]
    [itex]\Delta x=\sqrt{3.08\cdot 10^0\ m^2}[/itex]
    [itex]\Delta x = 1.75\cdot 10^0\ m[/itex]

    Unfortunately, [itex]1.75\ m[/itex] is not the correct answer. Perhaps I am not setting up Bernoulli's equation correctly for this problem.
     
  2. jcsd
  3. Apr 3, 2016 #2

    SteamKing

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    Staff Emeritus
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    Homework Helper

    No, the problem is not with the Bernoulli equation.

    You have:

    [itex]t=\sqrt{\frac{2h_0}{g}}[/itex]

    and

    [itex]v=\sqrt{2gh_{tank}}[/itex]

    when you combine these two equations:

    [itex]\Delta x=\sqrt{gh_{tank}} \sqrt{\frac{2h_0}{g}}[/itex]

    you dropped one of the factors of 2 under the radical sign for some reason.
     
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