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Knocking the bottom of the bottle.

  1. Jun 10, 2007 #1
    1. The problem statement, all variables and given/known data

    If vitreous bottle is full of water and we strike with a palm from the top to the bottleneck, we can knock out the bottom of the bottle. Why? How to evaluate the force which knocks out the bottom of the bottle?


    3. The attempt at a solution


    Because water has constant pressure:

    Pressure on the top = Force applied / neck area​
    =​
    Pressure on the bottom = Force bottom/ bottom area​
    Force bottom = Force top * bottom area / neck area​

    So we knock out the bottom of the bottle because the force gets magnified. Now I need to evaluate "Force top".

    Let's say that the mass of the palm is M, and it moves with speed v before strike. The strike takes time moment which is equal to t.

    So then this means that:

    Force top = M*v/t​

    And we get that:

    Force bottom = M*v/t * bottom area / neck area​

    So is the "Force bottom" evaluated correctly in this problem? Don't I miss anything?

    Thanks in advance.
     
  2. jcsd
  3. Jun 10, 2007 #2

    G01

    User Avatar
    Homework Helper
    Gold Member

    Its looks pretty good. Your pressure equation is right. Then you find the acceleration and force on the hand, which has to be equal to the force on the top of the bottle:

    For the hand:
    v_i=-v
    v_f=0
    t=t

    [tex]v_f=v_i+at[/tex]

    so [tex] a =\frac{-v_i}{t} =\frac{v}{t}[/tex]

    Then the force on the hand becomes:

    [tex]F=ma=\frac{mv}{t}[/tex]

    Which is equal to the force on the top of the bottle by Newton's Third Law.
    Yup looks good to me.
     
    Last edited: Jun 10, 2007
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