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    Find the flaw in the argument.



    (Answer later in the week.)
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    My first reaction:
    You applied the formula pressure=hdg to A, but this is only aplicable to columns of liquid which A isn't.

    I don't know whether it is right. It seems too simple...
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    Stephen Hawking is yum. :borat:
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    (Original post by Stonebridge)
    Find the flaw in the argument.


    hey great question but please dont give the answer yet i'm gonna give u my answer tomorrow -wednesday about 4:30 pm if u can wait please thanks its just i have a lot of work. thanks
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    i also agree with the first post. to explain it, it could be that some of the force is dissipated on the sides of the container because they are not straight
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    The forces could be the same and yet the pressure would still be different
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    the force pushing down on the table isnt the entire force, there is a reaction force from the container supporting some of the weight of the water i think so the base of the sloped container isnt supporting all the water because the total vertical force mg which has to be balanced is done so by the vertical components of the normal reaction as well as the reaction force at the bottom
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    When you consider the force/unit area you consider it at a specific height in the fluid, so the total force ends up being the same if you sum it over the height of fluid.
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    (Original post by Stonebridge)
    Find the flaw in the argument.



    (Answer later in the week.)

    The false assumption is that the force only acts on the base, where it actually acts on the side walls too.

    (need to revise this answer, I'm sure it's something along these lines though)
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    (Original post by trm90)
    When you consider the force/unit area you consider it at a specific height in the fluid, so the total force ends up being the same if you sum it over the height of fluid.
    This is what I was trying to get at!
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    (Original post by Stonebridge)
    Find the flaw in the argument.



    (Answer later in the week.)
    Yes the replies were mostly correct or very near.
    The flaw in the argument is the statement that it is just the force pA on the base that would equate to the weight of the one on the left.
    This would only equate to the weight of the mass of water directly above the base, and not include the mass above the sloping sides.
    Another way of looking at it, is that the pressure acts not only on the base, but also the sloping sides of the left container. This pressure, and the force on the sides, has a vertical component which will add the the overall downwards force.
    The pressure on the sides of the container on the right has no vertical component.

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    Additional Question: How would you measure the force on the sloping sides? :P
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    (Original post by BrilliantMinds)
    Additional Question: How would you measure the force on the sloping sides? :P
    Not directly, as the weight of the two containers is the same. But this would work...

    Spoiler:
    Show

    Force on sides of A = hdg(AB - AA)

    where you would measure the difference in the area of the bases of the two containers.

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    (Original post by trm90)
    When you consider the force/unit area you consider it at a specific height in the fluid, so the total force ends up being the same if you sum it over the height of fluid.
    That's exactly what I was thinking, since the area changes constantly in the first container.
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    (Original post by Stonebridge)
    Not directly, as the weight of the two containers is the same. But this would work...

    Spoiler:
    Show

    Force on sides of A = hdg(AB - AA)

    where you would measure the difference in the area of the bases of the two containers.

    That would only give the downward component. Divide that by cos theta, where theta = angle of inclination of side with bottom surface, and you get the full force. (Theoretical)

    You can poke a hole, from the range calculate backwards the acceleration then the mass flow rate and finally the force.
    (Experimental)
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    The side is slanted at an angle \theta. Consider poking a hole of area A at height h from the base.


    http://img852.imageshack.us/i/waterspray.jpg/
     F = v \frac{dm}{dt}

    and

     \frac{dm}{dt}= pAv

    so F = pAv^2

    Now to find v in terms of x. Consider a water molecule undergoing projectile motion.

    x = vT \cos \theta where T = time of flight

    From conservation of energy,
    \sqrt {2gh} = gT

    so x = v \cos \theta \sqrt{2h/g}

    Finally,

    F = \frac {pAgx^2}{2h \cos^2 \theta}

    Therefore, it can be seen that F is proportional to \frac{x^2}{h}.
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    (Original post by BrilliantMinds)
    That would only give the downward component.
    The original question was about the weight of the container and the downward forces. I assumed that's also what you were referring to.
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    I wonder, is it possible to find the total force using integration?

    I'm trying to construct an equation in the form:

    F = \displaystyle \int_0^H P(h) dA and to integrate the area elements along the height but I can't find a suitable expression for the area element. I was thinking a bunch of discs with an infintesimal thickness dh and their radius increases from r = r' (radius of bottom of container) to r = R (radius at top of container), but couldn't get anywhere.
 
 
 
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