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Given a pool with dimensions $$\ell \times w \times h \, ,$$ I am trying to derive an equation that will yield the force by the water on the sides of the pool, namely $$\ell\times h \quad \mathrm{or} \quad w \times h \, .$$ For the side of the pool with dimensions $\ell \times h$, I started by using the familiar equation for pressure $$F = PA \, .$$ Plugging in the expression for hydrostatic pressure for $P$ gives $$F = \rho ghA =\rho gh(\ell \times h) = \boxed{\rho g \ell h^2} \, .$$ Is my reasoning, and corresponding solution correct?

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  • $\begingroup$ Hydrostatic pressure changes with height. You have just multiplied by area, which means that you have assumed it to be constant. Instead, you should integrate over the area. You'll get an extra 1/2 term for the force. $\endgroup$
    – Goobs
    Sep 15, 2015 at 4:21

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As @Goobs says, the pressure force is $0$ at the top of the water line and increases to $\rho~g~y~dA$ on a surface of area $dA$ at depth $y$. Since this pressure increases linearly from $0$ to $\rho~g~y$ the average force on the wall is the average of the start and end: so, it is half of this value, and the total pressure is $\frac 12 \rho g h (h \ell).$

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  • $\begingroup$ Would this be correct? $\int dF = \int_0^H\rho g A \,\,dh = \rho g\ell\int_0^H h\,\,dh = \boxed{\frac{1}{2}\rho g H^2}$ $\endgroup$
    – rgarci0959
    Sep 15, 2015 at 4:51
  • $\begingroup$ Yes. For bonus points you would write it as $\int dA~\rho~g~h$ to start with, as that's one of those forces that you "know" is correct (to get the net force in some direction, sum all the little forces in that direction). $\endgroup$
    – CR Drost
    Sep 15, 2015 at 5:03

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