Equating $\frac{dp}{dt}$ and $F$ at a hydraulic jump?

A journal I am reading about circular hydraulic jumps provides the equation and I quote:

$H$ Depth after hydraulic jump, $h$ depth before hydraulic jump, $V$ velocity after hydraulic jump, $v$ velocity before hydraulic jump, $R$ radius of hydraulic jump.

The condition for conservation of momentum can be written as $$\frac{dp}{dt} = 2\pi R\rho HV^2 - 2\pi R\rho hv^2 = F_1-F_2$$ where $$F_1 = 2\pi R \rho g \int_0^hx\cdot dx = \rho g\pi Rh^2$$ $$F_2 = 2\pi R \rho g \int_0^Hx\cdot dx = \rho g\pi RH^2$$

I understand that units for $F$ and $\frac{dp}{dt}$ are both $kg\ m\ s^{-2}$, and that they are equivalent.

But I don't know how, in this example, $F_1$ and $F_2$ are derived i.e. where $g$ and the integrals come from. Could somebody please explain?

Page 4 from: On the circular hydraulic jump

$F=\frac{dp}{dt}$ is, in some sense, a more fundamental expression of Newton's law than $F=ma$ because $F=ma$ doesn't allow for situations with changing mass. You can easily derive the second from from the former by using $p=mv$ which gives $F=m\frac{dv}{dt}=ma$.

Your question seems to be more about how to derive the expressions for the force at the boundary of the hydraulic jump. To do this, think about the hydraulic jump as a wall; on one side of the wall the height of the water is $h$ and on the other it is $H$.

Now consider an infinitesimal square in this wall at depth $x$; the pressure exerted on this square is given by $$P=\rho\ g\ x.$$ The force exerted on the square is calculated by multiplying by the surface area of the infinitesimal patch $$F_{inf}=\rho\ g\ x\ dx\ dy.$$ Now you just have to sum up (with integration) all of the patches on the wall. The y direction is easy \begin{align} F_{tot}&=\rho\ g\int_{0}^{h}\left(\int_0^{2\pi R}dy\right) xdx\\ &=\rho\ g\ 2\pi R\int_0^{h}x dx\\ &=\pi R\ \rho\ g\ h^2 \end{align} • Thank you for the outstandingly clear explanation! I had wrongly assumed that the equations for $F_{tot}$ were derived from the momentum conservation equation; but instead are quite separate. Would you mind recreating this answer for my identical question on the physics stackexchange site (which has yet to produce such a good answer)? – MadCommy Sep 21 '15 at 13:30
• @MadCommy I'm glad that you found my answer helpful, you can accept it if you feel that it fully answered your question. Cross-posting questions to multiple different sites is generally frowned upon because it ends up fragmenting the question. I won't repost my answer at Physics.SE, but I have added links to both questions so that others are aware of both. In the future please try to choose the appropriate site and only post a single question. – Chris Mueller Sep 21 '15 at 14:05
• Thanks for the advice; I wasn't sure if my question was more suited for physics or engineering as, while being an engineering topic, I was researching it for my high-school extended essay in physics. – MadCommy Sep 21 '15 at 14:18
• No problem. I would say that fluid mechanics questions are generally better received here at Engineering.SE. Physicists, surprisingly, don't study fluid mechanics as part of the standard curriculum. – Chris Mueller Sep 21 '15 at 14:20
• Why are both $V$ and $v$ independent of the depth $x$ while the pressure is? – Hans Apr 30 at 7:52