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

Question

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

Answer 1

$F=\frac{dp}{dt}$ 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 $$ \begin{align} F_{tot}&=\rho\ g\int_{0}^{h}\left(\int_0^{2\pi R}dy\right) xdx\\ &=\rho\ g\ 2\pi R\int_0^{h}x dx\\ &=\pi R\ \rho\ g\ h^2 \end{align} $$