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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 $\rm kg\,m\,s^{-2}$, 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 http://adatbank.transindex.ro/vendeg/htmlk/pdf5017.pdf
AJP Journal

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  • $\begingroup$ It would probably be useful to provide the link to the journal article (or at least the author, title & year so an interested party my be able to search for it). $\endgroup$
    – Kyle Kanos
    Commented Sep 21, 2015 at 11:47
  • $\begingroup$ Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Commented Sep 21, 2015 at 12:38
  • $\begingroup$ Cross-posted to Engineering: engineering.stackexchange.com/q/5490 $\endgroup$ Commented Sep 21, 2015 at 13:57
  • $\begingroup$ I'm voting to close this question because it was crossposted. $\endgroup$
    – Qmechanic
    Commented Feb 25, 2017 at 17:19

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