 Solution

Manning's uniform-flow law implies the friction slope $S_f=n^2{u}^2/R^{4/3}$ for positive downstream flow. In a wide rectangular section $A\simeq wh$ and $R\simeq h$, so $\tau l/(\rho A)=g{n}^2{u}^2/h^{4/3}$.

Scale distance with $L$, depth with $h_0$, velocity with $u_0=h_0^{2/3}{S}^{1/2}/n$, and time with $L/u_0$. Divide momentum by $gS$ to obtainThe factor $ϵ$ compares depth change over $L$ with the mean bed drop $SL$, rather than with $L$ alone. Its product with $F^2$ measures inertial acceleration relative to bed-slope forcing.