The side wall of an open tank (wide) is connected to a tube whose cross –section is narrowing from $S_1$ to $S_2.$ The water level is $h$ higher. Neglecting viscosity, find the horizontal component of force tending to pull the tube out of the tank

 

$$\begin{gathered} \frac{R}{L_2}=\frac{r}{L_1}=\frac{y}{x} \\ y=r+\frac{R-r}{L_2-L_1}(x-L_1) \end{gathered}$$ 

Applying equation of continuity

$\pi R^2{\upsilon }_1=\pi r^2{\upsilon }_2=\pi y^2\upsilon \text{'};$  ${\upsilon }_{efflux}=\sqrt{2gh}$

$P_X+\frac{1}{2}\rho \upsilon {\text{'}}^2=P_{atm}+\frac{1}{2}\rho {\upsilon }_{efflux}^2$

Force on the nozzle tending to pull the tube is $F={\displaystyle \int \left(P_x-P_0\right)}\sin \theta 2\pi ydz$

$dz=dy\sec \theta$ and $\tan \theta =\frac{R-r}{L_2-L_1}$

$\therefore$ $F={\displaystyle \int \frac{1}{2}\left(v^2-u^2\right)\rho 2\pi y\frac{R-r}{L_2-L_1}dy}$

$=\rho \pi {\upsilon }^2{\displaystyle \underset{r}{\overset{R}{\int }}\left(1-\frac{r^4}{y^4}\right)ydy}$

$=\rho gh\left[\frac{\pi \left(R^2-r^2\right)}{R^2}\right]$