A smooth gate is kept in equilibrium by applying a horizontal force. What is the value of y so that no horizontal reaction force acts at the pivot?

The thrust force due to pressure act at center of gravity'

$\therefore F_{th}=PA=\left(\rho g\frac{h}{2}\right)\left(hw\right) \left[here w=width of gate\right]$

For translational equilibrium : 

 $$\begin{gathered} F_{th}+R_x=F \\ If horizontal reaction force is absent, R_x=0 \\ \Rightarrow F_{th}=F \end{gathered}$$

For rotational equilibrium about hinge point :

 $$\begin{gathered} Torque due to F_{th}=Torque due to F \\ \Rightarrow {\int }_0^h\left(\rho g\left(h-y\text{'}\right)\right)\left(wdy\text{'}\right)\left(y\text{'}\right)=\left(F\right)\left(y\right) \\ \Rightarrow \rho gw{\int }_0^h\left(hy\text{'}-y{\text{'}}^2\right)dy\text{'}=\left(F\right)\left(y\right) \\ \Rightarrow \rho gw\left(\frac{h}{6}\right)=\left(F\right)\left(y\right) \\ \Rightarrow \left(\rho g\frac{h}{2}\right)\left(hw\right) \left(\frac{h}{3}\right)=\left(F\right)\left(y\right) \\ \Rightarrow \left(F_{th}\right)\left(\frac{h}{3}\right)=\left(F\right)\left(y\right) \\ \Rightarrow y=\frac{h}{3} \end{gathered}$$

The center of liquid thrust force passes through $y=\frac{h}{3}$