I need your help on this fluid mechanics problem. Problem Statement: An open tank has a vertical partition and contains gasoline (ρ=700 kg/m3 ) to the right at a depth of 4m as sketched. The partition contains a rectangular gate that is 4m high and 2m wide and is hinged at the bottom with a stop at the top against outflow of gasoline. Water is slowly added to the left side of the tank that was originally empty. At what depth h will the gate start to open? Diagram: FBD: So my friend already worked out this analysis for me: FRg = γghcgAg where g refers to gasoline.. FRg = (700 kg/m^3)(9.81 m/s^2)(2m)(4m * 2 m) = 110*10^3 N = 110 kN FRw = γwhcwAw where w refers to water... FRw = (9.80*103 N/m^3)(h/2)(2m * h) where h is depth of the water... FRw = (9.80*103)h^2 For Equilibrium, ΣMH = 0 so that, FRw*lw = FRg*lg with lw = h/3 and lg = 4/3 m Thus, (9.80*10^3)(h^2)(h/3) = (110*10^3 N)(4/3 m) = h = 3.55 m This all makes sense to me except one part. Where does "with lw = h/3 and lg = 4/3m" come from? How are these moment arms derived? Thanks to anyone who can help.