The cross section of a 0.5 m wide vertical gate holding water and oil is shown in the figure. The unit weights of water and oil are 10 kN/m3 and 7.5 kN/m3, respectively.
(Figure not to scale)
The horizontal hydrostatic force (in kN) acting on the vertical gate is (rounded off to two decimal places).
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Split the gate into the oil zone (triangular pressure) and the water zone (trapezoidal pressure, since the oil above adds a constant surcharge pressure on top of the water's own hydrostatic pressure), then add the two forces.
Step 1: Convert the oil layer into an equivalent depth of water. A common trick for layered fluids is to replace the top fluid by an equivalent extra depth of the bottom fluid that would produce the same pressure at the interface. The oil produces a pressure of $\gamma_{oil} h_{oil} = 7.5 \times 0.5 = 3.75$ kPa at the interface. Expressed as an equivalent depth of water: \[ h_{eq} = \frac{\gamma_{oil}\,h_{oil}}{\gamma_{water}} = \frac{3.75}{10} = 0.375 \text{ m} \] So the water zone behaves as if it starts $0.375$ m higher than it really does, i.e. as if we had a $1.375$ m deep water column measured from the equivalent top.
Step 2: Get the pressure prism for the water zone using this equivalent depth. Pressure at the real top of water $= \gamma_{water} h_{eq} = 10 \times 0.375 = 3.75$ kPa (matches the surcharge found from the oil directly, confirming the equivalence). Pressure at the bottom $= \gamma_{water}(h_{eq}+h_{water}) = 10 \times 1.375 = 13.75$ kPa.
Step 3: Compute the water force as a trapezoid area geometrically. Split the trapezoid into a rectangle (from the constant $3.75$ kPa carried through the full 1.0 m) plus a triangle (from the water's own weight buildup of $10$ kPa over 1.0 m): \[ F_{water} = (3.75 \times 1.0 \times 0.5) + \left(\frac{1}{2}\times 10 \times 1.0^2 \times 0.5\right) = 1.875 + 2.5 = 4.375 \text{ kN} \]
Step 4: Add the oil force found the same way. \[ F_{oil} = \frac{1}{2}(7.5)(0.5)^2(0.5) = 0.46875 \text{ kN} \]
Step 5: Total. \[ F_{total} = 0.46875 + 4.375 = 4.84375 \text{ kN} \approx 4.84 \text{ kN} \] This rectangle-plus-triangle decomposition gives the identical trapezoid area as the average-pressure method, just built up piece by piece instead of averaged directly. \[ \boxed{F = 4.84 \text{ kN}} \]
