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: Understanding the Question. Oil sits on top of water against a vertical gate, 0.5 m of oil above 1.0 m of water (oil floats because it is lighter, 7.5 kN/m3 against 10 kN/m3). We need the total horizontal hydrostatic force on the 0.5 m wide gate, taking into account that the water is also pressed on from above by the weight of the oil layer (a surcharge).
Step 2: Force from the oil layer. Inside the oil, pressure builds up from 0 at the free surface to $\gamma_{oil}\,h_{oil}$ at the oil-water interface. This gives a triangular pressure diagram, so the force per unit width is the usual $\tfrac{1}{2}\gamma h^2$: \[ F_{oil} = \frac{1}{2}\gamma_{oil} h_{oil}^2 \times b = \frac{1}{2}(7.5)(0.5)^2 \times 0.5 = \frac{1}{2}(7.5)(0.25)(0.5) = 0.46875 \text{ kN} \]
Step 3: Force from the water layer, including the oil surcharge. At the top of the water, pressure is not zero, it already carries the oil's pressure: $p_{top} = \gamma_{oil} h_{oil} = 7.5 \times 0.5 = 3.75$ kPa. At the bottom of the water (bottom of gate), pressure is $p_{bottom} = 3.75 + \gamma_{water} h_{water} = 3.75 + 10(1.0) = 13.75$ kPa. This is a trapezoidal pressure diagram, so use the average pressure times the area: \[ F_{water} = \left(\frac{3.75+13.75}{2}\right)\times (1.0 \times 0.5) = 8.75 \times 0.5 = 4.375 \text{ kN} \]
