A uniform horizontal gate of length 2.4 m and mass 90 kg is hinged at one end. Its centre is 1.8 m below liquid of density 850 kg/m³. Liquid gives 3.6 kN downward at 1.5 m from hinge; a vertical cable acts at far end. Take g = 9.81 m/s². (a) Find gauge pressure at centre. (b) Find cable tension. (c) Find hinge vertical force. (d) Explain centre of pressure and complete equilibrium.

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Gauge pressure 15.0 kPa; cable tension 2.69 kN; hinge force 1.79 kN upward; both resultant force and moment must vanish. Gauge pressure p = ρgh = 850 × 9.81 × 1.8 = 1.50 × 10⁴ Pa = 15.0 kPa. Atmospheric pressure cancels if it acts on both sides. Gate weight is 90 × 9.81 = 883 N at 1.2 m. Taking moments about hinge: T(2.4) = 883(1.2) + 3600(1.5), so T = 2692 N = 2.69 kN. Vertical balance gives hinge force + 2692 − 883 − 3600 = 0, hence hinge force = 1791 N = 1.79 kN upward. Pressure increases with depth, so deeper areas contribute more strongly and the liquid resultant need not act at the geometric centre; its line is the centre of pressure. Complete static equilibrium requires vector sum of forces zero and sum of moments about any point zero. Moment balance alone does not prevent translation.

Submerged gate in equilibrium

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