A small horizontal hatch of area 0.025 m² lies 4 m below a liquid surface. Liquid density is 1000 kg/m³, g = 10 N/kg, and absolute pressure at the surface is 100000 Pa. Gas inside the sealed chamber has absolute pressure 140000 Pa. Assume uniform pressure across the small hatch, ignore hatch weight, and use liquid pressure increase = ρgh and force = pressure × area. Find liquid gauge pressure at the hatch, absolute outside pressure, and the resultant pressure force with direction.

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Gauge pressure 40000 Pa; outside absolute pressure 140000 Pa; resultant force 0 N, zero resultant. Gauge pressure here means the increase relative to the liquid's surface pressure. Calculate ρgh = 1000 × 10 × 4 = 40000 Pa, using vertical depth in metres. Absolute outside pressure includes the pressure already present at the surface: 100000 + 40000 = 140000 Pa. The liquid contribution does not replace the surface contribution. Pressure on the inside pushes in the opposite direction. The signed outside-minus-inside difference is 140000 − 140000 = 0 Pa; this difference is what determines the resultant. Multiply the difference magnitude by hatch area: 0 × 0.025 = 0 N. The result is zero resultant because the two absolute pressures are equal. Both surfaces can experience large forces even when the resultant is small or zero. This is why using liquid gauge pressure alone would be wrong whenever the sealed internal pressure differs from surface pressure.

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