A small horizontal hatch of area 0.008 m² lies 8 m below a liquid surface. Liquid density is 1000 kg/m³, g = 10 N/kg, and absolute pressure at the surface is 99000 Pa. Gas inside the sealed chamber has absolute pressure 99000 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 80000 Pa; outside absolute pressure 179000 Pa; resultant force 640 N, inward. Gauge pressure here means the increase relative to the liquid's surface pressure. Calculate ρgh = 1000 × 10 × 8 = 80000 Pa, using vertical depth in metres. Absolute outside pressure includes the pressure already present at the surface: 99000 + 80000 = 179000 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 179000 − 99000 = 80000 Pa; this difference is what determines the resultant. Multiply the difference magnitude by hatch area: 80000 × 0.008 = 640 N. The result is inward because outside pressure exceeds inside pressure. 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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