A small horizontal panel of area 0.032 m² is at depth 2.4 m in water of density 1000 kg/m³. Air at pressure 100000 Pa is above the water and also inside the equipment beneath the panel. Take g = 10 N/kg. Treat pressure as uniform over the panel. (a) Calculate the water's absolute pressure there. (b) Calculate the net force due to pressure on the panel. (c) Explain whether a wider tank changes the answer.

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Accepted Answer

Absolute pressure 124000 Pa; net force 768 N downward; tank width has no effect at the same depth. The pressure increase due to the water column is Δp = ρgh = 1000 × 10 × 2.4 = 24000 Pa. This is the excess above the pressure at the free surface, not the absolute pressure. Add the surface air pressure to obtain the external absolute pressure: p = 100000 + 24000 = 124000 Pa. Liquid pressure acts perpendicular to the panel. The internal air also pushes on the opposite face. Therefore the net pressure is only 24000 Pa, giving F = ΔpA = 24000 × 0.032 = 768 N downward. At fixed depth, density and gravitational field, liquid pressure does not depend on the tank's width or total volume. A wider tank contains more water but does not change the height of water responsible for the pressure at this panel.

Underwater inspection panel 7

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