Calculate pressure force from both sides

A zero gauge reading can hide two large forces

At the free surface the model’s gauge pressure is zero. Predict whether this means a panel there experiences no pressure force on either face. Gauge pressure describes a difference from a reference; it does not erase the physical pressure represented by that reference.

The experiment uses a thin horizontal panel of area 0.020 m². Liquid pushes on its upper face. A separate gas compartment beneath it is held at an absolute pressure of 100 kPa, equal to the pressure at the liquid’s free surface. The panel remains supported at the selected depth. Its weight, thickness, deformation and motion are omitted. The box in the diagram identifies the compartment; its width does not determine the stated area.

Distinguish the pressure references

Absolute pressure uses a vacuum reference. Gauge pressure here means liquid absolute pressure minus the specified 100 kPa reference. p_absolute = p_gauge + p_reference. Both are legitimate readings when their reference is stated. OpenStax measuring pressure distinguishes these conventions and the role of a reference pressure.

At depth 2 m and density 1000 kg/m³, the liquid has 20 kPa gauge pressure and 120 kPa absolute pressure. The compartment gas has 100 kPa absolute pressure and zero gauge pressure relative to the same reference. The pressure difference across the panel is 20 kPa.

If the compartment pressure were changed independently, the net panel load would change even with the same liquid depth and density. The current widget fixes this compartment pressure. Treating it as an arbitrary unknown would make gauge pressure alone insufficient to determine the net load.

Calculate each face force

For uniform pressure on a flat face, F = pA, with p in Pa and A in m², gives force magnitude in N. Pressure pushes normal to each face: the liquid force is downward, while the gas force on the lower face is upward.

At the default setting, the liquid force is 120000 Pa × 0.020 m² = 2400 N downward. The gas force is 100000 Pa × 0.020 m² = 2000 N upward. Choosing downward positive, the net pressure force is 2400 − 2000 = 400 N downward.

The equivalent calculation uses the pressure difference: (120000 − 100000) × 0.020 = 400 N. Here gauge pressure times area gives the same answer because the opposite face is at the gauge reference pressure. That shortcut is conditional, not a universal rule for every two-sided surface.

Check the force ledger before reading the summary. A correct liquid force of 2400 N is still an incorrect net-force answer if the opposing gas force is omitted. Pressure and force also have different units: a result in kPa cannot be called a force in newtons.

Use the surface limit as a check

Set depth to zero. Both absolute pressures are 100 kPa. Each face experiences a force of magnitude 2000 N, but the forces oppose and the net pressure force is zero. This example distinguishes zero resultant force from absent interactions.

Now increase depth to 4 m at density 1000 kg/m³. Liquid absolute pressure becomes 140 kPa, the downward liquid force 2800 N and the net pressure force 800 N. The upward gas force remains 2000 N. Capture both depths and identify exactly which force changes.

At fixed depth, increasing density changes the liquid pressure and force but not the prescribed gas pressure. The hypothetical liquid column above a panel-sized area has weight ρAhg, equal to the net pressure force in this specific arrangement. That column weight explains the pressure increment; it is not an additional load to add again to the already calculated liquid pressure force.

Separate load from motion and strength

Because the panel is held in place, its support must supply a force opposing the net pressure force, in addition to any omitted weight in a real design. A net pressure load does not by itself give the panel’s acceleration or establish whether it will bend or fail. Those questions require mass, material properties, geometry and a support model.

The horizontal-panel condition makes pressure uniform over the face at a single depth. A tall vertical panel spans different depths; its pressure varies across the area. One must account for that variation to find total force and its point of action. Multiplying the deepest pressure by the full area generally overestimates the resultant. This lab does not calculate bending moments or a center of pressure.

Transfer to a different compartment pressure

A horizontal panel has area 0.030 m², liquid absolute pressure 150 kPa above and gas absolute pressure 110 kPa below. Calculate each face force and the net pressure force. A nearby gauge is referenced to 100 kPa; explain why its reading times area would give the wrong net load.

Check the pressure force transfer

The liquid pushes down with 150000 × 0.030 = 4500 N; the gas pushes up with 110000 × 0.030 = 3300 N. Net pressure force is 1200 N downward. The gauge reads 50 kPa, but multiplying that by area gives 1500 N because its 100 kPa reference is not the actual 110 kPa pressure on the other face. Use the actual pressure difference of 40 kPa.