Fluid statics: the gate, the gauge and the oil film

Why the latch sits low

• A tank gate does not feel its load at its middle. Pressure in a fluid at rest grows with depth, $p = p_0 + \rho g h$, so the lower half of the gate is pushed harder than the upper half.
• The resultant is $F = \rho g h_c A$, the pressure at the centroid depth times the area, but it acts at the centre of pressure 压力中心, which is always below the centroid. That is where the latch must hold.

The tank gate

A height becomes a pressure

• A mercury U-tube (SG 13.6) across a fitting reads $\Delta h = 120$ mm. The water above the mercury cancels on both sides except over $\Delta h$, so $\Delta p = (\rho_m - \rho) g \Delta h = \mathbf{14.8}$ kPa.
• Using $\rho_m$ alone forgets that water column and overstates the reading by about 8%.

When the fluid moves: viscosity 黏度

• Statics ignores viscosity, but motion does not: $\tau = \mu\,du/dy$. A plate sliding at $2\ \mathrm{m/s}$ on a $0.5\ \mathrm{mm}$ oil film with $\mu = 0.29\ \mathrm{Pa\,s}$ feels $\tau = \mathbf{1160}$ Pa.
• The kinematic viscosity 运动黏度 $\nu = \mu/\rho$ is the one that appears in Reynolds numbers 雷诺数: $1.004 \times 10^{-6}\ \mathrm{m^2/s}$ for water.

The same law floats things

• Pressure that grows with depth pushes harder on a body's underside than its top. The net is the buoyancy 浮力 force, $\rho g V$: the weight of the displaced fluid (Archimedes).
• A 0.2 m³ float of 50 kg, held fully under water, feels $998 \times 9.81 \times 0.2 = 1958$ N up against 491 N of weight: its tether must hold $\mathbf{1468}$ N.

Worked example: hinge on top, latch below

• Hinge the 2 m gate along its top edge, 1.5 m deep, and latch it at the bottom. The 58.7 kN acts $2.633 - 1.5 = 1.133$ m below the hinge.
• Moments about the hinge: $R_{latch} \times 2 = 58.7 \times 1.133$, so $R_{latch} = \mathbf{33.3}$ kN, and the hinge takes the other 25.5 kN. The centre of pressure is what loads the latch more.

Three statics slips

Recap