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Equilibrium of a Fluid Element

A fluid element at rest is in equilibrium if the net force and net torque acting on it are zero, satisfying Newton’s laws.
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The statement of the theorem

Let Ω\Omega be a continuous, connected region in R3\mathbb{R}^3 representing the fluid domain, and let ρ(x⃗)\rho(\vec{x}) be the density function. For a differential fluid element dVdV located at x⃗∈Ω\vec{x} \in \Omega to be in static equilibrium, the net force acting on it must vanish. This condition is mathematically expressed by the balance of forces, which states that the gradient of the pressure field pp must exactly counteract the body force per unit volume f⃗b\vec{f}_b. Specifically, the governing equation is:\n\n∇p+f⃗b=0⃗\nabla p + \vec{f}_b = \vec{0}\n\nIf the body force is solely due to gravity, f⃗b=ρg⃗\vec{f}_b = \rho \vec{g}, leading to the fundamental differential equation of fluid statics:\n\n∇p=−ρg⃗\nabla p = -\rho \vec{g}\n\nFurthermore, the condition of static equilibrium requires that the velocity field v⃗\vec{v} is identically zero throughout Ω\Omega, i.e., v⃗=0⃗\vec{v} = \vec{0}, ensuring that the inertial terms vanish.