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Creeping flow past a falling sphere in a fluid (e.g., a droplet of fog falling through the air): streamlines, drag force F d and force by gravity F g. At terminal (or settling) velocity , the excess force F e due to the difference between the weight and buoyancy of the sphere (both caused by gravity [ 7 ] ) is given by:
Viscous drag of fluid in a pipe: Drag force on the immobile pipe decreases fluid velocity relative to the pipe. [ 4 ] [ 5 ] In the physics of sports, drag force is necessary to explain the motion of balls, javelins, arrows, and frisbees and the performance of runners and swimmers. [ 6 ]
In fluid dynamics, the drag equation is a formula used to calculate the force of drag experienced by an object due to movement through a fully enclosing fluid. The equation is: F d = 1 2 ρ u 2 c d A {\displaystyle F_{\rm {d}}\,=\,{\tfrac {1}{2}}\,\rho \,u^{2}\,c_{\rm {d}}\,A} where
The downward force of gravity (F g) equals the restraining force of drag (F d) plus the buoyancy. The net force on the object is zero, and the result is that the velocity of the object remains constant. Terminal velocity is the maximum speed attainable by an object as it falls through a fluid (air is the most common example).
Similarly, the force is proportional to the square of the mass of the object. One of these terms is from the gravitational force between the object and the wake. The second term is because the more massive the object, the more matter will be pulled into the wake. The force is also proportional to the inverse square of the velocity.
A set of equations describing the trajectories of objects subject to a constant gravitational force under normal Earth-bound conditions.Assuming constant acceleration g due to Earth's gravity, Newton's law of universal gravitation simplifies to F = mg, where F is the force exerted on a mass m by the Earth's gravitational field of strength g.
The equation of motion for Stokes flow can be obtained by linearizing the steady state Navier–Stokes equations.The inertial forces are assumed to be negligible in comparison to the viscous forces, and eliminating the inertial terms of the momentum balance in the Navier–Stokes equations reduces it to the momentum balance in the Stokes equations: [1]
To convert the velocity as a function of time to a particle velocity distribution as a function of distance, let's assume a 1-dimensional velocity jump in the direction. Let's assume x = 0 {\displaystyle x=0} is positioned where the shock wave is, and then integrate the previous equation to get: