Physics & Mechanics

Stokes' Law Calculator

Calculate the drag force acting on a spherical object moving through a viscous fluid. Perfect for fluid dynamics calculations.

Pa·s
m
m/s
Drag Force
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Sinking in Syrup

Stokes' law is a mathematical equation that calculates the drag force exerted on a small spherical object moving slowly through a viscous fluid. Derived by George Gabriel Stokes in 1851, the law is essential for understanding how small particles behave when settling out of a suspension.

When a particle falls through a fluid, gravity pulls it down, while buoyancy and viscous drag push it back up. As the particle accelerates, the drag force increases until it perfectly balances the downward gravitational pull. At this point, the particle stops accelerating and falls at a constant speed, known as the terminal velocity.

Industrial and Natural Applications

  • Raindrops and Fog: Stokes' law explains why tiny water droplets in fog remain suspended in the air almost indefinitely (their terminal velocity is near zero), while larger, heavier raindrops fall quickly to the ground.
  • Centrifuges: Laboratories use centrifuges to artificially increase the 'g-force' on suspended biological cells, allowing them to overcome viscous drag and separate out of the liquid plasma much faster than gravity alone would permit.
  • Brewing: When beer ferments, the yeast cells slowly sink to the bottom of the vat over weeks. Brewers use Stokes' law to predict how long 'flocculation' will take to clear the beer.

The Formula

Fd=6πμrv\begin{aligned} F_d = 6 \cdot \pi \cdot \mu \cdot r \cdot v \end{aligned}

Where:
FdF_d=
Frictional Drag Force (Newtons)
μ\mu=
Dynamic viscosity of the fluid (Pa·s)
r=
Radius of the spherical object (meters)
v=
Flow velocity relative to the object (m/s)

Example Calculation

Calculate the drag force on a tiny glass bead with a radius of $0.001 , ext{m}$ (1 mm) falling at $0.05 , ext{m/s}$ through thick motor oil (viscosity $\mu = 0.5 , ext{Pa}\cdot\text{s}$).

  1. Multiply $6 \cdot \pi \cdot \mu \cdot r \cdot v$: $6 \cdot \pi \cdot 0.5 \cdot 0.001 \cdot 0.05 \approx 0.00047 , ext{N}$.

The thick oil exerts a gentle drag force of $0.00047 , ext{Newtons}$ upward against the sinking bead. If this force equals the bead's weight (minus buoyancy), it has reached its terminal velocity.

Frequently Asked Questions

No, the standard equation $6 \pi \mu r v$ is derived strictly for perfect spheres. If the falling object is a flat disk, a jagged rock, or an aerodynamic teardrop, different drag coefficients must be applied because the fluid flows around them entirely differently.

Stokes' law assumes 'creeping flow' (where the Reynolds number is very low, usually $<0.1$). In this state, viscous forces completely dominate over inertial forces, and the fluid smoothly wraps around the sphere. At higher speeds, chaotic wakes and turbulent eddies form behind the object, drastically increasing the drag beyond what this simple formula predicts.

You set the drag force ($F_d$) plus the buoyant force equal to the gravitational weight of the sphere. Solving for velocity ($v$) will yield the exact terminal velocity equation: $v = \frac{2}{9} \frac{(\rho_{particle} - \rho_{fluid})}{\mu} g r^2$.