Stokes' Drag
Drag on a small sphere in viscous flow.
Formula
Variables
How it works
Stokes' drag gives the resistive force on a small sphere moving slowly through a viscous fluid, and it is linear in speed because the flow around the sphere is smooth and dominated by viscosity rather than inertia. The factor 6\pi\mu r comes from solving the low-Reynolds-number flow equations around a sphere with no-slip at the surface, so the formula is only valid when the fluid motion is creeping and laminar. It is especially useful because it links an easily measured terminal force to fluid viscosity and particle size.
When to use it
Use this for tiny spherical particles, droplets, or bubbles moving slowly in a thick fluid, especially when the Reynolds number is much less than 1. It is common in sedimentation, falling-ball viscometers, aerosol physics, and colloid dynamics.
Worked example, step by step
A steel sphere of radius r = 0.80 mm moves through glycerin at speed v = 0.12 m/s. The viscosity is μ = 1.50 Pa·s. Step 1: Convert radius to metres: r = 0.80 mm = 8.0 × 10^-4 m Step 2: Substitute into Stokes' drag: F = 6πμrv F = 6π(1.50)(8.0 × 10^-4)(0.12) Step 3: Calculate the product inside: (1.50)(8.0 × 10^-4)(0.12) = 1.44 × 10^-4 Step 4: Multiply by 6π: F = 6π(1.44 × 10^-4) ≈ 2.71 × 10^-3 N Final answer: F ≈ 2.7 × 10^-3 N
Common mistakes to avoid
- Using the formula for large or fast objects where the flow is not creeping and the drag is no longer linear in v
- Forgetting to convert the sphere radius to metres before substituting into the formula
- Mixing up viscosity units, such as using mPa·s as if it were Pa·s
Did You Know?
Stokes's drag law explains why fine dust and fog droplets fall so slowly they seem to hang in the air.
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