When a small sphere moves slowly through a fluid, it feels a viscous drag that opposes its motion. Stokes’ law tells you exactly how big that drag is — and it depends on three things: the fluid’s viscosity, the sphere’s radius, and its speed.
📘 What you need to know
Viscous drag is the frictional force between an object and a fluid that opposes their relative motion (in air, this is air resistance).
Stokes’ law:Fd = 6πηrv.
η = fluid viscosity (Pa s), r = sphere radius (m), v = speed through the fluid (m s⁻¹).
Viscosity measures how much a fluid resists flowing — water has a low viscosity, honey a high one.
Drag grows with speed, size and viscosity — bigger, faster spheres in thicker fluids feel more drag.
Viscous drag and Stokes’ law
Viscous drag is a friction-type force that appears whenever an object moves through a fluid (a liquid or a gas). For a small sphere moving slowly, its size is given by Stokes’ law:
Stokes’ lawFd = 6πηrv
The drag always acts opposite to the sphere’s motion through the fluid, and you can see from the formula that doubling the speed, the radius, or the viscosity each doubles the drag.
Drag depends on the sphere’s radius and speed and on the fluid’s viscosity — and always opposes the motion.
What viscosity means
The viscosityη is a property of the fluid — essentially how “thick” it is, or how strongly it resists flowing. A fluid with low viscosity (like water) pours easily; one with high viscosity (like honey or ketchup) flows slowly. At a given temperature, the rate of flow is inversely proportional to the viscosity.
Higher viscosity means the fluid resists flowing more — which also means more drag on anything moving through it.
Quick reference:Fd = 6πηrv — drag opposes motion and rises with viscosity η, radius r and speed v. Valid for a small sphere moving slowly through a fluid.
Worked examples
WE 1
Speed of a falling stone
A spherical stone of volume 2.7 × 10⁻⁴ m³ falls through air and, at one instant, feels a drag of 3.0 mN. Air has a viscosity of 1.81 × 10⁻⁵ Pa s. Find its speed at that instant.
First get the radius from V = (4/3)πr³r = ∛(3V ÷ 4π) = ∛(3 × 2.7×10⁻⁴ ÷ 4π)r ≈ 0.040 mRearrange Stokes’ law: v = Fd ÷ (6πηr)v = 3.0×10⁻³ ÷ (6π × 1.81×10⁻⁵ × 0.040)v ≈ 220 m s⁻¹
WE 2
Drag on a small sphere
A sphere of radius 2.0 mm moves at 0.50 m s⁻¹ through oil of viscosity 0.20 Pa s. Calculate the viscous drag on it.
Use Fd = 6πηrv (with r = 0.0020 m)Fd = 6π × 0.20 × 0.0020 × 0.50Fd ≈ 3.8 × 10⁻³ Nabout 3.8 mN, opposing the motion
💡 Top tips
Use the radius, not the diameter, in Fd = 6πηrv — and convert mm to m.
If you’re given a volume, find r from V = (4/3)πr³ first.
Drag opposes motion, so draw Fd pointing against the velocity.
Stokes’ law is for small spheres moving slowly (smooth, laminar flow) — it won’t apply to fast or large objects.
Viscosity depends on temperature — warm a fluid and it usually flows more easily.
⚠ Common mistakes
Forgetting the 6π or dropping a factor when rearranging the formula.
Using diameter as radius, or leaving the radius in mm.
Treating viscosity as a property of the object. It belongs to the fluid.
Applying Stokes’ law to large, fast objects, where the flow is turbulent and the law breaks down.
As a falling sphere speeds up, the drag grows until it balances the other forces and the sphere stops accelerating — its terminal velocity. We combine Stokes’ law with weight and upthrust to find that speed in the next topic. Up next: Buoyancy.
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