A-Level — physicsMedium4 min read

Friction and Drag Forces

Static and dynamic friction; drag and terminal velocity; Stokes' law

Tutor AI Team
Tutor AI Team

# Friction and Drag Forces — A-Level Physics

Friction and drag are resistive forces that oppose motion. While sometimes treated as nuisances, they are essential — without friction, you couldn't walk, drive, or write. At A-Level, you need a quantitative understanding of these forces.

1. Friction

Static Friction

Acts on objects that are not moving (or about to move). It adjusts to match the applied force up to a maximum.

fs≤μsNf_s \leq \mu_s N

Where:

  • fsf_s = static friction force (N)
  • μs\mu_s = coefficient of static friction
  • NN = normal reaction force (N)

Dynamic (Kinetic) Friction

Acts on objects that are sliding.

fk=μkNf_k = \mu_k N

Where μk\mu_k = coefficient of kinetic friction.

Key fact: μs>μk\mu_s > \mu_k (it's harder to start moving than to keep moving).

Properties of Friction

  • Acts parallel to the contact surface
  • Opposes the motion (or tendency to move)
  • Independent of surface area (approximately)
  • Independent of speed (approximately, for solid surfaces)
  • Depends on the nature of the surfaces and the normal force

2. Drag Forces in Fluids

When an object moves through a fluid (liquid or gas), it experiences a resistive force called drag.

Factors Affecting Drag

  • Speed — drag increases with speed
  • Cross-sectional area — larger area → more drag
  • Shape — streamlined shapes reduce drag
  • Fluid density — denser fluid → more drag
  • Fluid viscosity — more viscous → more drag

At Low Speeds (Viscous Drag)

For small, slow-moving objects in viscous fluids: Fdrag∝vF_{\text{drag}} \propto v

Stokes' Law

For a sphere moving at low speed through a viscous fluid:

F=6πηrv\boxed{F = 6\pi\eta rv}

Where:

  • FF = drag force (N)
  • η\eta = dynamic viscosity of the fluid (Pa·s)
  • rr = radius of the sphere (m)
  • vv = velocity (m/s)

Conditions for Stokes' Law to apply:

  • Small sphere
  • Slow speed (laminar flow)
  • Viscous fluid
  • Sphere is smooth

At Higher Speeds (Turbulent Drag)

Fdrag∝v2F_{\text{drag}} \propto v^2

This applies to larger objects at higher speeds (cars, skydivers, etc.).

3. Terminal Velocity

When an object falls through a fluid:

  1. Weight acts downwards
  2. Drag (+ upthrust) acts upwards
  3. As speed increases, drag increases
  4. When drag + upthrust = weight → resultant = 0 → terminal velocity

Terminal Velocity with Stokes' Law

For a sphere falling in a viscous fluid: mg=6πηrvt+upthrustmg = 6\pi\eta rv_t + \text{upthrust}

Upthrust = weight of displaced fluid = 43πr3ρfg\frac{4}{3}\pi r^3 \rho_f g

Weight = 43πr3ρsg\frac{4}{3}\pi r^3 \rho_s g

43πr3(ρs−ρf)g=6πηrvt\frac{4}{3}\pi r^3(\rho_s - \rho_f)g = 6\pi\eta r v_t

vt=2r2(ρs−ρf)g9η\boxed{v_t = \frac{2r^2(\rho_s - \rho_f)g}{9\eta}}

Worked Example: Friction on a Slope

Problem

A 5 kg block is on a slope at 30°. μs=0.6\mu_s = 0.6. Will it slide?

Force down slope: mgsin⁡30°=5(9.81)(0.5)=24.5mg\sin30° = 5(9.81)(0.5) = 24.5 N Max static friction: μsN=0.6×mgcos⁡30°=0.6×5(9.81)(0.866)=25.5\mu_s N = 0.6 \times mg\cos30° = 0.6 \times 5(9.81)(0.866) = 25.5 N

Since 24.5 < 25.5, the block does not slide. ✓

Solution

Worked Example: Stokes' Law

Problem

A steel ball (ρ=7800\rho = 7800 kg/m³, r=2r = 2 mm) falls through oil (η=0.5\eta = 0.5 Pa·s, ρ=900\rho = 900 kg/m³). Find the terminal velocity.

vt=2(0.002)2(7800−900)(9.81)9×0.5v_t = \frac{2(0.002)^2(7800 - 900)(9.81)}{9 \times 0.5} =2×4×10−6×6900×9.814.5= \frac{2 \times 4 \times 10^{-6} \times 6900 \times 9.81}{4.5} =0.54154.5=0.120 m/s= \frac{0.5415}{4.5} = 0.120 \text{ m/s}

Solution

Worked Example: Finding Viscosity

Problem

A ball bearing of radius 1.5 mm and density 7500 kg/m³ reaches terminal velocity of 0.08 m/s in glycerol (density 1260 kg/m³). Calculate the viscosity.

η=2r2(ρs−ρf)g9vt\eta = \frac{2r^2(\rho_s - \rho_f)g}{9v_t} =2(1.5×10−3)2(7500−1260)(9.81)9×0.08= \frac{2(1.5 \times 10^{-3})^2(7500 - 1260)(9.81)}{9 \times 0.08} =2×2.25×10−6×6240×9.810.72= \frac{2 \times 2.25 \times 10^{-6} \times 6240 \times 9.81}{0.72} =0.27550.72=0.383 Pa\cdotps= \frac{0.2755}{0.72} = 0.383 \text{ Pa·s}

Solution

5. Practice Questions

    1. A 3 kg block is pulled at constant speed across a surface by a 12 N force. Calculate μk\mu_k. (2 marks)
    1. State the conditions under which Stokes' Law is valid. (2 marks)
    1. A sphere of radius 3 mm (ρ=11,000\rho = 11{,}000 kg/m³) falls through water (η=10−3\eta = 10^{-3} Pa·s, ρ=1000\rho = 1000 kg/m³). Calculate the terminal velocity. Comment on whether Stokes' Law is valid. (4 marks)
    1. Explain why larger raindrops fall faster than smaller ones. (3 marks)

    Answers

    1. fk=μkNf_k = \mu_k N → 12=μk×3(9.81)12 = \mu_k \times 3(9.81) → μk=12/29.43=0.408\mu_k = 12/29.43 = 0.408.

Want to check your answers and get step-by-step solutions?

Get it on Google PlayDownload on the App Store

Summary

  • Static friction: fs≤μsNf_s \leq \mu_s N; Kinetic friction: fk=μkNf_k = \mu_k N
  • μs>μk\mu_s > \mu_k
  • Stokes' Law: F=6πηrvF = 6\pi\eta rv (low speed, small sphere, viscous fluid)
  • Terminal velocity: vt=2r2(ρs−ρf)g/(9η)v_t = 2r^2(\rho_s - \rho_f)g/(9\eta)
  • Drag ∝ vv at low speeds (Stokes); drag ∝ v2v^2 at high speeds (turbulent)
Tags:
#a-level#physics#mechanics#friction#drag#stokes-law#terminal-velocity

Related Topics

Ready to Ace Your A-Level physics?

Get instant step-by-step solutions to any problem. Snap a photo and learn with Tutor AI — your personal exam prep companion.

Get it on Google PlayDownload on the App Store