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CIE A-Level Maths Study Notes

3.1.1 Force Diagrams

Force diagrams are a fundamental aspect of physics, providing a visual representation of the forces acting on an object. These diagrams are crucial in understanding static situations, where forces are in equilibrium, meaning there is no net force causing motion.

Introduction to Force Diagrams

Purpose: Show force magnitude and direction on an object.


  • Object shown as a dot or shape.
  • Arrows for forces: length = magnitude, direction = force direction.
Force Diagram

Image courtesy of Varsity Tutors

Creating and Interpreting Force Diagrams


1. Represent object as a dot or shape.

2. Identify all forces on it.

3. Draw arrows from the object: length shows force size, direction shows force direction.

4. Label forces (e.g., F_gravity, F_tension).

Analyzing Diagrams: Check arrow balance. Balanced = static situation, net force = zero.

Types of Forces in Static Situations

  • Weight (Gravity): Downward force, mass x gravity acceleration (F_gravity = m*g).
  • Tension: Force in a stretched string/rope/wire.
  • Normal Reaction: Force from a surface against an object, perpendicular to surface.
  • Friction: Force opposing motion between two surfaces, parallel to surface.

Analysis of Forces

  • Weight and Gravity: Force due to gravity, downward, equals mass x gravity 9.81m/s29.81 m/s^2 on Earth.
  • Tension in Strings/Cables: Force along a string/rope/cable when pulled.
  • Normal Reaction Force: Surface force against an object, perpendicular to surface, equal and opposite to net pressing force.



  • Two boxes stacked: m1=10kg,m2=2kgm_1 = 10 kg, m_2 = 2 kg.
  • Find max horizontal force FF on bottom box to keep top box from sliding.
  • Coefficient of static friction μsμ_s between boxes = 0.2.
  • Ignore ground friction.


1. Static Friction Force (F_R12):

  • FR12=μs×m2×gF_R12 = μ_s × m_2 × g
  • FR12=0.2×2kg×9.81m/s2F_R12 = 0.2 × 2 kg × 9.81 m/s^2
  • FR12=3.924NF_R12 = 3.924 N

2. Max Force on Bottom Box (F_max):

  • Fmax=3.924NF_max = 3.924 N
  • Fmax=3.924NF_max = 3.924 N

3. Acceleration (a):

  • Total mass = m1+m2m_1 + m_2
  • a=Fm1+m2a = \frac{F}{m_1 + m_2}
  • $a = \frac{3.924 \, \text{N}}{10 \, \text{kg} + 2 \, \text{kg}} </li><li></li><li> a = 0.327 \, \text{m/s}^2 </li></ul><p><strong>Results:</strong></p><ul><li>Maxforcewithoutsliding=</li></ul><p><strong>Results:</strong></p><ul><li>Max force without sliding = 3.924 N</li><li>Accelerationofboxes=</li><li>Acceleration of boxes = 0.327 \, \text{m/s}^2 $
Force Diagram
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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