Photo AI

Last Updated Sep 26, 2025

Coefficient of Friction - F = ma Simplified Revision Notes

Revision notes with simplified explanations to understand Coefficient of Friction - F = ma quickly and effectively.

user avatar
user avatar
user avatar
user avatar
user avatar

305+ students studying

3.3.3 Coefficient of Friction - F = ma

In mechanics, the coefficient of friction plays a crucial role in determining the force of friction that opposes the motion of objects. When combined with Newton's second law, F=maF = ma, we can analyse the motion of objects on surfaces where friction is present.

1. Understanding Friction

  • Friction: Friction is a resistive force that acts against the relative motion of two surfaces in contact.
  • Coefficient of Friction ( μ\mu ): This is a dimensionless quantity that represents the ratio between the force of friction FfrictionF_{\text{friction}} and the normal force NN acting perpendicular to the surfaces.

μ=FfrictionN\mu = \frac{F_{\text{friction}}}{N}

  • Types of Friction:
    • Static Friction ( μs\mu_s ): The frictional force that must be overcome to start the motion.
    • Kinetic Friction ( μk\mu_k ): The frictional force acting on an object that is already in motion.

2. Force of Friction

The frictional force FfrictionF_{\text{friction}} is given by:

Ffriction=μNF_{\text{friction}} = \mu N

Where:

  • μ\mu is the coefficient of friction (static or kinetic).
  • NN is the normal force, which is often N=mgN = mg when the object is on a horizontal surface, with m m being the mass and gg the acceleration due to gravity.

3. Newton's Second Law: F=maF = ma with Friction

When friction is involved, Newton's second law is applied by accounting for the frictional force as part of the net force acting on the object.

  • Scenario 1: Object on a Horizontal Surface (with Friction)
    • Suppose a horizontal force FF is applied to an object of mass m m , causing it to accelerate.
    • The net force FnetF_{\text{net}} acting on the object is the applied force FF minus the frictional force FfrictionF_{\text{friction}} :

Fnet=FFfriction=maF_{\text{net}} = F - F_{\text{friction}} = ma

  • Substituting F_{\text{friction}} = $$\mu N and N=mgN = mg :

Fμmg=maF - \mu mg = ma

  • Solving for the acceleration aa :

a=Fμmgma = \frac{F - \mu mg}{m}

  • Scenario 2: Object on an Inclined Plane (with Friction)
    • Consider an object of mass mm on an inclined plane with angle θ\theta to the horizontal.
    • The forces acting on the object are:
    • Gravity: Acts downward with a force mgmg .
    • Normal Force: Acts perpendicular to the plane, N=mgcosθN = mg \cos \theta .
    • Frictional Force: Opposes motion, Ffriction=μN=μmgcosθF_{\text{friction}} = \mu N = \mu mg \cos \theta .
    • Component of Gravity along the Plane: mgsinθmg \sin \theta , which drives the object down the incline.
    • The net force along the plane is:

Fnet=mgsinθFfrictionF_{\text{net}} = mg \sin \theta - F_{\text{friction}}

Fnet=mgsinθμmgcosθF_{\text{net}} = mg \sin \theta - \mu mg \cos \theta

  • Applying Fnet=maF_{\text{net}} = ma :

mgsinθμmgcosθ=mamg \sin \theta - \mu mg \cos \theta = ma

  • Simplifying for acceleration a a :

a=g(sinθμcosθ)a = g (\sin \theta - \mu \cos \theta)

4. Example Problems

infoNote

Example 1: Object on a Horizontal Surface

  • Problem: A 10 kg box is pushed with a force of 60 N on a horizontal surface where the coefficient of kinetic friction μk\mu_k is 0.2. Find the acceleration of the box.
  • Solution:
  • Normal Force: N=mg=10×9.8=:highlight[98N]N = mg = 10 \times 9.8 = :highlight[98 \, \text{N}] .
  • Frictional Force: Ffriction=μkN=0.2×98=:highlight[19.6N]F_{\text{friction}} = \mu_k N = 0.2 \times 98 = :highlight[19.6 \, \text{N}] .
  • Net Force: Fnet=6019.6=:highlight[40.4N]F_{\text{net}} = 60 - 19.6 = :highlight[40.4 \, \text{N}] .
  • Acceleration: Using Fnet=maF_{\text{net}} = ma :

a=40.410=:success[4.04m/s2]a = \frac{40.4}{10} = :success[4.04 \, \text{m/s}^2]

infoNote

Example 2: Object on an Inclined Plane

  • Problem: A 5 kg block slides down a 30° inclined plane with a coefficient of kinetic friction μk\mu_k of 0.3. Find the acceleration of the block.
  • Solution:
  • Normal Force: N=mgcosθ=5×9.8×cos30°=:highlight[42.44N]N = mg \cos \theta = 5 \times 9.8 \times \cos 30° = :highlight[42.44 \, \text{N}] .
  • Frictional Force: Ffriction=μkN=0.3×42.44=:highlight[12.73N]F_{\text{friction}} = \mu_k N = 0.3 \times 42.44 = :highlight[12.73 \, \text{N}] .
  • Component of Gravity Along the Plane: mgsinθ=5×9.8×sin30°=:highlight[24.5N]mg \sin \theta = 5 \times 9.8 \times \sin 30° = :highlight[24.5 \, \text{N}] .
  • Net Force: Fnet=24.512.73=:highlight[11.77N]F_{\text{net}} = 24.5 - 12.73 = :highlight[11.77 \, \text{N}] .
  • Acceleration: Using Fnet=maF_{\text{net}} = ma :

a=11.775=:success[2.35m/s2]a = \frac{11.77}{5} = :success[2.35 \, \text{m/s}^2]

5. Summary

When friction is involved, Newton's second law F=maF = ma requires adjusting for the frictional force, which is determined by the coefficient of friction μ\mu and the normal force NN . Whether on a horizontal surface or an inclined plane, understanding how to incorporate friction into your calculations is crucial for accurately analysing the motion of objects.


Books

Only available for registered users.

Sign up now to view the full note, or log in if you already have an account!

500K+ Students Use These Powerful Tools to Master Coefficient of Friction - F = ma

Enhance your understanding with flashcards, quizzes, and exams—designed to help you grasp key concepts, reinforce learning, and master any topic with confidence!

60 flashcards

Flashcards on Coefficient of Friction - F = ma

Revise key concepts with interactive flashcards.

Try Maths Mechanics Flashcards

6 quizzes

Quizzes on Coefficient of Friction - F = ma

Test your knowledge with fun and engaging quizzes.

Try Maths Mechanics Quizzes

5 questions

Exam questions on Coefficient of Friction - F = ma

Boost your confidence with real exam questions.

Try Maths Mechanics Questions

27 exams created

Exam Builder on Coefficient of Friction - F = ma

Create custom exams across topics for better practice!

Try Maths Mechanics exam builder

15 papers

Past Papers on Coefficient of Friction - F = ma

Practice past papers to reinforce exam experience.

Try Maths Mechanics Past Papers

Other Revision Notes related to Coefficient of Friction - F = ma you should explore

Discover More Revision Notes Related to Coefficient of Friction - F = ma to Deepen Your Understanding and Improve Your Mastery

96%

114 rated

Further Forces & Newton's Laws

Resolving Forces & Inclined Planes

user avatar
user avatar
user avatar
user avatar
user avatar

208+ studying

199KViews

96%

114 rated

Further Forces & Newton's Laws

Coefficient of Friction

user avatar
user avatar
user avatar
user avatar
user avatar

325+ studying

181KViews

96%

114 rated

Further Forces & Newton's Laws

Coefficient of Friction - Inclined Planes

user avatar
user avatar
user avatar
user avatar
user avatar

344+ studying

193KViews

96%

114 rated

Further Forces & Newton's Laws

Coefficient of Friction - Harder Problems

user avatar
user avatar
user avatar
user avatar
user avatar

232+ studying

195KViews
Load more notes

Join 500,000+ A-Level students using SimpleStudy...

Join Thousands of A-Level Students Using SimpleStudy to Learn Smarter, Stay Organized, and Boost Their Grades with Confidence!

97% of Students

Report Improved Results

98% of Students

Recommend to friends

500,000+

Students Supported

50 Million+

Questions answered