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First teaching 2023

First exams 2025

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Frictional Forces (HL IB Physics)

Revision Note

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Frictional Forces

  • Frictional forces oppose the motion of an object
  • Frictional forces slow down the motion of an object 

 

  • When friction occurs, energy is transferred by heating
    • This raises the temperature (thermal energy) of the objects and their surroundings
    • The work done against frictional forces causes this rise in temperature
  • Fluid resistance or drag occurs when an object moves through a fluid (a gas or a liquid) 
    • The object collides with the particles in the liquid or gas
    • This slows down the motion of the object and causes heating of the object and the fluid
  • Surface friction occurs between two bodies that are in contact with one another
    • Imperfections in the surfaces of the objects in contact rub up against each other
    • Not only does this slow the object down but also causes an increase in thermal energy

Friction, IGCSE & GCSE Physics revision notes

The interface between the ground and the sled is bumpy which is the source of the frictional force

Static & Dynamic Friction

  • There are two kinds of surface friction to consider for IB DP Physics
    • Static friction occurs when a body is stationary on a surface
    • Dynamic friction occurs when a body is in motion on a surface, such as in the sledge example above
  • The surface frictional force always acts in a direction parallel to the plane of contact between a body and a surface
  • Both of these forms of friction depend on the normal reaction force, FN of one object sitting upon the other
  • Static friction will match any push or pull force that acts against it until it can no longer hold the two objects stationary
    • Static friction increases in magnitude until movement begins and dynamic friction occurs

  • For any given situation, static friction should reach a maximum value that is larger than that of dynamic friction
    • For a constant pushing force, dynamic friction will be a constant

  • This is because there are more forces at work keeping an object stationary than there are forces working to resist an object once it is in motion

Friction-1, downloadable IB Physics revision notes

The relationship between frictional forces and motion

  • The equation for static friction is given by:

F subscript f space less or equal than space µ subscript straight s F subscript straight N

  • Where:
    • Ff = frictional force (N)
    • μS = coefficient of static friction
    • FN = normal reaction force (N)

  • The coefficient of static friction is a number between 0 and 1 but does not include those numbers
    • It is a ratio of the force of static friction and the normal force
    • The larger the coefficient of static friction, the harder it is to move those two objects past one another

  • The equation for dynamic friction is given by:

F subscript f space equals space µ subscript straight d F subscript straight N

  • Where:
    • Ff = frictional force (N)
    • μd = coefficient of dynamic friction
    • FN = normal reaction force (N)

  • The coefficient of dynamic friction has similar properties to that of static friction
  • However:
    • dynamic friction has a definite force value for a given situation
    • static friction has an increasing force value for a given situation

Worked example

An 8.0 kg block sits on an incline of 20 degrees from the horizontal. It is stationary and does have a frictional force acting upon it.Friction-2-worked-example, downloadable IB Physics revision notes Determine the minimum possible value of the coefficient of static friction.

Answer:

Step 1: List the known quantities

  • Mass of the block, m = 8.0 kg
  • Angle between the slope and the horizontal, θ = 20°

Step 2: Determine the weight of the block

  • The weight will act directly downward and comes from the interaction of mass and acceleration due to gravity

Fg = mg

Fg = 8.0 × 9.81 = 78.48 N downwards

Step 3: Break the weight down into components based on the slope angle

friction-worked-example-ma

  • The component of the weight force that is parallel to the slope provides the force that moves the block down the slope
  • This component of the weight force is equal to the surface friction acting up the slope, F subscript f

F subscript f space equals space F subscript g sin space theta

F subscript f space equals space 78.48 space cross times space sin open parentheses 20 close parentheses space equals space 26.8 space straight N

 

  • The component of the weight force that is perpendicular to the slope has the same magnitude as the normal reaction force, F subscript N
     
  • F subscript N space equals space F subscript g cos space theta
  • F subscript N space equals space 78.48 space cross times space cos open parentheses 20 close parentheses space equals space 73.7 space straight N

Step 4: Use the equation of static friction to find the minimum value of the coefficient of static friction

  • The equation for static friction is:

F subscript f space less or equal than space µ subscript straight s F subscript straight N

 

  • Rearrange to make the coefficient of static friction the subject

straight mu subscript s space greater or equal than fraction numerator space F subscript f over denominator F subscript N end fraction

µ subscript s space greater or equal than space fraction numerator 26.8 space over denominator 73.7 end fraction

µ subscript s space greater or equal than space 0.36

Step 5: State the final answer

  • The coefficient for static friction must be 0.36 or greater for this situation

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Ashika

Author: Ashika

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.