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, F_N 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:

$F_{f} \leq µ_{s} F_{N}$

Where:
- F_f = frictional force (N)
- μ_S= coefficient of static friction
- F_N = 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:

$F_{f} = µ_{d} F_{N}$

Where:
- F_f = frictional force (N)
- μ_d= coefficient of dynamic friction
- F_N = 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

F_g = mg

F_g = 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, $F_{f}$

$F_{f} = F_{g} \sin θ$

$F_{f} = 78.48 \times \sin (20) = 26.8 N$

The component of the weight force that is perpendicular to the slope has the same magnitude as the normal reaction force, $F_{N}$
$F_{N} = F_{g} \cos θ$
$F_{N} = 78.48 \times \cos (20) = 73.7 N$

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:

$F_{f} \leq µ_{s} F_{N}$

Rearrange to make the coefficient of static friction the subject

$μ_{s} \geq \frac{F_{f}}{F_{N}}$

$µ_{s} \geq \frac{26.8}{73.7}$

$µ_{s} \geq 0.36$

Step 5: State the final answer

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

Frictional Forces (SL IB Physics)

Revision Note