Frictional Forces

Friction is a force that works in opposition to the motion of an object
It occurs between two solid bodies that are in contact with one another
- The opposition of friction slows down the motion of the object

When friction is present, energy is transferred in the form of heat
- This raises the temperature (thermal energy) of the object and its surroundings
- The work done against the frictional forces causes this rise in the temperature

Imperfections at the interface between the object and the surface bump into and 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 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 sled 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 the object sitting upon the other
Static friction will match any pushing 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 = surface 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 = dynamic 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
Yet, dynamic friction has a definite force value for a given situation
- Whereas the force of static friction has an increasing force value

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 downwards and comes from the interaction of mass and acceleration due to gravity

W = m × 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 that is parallel to the slope and provides a force moving the block down the slope is equal to the frictional force up the slope:

$F_{f} = W \sin (20)$

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

The component that is perpendicular to the slope and the same magnitude as the normal reaction force can be found from:
$F_{N} = 78.48 \times \cos (20) = 73.7 N$

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:

$F_{f} \leq µ_{s} F_{N}$

In this case, the F _f is the 26.8 N pushing the block down the slope
The F_N is the normal reaction force which has the same magnitude as the perpendicular component of the weight force which is 73.7 N
Therefore the value can be added and μ_Ssolved for:
Rearrange for μ_S

$µ_{s} \geq \frac{26.8}{73.7}$

$µ_{s} \geq 0.36$

Step 5: State the final answer

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

Frictional Forces (SL IB Physics)

Revision Note