While rocking a rocking chair listening to "2 rocking chairs" by Jon Bellion, it struck him; "what happens if this chair never stops rocking?". "How about engines in machines, imagine they ran endlessly without ever stopping. Eureka! I found it", Mr. Finicky Spins screamed in excitement and said, "everything needs a brake so that we don't break. We apply brakes to take a break, hence friction". In this exciting journey, you will learn about the equation, formula, measurement device as well as units of coefficient of friction. Let's rock without breaking!
What is the coefficient of friction?
The coefficient of friction, \(\mu\), is the ratio or quotient between the frictional force \((F)\) and normal reaction \((R)\).
This value gives you an idea of the ease with which movement occurs when two surfaces are in contact with each other.
When the coefficient of friction is high between materials it means there is more friction, hence, the resistance to movement between surfaces in contact is indeed high.
Meanwhile, when the coefficient of friction is low between materials it means there is less friction, hence, the resistance to movement between surfaces in contact is indeed low.
Also, the coefficient of friction is determined by the nature of the surfaces. Smoother surfaces will generally have less friction than rougher surfaces.
Before you proceed, it is beneficial to refresh your memory on frictional force and normal reaction.
What is frictional force?
The frictional force is that force which tends to resist or oppose the movement between objects or surfaces in contact. Before an object must begin motion on a surface, it must overcome the frictional force between both surfaces in contact.
Fig. 1. Description of frictional force.
What is a normal reaction?
The normal reaction often denoted as \(R\), is the force that counterbalances the weight of an object. It is equal to the weight, \(W\), of an object, however, it acts in an opposite direction. Since the weight of an object is a downward force impacted by the acceleration due to gravity, the normal reaction is an upward force.
Without the normal reaction, the weight from objects would make them to sink through the surfaces they are placed on.
Fig. 2. Image that describes normal reaction and weight.
Formula of coefficient of friction
Before determining the formula for the coefficient of friction, it is imperative to define the postulations of Charles-Augustin de Coulomb on friction in 1785. These postulations are:
1. The frictional force always resists the simultaneous movement which takes place between surfaces in contact.
2. The frictional force acts regardless of the relative speed of surfaces in contact and as such, the action of friction is not dependent on the rate at which the surfaces move.
3. However, the frictional force existing between surfaces in contact is dependent on the normal reaction between these surfaces as well as their level of roughness.
4. When sliding is not existing between surfaces in contact, the frictional force is said to be less than or equal to the product of the coefficient of friction and the normal reaction.
5. At the point sliding is to commence between surfaces in contact, the frictional force is described as 'limiting'. At this stage, the frictional force is equal to the product of the normal reaction and the coefficient of friction.
6. At the point where sliding is taking place, then frictional force is equal to the product of the normal reaction and the coefficient of friction.
From Coulomb's postulations, we can infer three instances that define the coefficient of friction. Such instances are:
No sliding
\[F≤µR\]
At the start of sliding
\[F=µR\]
During sliding
\[F=µR\]
Where \(F\) is the frictional force, \(R\) is the normal reaction and \(µ\) is the coefficient of friction.
Hence for an object moving in contact with a surface the coefficient of friction \(µ\) can thus be calculated with the formula \[µ=\frac{F}{R}\]
The unit of coefficient of friction
Knowing the units with which frictional force and normal reaction are measured, we can derive the unit used in measuring the coefficient of friction. Since both friction, \(F\), and normal reaction, \(R\), are measured in Newtons, \(N\), and the coefficient of friction is the quotient of friction and normal reaction, hence,
\[µ=\frac{N}{N}\]
Thus
\[µ=1\]
This means that the coefficient of friction has no unit.
Coefficient of friction measurement device
Based on Coulomb's research, he also stated that the coefficient of friction is a constant value or range of values between known surfaces in contact.
Now, the coefficient of friction is measured using the coefficient of friction testers. These measures the static and kinetic coefficient of friction (COF).
Below is a table that tells the coefficient of friction between certain surfaces in contact when they are static as well as when in motion.
| Material | Material of counter-surface | Static Coefficient of Friction | Kinetic Coefficient of Friction |
| Steel | Steel | 0.74 | 0.57 |
| Copper | Steel | 0.53 | 0.36 |
| Aluminum | Steel | 0.61 | 0.47 |
| Wood | Wood | 0.25 - 0.50 | 0.20 |
| Wood | Brick | 0.60 | 0.45 |
| Waxed wood | Dry snow | - | 0.040 |
| Waxed wood | Wet snow | 0.14 | 0.10 |
| Ice | Ice | 0.10 | 0.030 |
| Metal | lubricated metal | 0.15 | 0.060 |
| Rubber | Concrete | 1.0 | 0.8 |
| Glass | Glass | 0.94 | 0.40 |
| Teflon | Teflon | 0.040 | 0.040 |
| Joints | Joints with the synovial fluid in humans | 0.010 | 0.0030 |
Table 1. Coefficients of friction for different materials.
The Negative coefficient of friction
Generally, the frictional force increases as the weight of the object or load increases. However, in certain circumstances, with the decrease in load, there is a consequent increase in friction. This phenomenon is regarded as negative friction. A negative friction coefficient is seen to exist with minute masses of objects like those measured on nanoscales.
Equation of the coefficient of friction
Problems that involve the coefficient of friction would require the application of the formula of the coefficient of friction, forming some equations which are used to solve these problems.
Always recall that
\[µ=\frac{F}{R}\]
A rope is fitted to \(100\, \text{kg}\) mass of a rectangular block which is static on a plane surface. If the coefficient of friction existing between the block and plane is \(0.4\), determine the maximum force which can be exerted by pulling the rope without making the block move on the plane.
Solution:
Make a sketch of the information given to have a clearer picture.
Fig. 3. Determining the maximum force that keeps a block at rest.
Recall that the first inference from Coulomb's postulation explains the occasion of a body at rest. In this state, \[F≤µR\] This means that at this stage, the frictional force is less than or equal to the product of the normal reaction and the coefficient of friction.
The normal reaction is equivalent to the weight of the block although acting in an opposite direction.
The weight of the object, \(W\), is
\[W=mg\]
which is
\[W=100\times9.8\]
Hence, the weight of the object is \(980\, \text{N}\). This implies that
\[R=W=980\, \text{N}\]
The maximum force that can be applied to the body that would still keep it at rest would be so close to or equal to the frictional force. Hence, \[F≤µR\] which is
\[F≤0.4\times980\, \text{N}\]
thus,
\[F≤392\, \text{N}\]
This suggests that the maximum force applied on the rope fitted to the block which would still keep the block static is \(392\, \text{N}\).
Equation of coefficient of friction on an inclined plane
Imagine an object of mass \(m\) is placed on an inclined plane at an angle \(\theta\) to the horizontal. The following images below would guide you.
Fig. 4. Object on an inclined plane.
We see that the block is affected by the weight, normal reaction and friction from the above figure as it tends to slip down the inclined plane at an angle \(\theta\) to the horizontal.
Fig. 5. Defining the angle on an inclined plane using sum of angles in a triangle.
From the above, you can form a right triangle between the weight, \(mg\), and the horizontal. Hence, since the other angle is a right angle, the third angle is
\[180°-(90°+θ)=90°-θ\]
Fig. 6. Defining the angle of an inclined plane using opposite angles.
From the above diagram, we see that the angle formed between the frictional force, \(F\), and the weight is \(90°-θ\) because opposite angles are equal. The third angle in the initial right triangle is opposite to the angle formed by the frictional force and the weight.
Fig. 7. Defining the angle in an inclined plane using angles on a straight line.
From the above figure, we can determine the angle formed between the weight and normal reaction, since they all lie on the straight line of the inclined plane as \[180°-(90°+90°-θ)=θ\]
Recall that the sum of angles on a line is equal to \(180°\).
Fig. 8. Transformation from inclined plane to right triangle.
From the above, you should see that the inclined plane has finally been transformed into a right triangle. This would enable you to apply SOHCATOA to determine the relationship between the weight, normal reaction and friction. Thus,
\[F=mg\sin\theta\] while\[R=mg\cos\theta\]
Recall that \[µ=\frac{F}{R}\]
This means that the coefficient of friction can be derived through
\[µ=\frac{mg\sin\theta }{mg\cos\theta\ }\]
Therefore the equation of the coefficient of friction on an inclined plane is
\[µ=\tan\theta\]
Given that
\[\frac{\sin\theta}{\cos\theta}=\tan\theta\]
An object of mass \(30\, \text{kg}\) is placed on a slope \(38°\) to the horizontal. Find the coefficient of friction.
Solution:
Without much thinking, the coefficient of friction on an inclined plane is the tangent of the angle of inclination. Hence, \[µ=\tan38°\]
which is \[µ=0.78\]
Further examples on the coefficient of friction
To improve your competency in solving problems on the coefficient of friction, here are a few more examples.
A block of mass \(10\, \text{kg}\) is placed on a table and fitted on opposite sides by two springs attached to a \(5\, \text{kg}\) and \(12\, \text{kg}\) mass respectively. If blocks and tables have a standard coefficient of friction of \(0.4\), find the acceleration and tension in the springs.
Solution:
Make a diagram to have a clearer picture of what the question is saying.
Fig. 9. Determining the tension on springs using coefficient of friction.
Now, you need to determine the forces acting on the object on the table and indicate them with a diagram. Here you need to be very careful, note that because the \(12\, \text{kg}\) would pull more force than that of the \(5\, \text{kg}\) mass, thus the object is more likely to move towards the right.
However, this hypothesis of yours depends on if the force is greater than the frictional force, otherwise, the object would remain static on the table.
Hence, the frictional force is acting towards the right to prevent the tension pulled by the \(12\, \text{kg}\) mass.
Fig. 10. An Illustration of forces acting on a body pulled by springs attached to masses.
From the above diagram, you shall understand what happens at each point.
Don't fret, just start from the extreme ends, either left or right, and keep analysing the action of forces till you get to the opposite end.
From the extreme left, we see that the \(5\, \text{kg}\) mass applies a downward force, \(49\, N\), but the system above it causes tension, \(T_2\), which tends to move the mass upwards with an acceleration \(a\). This can thus be expressed as
\[T_2-49\, \text{N}=5\, \text{kg}\times a\]
This is because, in the end, the \(5\, \text{kg}\) mass is pulled up to move to an acceleration, \(a\).
Now, regarding the object on the table, you would observe that the tension, \(T_2\), tends to draw the object towards the left. Also, the frictional force acts towards the left since it tries to obstruct the rightward movement caused by the tension, \(T_1\), acting towards the right. This is expressed as
\[T_1-T_2-F=10\, \text{kg}\times a\]
This is because after the two leftward forces (i.e. \(T_2\) and \(F\) ) have tried to overcome the rightward force \(T_1\) and failed, it is expected that the object of mass \(10\, \text{kg}\) would move towards the right with an acceleration, \(a\).
When you look at the third mass at the left extreme, you would notice that the mass applies a downward force \(117.6\, \text{N}\), and it is being resisted by the upward tension on the spring, \(T_1\). Therefore, this can be expressed as
\[117.6\, \text{N}-T_1=12\, \text{kg}\times a\]
Due to the expectation that the downward force applied by the \(117.6\, \text{N}\) is meant to overpower that of the tension \(T_1\), then, the mass \(12\, \text{kg}\) should supposedly move with an acceleration, \(a\).
Now, we have three equations from the above explained.
These three equations are:
\[T_2-49\, \text{N}=5\, \text{kg}\times a\]
\[T_1-T_2-F=10\, \text{kg}\times a\]
\[117.6\, \text{N}-T_1=12\, \text{kg}\times a\]
Sum up all 3 equations, hence, \[T_2-49\, \text{N}+T_1-T_2-F+117.6\, \text{N}-T_1=5a+10a+12a\] which gives
\[68.6\, \text{N}-F=27a\]
Note that
\[F=µR\]
with
\[µ=0.4\]
and
\[R=W=98\, \text{N}\]
then,
\[F=0.4\times 98\, \text{N}\]
\[F=39.2\, \text{N}\]
Therefore, substitute the value of \(F\) into the equation and arrive at
\[68.6\, \text{N}-39.2\, \text{N}=27\times a\]
which is
\[27a=29.4\, \text{N}\]
Divide both sides by 27 to find the acceleration, \(a\), as
\[a=1.09\, \text{ms}^{-2}\]
To determine the tensions on the springs, \(T_1\) and \(T_2\), we substitute the earlier outlined equations.
Recall that
\[T_2-49\, \text{N}=5\, \text{kg} \times a\]
Therefore,
\[T_2-49\, \text{N}=5\, \text{kg}\times 1.09\, \text{ms}^{-2}\]
this gives
\[T_2-49\text{ N}=5.45\, \text{N}\]
Add \(49\, \text{N}\) to both sides of the equation to get our tension, \(T_2\), as
\[T_2=54.45\, \text{N}\]
Recall that
\[T_1-T_2-F=10\text{ kg} \times a\]
and \(F\) is \(39.2\, \text{N}\), \(a\) is \(1.09\, \text{ms}^{-2}\) and \(T_2\) is \(54.45\, \text{N}\).
Hence, substitute into the equation
\[T_1-54.45\, \text{N}-39.2\, \text{N}=10\, \text{kg}\times 1.09\, \text{ms}^{-2}\]
which gives
\[T_1-93.65\, \text{N}=10.9\, \text{N}\]
Add \(93.65\, \text{N}\) to both sides of the equation to get our tension, \(T_1\), as
\[T_1=104.55\, \text{N}\]
An individual stands immobile on the slope of a mountain and the coefficient of friction between the sole of his feet and the mountain surface is \(0.26\). If in the following year, there was a volcanic eruption that increased the coefficient of friction between the sole of his foot and the mountain by \(0.34\), by what angle has the slope of the mountain increased or decreased?
Solution:
To determine the angle made by the slope of the mountain, we recall that \[µ=\tan\theta\]
Hence the current slope of the mountain has an angle of
\[0.26=\tan\theta\]
Take the inverse to find \(\theta\)
\[\theta=\tan^{-1}(0.26)\]
Hence, the current slope of the mountain has an angle \[\theta=14.57°\]
However, the year after, the mountain experienced an eruption that increased the coefficient of friction by \(0.34\). Thus, the new coefficient of friction is
\[µ_{new}=0.26+0.34\]
which gives
\[µ_{new}=0.6\]
We need to determine the new angle of the slope of the mountain using
\[µ_{new}=\tan\theta\]
Thus,
\[0.6=\tan\theta\]
Take the inverse to find \(\theta\)
\[\theta=\tan^{-1}(0.6)\]
Hence, the new slope of the mountain has an angle
\[\theta=30.96°\]
The mountain slope had a previous angle of \(14.57°\), but upon the eruption it increased to \(30.96°\) by
\[30.96°-14.57°=16.39°\]
Therefore, the eruption increased the angle between the mountain slope by \(16.39°\).
Coefficient of Friction - Key takeaways
- Coefficient of friction, \(\mu\), is the ratio or quotient between the frictional force \((F)\) and normal reaction \((R)\).
- Frictional force is that force which tends to resist or oppose the movement between objects or surfaces in contact.
- For an object moving in contact with a surface the coefficient of friction \(µ\) can thus be calculated with the formula\[\mu=\frac{F}{R}\]
- The coefficient of friction has no unit.
- Negative friction occurs when the decrease in load brings a consequent increase in friction.
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