Have you ever wondered why roads get slippery during rainfall, making it more difficult for a car to stop? Turns out, it’s a direct consequence of the kinetic friction force, as dry asphalt creates a better grip between the tire and the road than wet asphalt, therefore reducing the stopping time of the vehicle.
Kinetic friction is a frictional force that is almost unavoidable in our daily lives. Sometimes it's a halt, but sometimes a necessity. It's there when we play football, use smartphones, walk, write, and do many other common activities. In real-life scenarios, whenever we are considering motion, kinetic friction will always accompany it. In this article, we'll develop a better understanding of what kinetic friction is and apply this knowledge to various example problems.
Kinetic Friction Definition
When you are trying to push a box, you will need to apply a certain amount of force. Once the box starts moving, it is easier to maintain the motion. From experience, the lighter the box, the easier it is to move it.
Let's picture a body resting on a flat surface. If a single contact force \(\vec{F}\) is applied to the body horizontally, we can identify four force components perpendicular and parallel to the surface as shown in the picture below.
Fig. 1 - If an object is placed on a horizontal surface and a horizontal force is applied, kinetic friction force will occur in the opposite direction of the motion and will be proportional to the normal force.
The normal force, \(\vec{F_\mathrm{N}}\), is perpendicular to the surface, and the friction force, \(\vec{F_\mathrm{f}}\),
is parallel to the surface. The friction force is in the opposite direction of the motion.
Kinetic friction is a type of friction force that acts on objects in motion.
It is denoted by \(\vec{F_{\mathrm{f, k}}}\) and its magnitude is proportional to the magnitude of the normal force.
This proportionality relation is quite intuitive, as we know from experience: the heavier the object, the harder it is to get it moving. On a microscopic level, greater mass equals greater gravitational pull; therefore the object will be closer to the surface, increasing the friction between the two.
Kinetic Friction Formula
The magnitude of kinetic friction force depends on the dimensionless coefficient of kinetic friction \(\mu_{\mathrm{k}}\) and the normal force \(\vec{F_\mathrm{N}}\) measured in newtons (\(\mathrm{N}\)). This relationship can be shown mathematically
$$ \vec{F}_{\mathrm{f,k}}=\mu_{\mathrm{k}} \vec{F_\mathrm{N}}.$$
Kinetic Friction Coefficient
The ratio of the kinetic friction force of contacting surfaces to the normal force is known as the coefficient of kinetic friction. It is denoted by \(\mu_{\mathrm{k}}\). Its magnitude depends on how slippery the surface is. Since it is the ratio of two forces, the coefficient of kinetic friction is unitless. In the table below, we can see the approximate coefficients of kinetic friction for some common combinations of materials.
| Materials | Coefficient of kinetic friction, \(\mu_{\mathrm{k}}\) |
| Steel on steel | \(0.57\) |
| Aluminum on steel | \(0.47\) |
| Copper on steel | \(0.36\) |
| Glass on glass | \(0.40\) |
| Copper on glass | \(0.53\) |
| Teflon on Teflon | \(0.04\) |
| Teflon on steel | \(0.04\) |
| Rubber on concrete (dry) | \(0.80\) |
| Rubber on concrete (wet) | \(0.25\) |
Now that we know the equation for calculating the kinetic friction force and have familiarized ourselves with the kinetic friction coefficient, let's apply this knowledge to some example problems!
Kinetic Friction Examples
To begin with, let's look at a simple case of directly applying the kinetic friction equation!
A car is moving at a uniform speed with the normal force of \(2000 \, \mathrm{N}\). If the kinetic friction applied on this car is \(400 \, \mathrm{N}\). Then compute the coefficient of the kinetic friction involved here?
Solution
In the example, the magnitudes of normal force and kinetic friction force are given. So, \(\vec{F}_{\mathrm{f,k}}=400 \, \mathrm{N}\) and \(F_\mathrm{N}= 2000 \, \mathrm{N}\). If we put these values in the kinetic friction formula
$$ \vec{F}_{\mathrm{f,k}}=\mu_{\mathrm{k}} \vec{F_\mathrm{N}},$$
we obtain the following expression
$$400 \, \mathrm{N} =\mu_{\mathrm{k}} \cdot 2000 \, \mathrm{N}, $$
which can be rearranged to find the coefficient of friction
$$ \begin{align} \mu_{\mathrm{k}} &= \frac{400\,\cancel{N}}{2000 \, \cancel{N}} \\ \mu_{\mathrm{k}}&=0.2.\end{align} $$
Now, let's look at a slightly more complicated example involving various forces acting on a box.
A \(200.0\, \mathrm{N}\) box needs to be pushed across a horizontal surface. Imagine dragging the rope up and \(30 ^{\circ}\) above the horizontal to move the box. How much force is required to maintain a constant velocity? Assume \(\mu_{\mathrm{k}}=0.5000\).
Fig. 2 - All the forces acting on the box - the normal force, weight, and a force at \(30 ^{\circ}\) to the horizontal surface. The kinetic friction force is in the opposite direction of the force.
Solution
In the example, it says we want to maintain a constant velocity. A constant velocity means that the object is in a state of equilibrium (i.e. the forces balance each other). Let's draw a free-body diagram to understand the forces better and look at the horizontal and vertical components.
Fig. 3 - Free-body diagram of the box. There are forces both in horizontal and vertical direction.
When we look at the perpendicular force components, upward forces should be equal to downward forces in magnitude.
Normal force does not always equal weight!
Now, we can write two separate equations. We'll use the fact that the sum of forces in the \(x\) and \(y\) directions, equal to zero. So, the horizontal forces are
$$ \sum F_\mathrm{x} = 0,$$
which, based on the free body diagram can be expressed as
$$ T \cdot \cos 30 ^{\circ} = F_{\mathrm{f,k}}=\mu_{\mathrm{k}} F_\mathrm{N}.$$
Vertical forces are also
$$ \sum F_\mathrm{y} = 0,$$
and give us the following equation
$$ F_\mathrm{N} + T \cdot \sin 30 ^{\circ} = w.$$
So \(F_\mathrm{N} = w - T \cdot \sin 30 ^{\circ}\). We can insert the \(F_\mathrm{N}\) value into the equation for the horizontal components
$$ \begin{align} T \cdot \cos 30 ^{\circ} &= \mu_\mathrm{k} (w - T \cdot \sin 30 ^{\circ} ) \\ T \cdot \cos 30 ^{\circ} &= \mu_\mathrm{k} w - \mu_\mathrm{k} \cdot \sin 30 ^{\circ} ), \end{align} $$
and gather and simplify all the like terms on the left-hand side
$$ \begin{align}T ( \cos 30 ^{\circ} + \mu_\mathrm{k} \cdot \sin 30 ^{\circ} ) &= \mu_\mathrm{k} w \\ T(\cos 30 ^{\circ} + \mu_\mathrm{k} \cdot \sin 30 ^{\circ}) &= \mu_\mathrm{k} w. \end{align} $$
Now we can plug in all the corresponding values and calculate the force \(T\):
$$ \begin{align} T &= \frac{\mu_\mathrm{k} w}{\cos 30 ^{\circ} + \mu_\mathrm{k} \cdot \sin 30 ^{\circ}} \\ T &= \frac{0.5000 \cdot 200.0 \, \mathrm{N}}{0.87 + 0.5000 \cdot 0.5} \\ T &= 89.29 \, \mathrm{N}. \end{align}$$
Finally, let's look at a similar example, only this time the box is placed on an inclined plane.
A box is sliding down at a constant velocity from an inclined plane that is at an angle \(\alpha\) with the horizontal. The surface has a coefficient of kinetic friction \(\mu_{\mathrm{k}}\). If the weight of the box is \(w\), find the angle \(\alpha\).
Fig. 4 - A box sliding down an inclined plane. It's moving at a constant velocity.
Let's look at the forces acting on the box in the figure below.
Fig. 5 - All the forces acting on a box sliding down an inclined plane. We can apply a new coordinate system to write the related equations.
If we attain new coordinates (\(x\) and \(y\)), we see that in the \(x\)-direction there is kinetic friction force and a horizontal component of weight. In the \(y\)-direction, there is the normal force and vertical component of weight. Since the box is moving at a constant velocity, the box is at equilibrium.
- For \(x\)-direction: \(w\cdot\sin\alpha=F_\mathrm{f,k} = \mu_{\mathrm{k}}F_\mathrm{N}\)
- For \(y\)-direction: \(F_\mathrm{N}=w\cdot\cos\alpha\)
We can insert the second equation into the first equation:
$$ \begin{align} w \cdot \sin\alpha & =\mu_\mathrm{k}w \cdot \cos\alpha \\ \cancel{w}\cdot\sin\alpha & =\mu_\mathrm{k} \cancel{w} \cdot \cos\alpha \\ \mu_\mathrm{k} & = \tan\alpha \end{align}$$
Then the angle \(\alpha\) is equal to
$$ \alpha = \arctan\mu_\mathrm{k}.$$
Static Friction vs Kinetic Friction
Altogether, there are two forms the coefficient of friction may take, kinetic friction being one of them. The other type is known as the static friction. As we have established by now, kinetic friction force is a type of frictional force acting on the objects that are in motion. So, what is the difference between static friction and kinetic friction exactly?
Static friction is a force that ensures that objects at rest relative to each other remain stationary.
In other words, kinetic friction applies to objects that are moving, meanwhile static friction is relevant for motionless objects.
The difference between the two types can be remembered directly from the vocabulary. While static means lacking in movement, kinetic means relating to or resulting from motion!
Mathematically, static friction \(F_\mathrm{f,s}\) looks very similar to kinetic friction,
$$ F_\mathrm{f,s} = \mu_\mathrm{s}F_\mathrm{N}$$
where the only difference is the use of a different coefficient \(\mu_\mathrm{s}\), which is the coefficient of static friction.
Let's look at an example, where an object experiences both types of friction.
A heavy box is resting on a table and remains stationary until some force is applied horizontally to slide it across the table. Because the surface of the table is quite bumpy, initially the box isn't moving, despite the applied force. As a result, the box is pushed even harder until, eventually, it starts moving across the table. Explain the different stages of the forces experienced by the box and plot friction versus the applied force.
Solution
- At first, no forces are applied to the box, so it only experiences the gravitational pull downwards and the normal force from the table pushing it upwards.
- Then, some pushing force \(F_\mathrm{p}\) is applied horizontally to the box. As a result, there will be resistance in the opposite direction, known as friction \(F_\mathrm{f}\).
- Considering the box is heavy and the surface of the table is bumpy, the box won't easily slide over, as both of these characteristics will affect friction.
The normal force and the roughness/smoothness of the surfaces involved are the main factors affecting friction.
- So, depending on the magnitude of the applied force, the box will remain stationary due to static friction \(F_\mathrm{f,s}\).
- With increasing applied force, eventually, \(F_\mathrm{p}\) and \(F_\mathrm{f,s}\) will be of the same magnitude. This point is known as the threshold of motion, and once reached, the box will start moving.
- Once the box starts moving, the friction force affecting the motion will be the kinetic friction \(F_\mathrm{f,k}\). It will become easier to maintain its motion, as the coefficient of friction for moving objects usually is less than that of stationary objects.
Graphically, all of these observations can be seen in the figure below.
Fig. 6 - Friction plotted as a function of the applied force.
Kinetic Friction - Key takeaways
- The kinetic friction force is a type of frictional force acting on the objects that are in motion.
- The magnitude of kinetic friction force depends on the coefficient of kinetic friction and the normal force.
- The ratio of the kinetic friction force of contacting surfaces to the normal force is known as the coefficient of kinetic friction.
- The equation used to calculate the coefficient of friction is \(\mu_{\mathrm{k}} = \frac{\vec{F}_{\mathrm{f,k}}}{\vec{F}_\mathrm{N}}\).
- The coefficient of kinetic friction depends on how slippery the surface is.
- Normal force does not always equal weight.
- Static friction, is a type of friction applied to stationary objects.
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