how to find friction force
When any two bodies come into contact, and they have relative motion between them, friction occurs. Friction can be define as a force that opposes the movement of two contacting surfaces, which slide relative to one another.
Friction is part of our daily lives and allows certain activities to take place. For example, walking and running requires friction, a vehicle accelerating or braking on the road, requires friction. The braking systems, clutch systems, found in vehicles, are based on the friction principles also.
The friction force depends on how smooth or rough the contacting surfaces are. At a microscopic level, even the smoothest surfaces have irregularities which come into contact and grip to each other. In the images below you can clearly see how the microstructure of the friction surface prevents movement between the two objects (solid bodies).
| Image: Contact between two surfaces | Image: Friction surface microstructure |
Depending on the state of the sliding surfaces, there are several types of friction:
- dry friction
- wet friction
Dry friction, as the name implies, occurs between two dry solid bodies, with no lubrication (oil, grease, etc.) between them. One example of dry friction is the braking system of a vehicle. Between the brake disc and the brake pad there is direct contact without any lubrication. Dry friction is also known as Coulomb friction, after the name of the French physicist Charles-Augustin de Coulomb who studied dry friction in depth.
Wet friction, occurs when there is a lubricant between the sliding surfaces. Examples of wet friction are: piston rings inside a cylinder, pumping elements inside a fuel pump, lock-up clutch of a torque converter, etc.
In this article we are going to focus on the characteristics of dry friction and how to calculate the friction force.
If we have a body on a rigid surface, and we apply a pushing force to it, the body will not move. This happens because the friction force between the body and the surface is opposite to the pushing force and keeps the body in place.
Image: Body equilibrium with friction
where:
Fp [N] – pushing force
Ff [N] – friction force
G [N] – body weight
N [-] – normal reaction
From the free body diagram, as long as the body is in equilibrium (does not move), we can write the force equilibrium equations for both axes (x and y).
Horizontal equilibrium:
\[\sum F_{x} = 0\]
\[F_{f}-F_{p} = 0 \tag{1}\]
\[F_{f}=F_{p} \tag{2}\]
Vertical equilibrium:
\[\sum F_{y} = 0\]
\[N-G = 0 \tag{3}\]
\[N = G \tag{4}\]
Equation (1) is true as long as the body doesn't move. The question is, with what force do we need to push the body, so that it starts moving?
The friction force is a reaction to the pushing force. If there is no pushing force, there is no friction force as well. If we keep pushing harder the body, it will eventually start moving. This happens because our pushing force became higher than the maximum friction force and the body is no longer in equilibrium.
The maximum value of the friction force, also called static friction force, depends on the normal force. The normal force is defined as the reaction force of the standing surface on the body, due to the body's weight force. If there was no normal reaction force to balance the weight force of the body, the body would sink into the standing surface.
Experimentally, it has been determined that this limiting static frictional force is directly proportional to the resultant normal force, and is calculated as:
\[\bbox[#FFFF9D]{F_{f} = \mu \cdot N} \tag{5}\]
where:
μ [-] – coefficient of friction
To answer the above question, the body will start to move when the pushing force is bigger than the friction force, which, from equations (4) and (5), means:
\[F_{p} \ge \mu \cdot G \tag{6}\]
We know that the body weight is calculated as:
\[G = m \cdot g \tag{7}\]
where:
m [kg] – vehicle mass
g [m/s2] – gravitational acceleration
Combining equations (6) and (7), gives the value of the pushing force function of the body mass and friction coefficient:
\[F_{p} \ge 9.81 \cdot \mu \cdot m \tag{8}\]
The friction coefficient depends on the relative movement between the friction surfaces. If there is no movement, the friction coefficient is called coefficient of static friction. If there is movement (speed) between the friction surfaces (bodies), the friction coefficient is called coefficient of kinetic friction.
Image: Coefficient of friction function of speed
where:
v [m/s] – relative speed between the two bodies in contact
μs [-] – coefficient of static friction
μk [-] – coefficient of kinetic friction
For two given bodies, the coefficient of friction is not fixed, it depends on the speed of relative movement of the bodies in contact. The maximum value of the coefficient is reached when there is no movement (zero speed) between the sliding surfaces. In this case the coefficient of kinetic friction becomes coefficient of static friction.
Mathematically, we ca define the coefficient of friction as:
\[ \begin{split}
\mu(v) = \left\{\begin{matrix}
\mu_{s} \text{ , if } v=0\\
\mu_{k}(v) \text{ , if } v>0
\end{matrix}\right.
\end{split} \]
In reality, the value of the kinetic coefficient of friction depends on the value of the relative speed of the friction surfaces. This happens because, the higher the relative friction speed, the higher the generated heat, the higher the changes of the material's properties. For simplicity, throughout this article, we are going to consider that the kinetic coefficient of friction is constant with speed.
Since the coefficient of friction depends on the relative speed of the sliding surfaces, the friction force will have the same dependency. The maximum friction force is achieved when the body is static, when it starts to move, the friction force drops along with the coefficient of friction.
Image: Friction force function of push force
where:
Ffmax [N] – maximum friction force
Ffs [N] – static friction force
Ffk [N] – kinetic friction force
Fp [N] – pushing force
x [m] – body (object) displacement
From the image above we can draw the following conclusions:
- in the static region, the friction force is directly proportional with the pushing force
- the maximum friction force is obtained in the static region
- when the pushing force exceeds the maximum static friction force, the body starts to move and the friction force drops
The maximum (static) friction force is calculated as:
\[F_{fmax} = \mu_{s} \cdot N \tag{9}\]
The maximum kinetic friction force is calculated as:
\[F_{fk} = \mu_{k} \cdot N \tag{10}\]
The coefficient of friction is determined experimentally, but there are some values available in physics books. The table below contains the values for the coefficient of friction compiled from the references [1], [2] and [3].
| Coefficients of friction | ||
|---|---|---|
| Material | μs [-] | μk [-] |
| Steel on steel | 0.74 | 0.57 |
| Steel on ice | 0.027 | 0.014 |
| Metal on ice | 0.03 – 0.05 | – |
| Aluminium on steel | 0.61 | 0.47 |
| Copper on steel | 0.53 | 0.36 |
| Copper on copper | 0.74 – 1.21 | – |
| Rubber on concrete | 1.00 | 0.80 |
| Wood on wood | 0.25 – 0.50 | 0.2 |
| Oak on oak (dry, along fibres) | 0.62 | 0.48 |
| Oak on oak (dry, perpendicular on fibres) | 0.54 | 0.34 |
| Glass on glass | 0.94 | 0.4 |
| Waxed wood on wet snow | 0.14 | 0.1 |
| Waxed wood on dry snow | – | 0.04 |
| Metal on metal (lubricated) | 0.15 | 0.06 |
| Ice on ice | 0.1 | 0.03 |
| Teflon on Teflon | 0.04 | 0.04 |
| Synovidal joints on humans | 0.01 | 0.003 |
| Bronze on bronze (slightly lubricated) | – | 0.2 |
| Cast iron on cast iron (slightly lubricated) | – | 0.15 |
| Cast iron on bronze (slightly lubricated) | – | 0.15 |
| Leather belt on cast iron (slightly lubricated) | 0.28 | – |
| Leather on wood | 0.20 – 0.50 | – |
| Leather on metal | 0.03 – 0.60 | – |
For a better understanding on how to calculate the friction force, let's go through some practical examples.
Example 1. A person is pushing a 50 kg aluminium crate on a steel floor. Calculate with what force does the person needs to push, for the crate to start moving. Once in motion, what is the required push force, to keep the crate sliding? What is the reduction in push force, when the crate starts moving?
Image: Pushing crate – friction force
From the definition of the problem, we can extract the input data as:
\[ \begin{split}
m &= 50 \text{ kg} \\
\mu_{s} &= 0.61\\
\mu_{k} &= 0.47\\
\end{split} \]
Step 1. Calculate the normal reaction force on the crate.
\[ \begin{split}
\sum F_{y} &= 0 \\
N – G &= 0\\
N &= G
\end{split} \]
Step 2. Calculate the static friction force using equation (5).
\[F_{fs} = \mu_{s} \cdot N = \mu_{s} \cdot G = \mu_{s} \cdot m \cdot g = 0.61 \cdot 50 \cdot 9.81 = 298.9 \text{ N}\]
Step 3. Calculate the push force from the equilibrium condition.
\[ \begin{split}
\sum F_{x} &= 0 \\
F_{p} – F_{f} &= 0\\
F_{p} &= F_{f}
\end{split} \]
For the crate to start moving, the push force needs to be higher than 298.9 N.
Step 4. Calculate the kinetic friction force.
\[F_{fk} = \mu_{k} \cdot N = \mu_{k} \cdot G = \mu_{k} \cdot m \cdot g = 0.47 \cdot 50 \cdot 9.81 = 230.535 \text{ N}\]
For the crate to keep moving, the push force needs to be higher than 230.535 N.
Step 5. The reduction in push force is calculated as:
\[\Delta F_{f} \text{ [%]} = \frac{\left | F_{fk} – F_{fs} \right |} {F_{fs}} \cdot 100 = 22.87 \text{ %}\]
As you can see, the push force needs to be higher to start moving the crate and a bit lower to keep the crate moving.
Example 2. A person is pulling a 50 kg aluminium crate on a steel floor, using a rope at a 30° angle with the horizontal. Calculate with what force does the person needs to pull, for the crate to start moving. Once in motion, what is the required pull force, to keep the crate sliding? What is the reduction in pull force, when the crate starts moving? Compared with the previous example, is it easier to pull or push a siding body (object)?
Image: Pulling crate – friction force
From the definition of the problem, we can extract the input data as:
\[ \begin{split}
m &= 50 \text{ kg} \\
\mu_{s} &= 0.61\\
\mu_{k} &= 0.47\\
\alpha &= 30^{\circ}
\end{split} \]
Step 1. Calculate the normal reaction force on the crate.
\[ \begin{split}
\sum F_{y} &= 0 \\
N + F_{py} – G &= 0\\
N + F_{p} \cdot \sin(\alpha) – G &= 0\\
N &= G – F_{p} \cdot \sin(\alpha)
\end{split} \]
Step 2. Calculate the static friction force using equation (5).
\[F_{fs} = \mu_{s} \cdot N = \mu_{s} \left ( G – F_{p} \cdot \sin(\alpha) \right )\]
Step 3. Calculate the static friction force from the horizontal force equilibrium.
\[ \begin{split}
\sum F_{x} &= 0 \\
F_{px} – F_{fs} &= 0\\
F_{p} \cdot \cos(\alpha) &= F_{fs}
\end{split} \]
Step 4. From step 2 and 3 we calculate the value of the (static) pull force to start moving the crate.
\[ \begin{split}
F_{ps} \cdot \cos(\alpha) &= \mu_{s} \cdot \left ( G – F_{p} \sin(\alpha) \right )\\
F_{ps} &= \frac{\mu_{s} \cdot m \cdot g}{\cos(\alpha) + \mu_{s} \cdot \sin(\alpha)}\\
F_{ps} &= 255.51 \text{ N}
\end{split} \]
For the crate to start moving, the pull force needs to be higher than 255.51 N.
We can also calculate the static friction force as:
\[F_{fs} = F_{ps} \cdot \cos(\alpha) = 221.28 \text{ N}\]
Step 5. Calculate the kinetic pull force.
\[F_{pk} = \frac{\mu_{k} \cdot m \cdot g}{\cos(\alpha) + \mu_{k} \cdot \sin(\alpha)} = 209.38 \text{ N}\]
For the crate to keep moving, the push force needs to be higher than 209.38 N.
We can also calculate the kinetic friction force as:
\[F_{fk} = F_{pk} \cdot \cos(\alpha) = 181.33 \text{ N}\]
Step 6. The reduction in pull force is calculated as:
\[\Delta F_{p} \text{ [%]} = \frac{\left | F_{pk} – F_{ps} \right |} {F_{ps}} \cdot 100 = 18.05 \text{ %}\]
As you can see, the pull force needs to be higher to start moving the crate and a bit lower to keep the crate moving.
Step 7. Compared with the push force, the pull force needs to be smaller to get the crate moving. The difference in pushing and pulling force for the static and kinematic regions are:
\[ \begin{split}
\Delta F_{ps} \text{ [%]} &= \frac{\left | 255.507 – 298.9 \right |} {298.9} \cdot 100 = 14.52 \text{ %}\\
\Delta F_{pk} \text{ [%]} &= \frac{\left | 209.38 – 230.535 \right |} {230.535} \cdot 100 = 9.18 \text{ %}
\end{split} \]
This is due to the effect of the angle of the rope, which reduces the normal reaction on the crate, hence the friction force. This decrease in normal force on the crate is however translated into a vertical force into the shoulder of the person, pressing down. This vertical force is equal with:
\[ \begin{split}
\text{static: }F_{ps} \cdot \sin(\alpha) &= 127.75 \text{ N}\\
\text{kinetic: }F_{pk} \cdot \sin(\alpha) &= 104.69 \text{ N}
\end{split} \]
Example 3. For the bodies in the figure below, find the value of the body mass m2, for which the body mass m1 = 30 kg starts to move upwards. The body 1 and the ramp are made from metal, with the sliding surfaces lubricated. The angle of the ramp is 45°.
Image: Diagram for friction force exercise
From the definition of the problem, we can extract the input data as:
\[ \begin{split}
m_{1} &= 30 \text{ kg} \\
\mu_{s} &= 0.15\\
\alpha &= 45^{\circ}
\end{split} \]
Step 1. Calculate the value of the pulling force Fp1 [N].
Horizontal equilibrium:
\[ \begin{split}
\sum F_{x} &= 0 \\
F_{p1} – F_{f1} – G_{1} \cdot \sin(\alpha) &= 0\\
F_{p1} &= F_{f1} + G_{1} \cdot \sin(\alpha)
\end{split} \]
Vertical equilibrium:
\[ \begin{split}
\sum F_{y} &= 0 \\
N – G_{1} \cdot \cos(\alpha) &= 0\\
N &= G_{1} \cdot \cos(\alpha)
\end{split} \]
The static friction force Ff1 [N] is calculated as:
\[F_{f1} = \mu_{s} \cdot N = \mu_{s} \cdot G_{1} \cdot \cos(\alpha)\]
Now we can get the expression of the pulling force function of the known parameters:
\[ \begin{split}
F_{p1} &= G_{1} \cdot \left ( \mu_{s} \cdot \cos(\alpha) + \sin(\alpha) \right )\\
&= m_{1} \cdot g \cdot \left ( \mu_{s} \cdot \cos(\alpha) + \sin(\alpha) \right )\\
&= 185.38 \text{ N}
\end{split} \]
Step 2. Calculate the value of the pulling force Fp2 [N].
Vertical equilibrium:
\[ \begin{split}
\sum F_{y} &= 0 \\
F_{p2} – G_{2} &= 0\\
F_{p2} &= G_{2}\\
F_{p2} &= m_{2} \cdot g
\end{split} \]
Step 3. Calculate m2 [kg].
The tension in the rope (wire) linking the two bodies is equal on both sides. From this condition we can calculate m2 as:
\[ \begin{split}
F_{p2} &= F_{p1}\\
m_{2} \cdot 9.81 &= 185.38 \text{} \\
m_{2} &= 18.9 \text{ kg}
\end{split} \]
If the body mass m2 is heavier than 18.9 kg, the body mass m1 will start sliding upwards onto the ramp.
Friction Force Calculator
The calculator below gives the values of the pulling forces (static and kinematic) and friction forces (static and kinematic) for a rigid body of mass m [kg] pulled on a surface with μs [-] and μk [-], with a pulling force angle α [°].
Image: Friction force diagram for calculator
The equations used to calculate the forces are based on the expressions from Example 2. For the equations to make sense, the mass value is limited to positive values only, the friction coefficient between [0, 2] and the pull force angle between [0°, 90°].
| m [kg] = | μs [-] = | μk [-] = | α [°] = |
| Static Pull Force, Fps [N] = | Static Friction Force, Ffs [N] = | ||
| Kinetic Pull Force, Fpk [N] = | Kinetic Friction Force, Ffk [N] = | ||
The results are quite interesting. They show that the optimum angle to have the minimum pulling force is 31° (static) and 25° (kinetic). Also, as expected, the friction forces are maximum at 0° angle and null at 90° angle.
References:
[1] R.C. Hibbeler, Engineering Mechanics – Statics, 14th Edition, Pearson, 2017.
[2] M. Radoi, E. Deciu, Mecanica, Editia a II-a revizuita, Editura Didactica si Pedagogica, Bucuresti, 1981.
[3] Physics : For Scientists and Engineers 6TH EDITION by Raymond A. Serway and John W. Jr. Jewett. Brooks/Cole Publishing Co.,2004.
how to find friction force
Source: https://x-engineer.org/undergraduate-engineering/physics/solid-mechanics/how-to-calculate-friction-force/
Posted by: graceevente1966.blogspot.com
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