6.13: Forces and constaints
In this section we discuss the properties of some of the most relevant forces and constraints that can occur in dynamics, as have been experimentally approximately determined using Newton’s laws.
6.13.1 Gravitational force
The gravitational force is important, because it acts on every point mass, and because it is the most prominent force that acts over a distance. In this textbook we focus on the dynamics of objects on or near the surface of the earth, and therefore we can often assume a constant distance to the earth’s centre. In this case the gravitational force vector on a mass \(m_{i}\) can be described by
\[\overrightarrow{\boldsymbol{F}}_{i, g}=m_{i} \overrightarrow{\boldsymbol{g}} \tag{6.27} \label{6.27}\]
where \(\overrightarrow{\boldsymbol{g}}\) is the gravitational acceleration vector with magnitude \(g=|\overrightarrow{\boldsymbol{g}}| \approx\) \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) that points downward in the direction of the centre of the earth.
6.13.2 Contact force
At a point \(C\) where the surfaces of two objects touch, the atoms in the objects can generate a contact force vector \(\overrightarrow{\boldsymbol{F}}_{C}\). One can define a contact plane through \(C\) that is tangential to the surface of the objects. Then, like shown in Figure 6.3, the component of the contact force \(\overrightarrow{\boldsymbol{F}}_{C}\) perpendicular to this plane is called the normal force \(\overrightarrow{\boldsymbol{F}}_{N}\), and the component parallel to the contact plane is the called the friction force \(\overrightarrow{\boldsymbol{F}}_{f}\). So, it always holds that \(\overrightarrow{\boldsymbol{F}}_{C}=\overrightarrow{\boldsymbol{F}}_{N}+\overrightarrow{\boldsymbol{F}}_{f}\), and it is allowed to draw and treat the contact force as a single vector \(\overrightarrow{\boldsymbol{F}}_{C}\). However, since the expressions for \(\overrightarrow{\boldsymbol{F}}_{N}\) and \(\overrightarrow{\boldsymbol{F}}_{f}\) are different, it can be useful to split them. Let us now discuss the normal forces and friction forces separately.
6.13.3 Normal force
Normal forces are the result of repulsive Coulomb forces and quantum mechanical effects. By definition, the normal force vector \(\overrightarrow{\boldsymbol{F}}_{N}\) points perpendicular to the contact plane. Normal forces can be calculated by combining Newton’s laws with constraint equations. If the contact surface is flat, the constraint equations for the motion components normal to the surface simplify to \(v_{n}=0\) and \(a_{n}=0\), otherwise one needs to use Equation 5.86. An example of how to determine the normal force \(F_{N}\) from the EoM was given in Equation 6.10 and Equation 6.17, where the movement of block \(A\) was constrained by the slope.
6.13.4 Friction force
The friction force is the component of the contact force tangential to the contact plane. We distinguish static or kinetic friction forces, which are characterised by a static \(\mu_{s}\) and kinetic \(\mu_{k}\) friction coefficient with \(\mu_{s} \geq \mu_{k}>0\). We now consider the friction force between two points \(A\) and \(B\) on different objects that are in contact.
Static friction
The static friction acting on point \(A\) is best described by this constraint equation:
\[\left\{\begin{array}{cl} a_{A / B, t}=0 & \text { if }\left|\overrightarrow{\boldsymbol{F}}_{A, f s}\right| \leq \mu_{s}\left|F_{A, N}\right| \text { and } \overrightarrow{\boldsymbol{v}}_{A / B}=\overrightarrow{\mathbf{0}} \tag{6.28} \label{6.28}\\[4pt] & \text { otherwise use kinetic friction } \end{array}\right.\]
The first condition for static friction is that the points \(A\) and \(B\) are not moving relative to each other \(\overrightarrow{\boldsymbol{v}}_{A / B}=0\). In that case Equation 6.28 states that the tangential component of acceleration of point \(A\) is zero \(\left(a_{A / B, t}=0\right)\) as long as the friction force \(\left|\overrightarrow{\boldsymbol{F}}_{A, f_{s}}\right|\) needed to satisfy that constraint equation does not exceed the maximum static friction force \(F_{A, f s, \max }=\mu_{s}\left|F_{A, N}\right|\), where \(F_{A, N}\) is the normal force on \(A\). In practice the EoM in the presence of the constraint \(a_{A / B, t}=0\) needs to be solved to determine the friction force \(\overrightarrow{\boldsymbol{F}}_{A, f s}\) after which it is checked if the static friction condition holds. If it does not hold, point \(A\) will accelerate and kinetic friction can be used.
Kinetic friction
When the points \(A\) and \(B\) slide or slip along each other, a kinetic friction force acts on \(A\). The kinetic friction force vector \(\overrightarrow{\boldsymbol{F}}_{A, f k}\) is tangential to the plane of contact and points opposite to the direction of the relative velocity vector:
\[\overrightarrow{\boldsymbol{F}}_{A, f k}= \begin{cases}-\mu_{k}\left|\overrightarrow{\boldsymbol{F}}_{A, N}\right| \hat{\boldsymbol{v}}_{A / B} & \text { if } \overrightarrow{\boldsymbol{v}}_{A / B} \neq 0 \text { or }\left|\overrightarrow{\boldsymbol{F}}_{A, f s}\right|>\mu_{s}\left|F_{A, N}\right| \tag{6.29} \label{6.29}\\[4pt] & \text { otherwise use static friction }\end{cases}\]
Solving problems with friction
If it is not known whether one should deal with static or kinetic friction, and/or if the direction of the velocity is not known in a kinetic friction situation, the following procedure can be used to solve the problem step by step:
- Determine the normal contact force \(F_{N}\) using constraint equations and the EoMs, like in Equation 6.17.
- Assume that the relative velocity is zero \(\overrightarrow{\boldsymbol{v}}_{A / B}=\overrightarrow{\mathbf{0}}\) and use the constraint equation \(a_{A / B, t}=0\) to determine the static friction force \(\overrightarrow{\boldsymbol{F}}_{A, f_{s}}\) from the EoMs.
- Check if \(\left|\overrightarrow{\boldsymbol{F}}_{A, f_{s}}\right| \leq \mu_{s} F_{N}\). If that is true, the static friction assumption is correct.
- Otherwise use kinetic friction. Guess a direction for the velocity \(\hat{\boldsymbol{v}}_{A / B}\) and use it to determine \(\overrightarrow{\boldsymbol{F}}_{A, f k}\) using Equation 6.29.
- Then solve the EoMs to determine if the direction of the velocity \(\overrightarrow{\boldsymbol{v}}_{A / B}\) matches the guessed direction of \(\hat{\boldsymbol{v}}_{A / B}\).
- If not, repeat from step 4 for the other direction of \(\hat{\boldsymbol{v}}_{A / B}\).
For friction of a rolling wheel with the ground two conditions can occur. If there is no slip, only static friction is present and \(a_{A / B, t}=0\). While in the case of slip or sliding, one deals with kinetic friction forces. Gearwheels always behave according to the no slip condition, since their teeth prevent slipping. Finally, a special situation are so-called frictionless contact problems, which have \(\mu_{s}=\mu_{k}=0\). For these cases the friction force is always zero \(\overrightarrow{\boldsymbol{F}}_{f}=\overrightarrow{\mathbf{0}}\), and the contact force is always normal to the contact plane.
6.13.5 Constraint equations on points
Some examples of common constraint equations that limit the motion of a point \(A\) in 3,2 or 1 dimensions are:
- A joint that fixes a point \(A\) to a certain other point \(P\), fixing it along all 3 coordinate directions:
\(\overrightarrow{\boldsymbol{r}}_{A}=\overrightarrow{\boldsymbol{r}}_{P}\)
- A point \(A\) is constrained to move along a slider joint, collar or rail with path curve \(\overrightarrow{\boldsymbol{r}}_{s}(s)\), constraining it along 2 coordinate axes:
\(\overrightarrow{\boldsymbol{r}}_{A}=\overrightarrow{\boldsymbol{r}}_{s}(s)\).
- Motion over a surface, for instance because the gravitational force ensures object \(A\) does not move in the \(z\) direction, constraining motion along 1 coordinate direction:
\(z_{A}(t)=0\)
- Angular constraints can fix certain rotation angles, e.g. \(\phi(t)=\phi_{0}\), they can result in moments on the constrained object.
The contact forces and moments resulting from these constraint equations can be analyzed by combining them with the EoMs.
6.13.6 Dynamics of massless mechanisms
When a mechanism \(m\) has zero mass, so \(m_{m}=0\), application of Newton’s second law tells us that the sum of external forces that act on the mechanism is zero, since otherwise its acceleration would be infinite:
\[\sum \overrightarrow{\boldsymbol{F}}_{\mathrm{ext}}=m_{m} \overrightarrow{\boldsymbol{a}}_{m}=\overrightarrow{\mathbf{0}} \tag{6.30} \label{6.30}\]
This means that the sum of the forces on the mechanism is always zero, just like in statics. Therefore you can use all methods you have learned in statics courses to determine the positions and forces of the massless elements in the mechanism. Unless it is explicitly mentioned they have mass, you may assume all mechanisms in this textbook to be massless.
6.13.7 Two-force members
A massless mechanism on which only two external forces \(\overrightarrow{\boldsymbol{F}}_{A}\) and \(\overrightarrow{\boldsymbol{F}}_{B}\) are acting is called a two-force member. Then we get from Equation 6.30 that \(\overrightarrow{\boldsymbol{F}}_{A}+\overrightarrow{\boldsymbol{F}}_{B}=\overrightarrow{\boldsymbol{0}}\) and:
\[\overrightarrow{\boldsymbol{F}}_{A}=-\overrightarrow{\boldsymbol{F}}_{B} \tag{6.31} \label{6.31}\]
We see from Equation 6.30 that because the mass of the mechanism is zero \(\left(m_{m}=0\right)\), its acceleration \(\overrightarrow{\boldsymbol{a}}_{m}\) can take any value, which means that the mechanism can translate extremely fast, such that it will instantaneously assume a position that ensures \(\sum \overrightarrow{\boldsymbol{F}}=\overrightarrow{\mathbf{0}}\). Note that, as will become clearer later, the same argumentation holds for rotations, such that the mechanism also instantaneously assumes a state where the sum of external moments is zero \(\sum \overrightarrow{\boldsymbol{M}}=\overrightarrow{\mathbf{0}}\), which is only possible if the two forces \(\overrightarrow{\boldsymbol{F}}_{A}\) and \(\overrightarrow{\boldsymbol{F}}_{B}\) are collinear. We will now discuss a few important massless mechanisms, show in Figure 6.7.
6.13.8 Rods and ropes
Probably the simplest two-force members are rods and ropes. Since these mechanisms have a fixed length \(L\) their operation is governed by this constraint equation:
\[\left|\overrightarrow{\boldsymbol{r}}_{A}-\overrightarrow{\boldsymbol{r}}_{B}\right|=L \tag{6.32} \label{6.32}\]
Ropes are flexible, therefore they can only handle tensile (’pulling’) forces that are parallel to the rope. The absolute value of those force vectors is called the tension in the rope and Equation 6.32 only holds if the rope is experiencing tensile force. If ropes are bent (e.g. by pulleys) they can be dealt with as discussed in 5.2.4. Since rods are rigid, they can handle also compressive forces and can transfer moments.
As shown in second FBD in Figure 6.7 the sum of the forces on a rope is not zero anymore \(\left(\overrightarrow{\boldsymbol{F}}_{A}+\overrightarrow{\boldsymbol{F}}_{B} \neq 0\right)\) if there is mass connected to the rope, because part of the force \(\overrightarrow{\boldsymbol{F}}_{A}\) is needed to accelerate the mass. So, be careful, \(\overrightarrow{\boldsymbol{F}}_{A}=-\overrightarrow{\boldsymbol{F}}_{B}\) can only be applied when no masses are accelerated by the forces.
6.13.9 Spring
A linear spring is a mechanism whose length extends by a distance \(\Delta L\) that is proportional to the external pulling force (see Figure 6.7). When there is no external force \(\left(\overrightarrow{\boldsymbol{F}}_{A}=\overrightarrow{\boldsymbol{F}}_{B}=\overrightarrow{\mathbf{0}}\right)\), it has a relaxed length \(L=L_{0}\), and when a force is applied its length increases by a distance \(\Delta L=L-L_{0}\) :
\[\overrightarrow{\boldsymbol{F}}_{k}=k \Delta L \hat{\boldsymbol{s}} \tag{6.33} \label{6.33}\]
Here \(k\) is the stiffness or spring constant of the spring, with unit \(\mathrm{N} / \mathrm{m}\), which normally has a positive value. The direction of the force can be indicated using a unit vector \(\hat{\boldsymbol{s}}\) that points outward parallel to the spring at point \(A\).
6.13.10 Damper
A linear damper is a mechanism whose length changes at a rate that is proportional to the externally applied pulling force (see Figure 6.7).
\[\overrightarrow{\boldsymbol{F}}_{c}=c \frac{\mathrm{d} L}{\mathrm{~d} t} \hat{\boldsymbol{s}} \tag{6.34} \label{6.34}\]
where \(c\) is the damping constant (unit \(\mathrm{N} \cdot \mathrm{s} / \mathrm{m}\) ), which is always positive. If point \(B\) is fixed, the velocity of point \(A\) is in the same direction as the applied force vector.
It is important to note that besides extending, springs and dampers can also be compressed, such that the distance \(\Delta L\) and time derivative \(\dot{L}\) become negative and the forces change sign. Secondly we note that we have discussed in Figure 6.7 the effect of external forces \(\overrightarrow{\boldsymbol{F}}_{A}\) and \(\overrightarrow{\boldsymbol{F}}_{B}\) generated on the mechanisms. But if one e.g. connects a point mass \(m\) at point \(A\) to a spring, the force generated by the mechanism is of opposite sign \(\overrightarrow{\boldsymbol{F}}_{m, k}=-k \Delta L \hat{\boldsymbol{s}}\), where the minus sign arises from Newton’s third law.
6.13.11 Other forces
There are many other types of forces, like thermodynamic, aerodynamic, fluid dynamic, electromagnetic, electrostatic and material forces. It carries too far here to discuss them all, and the methodology discussed in this chapter can be applied for all types of forces once the functional form is known.