6.11: Equations of motion in kinetics
The goal of solving the equations of motion is to obtain the position vectors \(\overrightarrow{\boldsymbol{r}}_{i}(t)\) of all objects \(i\) at all times \(t\) for given initial conditions. Solving these equations is part of kinematics and was already discussed in Ch. 5. We will briefly repeat the procedure here for the example from the previous section.
Since Newton’s second law gives us an expression for the acceleration \(\overrightarrow{\boldsymbol{a}}(t)\), the main challenge in solving the EoMs is to determine the time dependent position vector \(\overrightarrow{\boldsymbol{r}}(t)\) of an object from the acceleration \(\overrightarrow{\boldsymbol{a}}(t)\). Often this can be done using the kinematic integration techniques that we have discussed in the previous chapter. In some cases more complex differential equations, that cannot be solved by integration appear, as will be discussed in Ch. 13.
As an example we consider the block \(A\) in Figure 6.4. We know from the constraint equations that \(a_{A, n}=0\) and \(a_{A, b}=0\). So, we only need to determine the motion \(s_{A}(t)\) along the path coordinate. For a certain initial condition \(s_{A}(0)=s_{0}\) and \(\dot{s}_{A}(0)=v_{0}\), we integrate the equation of motion twice to obtain the motion as follows 6.21 :
\[\begin{align} & v_{A}(t)=v_{0}+\int_{0}^{t} a_{0} \mathrm{~d} t=v_{0}+a_{0} t \tag{6.22} \label{6.22}\\[4pt] & s_{A}(t)=s_{0}+\int_{0}^{t}\left(v_{0}+a_{0} t\right) \mathrm{d} t=\frac{1}{2} a_{0} t^{2}+v_{0} t \tag{6.23} \label{6.23}\end{align}\]
where \(a_{0}\) was obtained from the EoM in Equation 6.21.
6.11.2 Determining forces from the EoM
Besides using the EoM to determine the motion from known forces, it is also possible to substitute knowledge of the motion into the equation of motion, to determine unknown forces using the EoM. Essentially, this method follows directly from Newton’s second law \(\sum \overrightarrow{\boldsymbol{F}}=m \overrightarrow{\boldsymbol{a}}\), where the sum of forces can be determined if the mass \(m\) and acceleration \(\overrightarrow{\boldsymbol{a}}\) are known. Note that since the
sum of forces is obtained, if more than one force acts on the mass, this equation allows only obtaining one unknown force vector or 3 scalar force components.
An example of how forces can be determined from knowledge of the acceleration was already given in Example 6.1, where we determined in Equation 6.17 the contact force \(F_{N}\) by combining the knowledge that the acceleration component perpendicular to the slope was zero \(\left(a_{A, n}=0\right)\) and the knowledge of the mass and gravitational force. This is a powerful way to study and quantify forces.