Question 5.12b: A Lever Mechanism Consider the system shown in Figure 5.77, ...
A Lever Mechanism
Consider the system shown in Figure 5.77, in which a lever arm has a force applied on one side and a spring-damper combination on the other side with a suspended mass. When \theta=0 and f=0, the system is at static equilibrium. Assume that the lever arm can be approximated as a uniform slender rod. Draw the free-body diagram of the lever arm and the suspended mass. Derive the differential equations of motion for small angles \theta.
This is a mixed system with two masses. The suspended block undergoes translational motion along the vertical direction, and the motion of the rod is pure rotation about point \mathrm{O} that is fixed. We choose the displacement of the block y and the angular displacement of the rod \theta as the generalized coordinates. Note that y is measured from the static equilibrium position, which is set as the origin. The free-body diagrams and the kinematic diagram of the system are shown in Figure 5.78.
At static equilibrium, for the block, we have
\begin{gathered} +\uparrow: \quad \sum F_{y}=0, \\ k \delta_{\mathrm{st}}-m g=0, \\ k \delta_{\mathrm{st}}=m g, \end{gathered}
and for the arm,
\begin{gathered} +\curvearrowleft: \sum M_{\mathrm{O}}=0, \\ M g \cdot \frac{L}{4}-k \delta_{\mathrm{st}} \cdot \frac{L}{4}=0, \\ M g=k \delta_{\mathrm{st} \prime} \end{gathered}
where \delta_{\mathrm{st}} is the static deformation of the spring.
When a force f is applied on one side of the rod, the deformation of the spring caused by the rotation of the rod can be approximated as L \theta / 4 for small angles \theta. Assuming that the block and the rod are displaced in their positive directions and L \theta / 4>y>0, the spring is in tension and the magnitude of the spring force is f_{\mathrm{k}}=k\left(L \theta / 4-y+\delta_{\mathrm{st}}\right). The magnitude of the damping force is f_{\mathrm{b}}=b L \dot{\theta} / 4. The free-body diagram and the kinematic diagram of the system are shown in Figure 5.78b. Note that the acceleration components m \ddot{y} and I_{\mathrm{O}} \ddot{\theta} are shown together with the forces in the free-body diagram. For complex mechanical systems, we recommend to draw a separate kinematic diagram.
For the block (translation only), applying the force equation in the y direction gives
\begin{gathered} +\uparrow y: \sum F_{y}=m a_{C y} \\ k\left(\frac{L}{4} \theta-y+\delta_{\mathrm{st}}\right)-m g=m \ddot{y} . \end{gathered}
For the rod (rotation only), applying the moment equation about the fixed point \mathrm{O} gives
\begin{gathered} +\curvearrowleft: \sum M_{\mathrm{O}}=I_{\mathrm{O}} \alpha, \\ f \cdot \frac{3 L}{4} \cos \theta+M g \cdot \frac{L}{4} \cos \theta-k\left(\frac{L}{4} \theta-y+\delta_{\mathrm{st}}\right) \frac{L}{4} \cos \theta-b \frac{L}{4} \dot{\theta} \cdot \frac{L}{4} \cos \theta=I_{\mathrm{O}} \ddot{\theta}, \end{gathered}
where I_{\mathrm{O}} can be obtained using parallel-axis theorem,
I_{\mathrm{O}}=\frac{1}{12} M L^{2}+M\left(\frac{L}{4}\right)^{2}=\frac{7}{48} M L^{2} .
For small angular motions, \cos \theta \approx 1. Introducing the static equilibrium conditions and rearranging the two equations gives
\begin{gathered} m \ddot{y}+k y-\frac{k L}{4} \theta=0 \\ \frac{7}{48} M L^{2} \ddot{\theta}+\frac{b L^{2}}{16} \dot{\theta}+\frac{k L^{2}}{16} \theta-\frac{k L}{4} y=\frac{3 f L}{4} . \end{gathered}
