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 θ = 0 and f = 0, the system is at static equilibrium. Assume that the lever arm can be approximated as a

Question

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 [latex] heta=0[/latex] and [latex]f=0[/latex], 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 [latex] heta[/latex].

Accepted Answer

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 [latex]\mathrm{O}[/latex] that is fixed. We choose the displacement of the block [latex]y[/latex] and the angular displacement of the rod [latex] heta[/latex] as the generalized coordinates. Note that [latex]y[/latex] 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

[latex] \begin{gathered} +\uparrow: \quad \sum F_{y}=0, \ k \delta_{\mathrm{st}}-m g=0, \ k \delta_{\mathrm{st}}=m g, \end{gathered} [/latex]

and for the arm,

[latex] \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} [/latex]

where [latex]\delta_{\mathrm{st}}[/latex] is the static deformation of the spring.

When a force [latex]f[/latex] is applied on one side of the rod, the deformation of the spring caused by the rotation of the rod can be approximated as [latex]L heta / 4[/latex] for small angles [latex] heta[/latex]. Assuming that the block and the rod are displaced in their positive directions and [latex]L heta / 4>y>0[/latex], the spring is in tension and the magnitude of the spring force is [latex]f_{\mathrm{k}}=k\left(L heta / 4-y+\delta_{\mathrm{st}} ight)[/latex]. The magnitude of the damping force is [latex]f_{\mathrm{b}}=b L \dot{ heta} / 4[/latex]. The free-body diagram and the kinematic diagram of the system are shown in Figure 5.78b. Note that the acceleration components [latex]m \ddot{y}[/latex] and [latex]I_{\mathrm{O}} \ddot{ heta}[/latex] 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 [latex]y[/latex] direction gives

[latex] \begin{gathered} +\uparrow y: \sum F_{y}=m a_{C y} \ k\left(\frac{L}{4} heta-y+\delta_{\mathrm{st}} ight)-m g=m \ddot{y} . \end{gathered} [/latex]

For the rod (rotation only), applying the moment equation about the fixed point [latex]\mathrm{O}[/latex] gives

[latex] \begin{gathered} +\curvearrowleft: \sum M_{\mathrm{O}}=I_{\mathrm{O}} \alpha, \ f \cdot \frac{3 L}{4} \cos heta+M g \cdot \frac{L}{4} \cos heta-k\left(\frac{L}{4} heta-y+\delta_{\mathrm{st}} ight) \frac{L}{4} \cos heta-b \frac{L}{4} \dot{ heta} \cdot \frac{L}{4} \cos heta=I_{\mathrm{O}} \ddot{ heta}, \end{gathered} [/latex]

where [latex]I_{\mathrm{O}}[/latex] can be obtained using parallel-axis theorem,

[latex]I_{\mathrm{O}}=\frac{1}{12} M L^{2}+M\left(\frac{L}{4} ight)^{2}=\frac{7}{48} M L^{2} .[/latex]

For small angular motions, [latex]\cos heta \approx 1[/latex]. Introducing the static equilibrium conditions and rearranging the two equations gives

[latex] \begin{gathered} m \ddot{y}+k y-\frac{k L}{4} heta=0 \ \frac{7}{48} M L^{2} \ddot{ heta}+\frac{b L^{2}}{16} \dot{ heta}+\frac{k L^{2}}{16} heta-\frac{k L}{4} y=\frac{3 f L}{4} . \end{gathered} [/latex]

Question 5.12b: A Lever Mechanism Consider the system shown in Figure 5.77, ...

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