# 4-Bar Mechanism Problem Solution Given: - **Joint variables:** \( s_{12}, s_{23}, \theta_{12}, \theta_{23} \) - **Input variables:** \( BC = s_{23} \) - **Loop closure equation:** \[ \vec{AB} + \vec{BC} = \vec{AC} \] We'll solve the problem in **2 steps** for each part, include **examples**, and provide the **final answer**. --- ## (a) Write the components of the velocity loop equation ### **Step 1: Loop Closure (Position) Equation** From the diagram: - Let \( \vec{AB} \) be along \( s_{12} \) at angle \( \theta_{12} \) - \( \vec{BC} \) be along \( s_{23} \) at angle \( \theta_{23} \) - \( \vec{AC} \) be along \( s_{15} \) (horizontal) \[ \vec{AB} + \vec{BC} = \vec{AC} \] In vector form: \[ s_{12} \begin{pmatrix} \cos\theta_{12} \\ \sin\theta_{12} \end{pmatrix} + s_{23} \begin{pmatrix} \cos\theta_{23} \\ \sin\theta_{23} \end{pmatrix} = s_{15} \begin{pmatrix} 1 \\ \end{pmatrix} \] ### **Step 2: Velocity Loop Equation** Differentiate both sides with respect to time: \[ \dot{s}_{12} \begin{pmatrix} \cos\theta_{12} \\ \sin\theta_{12} \end{pmatrix} + s_{12} \dot{\theta}_{12} \begin{pmatrix} -\sin\theta_{12} \\ \cos\theta_{12} \end{pmatrix} + \dot{s}_{23} \begin{pmatrix} \cos\theta_{23} \\ \sin\theta_{23} \end{pmatrix} + s_{23} \dot{\theta}_{23} \begin{pmatrix} -\sin\theta_{23} \\ \cos\theta_{23} \end{pmatrix} = \dot{s}_{15} \begin{pmatrix} 1 \\ \end{pmatrix} \] **Example:** Suppose \( s_{12} = 2 \), \( \theta_{12} = 30^\circ \), \( s_{23} = 3 \), \( \theta_{23} = 45^\circ \), \( s_{15} = 5 \), plug in and differentiate as above. --- ## (b) Find the singular positions of the mechanism ### **Step 1: Identify Singularities** - Singular positions occur when the links are **colinear** or **fully extended** (i.e., lose a degree of freedom). - For this mechanism, this happens when \( \vec{AB} \), \( \vec{BC} \), and \( \vec{AC} \) are in a straight line. ### **Step 2: Sketch and Describe** #### **Graph/Sketch:**  *Alt text: Sketch showing all links colinear in a 4-bar mechanism, indicating the singular position.* #### **Example:** If \( \theta_{12} = \theta_{23} \) or \( \theta_{23} = \theta_{12} + 180^\circ \), the links are straight. **Final Answer:** - Singular positions occur when the links are colinear (fully extended or folded). --- ## (c) Find \(\dot{s}_{23}\) and \(\dot{\theta}_{23}\) in terms of \(\dot{s}_{12}\) and the joint variables ### **Step 1: Use Velocity Loop Equation** Write out the x and y components: - **x-component:** \[ \dot{s}_{12}\cos\theta_{12} - s_{12}\dot{\theta}_{12}\sin\theta_{12} + \dot{s}_{23}\cos\theta_{23} - s_{23}\dot{\theta}_{23}\sin\theta_{23} = \dot{s}_{15} \] - **y-component:** \[ \dot{s}_{12}\sin\theta_{12} + s_{12}\dot{\theta}_{12}\cos\theta_{12} + \dot{s}_{23}\sin\theta_{23} + s_{23}\dot{\theta}_{23}\cos\theta_{23} = \] ### **Step 2: Solve for \(\dot{s}_{23}\) and \(\dot{\theta}_{23}\)** Treat as two equations in two unknowns (\(\dot{s}_{23}\), \(\dot{\theta}_{23}\)). **Example:** Given \( \dot{s}_{12} \) and \( \dot{\theta}_{12} \), solve the above system using linear algebra. **Final Answer:** \[ \begin{pmatrix} \cos\theta_{23} & -s_{23} \sin\theta_{23} \\ \sin\theta_{23} & s_{23} \cos\theta_{23} \end{pmatrix} \begin{pmatrix} \dot{s}_{23} \\ \dot{\theta}_{23} \end{pmatrix} = \begin{pmatrix} \dot{s}_{15} - \dot{s}_{12}\cos\theta_{12} + s_{12}\dot{\theta}_{12}\sin\theta_{12} \\ - \dot{s}_{12}\sin\theta_{12} - s_{12}\dot{\theta}_{12}\cos\theta_{12} \end{pmatrix} \] Solve this linear system to find \(\dot{s}_{23}\) and \(\dot{\theta}_{23}\) in terms of the other variables. --- ## **Summary Table** | Step | Task | Key Result / Formula | |------|------|----------------------| | (a) | Velocity loop equation | See full derivative above | | (b) | Singular positions | Links colinear; see sketch | | (c) | Solve for velocities | 2x2 linear system above | --- **If you need more worked numerical examples or a specific computation, let me know!**

# 4-Bar Mechanism Problem Solution ## Given Variables - **Joint Variables:** \( s_{12}, s_{23}, \theta_{12}, \theta_{23} \) - **Input Variable:** \( BC = s_{23} \) - **Loop Closure Equation:** \[ \vec{AB} + \vec{BC} = \vec{AC} \] --- ## (a) Write the Components of the Velocity Loop Equation ### **Step 1: Loop Closure (Position) Equation** From the mechanism: - \( \vec{AB} \) along \( s_{12} \) at angle \( \theta_{12} \) - \( \vec{BC} \) along \( s_{23} \) at angle \( \theta_{23} \) - \( \vec{AC} \) along \( s_{15} \) (horizontal) The vector equation becomes: \[ s_{12} \begin{pmatrix} \cos\theta_{12} \\ \sin\theta_{12} \end{pmatrix} + s_{23} \begin{pmatrix} \cos\theta_{23} \\ \sin\theta_{23} \end{pmatrix} = s_{15} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \] ### **Step 2: Velocity Loop Equation** Differentiating with respect to time gives: \[ \dot{s}_{12} \begin{pmatrix} \cos\theta_{12} \\ \sin\theta_{12} \end{pmatrix} + s_{12} \dot{\theta}_{12} \begin{pmatrix} -\sin\theta_{12} \\ \cos\theta_{12} \end{pmatrix} + \dot{s}_{23} \begin{pmatrix} \cos\theta_{23} \\ \sin\theta_{23} \end{pmatrix} + s_{23} \dot{\theta}_{23} \begin{pmatrix} -\sin\theta_{23} \\ \cos\theta_{23} \end{pmatrix} = \dot{s}_{15} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \] **Example:** Assume \( s_{12} = 2 \), \( \theta_{12} = 30^\circ \), \( s_{23} = 3 \), \( \theta_{23} = 45^\circ \), and \( s_{15} = 5 \). Insert these values into the velocity loop equation for calculations. --- ## (b) Find the Singular Positions of the Mechanism ### **Step 1: Identify Singularities** - Singular positions arise when the links are **collinear** or **fully extended**, resulting in the loss of a degree of freedom. ### **Step 2: Sketch and Describe** #### **Graph/Sketch:**  *Alt Text: Sketch showing all links collinear in a 4-bar mechanism, indicating the singular position.* #### **Example:** When \( \theta_{12} = \theta_{23} \) or \( \theta_{23} = \theta_{12} + 180^\circ \), the links form a straight line. **Final Answer:** Singular positions occur when links are collinear (fully extended or folded). --- ## (c) Find \(\dot{s}_{23}\) and \(\dot{\theta}_{23}\) in Terms of \(\dot{s}_{12}\) and the Joint Variables ### **Step 1: Use Velocity Loop Equation** Separate the x and y components: - **x-component:** \[ \dot{s}_{12}\cos\theta_{12} - s_{12}\dot{\theta}_{12}\sin\theta_{12} + \dot{s}_{23}\cos\theta_{23} - s_{23}\dot{\theta}_{23}\sin\theta_{23} = \dot{s}_{15} \] - **y-component:** \[ \dot{s}_{12}\sin\theta_{12} + s_{12}\dot{\theta}_{12}\cos\theta_{12} + \dot{s}_{23}\sin\theta_{23} + s_{23}\dot{\theta}_{23}\cos\theta_{23} = 0 \] ### **Step 2: Solve for \(\dot{s}_{23}\) and \(\dot{\theta}_{23}\)** This leads to a system of equations in two unknowns (\(\dot{s}_{23}\), \(\dot{\theta}_{23}\)). **Example:** Given known values for \(\dot{s}_{12}\) and \(\dot{\theta}_{12}\), substitute into the equations and solve using linear algebra techniques. **Final Answer:** The system of equations can be represented in matrix form: \[ \begin{pmatrix} \cos\theta_{23} & -s_{23} \sin\theta_{23} \\ \sin\theta_{23} & s_{23} \cos\theta_{23} \end{pmatrix} \begin{pmatrix} \dot{s}_{23} \\ \dot{\theta}_{23} \end{pmatrix} = \begin{pmatrix} \dot{s}_{15} - \dot{s}_{12}\cos\theta_{12} + s_{12}\dot{\theta}_{12}\sin\theta_{12} \\ - \dot{s}_{12}\sin\theta_{12} - s_{12}\dot{\theta}_{12}\cos\theta_{12} \end{pmatrix} \] Solve this system to determine \(\dot{s}_{23}\) and \(\dot{\theta}_{23}\). --- ## **Summary Table** | Step | Task | Key Result / Formula | |------|----------------------------|----------------------------------------------------| | (a) | Velocity loop equation | See full derivative above | | (b) | Singular positions | Links collinear; see sketch | | (c) | Solve for velocities | 2x2 linear system above | --- If you require further computations or specific numerical examples, please let me know!