Let's solve **Problem 2.17** step-by-step for **Figure P2.17**. --- ## (a) **Free-Body Diagrams and Differential Equations** ### **Free-Body Diagrams** #### **Mass \(M_1\):** - **Upward forces:** Spring force from top spring \(K(x_1)\) - **Downward forces:** Spring force from bottom spring \(K(x_1 - x_3)\), Damping \(B_1 \dot{x}_1\), Weight \(M_1 g\) #### **Mass \(M_2\):** - **Upward force:** Spring force from top spring \(K(x_2)\) - **Downward forces:** Spring force from bottom spring \(K(x_2 - x_3)\), Damping \(B_2 \dot{x}_2\), Weight \(M_2 g\) #### **Mass \(M_3\):** - **Upward forces:** Spring forces from both upper springs (\(K(x_1 - x_3)\) from left, \(K(x_2 - x_3)\) from right) - **Downward forces:** Weight \(M_3 g\), External force \(f_a(t)\) --- ### **Equations of Motion (Newton's Second Law)** #### **For \(M_1\):** \[ M_1 \ddot{x}_1 = -K x_1 - B_1 \dot{x}_1 + K(x_3 - x_1) - M_1 g \] \[ M_1 \ddot{x}_1 + B_1 \dot{x}_1 + 2K x_1 - K x_3 = -M_1 g \] #### **For \(M_2\):** \[ M_2 \ddot{x}_2 = -K x_2 - B_2 \dot{x}_2 + K(x_3 - x_2) - M_2 g \] \[ M_2 \ddot{x}_2 + B_2 \dot{x}_2 + 2K x_2 - K x_3 = -M_2 g \] #### **For \(M_3\):** \[ M_3 \ddot{x}_3 = -K(x_3 - x_1) - K(x_3 - x_2) - M_3 g + f_a(t) \] \[ M_3 \ddot{x}_3 + 2K x_3 - K x_1 - K x_2 = f_a(t) - M_3 g \] --- ## (b) **Static Equilibrium (Constant Elongation of Each Spring, \(f_a(t) = \))** At equilibrium, \(\ddot{x}_i = \), \(\dot{x}_i = \), \(f_a(t) = \): ### **Write Static Equations** 1. \(2K x_1 - K x_3 = -M_1 g\) 2. \(2K x_2 - K x_3 = -M_2 g\) 3. \(-K x_1 - K x_2 + 2K x_3 = -M_3 g\) ### **Solve the System** #### **Step 1:** Express \(x_1\) and \(x_2\) in terms of \(x_3\): From (1): \(2K x_1 - K x_3 = -M_1 g\) \(\rightarrow x_1 = \frac{K x_3 - M_1 g}{2K}\) From (2): \(2K x_2 - K x_3 = -M_2 g\) \(\rightarrow x_2 = \frac{K x_3 - M_2 g}{2K}\) #### **Step 2:** Substitute into (3): \[ -K x_1 - K x_2 + 2K x_3 = -M_3 g \] Substitute \(x_1\) and \(x_2\): \[ -K \left(\frac{K x_3 - M_1 g}{2K}\right) - K \left(\frac{K x_3 - M_2 g}{2K}\right) + 2K x_3 = -M_3 g \] \[ -\frac{K x_3 - M_1 g}{2} - \frac{K x_3 - M_2 g}{2} + 2K x_3 = -M_3 g \] \[ -\frac{K x_3 - M_1 g + K x_3 - M_2 g}{2} + 2K x_3 = -M_3 g \] \[ -\frac{2K x_3 - (M_1 + M_2)g}{2} + 2K x_3 = -M_3 g \] \[ -(K x_3 - \frac{(M_1 + M_2)g}{2}) + 2K x_3 = -M_3 g \] \[ -K x_3 + \frac{(M_1 + M_2)g}{2} + 2K x_3 = -M_3 g \] \[ K x_3 + \frac{(M_1 + M_2)g}{2} = -M_3 g \] \[ K x_3 = -M_3 g - \frac{(M_1 + M_2)g}{2} \] \[ x_3 = -\frac{M_3 g}{K} - \frac{(M_1 + M_2)g}{2K} \] #### **Step 3:** Plug \(x_3\) into \(x_1\) and \(x_2\): \[ x_1 = \frac{K x_3 - M_1 g}{2K} \] \[ x_1 = \frac{K \left(-\frac{M_3 g}{K} - \frac{(M_1 + M_2)g}{2K}\right) - M_1 g}{2K} \] \[ x_1 = \frac{-M_3 g - \frac{(M_1 + M_2) g}{2} - M_1 g}{2K} \] \[ x_1 = \frac{-M_3 g - M_1 g - \frac{(M_1 + M_2)g}{2}}{2K} \] \[ x_1 = \frac{-M_3 g - M_1 g - \frac{M_1 g}{2} - \frac{M_2 g}{2}}{2K} \] \[ x_1 = \frac{-M_3 g - \frac{3M_1 g}{2} - \frac{M_2 g}{2}}{2K} \] Similarly, for \(x_2\): \[ x_2 = \frac{K x_3 - M_2 g}{2K} \] \[ x_2 = \frac{-M_3 g - \frac{(M_1 + M_2) g}{2} - M_2 g}{2K} \] \[ x_2 = \frac{-M_3 g - M_2 g - \frac{M_1 g}{2} - \frac{M_2 g}{2}}{2K} \] \[ x_2 = \frac{-M_3 g - \frac{3M_2 g}{2} - \frac{M_1 g}{2}}{2K} \] --- ## **Final Answers** ### **(a) Differential Equations:** \[ \boxed{ \begin{aligned} &M_1 \ddot{x}_1 + B_1 \dot{x}_1 + 2K x_1 - K x_3 = -M_1 g \\ &M_2 \ddot{x}_2 + B_2 \dot{x}_2 + 2K x_2 - K x_3 = -M_2 g \\ &M_3 \ddot{x}_3 + 2K x_3 - K x_1 - K x_2 = f_a(t) - M_3 g \end{aligned} } \] ### **(b) Static Equilibrium (Elongations):** \[ \boxed{ \begin{aligned} x_1 &= \frac{-M_3 g - \frac{3M_1 g}{2} - \frac{M_2 g}{2}}{2K} \\ x_2 &= \frac{-M_3 g - \frac{3M_2 g}{2} - \frac{M_1 g}{2}}{2K} \\ x_3 &= -\frac{M_3 g}{K} - \frac{(M_1 + M_2)g}{2K} \end{aligned} } \] --- **Let me know if you need the solution for Figure P2.18 as well!**