Sure! Let's break down the problem and find the **governing equation** for the system. --- ## **Problem Recap** - **Rod**: Ideal (no mass, no stiffness, no damping) - **Pivot**: Point P - **Weight**: \( mg = 5N \) at the end of the rod - **Spring**: Spring constant \( k = 400 \, \text{N/m} \) - **Displacement**: \( x \) (measured from equilibrium) - **Goal**: Find the governing equation. --- ## **Step 1: System Diagram and Variables** Let's label: - **Rod length:** \( l \) - **Displacement at end:** \( x \) - **Spring attached at distance \( d \) from pivot** When the weight moves down by \( x \), the spring also stretches by an amount **proportional to the arc length** at its attachment point. --- ## **Step 2: Kinematics** Assume small angles (\( \theta \)), so that: - Displacement at rod end: \( x = l \theta \) ⇒ \( \theta = \frac{x}{l} \) - Displacement at spring point: \( x_s = d \theta = d \frac{x}{l} \) --- ## **Step 3: Forces and Torques** ### **Forces Acting** 1. **Spring Force:** \( F_{\text{spring}} = -k x_s = -k \frac{d}{l} x \) (restoring force) 2. **Weight:** \( mg \) (acts downward at rod end) ### **Torques about Pivot P** - **Spring Torque:** \( \tau_{\text{spring}} = -k x_s \cdot d = -k \frac{d^2}{l} x \) - **Weight Torque:** \( \tau_{\text{weight}} = -mg \cdot l \cdot \sin\theta \approx -mg \cdot l \cdot \theta = -mg \cdot x \) - For small angles, \( \sin\theta \approx \theta \). #### **Total Restoring Torque** \[ \tau_{\text{total}} = -mg \cdot x - k \frac{d^2}{l} x \] --- ## **Step 4: Equation of Motion** Since the rod is massless and ideal, **no inertia term**. The sum of torques must be zero in static equilibrium, or if you assume small oscillations, you would write: \[ (mg) \cdot x + k \frac{d^2}{l} x = \] Or, more generally (if considering dynamics and adding mass \( m_{\text{eff}} \) at the end): \[ m_{\text{eff}} \ddot{x} + mg x + k \frac{d^2}{l} x = \] But since the rod is massless and only the weight at the end matters, if considering only static or quasi-static equilibrium: \[ mg x + k \frac{d^2}{l} x = \] --- ## **Step 5: Substitute Values** Given: - \( mg = 5 \) N - \( k = 400 \) N/m - \( d \), \( l \): Distances from diagram The **governing equation** is: \[ 5x + 400 \frac{d^2}{l} x = \] Or, factoring \( x \): \[ \boxed{ \left[5 + 400 \frac{d^2}{l} \right] x = } \] This is the equation that governs the equilibrium (or motion for small perturbations) of the system. --- ### **Summary Table** | Term | Physical Meaning | |-----------|----------------------------------| | \( 5x \) | Gravitational restoring force | | \( 400 \frac{d^2}{l} x \) | Spring restoring force | --- **If you have the values for \( d \) and \( l \), you can substitute them for a numerical answer.** Let me know if you'd like to see the dynamic equation (with mass, if included), or if you want a step-by-step for a specific value!