Let's solve the problem step by step: --- ### **Step 1: Find the Vertical Displacement ($\delta y_A$) of Point A due to Small Change $\delta\theta$** When the angle $\theta$ increases by $\delta\theta$, the vertical position of point A changes. - The length of each rod is 1 m. - The vertical height of A from the base is: \( y_A = 1 \cdot \sin\theta \) A small change in $\theta$ gives a change in $y_A$: \[ \delta y_A = \frac{d}{d\theta} (1 \cdot \sin\theta)\, \delta\theta = \cos\theta \, \delta\theta \] **For $\theta = 30^\circ = \pi/6$:** \[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \] \[ \delta y_A = \frac{\sqrt{3}}{2} \, \delta\theta \] **Explanation:** This step finds how much the vertical position of point A changes for a small change in angle, which is needed because the weight does work as point moves vertically. --- ### **Step 2: Calculate the Virtual Work Done by the Weight ($\delta U_w$)** The virtual work done by the weight \( w \) is: \[ \delta U_w = w \cdot (\text{vertical displacement of A}) = w \, \delta y_A \] Given \( w = 200\, \text{N} \): \[ \delta U_w = 200 \cdot \frac{\sqrt{3}}{2} \, \delta\theta \approx 200 \cdot .866 \, \delta\theta = 173.2\, \delta\theta \] Since the weight acts downward and $\delta y_A$ is upward for positive $\delta\theta$, the work **done by the weight is negative**: \[ \delta U_w = -173\, \delta\theta \] **Explanation:** The weight does negative work as point A moves upward (since the force is downward and the displacement is upward). --- ## **Final Answer** **The virtual work done by the weight $w$ is:** \[ \boxed{\delta U_w = -173\, \delta\theta} \] **(Units: N·rad, option 5)**