Let's solve part (B) as requested, in **2 steps** with explanation after each step and the **final answer at the end**. --- ## (B) If H = 4V, what is the magnitude and direction of the net force at the hinge (in terms of mg)? ### **Step 1: Force and Torque Equilibrium** **Explanation:** - The lamp is in **static equilibrium**: - The sum of forces in both the x and y directions equals zero. - The sum of torques about any point (choose the hinge) equals zero. **Forces acting:** - Tension, **T**, in the rod (direction as shown). - Weight, **mg**, acting downward at the lamp. - Hinge forces (**V** vertically, **H** horizontally). **Take torques about the hinge:** - The torque due to **mg** is \( mg \cdot H \) (since it is perpendicular to the rod, distance is \( H \)). - The torque due to **T** is \( T \cdot V \) (since the tension acts perpendicular to the rod at distance \( V \) from the hinge). Set the net torque to zero (counterclockwise positive): \[ T \cdot V = mg \cdot H \] Given \( H = 4V \): \[ T \cdot V = mg \cdot 4V \implies T = 4mg \] --- ### **Step 2: Find Net Force at the Hinge** **Explanation:** - At the hinge, the rod exerts force components \( V \) (vertical) and \( H \) (horizontal). - The vertical force at the hinge must balance the vertical forces: - Upward: \( V \) (from hinge), \( T \) (from tension) - Downward: \( mg \) - The horizontal force at the hinge must balance the horizontal forces: - Only \( T \) acts horizontally to the right, so hinge must pull to the left with force \( H \). **Vertical equilibrium:** \[ V + T_y = mg \] But since \( T \) is horizontal (from diagram), \( T_y = \): \[ V = mg \] **Horizontal equilibrium:** \[ H = T = 4mg \] **Net force at the hinge:** \[ f = \sqrt{H^2 + V^2} = \sqrt{(4mg)^2 + (mg)^2} = \sqrt{16m^2g^2 + m^2g^2} = \sqrt{17}mg \] **Direction:** \[ \theta = \arctan\left(\frac{V}{H}\right) = \arctan\left(\frac{mg}{4mg}\right) = \arctan\left(\frac{1}{4}\right) \approx 14^\circ \text{ above the horizontal (to the left)} \] --- ## **Final Answer** - **Magnitude**: \( \boxed{\sqrt{17}\,mg} \) - **Direction**: \( \boxed{14^\circ} \) above the horizontal, pointing to the left.

Let's solve all parts of the problem step by step. --- # Problem Breakdown ## (A) Complete the Extended Force Body Diagram ### **Step 1: Identify Forces** **Forces Acting on the Lamp:** - **Weight (mg)**: Acts downward at the center of mass of the lamp. - **Tension (T)**: Acts along the rod towards the hinge. - **Hinge Reaction Forces (H, V)**: - **H**: Horizontal component (to the left). - **V**: Vertical component (upward). ### **Step 2: Draw Forces on the Diagram** 1. **Weight (mg)**: Draw an arrow pointing downwards from the lamp's center of mass. 2. **Tension (T)**: Draw an arrow along the rod towards the hinge. 3. **Hinge Forces**: - Draw an arrow pointing left for **H**. - Draw an arrow pointing upward for **V**. **Note:** Ensure that the lengths of the arrows are proportional to the magnitudes of the forces. --- ## (B) If H = 4V, Find the Magnitude and Direction of the Net Force at the Hinge (in terms of mg) ### **Step 1: Force and Torque Equilibrium** **Equilibrium Conditions:** - The system is in static equilibrium, so: - Sum of forces in both x and y directions = 0 - Sum of torques about the hinge = 0 **Torque Calculation:** The torque about the hinge due to the weight of the lamp and the tension in the rod must balance. Take torques about the hinge: 1. **Torque due to mg**: \[ \tau_{mg} = mg \cdot H \] 2. **Torque due to T**: \[ \tau_T = T \cdot V \] Set them equal for equilibrium: \[ T \cdot V = mg \cdot H \] Substituting \( H = 4V \): \[ T \cdot V = mg \cdot 4V \] Dividing both sides by \( V \) (assuming \( V \neq 0 \)): \[ T = 4mg \] --- ### **Step 2: Find Net Force at the Hinge** **Vertical and Horizontal Equilibrium:** 1. **Vertical Forces**: \[ V + T_y = mg \quad \text{(where \( T_y = 0 \) since T is horizontal)} \] Thus: \[ V = mg \] 2. **Horizontal Forces**: \[ H = T \] Substituting \( T \): \[ H = 4mg \] **Net Force at the Hinge**: \[ f = \sqrt{H^2 + V^2} = \sqrt{(4mg)^2 + (mg)^2} \] Calculating: \[ f = \sqrt{16m^2g^2 + m^2g^2} = \sqrt{17m^2g^2} = \sqrt{17}mg \] **Direction**: \[ \theta = \arctan\left(\frac{V}{H}\right) = \arctan\left(\frac{mg}{4mg}\right) = \arctan\left(\frac{1}{4}\right) \approx 14^\circ \text{ above the horizontal (to the left)} \] --- ## **Final Answers** - **Magnitude of Net Force**: \( \boxed{\sqrt{17}\,mg} \) - **Direction**: \( \boxed{14^\circ} \) above the horizontal, pointing to the left.