Let's solve both parts step by step: --- ## **a. Find \( F_B \) for Static Equilibrium** ### **Step 1: Free-Body Diagram and Force Components** - **Forces acting:** - \( 30\,\text{N} \) at A, \( 30\,\text{N} \) at C, \( F_B \) at B. - **Angles:** - \( 30\,\text{N} \) at A is at \( 45^\circ \) from the horizontal. - All other forces are shown in the diagram. ### **Step 2: Sum of Forces** #### **Horizontal (\( x \)) Direction** \[ \sum F_x = \] - \( 30\,\text{N} \) at A: \( 30\cos(45^\circ) \) right - \( F_B \): \( F_B\cos(45^\circ) \) right - \( 30\,\text{N} \) at C: \( 30 \) left So, \[ 30\cos(45^\circ) + F_B\cos(45^\circ) - 30 = \] \[ (30 + F_B)\cos(45^\circ) = 30 \] \[ (30 + F_B)\left(\frac{\sqrt{2}}{2}\right) = 30 \] \[ 30 + F_B = 30 \times \frac{2}{\sqrt{2}} \] \[ 30 + F_B = 30\sqrt{2} \] \[ F_B = 30(\sqrt{2} - 1) \] #### **Vertical (\( y \)) Direction** \[ \sum F_y = \] - \( 30\,\text{N} \) at A: \( 30\sin(45^\circ) \) up - \( F_B \): \( F_B\sin(45^\circ) \) up \[ 30\sin(45^\circ) + F_B\sin(45^\circ) = \] \[ (30 + F_B)\sin(45^\circ) = \] \[ 30 + F_B = \] \[ F_B = -30\,\text{N} \] --- ### **Step 3: Check Consistency** From \( x \)- and \( y \)-directions, both must be satisfied. But from \( x \)-direction: \( F_B = 30(\sqrt{2} - 1) \approx 12.43\,\text{N} \) From \( y \)-direction: \( F_B = -30\,\text{N} \) **Contradiction: Therefore, static equilibrium is NOT possible.** --- ## **b. Find \( F_B \) for Static Equilibrium** ### **Step 1: Free-Body Diagram and Force Components** - \( 30\,\text{N} \) at A, at \( 30^\circ \) - \( F_B \) at B (at a given angle) - \( 30\,\text{N} \) at C (horizontal left) ### **Step 2: Force Equilibrium** #### **Horizontal (\( x \)) Direction** \[ \sum F_x = \] - \( 30\cos(30^\circ) \) right (A) - \( F_B\cos(30^\circ) \) right (B) - \( 30 \) left (C) \[ 30\cos(30^\circ) + F_B\cos(30^\circ) - 30 = \] \[ (30 + F_B)\cos(30^\circ) = 30 \] \[ 30 + F_B = \frac{30}{\cos(30^\circ)} \] \[ 30 + F_B = \frac{30}{.866} \approx 34.64 \] \[ F_B \approx 4.64\,\text{N} \] #### **Vertical (\( y \)) Direction** \[ \sum F_y = \] - \( 30\sin(30^\circ) \) up (A) - \( F_B\sin(30^\circ) \) up (B) \[ 30\sin(30^\circ) + F_B\sin(30^\circ) = \] \[ (30 + F_B)\sin(30^\circ) = \] \[ 30 + F_B = \] \[ F_B = -30\,\text{N} \] --- ### **Step 3: Check Consistency** From \( x \) direction: \( F_B \approx 4.64\,\text{N} \) From \( y \) direction: \( F_B = -30\,\text{N} \) **Contradiction: Therefore, static equilibrium is NOT possible.** --- ## **Final Answers** ### **a. Not possible.** ### **b. Not possible.** ### **Explanation:** In both cases, the force equilibrium equations in \( x \) and \( y \) directions give inconsistent values for \( F_B \), so it is not possible to achieve static equilibrium with the given forces and configurations.