Let's break down the problem step by step: ### 1. **Identify Loads and Reactions** #### **Loads:** - **Uniformly distributed load (UDL):** \( w = 8\, \text{kN/m} \) over \( 8\, \text{m} \) Total load = \( 8 \times 8 = 64\, \text{kN} \) Acts at the center of the 8 m section, i.e., 4 m from A. - **Point load due to 50 kg mass:** Weight \( W = mg = 50 \times 9.81 = 490.5\, \text{N} = 0.491\, \text{kN} \) This acts vertically downward at the right end (distance = 12 m from A). The cable forms an angle of \( 120^\circ \) with the beam, so the force in the cable (\( T \)) is: - Vertical component \( T_v = T \sin(120^\circ) \) - Horizontal component \( T_h = T \cos(120^\circ) \) #### **Reactions:** - **At A (fixed support):** \( A_x \), \( A_y \), and moment \( M_A \) - **At B (roller):** \( B_y \) (vertical only) --- ### 2. **Equilibrium Equations** #### **Sum of Forces in X direction:** \[ \sum F_x = 0: \quad A_x + T \cos(120^\circ) = 0 \] #### **Sum of Forces in Y direction:** \[ \sum F_y = 0: \quad A_y + B_y - 64 - T \sin(120^\circ) = 0 \] #### **Sum of Moments about A:** Take moments about point A (clockwise positive): - UDL: \( 64\, \text{kN} \) acts at 4 m from A. - \( B_y \) acts at 10 m from A. - The vertical component of the cable (\( T_v \)) acts at 12 m from A. \[ \sum M_A = 0: \quad -64 \times 4 + B_y \times 10 + [T \sin(120^\circ)] \times 12 = 0 \] --- ### 3. **Calculate Cable Force \( T \):** \[ T = 0.491\, \text{kN} \] \[ \sin(120^\circ) = 0.866, \quad \cos(120^\circ) = -0.5 \] \[ T_v = T \sin(120^\circ) = 0.491 \times 0.866 = 0.425\, \text{kN} \] \[ T_h = T \cos(120^\circ) = 0.491 \times (-0.5) = -0.246\, \text{kN} \] --- ### 4. **Solve for Unknowns** #### **Moment about A:** \[ -64 \times 4 + B_y \times 10 + 0.425 \times 12 = 0 \] \[ -256 + 10B_y + 5.1 = 0 \] \[ 10B_y = 256 - 5.1 = 250.9 \] \[ \boxed{B_y = 25.09\, \text{kN}\ \uparrow} \] --- #### **Sum of Vertical Forces:** \[ A_y + B_y - 64 - 0.425 = 0 \] \[ A_y + 25.09 - 64 - 0.425 = 0 \] \[ A_y = 64 + 0.425 - 25.09 = 39.335\, \text{kN} \] \[ \boxed{A_y = 39.34\, \text{kN}\ \uparrow} \] --- #### **Sum of Horizontal Forces:** \[ A_x + T_h = 0 \] \[ A_x - 0.246 = 0 \] \[ \boxed{A_x = 0.246\, \text{kN}\ \rightarrow} \] --- ### **Final Highlighted Answers** #### **Reactions at Supports:** - **At A:** - \( \boxed{A_x = 0.25\ \text{kN}\ \rightarrow} \) - \( \boxed{A_y = 39.34\ \text{kN}\ \uparrow} \) - **At B:** - \( \boxed{B_y = 25.09\ \text{kN}\ \uparrow} \) Let me know if you need the moment at A or a summary table!