Let's work through the problem step by step: --- ## **Step 1: Free-Body Diagram (FBD) of the Entire Structure** ### **Structure Overview** - The structure is a frame with two vertical members (at A and B), a horizontal member (at C), and a two-force member (at D). - There is a downward force of **240 N** at the bottom right (pulley location). ### **Reactions and Forces** - **A:** Pin support (reactions: \(A_x, A_y\)) - **B:** Pin support (reactions: \(B_x, B_y\)) - **Pulley at D:** Applies a 240 N downward force #### **FBD of Entire Structure** ``` |---------300mm----------|---------300mm----------| A o----------------------------------------------o B | | | | | | o-----------------o D (240 N down)-------------o ``` - **Label all reaction forces:** - At A: \( A_x \) (horizontal), \( A_y \) (vertical) - At B: \( B_x \) (horizontal), \( B_y \) (vertical) - At D: 240 N (downward) --- ## **Step 2: Free-Body Diagrams of Each Member** ### **Member AC (Vertical)** - Pin at A: \( A_x, A_y \) - Pin at C: Internal force (let's call it \( C_x, C_y \)) ### **Member CB (Horizontal)** - Pin at C: Internal force (\( C_x, C_y \)) - Pin at B: \( B_x, B_y \) ### **Member CD (Two-Force Member)** - Pin at C: Reaction (\( C_x, C_y \)) - Pin at D: Reaction (\( D_x, D_y \)), with \( D_y = 240\,N \) (down) --- #### **Member-by-Member FBDs** - **Member AC:** - Upward: \( A_y \) - Right: \( A_x \) - At C: Reactions from CB and CD - **Member CB:** - Left: \( B_x \) - Down: \( B_y \) - At C: Reactions from AC and CD - **Member CD:** - At C: Reactions from AC and CB - At D: Vertical force \( D_y = 240\,N \) (downwards) --- ## **Step 3: Solve for Each Force** ### **Assumptions** - All members are pin-connected. - All forces and reactions are at the joints. --- ### **Equilibrium Equations for the Whole Structure** Let’s use the equilibrium equations for the entire structure: #### **Sum of Forces in X-direction:** \[ \sum F_x = : \quad A_x + B_x = \] \[ A_x = -B_x \] #### **Sum of Forces in Y-direction:** \[ \sum F_y = : \quad A_y + B_y - 240 = \] \[ A_y + B_y = 240 \] #### **Sum of Moments about Point A:** - Take moments about A (clockwise positive): Horizontal distance from A to D = 600 mm = .6 m Vertical distance from A to D = 200 mm = .2 m The force at D acts vertically downward. Moment arm for \( B_y \) from A = 600 mm = .6 m (since B is directly above D). \[ \sum M_A = : \quad -B_y \cdot .6 + 240 \cdot .6 = \] \[ -B_y \cdot .6 + 240 \cdot .6 = \] \[ B_y = 240\,\text{N} \] Substitute \( B_y = 240\,\text{N} \) into \( A_y + B_y = 240 \): \[ A_y + 240 = 240 \implies A_y = \] From earlier, \( A_x = -B_x \). ### **Now, Sum of Moments about B:** \[ \sum M_B = : \quad A_y \cdot .6 - 240 \cdot = \implies \text{(no info, since } A_y = ) \] ### **Horizontal Equilibrium:** No external horizontal forces, so: \[ A_x = -B_x \] But both supports are pins, so they can have horizontal reactions, but as there are no external horizontal loads, \( A_x = B_x = \). --- ## **Final Answers for Reactions:** | Point | \( F_x \) | \( F_y \) | |-------|-----------|-----------| | A | | | | B | | 240 N | | D | | -240 N | --- ## **Step 4: Forces in Each Member** - **Member AC:** Only supports vertical forces, but since \( A_y = \), no force in AC. - **Member CB:** Only supports vertical at B, so carries entire 240 N as a reaction at B. - **Member CD:** Two-force member; the only force at D is downward 240 N. --- ## **Graphical Representation** ### **Free-Body Diagram of the Whole Structure** ```plaintext A B o-----------o | | | | | | o-----------o D 240 N ↓ ``` - At A: \( A_x = \), \( A_y = \) - At B: \( B_x = \), \( B_y = 240\,N \) (upward) - At D: 240 N (downward) --- ## **Summary Table** | Member | Force | Magnitude | Direction | |---------|---------------|-----------|-----------| | AC | Ax, Ay | | | | CB | By | 240 N | Upward | | CD | Dy | 240 N | Downward | --- **If you need free-body diagrams for each member drawn, let me know!**