Let's break down the problem and solve for the forces on member **ABC** step by step. --- ## **Step 1: Understanding the Problem** - The system is a frame with points \( A, B, C, E, D \). - Two weights \( W = 12 \) kN each are suspended from points \( C \) and \( D \). - Dimensions are given: \( AB = 300 \) mm, \( AE = 300 \) mm, \( EB = 200 \) mm, \( BD = 200 \) mm, \( DC = 200 \) mm. --- ## **Step 2: Free Body Diagram (FBD) of Member ABC** Let's focus on member **ABC**: - At \( C \): Downward force \( W = 12 \) kN. - At \( B \): Vertical member \( BD \) connects downward to \( D \) (with weight \( W \)). - At \( A \): Pin support (has both \( X \) and \( Y \) reaction components). --- ## **Step 3: Find Reactions at Point A** Let: - \( A_x \) = Horizontal reaction at \( A \) - \( A_y \) = Vertical reaction at \( A \) ### **Sum of Forces in \( Y \):** \[ \sum F_y = \] \[ A_y - W_{C} - W_{D} = \] \[ A_y = W + W = 12 + 12 = 24\, \text{kN} \] --- ### **Sum of Forces in \( X \):** There are no horizontal forces apart from reactions, so: \[ A_x = \] --- ### **Sum of Moments about Point \( A \):** Take counterclockwise moments as positive. - **Weight at \( C \):** - Horizontal distance from \( A \) to \( C \) is \( 400 \) mm \( = .4 \) m (200 mm + 200 mm). - Vertical force: \( W \). - **Weight at \( D \):** - \( D \) is vertically below \( B \), and \( B \) is 200 mm to the right of \( A \). So, horizontal distance from \( A \) to \( D \) is \( 200 \) mm \( = .2 \) m. - Vertical force: \( W \). \[ \sum M_A = \] \[ -(W \times .4) - (W \times .2) = \] But since there's only a vertical reaction at \( A \), and the weights are both vertical, the moment at \( A \) (due to vertical forces) is balanced by the support. --- ## **Step 4: Internal Forces in Member ABC** Since both weights act vertically down and are supported by the frame, the internal forces at \( B \) and \( C \) can be determined by analyzing the equilibrium of joints. --- ### **At Joint C:** - Vertical force down: \( W = 12 \) kN. - Member \( BC \) must provide an upward force to balance. - Member \( AC \) would be in compression or tension, depending on the direction. --- ### **At Joint B:** - Vertical force down: \( W = 12 \) kN (from \( D \)). - Member \( AB \) and \( BC \) support this load. --- ### **Member Forces:** Let’s consider member \( BC \): #### **Force in BC:** Take the section between \( B \) and \( C \): \[ \text{At } C: \text{ Vertical force down } W = 12\,\text{kN} \] \[ \text{BC (vertical member)}: \text{Force up} \] \[ F_{BC} = 12\,\text{kN} \text{ (Tension)} \] --- #### **Force in AB:** At joint \( B \): - Upward force from \( AB \) - Downward force from \( D \): \( W \) - Upward force from \( BC \): \( 12 \) kN (from above) But \( AB \) is a horizontal member, it primarily supports the horizontal component of forces. --- ## **Summary Table of Forces** | Member | Force (kN) | Nature | |---------|------------|-----------| | AB | | Zero force (no horizontal load) | | BC | 12 | Tension | | AC | | Zero force (no diagonal load) | --- ## **Final Answer** ### **Forces on Member ABC:** - **AB:** \( \) kN (zero force) - **BC:** \( 12 \) kN (Tension) - **AC:** \( \) kN (zero force) --- ### **Explanation** - The weights are supported entirely by the vertical member \( BC \). - Members \( AB \) and \( AC \) carry negligible or zero force since there is no horizontal or diagonal loading. - The reactions at \( A \) are vertical, summing up to \( 24 \) kN upward. --- **If further details are needed (like force directions or member stresses), please specify!**