Let's solve the problem step by step: --- ## **1. Free-Body Diagram for Bar AB** Bar AB is a two-force member with: - Pin at **A** (reaction: \( A_x, A_y \)) - Pin at **B** (reaction: \( B_x, B_y \)) - Two downward forces of 90 N each at 100 mm and 200 mm from A. **Diagram:** ``` A (pin) |-------------------|-------------------| <--100mm--|--100mm--|--100mm--> (200mm total) ↓90N ↓90N ``` - At A: \( A_x \) (horizontal), \( A_y \) (vertical) - At B: \( B_x \) (horizontal), \( B_y \) (vertical) - At 100 mm from A: \( 90\,N \) downward - At 200 mm from A (at B): \( 90\,N \) downward --- ## **2. Free-Body Diagram for Bar DB** Bar DB is pinned at D and B. The bar is 100 mm long, vertical. **Diagram:** ``` D (pin) | | | B (pin) ``` - At D: \( D_x, D_y \) (reactions at D) - At B: \( B_x', B_y' \) (reactions at B, equal and opposite to those on AB) --- ## **3. Equilibrium Equations** ### **For Bar AB:** Sum of forces in x and y, sum of moments about A. #### \(\Sigma F_x = \): \[ A_x + B_x = \] #### \(\Sigma F_y = \): \[ A_y + B_y - 90 - 90 = \implies A_y + B_y = 180 \] #### \(\Sigma M_A = \): (Take counterclockwise as positive, moments due to forces at distances from A) - 90 N at 100 mm: \(90 \times 100\) - 90 N at 200 mm: \(90 \times 200\) - B's forces at 200 mm: \(B_y \times 200\) (vertical), \(B_x \times \) (horizontal, passes through A, no moment) \[ -90 \times 100 - 90 \times 200 + B_y \times 200 = \] \[ -900 - 18000 + 200B_y = \] \[ -27000 + 200 B_y = \] \[ B_y = \frac{27000}{200} = 135\,N \] From \(\Sigma F_y\): \[ A_y + 135 = 180 \implies A_y = 45\,N \] From \(\Sigma F_x\): \[ A_x + B_x = \implies A_x = -B_x \] --- ### **For Bar DB:** Bar DB is in equilibrium (assume DB is a two-force member, so the force is along DB). - At D: \( D_x, D_y \) - At B: \( B_x', B_y' \) (should be equal and opposite to B_x, B_y from AB) #### \(\Sigma F_x = \): \[ D_x + B_x' = \] #### \(\Sigma F_y = \): \[ D_y + B_y' = \] Since at B, the sum of forces from both bars must be zero, and B_x', B_y' are equal and opposite to B_x, B_y from AB. --- ## **4. Solve for Reactions at A, B, and D** ### **At B:** From above: \[ B_y = 135\,N \quad (upward) \] \[ B_x: \text{Let us find B_x} \] But bar DB is at a slope, so the force in DB (call it F_DB) acts along DB. The geometry: - Horizontal distance from D to B: 100 mm - Vertical distance: 100 mm So, angle \( \theta = 45^\circ \). \[ \sin\theta = \cos\theta = \frac{100}{\sqrt{100^2 + 100^2}} = \frac{100}{141.42} = .707 \] Force in DB (F_DB) at B: - \( B_x = F_{DB} \cos(45^\circ) \) - \( B_y = F_{DB} \sin(45^\circ) \) But from above, \( B_y = 135\,N \). So, \[ B_y = F_{DB} \times .707 = 135 \implies F_{DB} = \frac{135}{.707} = 191\,N \] \[ B_x = F_{DB} \times .707 = 191 \times .707 = 135\,N \] But which direction is B_x? Since A_x = -B_x, and A_x is at A, let's check the sense. ### **At A:** \[ A_x = -B_x = -135\,N \] \[ A_y = 45\,N \] ### **At D:** At D (reactions from bar DB): \[ D_x = -B_x = -135\,N \] \[ D_y = -B_y = -135\,N \] (acts up and to the left) --- ## **Summary Table** | Point | \( F_x \) (N) | \( F_y \) (N) | |-------|:-------------:|:-------------:| | A | -135 | 45 | | B | 135 | 135 | | D | -135 | -135 | --- ## **Final Answers** 1. **Free-body diagrams:** - For AB: Show A reactions (\(A_x, A_y\)), B reactions (\(B_x, B_y\)), two downward 90 N forces. - For DB: Show D reactions (\(D_x, D_y\)), B reactions (\(B_x', B_y'\)), both along DB's axis. 2. **Equilibrium Equations:** - For AB: \[ \sum F_x = : \quad A_x + B_x = \] \[ \sum F_y = : \quad A_y + B_y - 180 = \] \[ \sum M_A = : \quad -90 \times 100 - 90 \times 200 + B_y \times 200 = \] - For DB: \[ \sum F_x = : \quad D_x + B_x' = \] \[ \sum F_y = : \quad D_y + B_y' = \] 3. **Reactions:** - At A: \(A_x = -135\,N\), \(A_y = 45\,N\) - At B: \(B_x = 135\,N\), \(B_y = 135\,N\) - At D: \(D_x = -135\,N\), \(D_y = -135\,N\) --- ### **Boxed Final Answers** \[ \boxed{ \begin{aligned} A_x &= -135\,\text{N} \quad & A_y &= 45\,\text{N} \\ B_x &= 135\,\text{N} \quad & B_y &= 135\,\text{N} \\ D_x &= -135\,\text{N} \quad & D_y &= -135\,\text{N} \end{aligned} } \] If you need the diagrams sketched or a clearer layout, let me know!

# Step-by-Step Solution for the Given Problem ## 1. Free-Body Diagram for Bar AB ### Description - **Bar AB** is a two-force member supported by pins at **A** and **B**. - **Forces Acting:** - Two downward forces of \(90\,N\) each, located at \(100\,mm\) and \(200\,mm\) from point **A**. ### Diagram Representation ``` A (pin) |-------------------|-------------------| <--100mm--|--100mm--|--100mm--> (200mm total) ↓90N ↓90N ``` - Reactions at A: \(A_x\) (horizontal), \(A_y\) (vertical) - Reactions at B: \(B_x\) (horizontal), \(B_y\) (vertical) --- ## 2. Free-Body Diagram for Bar DB ### Description - **Bar DB** is pinned at **D** and **B** and is vertical. ### Diagram Representation ``` D (pin) | | | B (pin) ``` - Reactions at D: \(D_x, D_y\) - Reactions at B: \(B_x', B_y'\) (equal and opposite to those on AB) --- ## 3. Equilibrium Equations ### For Bar AB 1. **Sum of Forces in x-direction** (\(\Sigma F_x\)): \[ A_x + B_x = 0 \] 2. **Sum of Forces in y-direction** (\(\Sigma F_y\)): \[ A_y + B_y - 90 - 90 = 0 \implies A_y + B_y = 180 \] 3. **Sum of Moments about point A** (\(\Sigma M_A\)): \[ -90 \times 100 - 90 \times 200 + B_y \times 200 = 0 \] \[ -900 - 18000 + 200B_y = 0 \] \[ 200B_y = 18900 \implies B_y = \frac{18900}{200} = 94.5\,N \] ### Substitute \(B_y\) into \(\Sigma F_y\): \[ A_y + 94.5 = 180 \implies A_y = 180 - 94.5 = 85.5\,N \] ### For Bar DB 1. **Sum of Forces in x-direction** (\(\Sigma F_x\)): \[ D_x + B_x' = 0 \] 2. **Sum of Forces in y-direction** (\(\Sigma F_y\)): \[ D_y + B_y' = 0 \] Since \(B_y' = B_y\) and \(B_x' = B_x\). --- ## 4. Solve for Reactions at A, B, and D ### At B - From equilibrium equations: \[ B_y = 94.5\,N \] - Assuming \(B_x\) is horizontal. ### At A - From x-direction equilibrium: \[ A_x + B_x = 0 \implies A_x = -B_x \] ### At D - Use the relationship between reactions: \[ D_x = -B_x, \quad D_y = -B_y \] ### Final Values Assuming \(B_x = 0\) (as vertical), we can calculate: 1. **At A**: \[ A_x = 0, \quad A_y = 85.5\,N \] 2. **At B**: \[ B_x = 0, \quad B_y = 94.5\,N \] 3. **At D**: \[ D_x = 0, \quad D_y = -94.5\,N \] --- ## Summary Table | Point | \( F_x \) (N) | \( F_y \) (N) | |-------|:-------------:|:-------------:| | A | 0 | 85.5 | | B | 0 | 94.5 | | D | 0 | -94.5 | --- ## Final Answers \[ \boxed{ \begin{aligned} A_x &= 0 \quad & A_y &= 85.5\,\text{N} \\ B_x &= 0 \quad & B_y &= 94.5\,\text{N} \\ D_x &= 0 \quad & D_y &= -94.5\,\text{N} \end{aligned} } \] This concludes the analysis of the forces acting on the bars AB and DB. If you require further clarification or additional information, please let me know!