Let's solve the problem step by step. --- ## **Step 1: Understand the Problem** Given: - Bent rod **ABEF** is supported at **C** and **D** (bearings allow reactions in \(j\) and \(k\) directions). - Supported by wire **AH**. - \(AB = 250 \text{ mm}\), \(BE = 300 \text{ mm}\), \(EF = 250 \text{ mm}\), \(FA = 50 \text{ mm}\). - Load \(W = 520 \text{ N}\) acts downward at \(F\). - Bearing at **D** does **not** exert axial thrust (\(\vec{i}\) direction reaction at \(D\) is zero). --- ## **Step 2: Free-Body Diagram and Coordinates** Let's assign coordinates to key points: - \(A: (, 250, 50)\) - \(B: (, 250, )\) - \(C: (, , )\) - \(E: (250, , )\) - \(F: (250, , 50)\) - \(H: (, 500, )\) --- ## **Step 3: Write Equilibrium Equations** **Sum of Forces:** \[ \sum \vec{F} = \] In \(j\) and \(k\) directions (since bearings cannot resist in \(i\)). Let: - Reaction at \(C = C_j \mathbf{j} + C_k \mathbf{k}\) - Reaction at \(D = D_j \mathbf{j} + D_k \mathbf{k}\) - Wire tension at \(A\) (let's call it \(T_{AH}\)), having only a \(j\) component since it runs in the y-direction. - Load at \(F: -520\mathbf{k}\) #### **Force Equilibrium in \(j\) and \(k\):** \[ C_j + D_j + T_{AH} = \tag{1} \] \[ C_k + D_k - 520 = \tag{2} \] --- ## **Step 4: Moment Equilibrium about Point \(C\)** Take moments about point \(C\): - Position vectors from \(C\) to each force application point: - \(E_C = (250, , )\) - \(F_C = (250, , 50)\) - \(A_C = (, 250, 50)\) - Let's write the moment due to each: ### **a) Load at F** \[ \vec{r}_{F/C} = (250, , 50) \] \[ \vec{F}_F = -520\mathbf{k} \] \[ \vec{M}_{F/C} = \vec{r}_{F/C} \times \vec{F}_F \] \[ = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 250 & & 50 \\ & & -520 \\ \end{vmatrix} = \mathbf{i}( \times -520 - 50 \times ) - \mathbf{j}(250 \times -520 - 50 \times ) + \mathbf{k}(250 \times - \times ) \] \[ = \mathbf{i}() - \mathbf{j}(-130,000) + \mathbf{k}() = 130,000\mathbf{j} \] ### **b) Reaction at D** \[ \vec{r}_{D/C} = (250, , ) \] \[ \vec{F}_D = D_j \mathbf{j} + D_k \mathbf{k} \] \[ \vec{M}_{D/C} = \vec{r}_{D/C} \times \vec{F}_D = (250, , ) \times (D_j \mathbf{j} + D_k \mathbf{k}) \] Compute: - \( (250, , ) \times (, D_j, ) = (, , 250 D_j) \) - \( (250, , ) \times (, , D_k) = (, -250 D_k, ) \) So, \[ \vec{M}_{D/C} = 250 D_j \mathbf{k} - 250 D_k \mathbf{j} \] ### **c) Tension at A (Wire AH)** \[ \vec{r}_{A/C} = (, 250, 50) \] \[ \vec{F}_{AH} = T_{AH}\mathbf{j} \] \[ \vec{M}_{A/C} = (, 250, 50) \times (, T_{AH}, ) \] \[ = \mathbf{i}(250 \times - 50 \times T_{AH}) - \mathbf{j}( \times - 50 \times ) + \mathbf{k}( \times T_{AH} - 250 \times ) \] \[ = -50 T_{AH} \mathbf{i} \] --- ### **Sum Moments about C:** \[ \sum \vec{M}_C = \] \[ \vec{M}_{F/C} + \vec{M}_{D/C} + \vec{M}_{A/C} = \] \[ (130,000\mathbf{j}) + [250 D_j \mathbf{k} - 250 D_k \mathbf{j}] + [-50 T_{AH} \mathbf{i}] = \] So, group components: - \(i:\) \(-50 T_{AH} = \implies T_{AH} = \) - \(j:\) \(130,000 - 250 D_k = \implies D_k = \frac{130,000}{250} = 520\, \text{N}\) - \(k: \) \(250 D_j = \implies D_j = \) --- ## **Step 5: Substitute Back into Force Equations** From above: - \(T_{AH} = \) - \(D_k = 520\, N\) - \(D_j = \) From the force equilibrium equations: \[ C_j + D_j = \implies C_j = -D_j = \] \[ C_k + D_k - 520 = \implies C_k + 520 - 520 = \implies C_k = \] --- ## **Step 6: Final Answers** - **Reaction at C:** \(\, \mathbf{j} + \, \mathbf{k}\) - **Reaction at D:** \(\, \mathbf{j} + 520\, \mathbf{k}\) --- ## **Final Answers (Fill in the Blanks)** \[ \boxed{ \begin{aligned} \text{The reaction at C is } & \quad ()\text{ N}\, \mathbf{j} + ()\text{ N}\, \mathbf{k} \\ \text{The reaction at D is } & \quad ()\text{ N}\, \mathbf{j} + (520)\text{ N}\, \mathbf{k} \\ \end{aligned} } \] --- **Summary Table:** | Location | \(j\) (N) | \(k\) (N) | |----------|-----------|-----------| | C | | | | D | | 520 | --- If you have any further questions, let me know!