Let's break down the solution step by step: --- ## **Step 1: Understanding the Problem** You are given: - The force at point \( E \) on the gear of shaft \( CD \): \[ (\vec{F}_E)_{CD} = (-92.\mathbf{i} - 368.\mathbf{j} + 801\mathbf{k}) \text{ lbf} \] - You need to determine the reactions at the supports \( A \) and \( B \): - \( A_z \): Reaction at \( A \) in the \( z \)-direction - \( A_y \): Reaction at \( A \) in the \( y \)-direction - \( B_z \): Reaction at \( B \) in the \( z \)-direction - \( B_y \): Reaction at \( B \) in the \( y \)-direction --- ## **Step 2: Free-Body Diagram and Equilibrium Equations** The shaft is in equilibrium, so: ### **Sum of Forces** \[ \sum F_y = : \quad A_y + B_y + F_{Ey} = \] \[ \sum F_z = : \quad A_z + B_z + F_{Ez} = \] ### **Sum of Moments About Point A** Take moments about point \( A \). Use the distances from \( A \): - \( B \): \( 3 \) in \( x \)-direction from \( A \) - \( E \): \( (3 + 6.5) = 9.5 \) in \( x \)-direction from \( A \) #### **Moment Due to Force at E** \[ \vec{r}_{AE} = 9.5\mathbf{i} \] \[ \vec{M}_A = \vec{r}_{AE} \times \vec{F}_E \] Calculate the cross product: \[ \vec{M}_A = (9.5\mathbf{i}) \times (-92.\mathbf{i} - 368.\mathbf{j} + 801\mathbf{k}) \] \[ = 9.5 \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & & \\ -92. & -368. & 801 \\ \end{matrix} \right| \] \[ = 9.5 \Big[ ()(801) - ()(-368.)\mathbf{i} - (1)(801)\mathbf{j} + (1)(-368.)\mathbf{k} \Big] \] \[ = 9.5 [\mathbf{i} - 801\mathbf{j} + 368.\mathbf{k}] \] \[ = -7609.5\mathbf{j} + 3496.\mathbf{k} \] #### **Moments Due to Reactions at B** - \( \vec{r}_{AB} = 3\mathbf{i} \) - Reactions at \( B \): \( B_y \mathbf{j} + B_z \mathbf{k} \) \[ \vec{M}_A^{(B)} = (3\mathbf{i}) \times (B_y\mathbf{j} + B_z\mathbf{k}) \] \[ = 3 [\mathbf{i} \times B_y\mathbf{j} + \mathbf{i} \times B_z\mathbf{k}] \] \[ = 3 [B_y \mathbf{k} - B_z \mathbf{j}] \] \[ = 3B_y\mathbf{k} - 3B_z\mathbf{j} \] --- ## **Step 3: Equilibrium of Moments About A** \[ \vec{M}_A = = \text{Moment due to } \vec{F}_E + \text{Moment due to B} \] \[ -7609.5\mathbf{j} + 3496.\mathbf{k} + [ -3B_z\mathbf{j} + 3B_y\mathbf{k} ] = \] Equate components: - \( \mathbf{j} \) component: \( -7609.5 - 3B_z = \Rightarrow B_z = -2536.5 \) lbf - \( \mathbf{k} \) component: \( 3496. + 3B_y = \Rightarrow B_y = -1165.3 \) lbf --- ## **Step 4: Use Force Equilibrium to Find \( A_y \) and \( A_z \)** Given \( F_{Ez} = 801 \) lbf and \( F_{Ey} = -368. \) lbf: #### **Sum of Forces in y:** \[ A_y + B_y + F_{Ey} = \] \[ A_y = -B_y - F_{Ey} \] \[ A_y = -(-1165.3) - (-368.) \] \[ A_y = 1165.3 + 368. \] \[ A_y = 1533.3 \text{ lbf} \] #### **Sum of Forces in z:** \[ A_z + B_z + F_{Ez} = \] \[ A_z = -B_z - F_{Ez} \] \[ A_z = -(-2536.5) - 801 \] \[ A_z = 2536.5 - 801 \] \[ A_z = 1735.5 \text{ lbf} \] --- ## **Final Answers** | Variable | Value (lbf) | |----------|-------------| | \( A_z \) | **1735.5** | | \( A_y \) | **1533.3** | | \( B_z \) | **-2536.5** | | \( B_y \) | **-1165.3** | **Boxed for your form:** The value of \( A_z \) is **1735.5** lbf. The value of \( A_y \) is **1533.3** lbf. The value of \( B_z \) is **-2536.5** lbf. The value of \( B_y \) is **-1165.3** lbf. --- ### **Summary Table** | Variable | Value (lbf) | |----------|-------------| | \( A_z \) | 1735.5 | | \( A_y \) | 1533.3 | | \( B_z \) | -2536.5 | | \( B_y \) | -1165.3 | --- **Let me know if you need the solution for the support reactions at C and D as well!**