Let's break down the solution step by step: ### Step 1: **Identify All Forces Acting on the Shaft** #### At A (Sheave) - 24 lb (in the negative z-direction) - 18 lb (in the positive y-direction) #### At C (Sheave) - \(P = 35.5 \) lb (in the negative y-direction) - \(T\) (in the negative z-direction) \(T = \frac{2.5 \text{ in.}}{2 \text{ in.}} \times 24 \text{ lb} = 30 \text{ lb}\) (Since tension is proportional to the radius, and the belt at A is 24 lb, the tension at C is calculated as above.) #### Bearings: - At B and D: Reactions in y and z directions: \(B_y, B_z, D_y, D_z\). ### Step 2: **Write Equilibrium Equations** #### (1) \(\sum F_y = 0\) \[ B_y + D_y + 18\,(-y \text{ at A}) - 35.5\,(-y \text{ at C}) = 0 \] \[ B_y + D_y + 18 - 35.5 = 0 \] \[ B_y + D_y = 17.5 \] #### (2) \(\sum F_z = 0\) \[ B_z + D_z - 24\,(-z \text{ at A}) - 30\,(-z \text{ at C}) = 0 \] \[ B_z + D_z - 24 - 30 = 0 \] \[ B_z + D_z = 54 \] --- ### Step 3: **Moment Equilibrium About D (\(\sum M_D = 0\))** Take moments about point D (using right-hand rule, positive y is up, z is out of page): #### For Each Force: - **At A (12 in from D):** - 24 lb in -z: \(12\,\text{in} \times 24\,\text{lb} = 288\,\text{in}\cdot\text{lb}\) (about y) - 18 lb in +y: \(12\,\text{in} \times 18\,\text{lb} = 216\,\text{in}\cdot\text{lb}\) (about z) - **At C (6 in from D):** - 35.5 lb in -y: \(6\,\text{in} \times 35.5\,\text{lb} = 213\,\text{in}\cdot\text{lb}\) (about z) - 30 lb in -z: \(6\,\text{in} \times 30\,\text{lb} = 180\,\text{in}\cdot\text{lb}\) (about y) - **At B (6 in from D):** - \(B_y, B_z\) at 6 in from D #### Write Moment Equations (using the cross product): \[ \sum M_D = 0 = r \times F \] For each force, \(r\) is the position vector from D to the point of application. \[ \begin{align*} \text{At A:} & \quad r_A = -12\,\hat{i} \\ & \quad F_{A,y} = 18\,\hat{j} \implies M_{A,18} = (-12\,\hat{i}) \times (18\,\hat{j}) = -216\,\hat{k} \\ & \quad F_{A,z} = -24\,\hat{k} \implies M_{A,24} = (-12\,\hat{i}) \times (-24\,\hat{k}) = -288\,\hat{j} \\ \text{At C:} & \quad r_C = -6\,\hat{i} \\ & \quad F_{C,y} = -35.5\,\hat{j} \implies M_{C,35.5} = (-6\,\hat{i}) \times (-35.5\,\hat{j}) = -213\,\hat{k} \\ & \quad F_{C,z} = -30\,\hat{k} \implies M_{C,30} = (-6\,\hat{i}) \times (-30\,\hat{k}) = -180\,\hat{j} \\ \text{At B:} & \quad r_B = -6\,\hat{i} \\ & \quad B_y\,\hat{j} \implies M_{B_y} = (-6\,\hat{i}) \times (B_y\,\hat{j}) = -6B_y\,\hat{k} \\ & \quad B_z\,\hat{k} \implies M_{B_z} = (-6\,\hat{i}) \times (B_z\,\hat{k}) = 6B_z\,\hat{j} \\ \end{align*} \] Sum up moments about D: - **j components:** \[ -288 - 180 + 6B_z = 0 \implies 6B_z = 468 \implies B_z = 78 \] - **k components:** \[ -216 - 213 - 6B_y = 0 \implies -6B_y = 429 \implies B_y = -71.5 \] --- ### Step 4: **Solve for D_y and D_z Using the Force Equations** Recall: \[ B_y + D_y = 17.5 \implies D_y = 17.5 - B_y = 17.5 - (-71.5) = 89 \] \[ B_z + D_z = 54 \implies D_z = 54 - B_z = 54 - 78 = -24 \] --- ### **Final Answers** \[ \boxed{ \begin{align*} \text{The reaction at } B &: (-71.5\,\hat{j} + 78\,\hat{k}) \text{ lb} \\ \text{The reaction at } D &: (89\,\hat{j} + (-24)\,\hat{k}) \text{ lb} \end{align*} } \] **Plug these into the answer boxes:** - The reaction at \( B \) is ( -71.5 ) lb j + ( 78 ) lb k. - The reaction at \( D \) is ( 89 ) lb j + ( -24 ) lb k. --- If you need further clarification or a free body diagram, let me know!