Let's solve the problem step by step. --- ## **Step 1: Write Position Vectors and Force Vector** ### **Coordinates** Let’s assign coordinates based on the diagram: - \( O = (0, 0, 0) \) - \( F = (13, 10, 0) \) - \( A = (8, 10, 9) \) - \( B = (0, 6, 9) \) ### **Force at F** The 280 lb force acts along \( \vec{OF} \): \[ \vec{OF} = (13, 10, 0) - (0, 0, 0) = 13\hat{i} + 10\hat{j} \] Magnitude of \( \vec{OF} \): \[ |\vec{OF}| = \sqrt{13^2 + 10^2} = \sqrt{169 + 100} = \sqrt{269} \] Unit vector along \( \vec{OF} \): \[ \hat{n} = \frac{13\hat{i} + 10\hat{j}}{\sqrt{269}} \] Force vector: \[ \vec{F} = 280 \times \hat{n} = 280 \times \frac{13\hat{i} + 10\hat{j}}{\sqrt{269}} \] \[ \vec{F} = \left( \frac{280 \times 13}{\sqrt{269}} \right)\hat{i} + \left( \frac{280 \times 10}{\sqrt{269}} \right)\hat{j} \] \[ \vec{F} \approx (221.86\hat{i} + 170.86\hat{j}) \text{ lb} \] --- ## **Step 2: Find Moment About Point A** The moment about point \(A\): \[ \vec{M}_A = \vec{r}_{AF} \times \vec{F} \] Where: - \(\vec{r}_{AF} = \vec{F} - \vec{A} = (13-8)\hat{i} + (10-10)\hat{j} + (0-9)\hat{k} = 5\hat{i} - 9\hat{k}\) So, \[ \vec{r}_{AF} = 5\hat{i} + 0\hat{j} - 9\hat{k} \] Now, calculate the cross product: \[ \vec{M}_A = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 0 & -9 \\ 221.86 & 170.86 & 0 \\ \end{vmatrix} \] \[ = \hat{i} \left| \begin{matrix} 0 & -9 \\ 170.86 & 0 \end{matrix} \right| - \hat{j} \left| \begin{matrix} 5 & -9 \\ 221.86 & 0 \end{matrix} \right| + \hat{k} \left| \begin{matrix} 5 & 0 \\ 221.86 & 170.86 \end{matrix} \right| \] \[ = \hat{i} (0 \cdot 0 - (-9) \cdot 170.86) - \hat{j} (5 \cdot 0 - (-9) \cdot 221.86) + \hat{k} (5 \cdot 170.86 - 0 \cdot 221.86) \] \[ = \hat{i} (1537.74) - \hat{j} (1996.74) + \hat{k} (854.3) \] **So,** \[ \boxed{ \vec{M}_A = 1537.74\,\hat{i} - 1996.74\,\hat{j} + 854.3\,\hat{k}\ \text{lb-in} } \] --- ## **Step 3: Find Moment About Point B** \[ \vec{r}_{BF} = \vec{F} - \vec{B} = (13-0)\hat{i} + (10-6)\hat{j} + (0-9)\hat{k} = 13\hat{i} + 4\hat{j} - 9\hat{k} \] Set up the cross product: \[ \vec{M}_B = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 13 & 4 & -9 \\ 221.86 & 170.86 & 0 \\ \end{vmatrix} \] \[ = \hat{i} (4 \times 0 - (-9) \times 170.86) - \hat{j} (13 \times 0 - (-9) \times 221.86) + \hat{k} (13 \times 170.86 - 4 \times 221.86) \] \[ = \hat{i} (1537.74) - \hat{j} (1996.74) + \hat{k} (2221.18 - 887.44) \] \[ = \hat{i} (1537.74) - \hat{j} (1996.74) + \hat{k} (1333.74) \] --- ## **Final Answers** ### **Moment about Point A:** \[ \boxed{ \vec{M}_A = 1537.74\,\hat{i} - 1996.74\,\hat{j} + 854.3\,\hat{k}\ \text{lb-in} } \] ### **Moment about Point B:** \[ \boxed{ \vec{M}_B = 1537.74\,\hat{i} - 1996.74\,\hat{j} + 1333.74\,\hat{k}\ \text{lb-in} } \]