Let's break down the problem step by step: --- ## **Step 1: Assign Coordinates** Let's assign coordinates to all relevant points using the diagram: - \( O \): Origin, \( (0, 0, 0) \) - \( F \): Point where the force is applied. - \( A \): \( (8, 0, 5) \) in inches - \( B \): \( (0, 6, 0) \) in inches ### **Locate Point F:** From O, move 13 in along \( x \), 10 in along \( y \), and 9 in along \( z \). - \( F = (13, 10, 9) \) in inches --- ## **Step 2: Write the Force Vector** The force is **downward along the z-axis**: \[ \vec{F} = -280\, \hat{k}\ \text{lb} \] (In vector form: \( \vec{F} = 0\, \hat{i} + 0\, \hat{j} -280\, \hat{k} \) lb) --- ## **Step 3: Position Vectors from A and B to F** ### **From A to F:** \[ \vec{r}_{AF} = \vec{r}_F - \vec{r}_A = (13 - 8)\, \hat{i} + (10 - 0)\, \hat{j} + (9 - 5)\, \hat{k} \] \[ \vec{r}_{AF} = (5\, \hat{i} + 10\, \hat{j} + 4\, \hat{k})\ \text{in} \] ### **From B to F:** \[ \vec{r}_{BF} = \vec{r}_F - \vec{r}_B = (13 - 0)\, \hat{i} + (10 - 6)\, \hat{j} + (9 - 0)\, \hat{k} \] \[ \vec{r}_{BF} = (13\, \hat{i} + 4\, \hat{j} + 9\, \hat{k})\ \text{in} \] --- ## **Step 4: Calculate Moments** The moment about a point is: \[ \vec{M} = \vec{r} \times \vec{F} \] ### **(a) Moment about A:** \[ \vec{M}_A = \vec{r}_{AF} \times \vec{F} \] Plug in: \[ \vec{r}_{AF} = 5\, \hat{i} + 10\, \hat{j} + 4\, \hat{k} \] \[ \vec{F} = 0\, \hat{i} + 0\, \hat{j} -280\, \hat{k} \] Compute the cross product: \[ \vec{M}_A = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 10 & 4 \\ 0 & 0 & -280 \end{vmatrix} \] - \( \hat{i} \) component: \( (10)(-280) - (4)(0) = -2800 \) - \( \hat{j} \) component: \( - [ (5)(-280) - (4)(0) ] = - [ -1400 ] = 1400 \) - \( \hat{k} \) component: \( (5)(0) - (10)(0) = 0 \) So, \[ \vec{M}_A = -2800\, \hat{i} + 1400\, \hat{j} + 0\, \hat{k}\ \text{lb-in} \] --- ### **(b) Moment about B:** \[ \vec{M}_B = \vec{r}_{BF} \times \vec{F} \] \[ \vec{r}_{BF} = 13\, \hat{i} + 4\, \hat{j} + 9\, \hat{k} \] \[ \vec{F} = 0\, \hat{i} + 0\, \hat{j} -280\, \hat{k} \] Compute the cross product: \[ \vec{M}_B = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 13 & 4 & 9 \\ 0 & 0 & -280 \end{vmatrix} \] - \( \hat{i} \) component: \( (4)(-280) - (9)(0) = -1120 \) - \( \hat{j} \) component: \( - [ (13)(-280) - (9)(0) ] = - [ -3640 ] = 3640 \) - \( \hat{k} \) component: \( (13)(0) - (4)(0) = 0 \) So, \[ \vec{M}_B = -1120\, \hat{i} + 3640\, \hat{j} + 0\, \hat{k}\ \text{lb-in} \] --- ## **Step 5: Final Answers** ### **Moment about Point A:** \[ \boxed{ \vec{M}_A = -2800\, \hat{i} + 1400\, \hat{j}\ \text{lb-in} } \] ### **Moment about Point B:** \[ \boxed{ \vec{M}_B = -1120\, \hat{i} + 3640\, \hat{j}\ \text{lb-in} } \] --- **These are the moments of the 280 lb force about points A and B, respectively.**

# Step-by-Step Solution to Find Moments ## **Step 1: Assign Coordinates** Let's assign coordinates to all relevant points based on the provided diagram: - **O**: Origin, \( (0, 0, 0) \) - **A**: \( (8, 0, 5) \) inches - **B**: \( (0, 6, 0) \) inches ### **Locate Point F:** From O, moving: - 13 inches along the \( x \)-axis - 10 inches along the \( y \)-axis - 9 inches along the \( z \)-axis Thus, the coordinates of point **F** are: - **F**: \( (13, 10, 9) \) inches --- ## **Step 2: Write the Force Vector** The force is acting downward along the \( z \)-axis: \[ \vec{F} = -280\, \hat{k}\ \text{lb} \] In vector form, this can be represented as: \[ \vec{F} = 0\, \hat{i} + 0\, \hat{j} - 280\, \hat{k}\ \text{lb} \] --- ## **Step 3: Position Vectors from A and B to F** ### **From A to F:** The position vector from point A to F is calculated as: \[ \vec{r}_{AF} = \vec{r}_F - \vec{r}_A = (13 - 8)\, \hat{i} + (10 - 0)\, \hat{j} + (9 - 5)\, \hat{k} \] \[ \vec{r}_{AF} = 5\, \hat{i} + 10\, \hat{j} + 4\, \hat{k}\ \text{in} \] ### **From B to F:** The position vector from point B to F is calculated as: \[ \vec{r}_{BF} = \vec{r}_F - \vec{r}_B = (13 - 0)\, \hat{i} + (10 - 6)\, \hat{j} + (9 - 0)\, \hat{k} \] \[ \vec{r}_{BF} = 13\, \hat{i} + 4\, \hat{j} + 9\, \hat{k}\ \text{in} \] --- ## **Step 4: Calculate Moments** The moment about a point is given by: \[ \vec{M} = \vec{r} \times \vec{F} \] ### **(a) Moment about A:** Calculating the moment about point A: \[ \vec{M}_A = \vec{r}_{AF} \times \vec{F} \] Substituting the values: \[ \vec{r}_{AF} = 5\, \hat{i} + 10\, \hat{j} + 4\, \hat{k} \] \[ \vec{F} = 0\, \hat{i} + 0\, \hat{j} - 280\, \hat{k} \] **Compute the cross product:** \[ \vec{M}_A = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 10 & 4 \\ 0 & 0 & -280 \end{vmatrix} \] Calculating each component: - \( \hat{i} \) component: \( (10)(-280) - (4)(0) = -2800 \) - \( \hat{j} \) component: \( -\left[(5)(-280) - (4)(0)\right] = 1400 \) - \( \hat{k} \) component: \( (5)(0) - (10)(0) = 0 \) Thus, the moment about A is: \[ \vec{M}_A = -2800\, \hat{i} + 1400\, \hat{j} + 0\, \hat{k}\ \text{lb-in} \] --- ### **(b) Moment about B:** Calculating the moment about point B: \[ \vec{M}_B = \vec{r}_{BF} \times \vec{F} \] Substituting the values: \[ \vec{r}_{BF} = 13\, \hat{i} + 4\, \hat{j} + 9\, \hat{k} \] \[ \vec{F} = 0\, \hat{i} + 0\, \hat{j} - 280\, \hat{k} \] **Compute the cross product:** \[ \vec{M}_B = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 13 & 4 & 9 \\ 0 & 0 & -280 \end{vmatrix} \] Calculating each component: - \( \hat{i} \) component: \( (4)(-280) - (9)(0) = -1120 \) - \( \hat{j} \) component: \( -\left[(13)(-280) - (9)(0)\right] = 3640 \) - \( \hat{k} \) component: \( (13)(0) - (4)(0) = 0 \) Thus, the moment about B is: \[ \vec{M}_B = -1120\, \hat{i} + 3640\, \hat{j} + 0\, \hat{k}\ \text{lb-in} \] --- ## **Step 5: Final Answers** ### **Moment about Point A:** \[ \boxed{ \vec{M}_A = -2800\, \hat{i} + 1400\, \hat{j}\ \text{lb-in} } \] ### **Moment about Point B:** \[ \boxed{ \vec{M}_B = -1120\, \hat{i} + 3640\, \hat{j}\ \text{lb-in} } \] **These values represent the moments of the 280 lb force about points A and B, respectively.**

# Step-by-Step Solution to Find Moments ## Step 1: Assign Coordinates We will first assign coordinates to all relevant points based on the provided diagram. - **O**: Origin, \( (0, 0, 0) \) - **A**: Point A located at \( (8, 0, 5) \) inches - **B**: Point B located at \( (0, 6, 0) \) inches ### Locate Point F Point F is determined by moving from the origin: - 13 inches along the \( x \)-axis - 10 inches along the \( y \)-axis - 9 inches along the \( z \)-axis Thus, the coordinates for point **F** are: - **F**: \( (13, 10, 9) \) inches --- ## Step 2: Write the Force Vector The applied force acts downward along the \( z \)-axis: \[ \vec{F} = -280\, \hat{k}\ \text{lb} \] In vector form, this is expressed as: \[ \vec{F} = 0\, \hat{i} + 0\, \hat{j} - 280\, \hat{k}\ \text{lb} \] --- ## Step 3: Position Vectors from A and B to F ### From A to F The position vector from A to F is calculated as: \[ \vec{r}_{AF} = \vec{r}_F - \vec{r}_A = (13 - 8)\, \hat{i} + (10 - 0)\, \hat{j} + (9 - 5)\, \hat{k} \] Thus: \[ \vec{r}_{AF} = 5\, \hat{i} + 10\, \hat{j} + 4\, \hat{k}\ \text{in} \] ### From B to F The position vector from B to F is: \[ \vec{r}_{BF} = \vec{r}_F - \vec{r}_B = (13 - 0)\, \hat{i} + (10 - 6)\, \hat{j} + (9 - 0)\, \hat{k} \] Thus: \[ \vec{r}_{BF} = 13\, \hat{i} + 4\, \hat{j} + 9\, \hat{k}\ \text{in} \] --- ## Step 4: Calculate Moments The moment about a point is given by the cross product: \[ \vec{M} = \vec{r} \times \vec{F} \] ### (a) Moment about A Calculating the moment about point A: \[ \vec{M}_A = \vec{r}_{AF} \times \vec{F} \] Substituting the values: \[ \vec{r}_{AF} = 5\, \hat{i} + 10\, \hat{j} + 4\, \hat{k} \] \[ \vec{F} = 0\, \hat{i} + 0\, \hat{j} - 280\, \hat{k} \] **Compute the cross product:** \[ \vec{M}_A = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 10 & 4 \\ 0 & 0 & -280 \end{vmatrix} \] Calculating each component: - \( \hat{i} \) component: \( (10)(-280) - (4)(0) = -2800 \) - \( \hat{j} \) component: \( -\left[(5)(-280) - (4)(0)\right] = 1400 \) - \( \hat{k} \) component: \( (5)(0) - (10)(0) = 0 \) Thus, the moment about A is: \[ \vec{M}_A = -2800\, \hat{i} + 1400\, \hat{j} + 0\, \hat{k}\ \text{lb-in} \] --- ### (b) Moment about B Calculating the moment about point B: \[ \vec{M}_B = \vec{r}_{BF} \times \vec{F} \] Substituting the values: \[ \vec{r}_{BF} = 13\, \hat{i} + 4\, \hat{j} + 9\, \hat{k} \] \[ \vec{F} = 0\, \hat{i} + 0\, \hat{j} - 280\, \hat{k} \] **Compute the cross product:** \[ \vec{M}_B = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 13 & 4 & 9 \\ 0 & 0 & -280 \end{vmatrix} \] Calculating each component: - \( \hat{i} \) component: \( (4)(-280) - (9)(0) = -1120 \) - \( \hat{j} \) component: \( -\left[(13)(-280) - (9)(0)\right] = 3640 \) - \( \hat{k} \) component: \( (13)(0) - (4)(0) = 0 \) Thus, the moment about B is: \[ \vec{M}_B = -1120\, \hat{i} + 3640\, \hat{j} + 0\, \hat{k}\ \text{lb-in} \] --- ## Step 5: Final Answers ### Moment about Point A \[ \boxed{ \vec{M}_A = -2800\, \hat{i} + 1400\, \hat{j}\ \text{lb-in} } \] ### Moment about Point B \[ \boxed{ \vec{M}_B = -1120\, \hat{i} + 3640\, \hat{j}\ \text{lb-in} } \] These results represent the moments of the 280 lb force about points A and B, respectively.