## Step-by-Step Solution ### **Given Data** - **Force in member AC**: \( F_{AC} = 240 \) kips - **We are to find:** Moment of this force **about point D**. --- ### **Step 1: Assign Coordinates** Let's assign coordinates to the points based on the diagram: - \( A = (, , ) \) - \( C = (25, , 17.5) \) - \( D = (80, , 60) \) --- ### **Step 2: Find the Force Vector \( \vec{F}_{AC} \)** The force acts **along AC** from A to C. #### **Direction Vector AC:** \[ \vec{r}_{AC} = (25-)\hat{i} + (-)\hat{j} + (17.5-)\hat{k} = 25\hat{i} + \hat{j} + 17.5\hat{k} \] #### **Magnitude of AC:** \[ |\vec{AC}| = \sqrt{25^2 + ^2 + 17.5^2} = \sqrt{625 + + 306.25} = \sqrt{931.25} \approx 30.52 \] #### **Unit Vector Along AC:** \[ \hat{u}_{AC} = \frac{1}{30.52} \langle 25, , 17.5 \rangle = \langle .819, , .574 \rangle \] #### **Force Vector:** \[ \vec{F}_{AC} = 240 \cdot \langle .819, , .574 \rangle = \langle 196.56, , 137.76 \rangle \text{ kips} \] --- ### **Step 3: Position Vector from D to A (\(\vec{r}_{DA}\))** \[ \vec{r}_{DA} = \vec{A} - \vec{D} = (-80)\hat{i} + (-)\hat{j} + (-60)\hat{k} = -80\hat{i} + \hat{j} -60\hat{k} \] --- ### **Step 4: Moment Calculation (\(\vec{M}_D = \vec{r}_{DA} \times \vec{F}_{AC}\))** Set up the cross product: \[ \vec{M}_D = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -80 & & -60 \\ 196.56 & & 137.76 \\ \end{vmatrix} \] Expand: \[ \vec{M}_D = \hat{i} \begin{vmatrix} & -60 \\ & 137.76 \\ \end{vmatrix} - \hat{j} \begin{vmatrix} -80 & -60 \\ 196.56 & 137.76 \\ \end{vmatrix} + \hat{k} \begin{vmatrix} -80 & \\ 196.56 & \\ \end{vmatrix} \] Calculate each component: - **i-component**: \( ()(137.76) - ()(-60) = \) - **j-component**: \( -\left[(-80)(137.76) - (196.56)(-60)\right] = -\left[-11020.8 + 11793.6\right] = -772.8 \) - **k-component**: \( (-80)() - ()(196.56) = \) So, \[ \vec{M}_D = \hat{i} - 772.8\hat{j} + \hat{k} \] --- ### **Step 5: Final Answer** \[ \boxed{ \vec{M}_D = -772.8\ \hat{j} \ \text{kips-in} } \] - **The moment of force AC about point D is \( 772.8 \) kips-in, acting in the negative y-direction.**

# Step-by-Step Solution ## Given Data - **Force in member AC**: \( F_{AC} = 240 \) kips - **Objective**: Find the moment of this force about point D. --- ## Step 1: Assign Coordinates Assume the coordinates based on the provided diagram (exact coordinates need to be inferred from the image): - \( A = (x_A, y_A, z_A) \) - \( C = (25, y_C, 17.5) \) - \( D = (80, y_D, 60) \) --- ## Step 2: Find the Force Vector \( \vec{F}_{AC} \) The force acts along line AC. ### Direction Vector AC \[ \vec{r}_{AC} = (25 - x_A)\hat{i} + (y_C - y_A)\hat{j} + (17.5 - z_A)\hat{k} \] ### Magnitude of AC \[ |\vec{AC}| = \sqrt{(25 - x_A)^2 + (y_C - y_A)^2 + (17.5 - z_A)^2} \] ### Unit Vector Along AC \[ \hat{u}_{AC} = \frac{1}{|\vec{AC}|} \langle 25 - x_A, y_C - y_A, 17.5 - z_A \rangle \] ### Force Vector \[ \vec{F}_{AC} = 240 \cdot \hat{u}_{AC} \] --- ## Step 3: Position Vector from D to A (\(\vec{r}_{DA}\)) \[ \vec{r}_{DA} = \vec{A} - \vec{D} = (x_A - 80)\hat{i} + (y_A - y_D)\hat{j} + (z_A - 60)\hat{k} \] --- ## Step 4: Moment Calculation (\(\vec{M}_D = \vec{r}_{DA} \times \vec{F}_{AC}\)) Set up the cross product: \[ \vec{M}_D = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x_A - 80 & y_A - y_D & z_A - 60 \\ F_{ACx} & F_{ACy} & F_{ACz} \\ \end{vmatrix} \] ### Calculate Each Component - **i-component**: \[ M_{D_x} = (y_A - y_D)F_{ACz} - (z_A - 60)F_{ACy} \] - **j-component**: \[ M_{D_y} = -\left[(x_A - 80)F_{ACz} - (z_A - 60)F_{ACx}\right] \] - **k-component**: \[ M_{D_z} = (x_A - 80)F_{ACy} - (y_A - y_D)F_{ACx} \] So, the moment vector is: \[ \vec{M}_D = M_{D_x}\hat{i} + M_{D_y}\hat{j} + M_{D_z}\hat{k} \] --- ## Step 5: Final Answer Based on the calculations: \[ \vec{M}_D = M_{D_x} \hat{i} - M_{D_y} \hat{j} + M_{D_z} \hat{k} \text{ (in kips-in)} \] - **The moment of force AC about point D is represented by the vector \( \vec{M}_D \)**, which can be computed once specific coordinates are assigned. If the coordinates for points A, C, and D were provided accurately, you could substitute them into the equations to find numerical values for \( M_{D_x} \), \( M_{D_y} \), and \( M_{D_z} \).

# Step-by-Step Solution ## Given Data - **Force in member AC**: \( F_{AC} = 240 \) kips - **Objective**: Find the moment of this force about point D. --- ## Step 1: Assign Coordinates Assuming hypothetical coordinates based on the diagram (exact coordinates should be inferred from the image): - Let \( A = (x_A, y_A, z_A) \) - \( C = (25, y_C, 17.5) \) - \( D = (80, y_D, 60) \) --- ## Step 2: Find the Force Vector \( \vec{F}_{AC} \) The force acts along line AC. ### Direction Vector AC \[ \vec{r}_{AC} = (25 - x_A)\hat{i} + (y_C - y_A)\hat{j} + (17.5 - z_A)\hat{k} \] ### Magnitude of AC \[ |\vec{AC}| = \sqrt{(25 - x_A)^2 + (y_C - y_A)^2 + (17.5 - z_A)^2} \] ### Unit Vector Along AC \[ \hat{u}_{AC} = \frac{1}{|\vec{AC}|} \langle 25 - x_A, y_C - y_A, 17.5 - z_A \rangle \] ### Force Vector \[ \vec{F}_{AC} = 240 \cdot \hat{u}_{AC} \] --- ## Step 3: Position Vector from D to A (\(\vec{r}_{DA}\)) \[ \vec{r}_{DA} = \vec{A} - \vec{D} = (x_A - 80)\hat{i} + (y_A - y_D)\hat{j} + (z_A - 60)\hat{k} \] --- ## Step 4: Moment Calculation (\(\vec{M}_D = \vec{r}_{DA} \times \vec{F}_{AC}\)) Set up the cross product: \[ \vec{M}_D = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x_A - 80 & y_A - y_D & z_A - 60 \\ F_{ACx} & F_{ACy} & F_{ACz} \\ \end{vmatrix} \] ### Calculate Each Component - **i-component**: \[ M_{D_x} = (y_A - y_D)F_{ACz} - (z_A - 60)F_{ACy} \] - **j-component**: \[ M_{D_y} = -\left[(x_A - 80)F_{ACz} - (z_A - 60)F_{ACx}\right] \] - **k-component**: \[ M_{D_z} = (x_A - 80)F_{ACy} - (y_A - y_D)F_{ACx} \] Thus, the moment vector is: \[ \vec{M}_D = M_{D_x}\hat{i} + M_{D_y}\hat{j} + M_{D_z}\hat{k} \] --- ## Step 5: Final Answer The moment of force AC about point D can be expressed as: \[ \vec{M}_D = M_{D_x} \hat{i} - M_{D_y} \hat{j} + M_{D_z} \hat{k} \text{ (in kips-in)} \] - **Numerical values for \( M_{D_x} \), \( M_{D_y} \), and \( M_{D_z} \)** can be computed once the specific coordinates for points A, C, and D are determined from the diagram. In conclusion, substitute the assigned coordinates into the equations for a complete solution.