Let's break down the solution step by step. --- # **1. Given Data** **Forces:** - \(\vec{F}_1 = \langle 15, -20, -22 \rangle \) lb - \(\vec{F}_2 = \langle 10, 25, 30 \angle \) lb **Coordinates (from the diagram):** - \(A = (, , )\) - \(B = (, 4, 4)\) - \(C = (, 4, )\) - \(D = (5, , )\) - \(E = (5, 1, )\) **Force application points (by observation):** - \(\vec{F}_1\) is applied at \(B = (,4,4)\) - \(\vec{F}_2\) is applied at \(E = (5,1,)\) --- # **2. Equivalent Force at Point A** The **equivalent force** at point \(A\) is the **vector sum** of all forces: \[ \vec{F}_{\text{eq}} = \vec{F}_1 + \vec{F}_2 \] \[ \vec{F}_{\text{eq}} = \langle 15, -20, -22 \rangle + \langle 10, 25, 30 \rangle = \langle 25, 5, 8 \rangle \ \text{lb} \] --- # **3. Equivalent Moment at Point A** The **moment** of a force about point \(A\) is: \[ \vec{M}_A = \vec{r}_B \times \vec{F}_1 + \vec{r}_E \times \vec{F}_2 \] where: - \(\vec{r}_B = \overrightarrow{AB} = (,4,4) - (,,) = \langle , 4, 4 \rangle\) - \(\vec{r}_E = \overrightarrow{AE} = (5,1,) - (,,) = \langle 5, 1, \rangle\) ### **Moment due to \(\vec{F}_1\) at \(B\):** \[ \vec{M}_1 = \vec{r}_B \times \vec{F}_1 \] \[ = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ & 4 & 4 \\ 15 & -20 & -22 \end{vmatrix} \] \[ = \mathbf{i} [4 \cdot (-22) - 4 \cdot (-20)] - \mathbf{j}[ \cdot (-22) - 4 \cdot 15] + \mathbf{k}[ \cdot (-20) - 4 \cdot 15] \] \[ = \mathbf{i} [ -88 + 80 ] - \mathbf{j}[ - 60 ] + \mathbf{k}[ - 60 ] \] \[ = \mathbf{i} [ -8 ] + \mathbf{j}[ 60 ] + \mathbf{k}[ -60 ] \] \[ = \langle -8, 60, -60 \rangle \text{ lb}\cdot\text{ft} \] --- ### **Moment due to \(\vec{F}_2\) at \(E\):** \[ \vec{M}_2 = \vec{r}_E \times \vec{F}_2 \] \[ = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & 1 & \\ 10 & 25 & 30 \end{vmatrix} \] \[ = \mathbf{i}[1 \cdot 30 - \cdot 25] - \mathbf{j}[5 \cdot 30 - \cdot 10] + \mathbf{k}[5 \cdot 25 - 1 \cdot 10] \] \[ = \mathbf{i}[30] - \mathbf{j}[150] + \mathbf{k}[125 - 10] \] \[ = \langle 30, -150, 115 \rangle \text{ lb}\cdot\text{ft} \] --- ### **Total Moment at \(A\):** \[ \vec{M}_A = \vec{M}_1 + \vec{M}_2 = \langle -8, 60, -60 \rangle + \langle 30, -150, 115 \rangle = \langle 22, -90, 55 \rangle \text{ lb}\cdot\text{ft} \] --- # **Summary of Answers (Cartesian Vectors)** ## **Equivalent Force at A:** \[ \boxed{ \vec{F}_{\text{eq}} = \langle 25, 5, 8 \rangle \ \text{lb} } \] ## **Equivalent Moment at A:** \[ \boxed{ \vec{M}_A = \langle 22, -90, 55 \rangle \ \text{lb}\cdot\text{ft} } \] --- # **4. Triple Scalar Product (Moment about Imaginary Axis)** Suppose the imaginary axis \(V\) is a unit vector \(\vec{e}_V\). The moment about this axis is: \[ M_V = \vec{e}_V \cdot \vec{M}_A \] If you specify the direction of the imaginary axis, use the result above for \(\vec{M}_A\). --- **Let me know if you need the moment about a specific axis or if you want a worked example for the triple scalar product!**