Let's break down the problem step-by-step: ### **Given Data:** - Position vector \(\vec{R}\) from \(A\) to \(B\): - \(R_x = -8 \, \text{ft}\) - \(R_y = 3 \, \text{ft}\) - Force vector \(\vec{F}\) at \(B\): - \(F_x = 36 \, \text{lb}\) - \(F_y = 15 \, \text{lb}\) We are to find the **moment of force \(\vec{F}\) about point \(A\)**. --- ### **Step 1: Write the vectors** \[ \vec{R} = R_x \hat{i} + R_y \hat{j} = (-8)\hat{i} + (3)\hat{j} \ \text{ft} \] \[ \vec{F} = F_x \hat{i} + F_y \hat{j} = (36)\hat{i} + (15)\hat{j} \ \text{lb} \] --- ### **Step 2: Moment formula (in 2D)** The moment of \(\vec{F}\) about point \(A\) (cross product): \[ M_A = \vec{R} \times \vec{F} \] In 2D: \[ M_A = (R_x F_y - R_y F_x) \] --- ### **Step 3: Substitute values** \[ M_A = (-8)(15) - (3)(36) \] \[ M_A = (-120) - (108) \] \[ M_A = -228 \ \text{lb} \cdot \text{ft} \] --- ### **Step 4: Direction** A negative sign means the moment is **clockwise**. --- ## **Final Answers for the Blanks** **x-component of the position vector \(R\):** \(-8\) ft **y-component of the position vector \(R\):** \(3\) ft **x-component of the force vector \(F\):** \(36\) lb **y-component of the force vector \(F\):** \(15\) lb **Moment about point A:** \(-228\) lb·ft (**Clockwise**) --- ### **Summary Table** | Quantity | Value | |-------------------------------------------|------------------| | x-component of position vector \(R\) | \(-8\) ft | | y-component of position vector \(R\) | \(3\) ft | | x-component of force vector \(F\) | \(36\) lb | | y-component of force vector \(F\) | \(15\) lb | | Moment about point \(A\) | \(-228\) lb·ft | | Moment direction | Clockwise | Let me know if you need more explanation!