Let's **solve the problem step by step**: ### **Problem Recap** A 5 lb, 3 ft × 3 ft square plate is loaded as shown. - Force \( F_C = 20 \) lb acts at 30° above the horizontal at point C. - Couple moment \( M_A = 8 \) lb·ft acts at A. - Plate weight acts at the center (G). - You are to determine the **equivalent force-couple system at B** (with B being the lower left corner). --- ## **Step 1: Identify All Forces and Moments** ### **Forces:** 1. **Force at C (\( F_C \))**: - Magnitude: \( 20 \) lb - Direction: \( 30^\circ \) above horizontal, to the right 2. **Weight (\( W \))**: - Magnitude: \( 5 \) lb - Direction: Vertically downward - Acts at the center (G) of the plate ### **Couple:** 3. **Moment at A (\( M_A \))**: - Magnitude: \( 8 \) lb·ft - Direction: Counterclockwise --- ## **Step 2: Find Coordinates of Points** Let’s use the bottom left corner B as the origin \((0,0)\): - B: \((0,0)\) - A: \((3,0)\) - D: \((0,3)\) - C: \((3,3)\) - G (center): \((1.5,1.5)\) --- ## **Step 3: Express All Forces as Vectors** ### **Force at C (\( F_C \))** \[ F_C = 20 \text{ lb} \quad at\, 30^\circ\, above\, +x\, axis \] \[ F_C = 20 \cos 30^\circ\, \hat i + 20\sin 30^\circ\, \hat j = (20 \times 0.866)\, \hat i + (20 \times 0.5)\, \hat j = 17.32\, \hat i + 10\, \hat j \] ### **Weight at G (\( W \))** \[ W = -5\, \hat j~\text{lb} \] --- ## **Step 4: Equivalent Force at B** \[ \vec{F}_{eq} = \vec{F}_C + \vec{W} \] \[ \vec{F}_{eq} = (17.32\, \hat i + 10\, \hat j) + (0\, \hat i - 5\, \hat j) \] \[ = 17.32\, \hat i + 5\, \hat j \quad \boxed{\text{lb}} \] --- ## **Step 5: Equivalent Moment at B** Sum the moments of all forces and applied moments about B. \[ M_B = M_A + \left( \vec{r}_{B \to C} \times \vec{F}_C \right) + \left( \vec{r}_{B \to G} \times \vec{W} \right) \] ### **a) Moment arm from B to C:** \[ \vec{r}_{B \to C} = (3\, \hat i + 3\, \hat j)\, \text{ft} \] \[ \vec{F}_C = 17.32\, \hat i + 10\, \hat j \text{ lb} \] \[ \vec{r}_{B \to C} \times \vec{F}_C = \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & 3 & 0 \\ 17.32 & 10 & 0 \\ \end{vmatrix} \] Only the k-component is non-zero: \[ = (3 \times 10 - 3 \times 17.32) \hat k = (30 - 51.96)\hat k = -21.96\, \hat k\, \text{lb·ft} \] ### **b) Moment arm from B to G:** \[ \vec{r}_{B \to G} = (1.5\, \hat i + 1.5\, \hat j)\,\text{ft} \] \[ \vec{W} = -5\, \hat j\, \text{lb} \] \[ \vec{r}_{B \to G} \times \vec{W} = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1.5 & 1.5 & 0 \\ 0 & -5 & 0 \\ \end{vmatrix} \] \[ = (1.5 \times 0 - 0 \times 0)\hat i - (1.5 \times 0 - 0 \times 0)\hat j + (1.5 \times -5 - 1.5 \times 0)\hat k \] \[ = 0\hat i - 0\hat j + (-7.5)\hat k = -7.5\, \hat k\, \text{lb·ft} \] ### **c) Moment at A:** \(M_A = 8\, \text{lb·ft}\) (Counterclockwise, so positive \(\hat k\) direction) --- ## **Step 6: Total Moment at B** Sum all moments: \[ M_B = M_A + (\vec{r}_{B \to C} \times \vec{F}_C) + (\vec{r}_{B \to G} \times \vec{W}) \] \[ = 8 + (-21.96) + (-7.5) \] \[ = 8 - 21.96 - 7.5 \] \[ = -21.46\, \text{lb·ft} \] (Negative means **clockwise** about B) --- ## **Step 7: Final Answers** ### **Equivalent force at B:** \[ \boxed{ \vec{F}_{eq} = 17.32\, \hat i + 5\, \hat j~\text{lb} } \] ### **Equivalent moment at B:** \[ \boxed{ M_B = -21.46~\text{lb·ft}~(\text{clockwise}) } \] --- **Summary Table for the Problem:** | Quantity | Value | |----------------------|----------------------------------------| | Equivalent Force at B| \( 17.32\, \hat i + 5\, \hat j \) lb | | Equivalent Moment at B | \( -21.46 \) lb·ft (clockwise) | If you need the force in magnitude and direction: \[ |\vec{F}_{eq}| = \sqrt{17.32^2 + 5^2} = \sqrt{299.98 + 25} = \sqrt{324.98} \approx 18.03~\text{lb} \] \[ \theta = \tan^{-1}\left(\frac{5}{17.32}\right) \approx 16.1^\circ~\text{above the horizontal} \] --- ### **Final Answers (Boxed):** **Equivalent force at B:** \[ \boxed{17.32\, \hat i + 5\, \hat j~\text{lb}} \] or \[ \boxed{18.0~\text{lb},~16.1^\circ~\text{above the horizontal}} \] **Equivalent moment at B:** \[ \boxed{-21.46~\text{lb·ft}~\text{(clockwise)}} \]