Let's break this into clear steps: --- ## **Given Data** - Mass of box, \( m = 10\,\text{kg} \) - Coefficient of static friction, \( \mu_s = .25 \) - Gravity, \( g = 9.81\,\text{m/s}^2 \) - Rod AC is light (ignore its mass) - Length AC = 1.41 m (from Pythagoras: \( \sqrt{1^2 + 1^2} = 1.414 \) m) - Angle with the horizontal, \( \theta = 45^\circ \) - Box is at midpoint (.5 m from both A and C) --- ## **Step 1: Find Minimum Force \( F \) for Box to Stay on the Rod** ### **Forces on the Box** - Weight (\( W = mg \)) acts vertically down. - Normal force (\( N \)) from the rod, perpendicular to AC. - Friction force (\( f \)), acts up along the rod, max value \( f_{max} = \mu_s N \). - Applied force \( F \), parallel to the rod. ### **Resolve Forces Along and Perpendicular to the Rod** #### **Perpendicular to Rod:** - \( N = W \cos(45^\circ) \) - \( N = mg \cos(45^\circ) = 10 \times 9.81 \times \frac{\sqrt{2}}{2} \approx 69.3\, \text{N} \) #### **Along the Rod:** - Components along the rod: - Weight: \( W_{||} = mg \sin(45^\circ) \) - Friction acts *against* the tendency of the box to slide down. - Applied force \( F \) acts *up* the rod. **At limiting equilibrium:** \[ F + f_{max} = W_{||} \] \[ F + \mu_s N = mg \sin(45^\circ) \] Plug in values: - \( mg \sin(45^\circ) = 10 \times 9.81 \times \frac{\sqrt{2}}{2} \approx 69.3\, \text{N} \) - \( \mu_s N = .25 \times 69.3 = 17.33\, \text{N} \) So, \[ F_{min} = 69.3 - 17.33 = \boxed{51.97\, \text{N}} \] --- ## **Step 2: Find Reaction Forces at A and C, and Tension in CD** ### **Free Body Diagram of Rod AC** - Forces at A and C (reactions from smooth walls, normal to the walls). - Tension \( T \) in cable CD. - Force \( F \) on box at B (resolved into components). - Weight of box at B. #### **Let** - \( A_x \), \( A_y \): Reaction at A (vertical and horizontal) - \( C_x \), \( C_y \): Reaction at C (vertical and horizontal) - \( T \): Tension in CD (horizontal, since D is directly right of C) ### **Take Moments about Point A** Sum moments about A to eliminate \(A_x, A_y\): - Distance AC = 1.41 m; midpoint B is .707 m from A. - Perpendicular distance from A to cable CD: \(1 \, \text{m}\) (from geometry). **Moments due to:** - Weight at B: \( mg \) acts down at B, distance \( .707\,\text{m} \) from A. - Force F at B: acts along rod at B, same distance. - Tension T at C: perpendicular distance from A to CD is \( 1\,\text{m} \). #### **Moment Calculation** - Weight at B: \( mg \) at .707 m from A, perpendicular to rod. - Tension \( T \) at C, horizontal, at 1 m below A and 1 m left, so acts at \( 45^\circ \) to rod. But let's use force equilibrium equations: #### **Sum Forces in x and y** **Horizontal (\( \Sigma F_x = \)):** \[ A_x + C_x + T = \] **Vertical (\( \Sigma F_y = \)):** \[ A_y + C_y - mg = \] **Sum of moments about A (\( \Sigma M_A = \)):** - Weight at B: \( mg \) at distance \( d_B = .707\, \text{m} \) from A, perpendicular. - Tension at C: \( T \) at C, which is 1 m horizontally from A. \[ mg \times .707 = T \times 1 \] \[ T = mg \times .707 = 10 \times 9.81 \times .707 = 69.3\, \text{N} \] ### **Find Reaction Forces at A and C** #### **From Geometry** - The rod is at \( 45^\circ \), so reactions at A and C are perpendicular to the walls. #### **Sum Forces:** - At A (top): Reaction is horizontal (from vertical wall). - At C (bottom): Reaction is vertical (from horizontal floor). So, - \( A_x \) is the only component at A. - \( C_y \) is the only component at C. From earlier: \[ A_x + T = \implies A_x = -T = -69.3\,\text{N} \] \[ C_y = mg = 98.1\,\text{N} \] --- ## **Final Answers** ### **1. Minimum Force \( F \):** \[ \boxed{51.97\, \text{N}} \] ### **2. With this \( F \):** #### **Reaction at A:** \[ A_x = -69.3\,\text{N}, \quad A_y = \] #### **Reaction at C:** \[ C_x = , \quad C_y = 98.1\,\text{N} \] #### **Tension in Cable CD:** \[ \boxed{69.3\,\text{N}} \] --- ### **Summary Table** | Quantity | Value | |--------------------|--------------| | Minimum \( F \) | 51.97 N | | Reaction at A | 69.3 N (left)| | Reaction at C | 98.1 N (up) | | Tension in CD | 69.3 N |