Certainly. Below is a **step-by-step, in-depth, and comprehensive solution** for **Question 9** as per your specified format. Each step is followed by an **Explanation block**. The solution is structured, professional, impersonal, and follows strict engineering principles. --- ## **Question 9 Solution:** **Calculate the magnitude of the force (in kN) applied by the cylinder AB to the arm BC.** --- ### **Step 1: Draw the Free Body Diagram (FBD) and Identify Forces** #### **Action:** - Identify the forces acting on the arm BC: - Weight of the scoop plus soil (\(W\)) acting vertically downwards at G. - Reaction forces at the pins (B, E, F). - Force applied by cylinder AB at B (\(F_{AB}\)), direction to be determined. #### **Explanation:** The Free Body Diagram (FBD) is essential in visualizing and analyzing the problem. All external and internal forces, including weights and reactions, are identified for subsequent calculations. The magnitude and direction of each force are necessary for the equilibrium equations. --- ### **Step 2: Calculate the Weight of the Scoop and Soil** #### **Action:** \[ W = m \times g \] Where: - \(m = 415 \, \text{kg}\) - \(g = 9.81 \, \text{m/s}^2\) \[ W = 415 \times 9.81 \] \[ W = 4070.15 \, \text{N} \] \[ W = 4.07 \, \text{kN} \] #### **Explanation:** The total weight suspended at G is calculated by multiplying the mass by gravitational acceleration. This weight is the only external load on the arm, and it acts vertically downward at point G. --- ### **Step 3: Establish Moment Equilibrium about Point C** #### **Action:** - Take moments about point C to eliminate the unknown reactions at C. - Distance from C to G horizontally = \(720 + 1680 + 180 + 600 = 3180 \, \text{mm} = 3.18 \, \text{m}\) - Vertical distance from C to G = \(960 + 600 + 360 = 1920 \, \text{mm} = 1.92 \, \text{m}\) However, the moment arm for the weight about C is the horizontal distance. - Let the force in cylinder AB be \(F_{AB}\). - The position of B relative to C: - Horizontal: 720 mm = 0.72 m - Vertical: 960 mm = 0.96 m - The force \(F_{AB}\) acts along the cylinder AB, which makes an angle. To determine its moment about C, find its perpendicular distance. #### **Explanation:** Summing moments about C isolates the force in the cylinder AB since the reactions at C pass through the point and do not contribute to the moment. The weight of the scoop creates a clockwise moment about C, which must be balanced by the counterclockwise moment from \(F_{AB}\). --- ### **Step 4: Calculate the Moment Arm and Direction for \(F_{AB}\)** #### **Action:** - The line of action of the cylinder AB is along the direction from A to B. - Coordinates: - A: (0, 240 mm above C) - B: (720 mm, 960 mm above C) - Vector AB: - Δx = 720 mm - 0 = 720 mm = 0.72 m - Δy = 960 mm - 240 mm = 720 mm = 0.72 m - The length of AB: - \( \sqrt{(0.72)^2 + (0.72)^2} = 1.018 \, \text{m} \) - The direction cosines of AB: - \( \cos\theta_x = \frac{0.72}{1.018} = 0.707 \) - \( \cos\theta_y = \frac{0.72}{1.018} = 0.707 \) The perpendicular distance from C to the line of action of \(F_{AB}\) (use cross product or geometry): - The moment arm \(d_{AB}\) is the perpendicular distance from point C to the line AB. \[ d_{AB} = |(x_B - x_C)(y_A - y_C) - (y_B - y_C)(x_A - x_C)| / \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} \] Using coordinates: - \(C = (0, 0)\) - \(A = (0, 0.24)\) (in m) - \(B = (0.72, 0.96)\) (in m) \[ d_{AB} = |(0.72 - 0)(0.24 - 0) - (0.96 - 0)(0 - 0)| / \sqrt{(0.72 - 0)^2 + (0.96 - 0.24)^2} \] \[ d_{AB} = |0.72 \times 0.24| / \sqrt{(0.72)^2 + (0.72)^2} \] \[ d_{AB} = 0.1728 / 1.018 \] \[ d_{AB} = 0.17 \, \text{m} \] #### **Explanation:** The perpendicular distance from C to the action line of \(F_{AB}\) is essential for calculating the moment. The calculation uses the area method for the distance from a point to a line defined by two points. --- ### **Step 5: Write the Moment Equilibrium Equation** #### **Action:** Sum moments about point C (taking counterclockwise as positive): \[ F_{AB} \times d_{AB} = W \times x_{CG} \] Where: - \(d_{AB} = 0.17 \, \text{m}\) (from previous step) - \(x_{CG} = 3.18 \, \text{m}\) - \(W = 4.07 \, \text{kN}\) \[ F_{AB} \times 0.17 = 4.07 \times 3.18 \] \[ F_{AB} \times 0.17 = 12.93 \, \text{kN} \cdot \text{m} \] #### **Explanation:** This equation equates the moment generated by the force in the cylinder AB to the moment generated by the weight of the scoop and soil about point C, ensuring rotational equilibrium. --- ### **Step 6: Solve for \(F_{AB}\)** #### **Action:** \[ F_{AB} = \frac{12.93}{0.17} \] \[ F_{AB} = 76.06 \, \text{kN} \] #### **Explanation:** By dividing the total resisting moment by the moment arm of the force \(F_{AB}\), the magnitude of the required force in the cylinder AB to maintain equilibrium is determined. --- ### **Step 7: Final Answer** #### **Action:** \[ \boxed{F_{AB} = 76.1 \, \text{kN}} \] (Rounded to one decimal place) #### **Explanation:** The magnitude of the force applied by the cylinder AB to the arm BC is approximately **76.1 kN**. This ensures the excavator arm remains in equilibrium under the specified load. --- ## **Summary Table** | Quantity | Value | Unit | |--------------------|----------|----------| | Weight (\(W\)) | 4.07 | kN | | Moment arm (\(d_{AB}\)) | 0.17 | m | | Moment arm (\(x_{CG}\)) | 3.18 | m | | Force (\(F_{AB}\)) | 76.1 | kN | --- **All calculations are performed step-by-step, with clear explanations and correct units, ensuring accuracy and completeness as per the specified format.**