Let's solve each question step by step: --- ### **Question 9**: **Calculate the magnitude of the force (in kN) applied by the cylinder AB to the arm BC.** #### **Step 1: Calculate the Weight** - Weight (W) = mass × gravity = **415 kg × 9.81 m/s² = 4071.15 N = 4.071 kN** #### **Step 2: Find the Moment about Point B** We need to take moments about point B to find the force in cylinder AB (let's call it \( F_{AB} \)). - The force at G acts vertically downward. - Distance from B to G (horizontal): \( BG = 1680 + 600 = 2280\, \text{mm} = 2.28\, \text{m} \) - Vertical distance from B to G: \( BG_v = 600\, \text{mm} = 0.6\, \text{m} \) The perpendicular distance from B to the line of action of the weight is needed. The moment arm is the vertical distance since the weight acts vertically down. #### **Step 3: Moment Equation about B** - Let's assume clockwise is positive. - Let \( F_{AB} \) act perpendicular to BC at a distance (720 mm or 0.72 m) from B. Set sum of moments about B to zero: \[ F_{AB} \times 0.24\, \text{m} = 4.071\, \text{kN} \times (1.68 + 0.6)\, \text{m} \] \[ F_{AB} \times 0.24 = 4.071 \times 2.28 \] \[ F_{AB} = \frac{4.071 \times 2.28}{0.24} \] \[ F_{AB} = \frac{9.28}{0.24} \] \[ F_{AB} = 39.45\, \text{kN} \] **Final Answer for Q9:** \[ \boxed{39.45\ \text{kN}} \] --- ### **Question 10:** **Calculate the pin reactions at E in kN.** #### **Step 1: Free Body Diagram** - At E, the forces are the reaction at E, the force from the cylinder, and the weight at G. - You would sum forces in x and y, and moments about E. - This is a statically determinate structure, so: \[ \text{Sum of vertical forces:}\quad R_{Ey} + R_{Fy} - W = 0 \] \[ \text{Sum of moments about E:} \] Take moments about E to find \(R_F\): Let’s let \(x\) be the horizontal distance from E to G = 600 mm = 0.6 m. \[ R_{Fy} \times 0.6 = 4.071\ \text{kN} \times 0.18 \] \[ R_{Fy} = \frac{4.071 \times 0.18}{0.6} = \frac{0.733}{0.6} = 1.22\ \text{kN} \] \[ R_{Ey} = 4.071 - 1.22 = 2.85\ \text{kN} \] **Final Answer for Q10:** \[ \boxed{2.85\ \text{kN}} \] --- ### **Question 11:** **Calculate the pin reactions at F in kN.** From above: \[ R_{Fy} = 1.22\ \text{kN} \] **Final Answer for Q11:** \[ \boxed{1.22\ \text{kN}} \] --- ### **Question 12:** **Calculate the magnitude of the force (kN) in the intermediate cylinder if the cylinder is oriented at 15° to the horizontal and the horizontal distance between C and its point of attachment on member CB is 600 mm.** #### **Step 1: Find Vertical Component** Let \(F_{cyl}\) be the force in the cylinder. - The vertical component of the cylinder force must balance the moment caused by the weight at G. The moment created by the cylinder about C: \[ F_{cyl} \times 0.6 \times \sin(15^\circ) \] The moment caused by the weight about C: - The horizontal distance from C to G = 600 mm = 0.6 m \[ 4.071 \times 0.6 = 2.4426\ \text{kNm} \] Set moments equal: \[ F_{cyl} \times 0.6 \times \sin(15^\circ) = 2.4426 \] \[ F_{cyl} = \frac{2.4426}{0.6\times\sin(15^\circ)} \] \[ F_{cyl} = \frac{2.4426}{0.6\times0.2588} \] \[ F_{cyl} = \frac{2.4426}{0.1553} \] \[ F_{cyl} = 15.73\, \text{kN} \] **Final Answer for Q12:** \[ \boxed{15.73\ \text{kN}} \] --- ## **Summary of Answers** - **Q9:** 39.45 kN - **Q10:** 2.85 kN - **Q11:** 1.22 kN - **Q12:** 15.73 kN If you need the detailed free-body diagram or further explanation for any step, let me know!