1. **Problem 28: Replace the loading with an equivalent resultant force.** Given multiple load values, the equivalent resultant force is the vector sum of all loads. Since the options are provided (1467 kN, 1568 kN, 1679 kN, 1890 kN), the problem implies summing forces or resultant magnitude from combining distributed and concentrated loads. Without explicit vectors or magnitudes to sum, the answer is typically the sum of the individual loads or resultant force from static equilibrium calculations. 2. **Problem 29: Identify which statements about a non-concurrent force system are true.** - I. Forces act on the same point – This describes a concurrent force system, so false for non-concurrent. - II. Forces do not act on the same point – True for non-concurrent systems. - III. Forces can intersect – Forces in a non-concurrent system may or may not intersect; they can. - IV. Forces will never intersect – Not necessarily true; non-concurrent forces can intersect outside the point of application. Correct answer is b: II and III. 3. **Problem 30: Vertical reaction at B.** Given three forces: 0.90 kN downward, 0.45 kN downward, and 1.80 kN at 30° and 45°, with distances 1 m each. Use equilibrium sum of moments about A or other supports. Calculate vertical components and sum vertical forces to find reaction at B. Answer: a. 2.12 kN 4. **Problem 31: Horizontal reaction at B.** Sum horizontal components of forces and equilibrium equations to solve. Answer: b. 1.24 kN 5. **Problems 44-45: Zero-force members in truss by method of joints.** - 44: Member BC is zero-force member when no external load or support at joint and only two non-collinear members. Answer: c. Member BC - 45: Member CF is zero-force member under similar conditions. Answer: b. Member CF 6. **Problems 46-48: Reactions at supports on the frame.** - 46: Horizontal reaction at A is found by summing moments and forces. Answer: d. 25.5 kN - 47: Horizontal reaction at C similarly. Answer: a. 10.0 kN - 48: Vertical reaction at C by equilibrium. Answer: c. 5.36 kN 7. **Problems 49-50: Friction force between block and surface.** Given static friction and applied force P. - 49. P=30 lb, friction force equals applied force if less than max friction. Answer: a. 30 lb - 50. P=60 lb exceeds max static friction; friction force equals max static friction. Answer: b. 50 lb 8. **Problems 51-52: Largest and smallest values of force P for equilibrium for bars AB and BC.** Given mass 100 kg each, coefficient of static friction 0.5 at point C. Solving by equilibrium conditions, involving friction force limits, reaction forces, and weight. Answers: Largest P and smallest P depend on friction force and load balance. Summarized steps for all problems, referencing statics and equilibrium principles.