Meriam, Kraige & Bolton — Equilibrium of Rigid Bodies: Problem 4_77

⚡ Mecademy AIENG정역학 · ch4  Problem Statement The wingnut of the collapsible bucksaw is tightened until the tension in rod is . Determine the force in the saw blade and the magnitude of the force supported by pin . Problem 4/77 (a) Force in the saw blade ● Calculation Process 1. Present Final Formula: Consider the equilibrium of the side member . Take the sum of moments about the pin to eliminate the unknown reaction force at that point. 2. Substitute Values: The tension in rod is acting at a vertical distance of above . The force in the saw blade is acting at a vertical distance of below . Both forces are horizontal. 3. Partial Operations: Solve for : 4. Final Calculation: ● Final Conclusion: The force in the saw blade is (tension). (b) Magnitude of the force supported by pin ● Calculation Process BAB200 N EFF C EF ACE C M = ∑ C 0 ABT = AB 200 N 80 mmCEFT EF 160 mmC T ⋅ AB (80 mm)−T ⋅ EF (160 mm)=0 T EF 200 N⋅80 mm=T ⋅ EF 160 mm 16000 N⋅mm=T ⋅ EF 160 mm T = EF N= 160 16000 100 N EF100 N FC 1. Present Final Formula: Use the equilibrium of forces in the horizontal and vertical directions for the side member . Let and be the components of the reaction force at pin . 2. Substitute Values: In the -direction (horizontal), assuming right is positive: (where is the compression from the crossbar ). In the -direction (vertical): Since the rod and blade are horizontal two-force members, they exert no vertical force. Thus, . 3. Partial Operations: Calculate the horizontal component: The