Meriam, Kraige & Bolton — Rigid Bodies: Equivalent Systems of Forces: Problem 3_12

⚡ Mecademy AIENG정역학 · ch3  Problem Statement To facilitate shifting the position of a lifting hook when it is not under load, the sliding hanger shown is used. The projections at A and B engage the flanges of a box beam when a load is supported, and the hook projects through a horizontal slot in the beam. Compute the forces at A and B when the hook supports a 300-kg mass. Problem 3/12 (a) Computation of the contact forces at A and B 1. Formula: First, we determine the weight of the supported mass: Then, apply the equilibrium equations for the hanger (treated as a rigid body): 2. Substitution: Using , the weight is: Taking moments about point (assuming clockwise is positive): Summing vertical forces (assuming upward is positive): 3. Calculation: Calculate the weight: Solve for the force at from the moment equation: W W=m⋅g M = ∑ A 0 F = ∑ y 0 g= 9.81m/s 2 W=300kg× 9.81m/s 2 A W⋅(0.4m)−F ⋅ B (0.6m)=0 F − A F − B W=0 W=2943N B F = B = 0.6m 2943N×0.4m 1962N (This force acts downward on the hanger as it presses against the underside of the top flange). Solve for the force at from the vertical force equation: (This force acts upward on the hanger as it rests on the top of the bottom flange). 4. Result: ● Final Conclusion: The contact force at projection is acting upward, and the contact force at projection is acting downward on the hanger. ✨ Final Answer Summary (a) (upward), (downward) Mecademy AI Solution · ENGProblem 3/12 A F = A F + B W=1962N+2943N=4905N F = A 4.91kN F = B 1.9