Engineering Mechanics: Statics 9th Edition · Rigid Bodies: Equivalent Systems of Forces · Problem 3_91
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Meriam, Kraige & Bolton — Rigid Bodies: Equivalent Systems of Forces: Problem 3_91
⚡ Mecademy AIENG정역학 · ch3 Problem Statement The tool shown is used for straightening twisted members as wooden framing is completed. If the force is applied to the handle as shown, determine the normal forces applied to the installed stud at points and . Ignore friction. Problem 3/91 (a) Normal forces at A and B ● Calculation Process 1. Present Final Formula: We consider the equilibrium of the tool. Let and be the magnitudes of the normal forces exerted by the stud on the tool at points and , respectively. By Newton's third law, these are equal to the forces the tool applies to the stud. Based on the geometry, the force and the reaction at (which must act downward on the tool to balance the moment) are in the same direction, while the reaction at acts upward on the tool. The equilibrium equations are: where is the horizontal distance from to , and is the horizontal distance from to . 2. Substitute Values: Substitute , , and into the moment equation: 3. Partial Operations: Solve for : Now solve for using the force equilibrium equation: P=30 lb AB N A N B A B PB A M = ∑ A 0⟹P⋅d − 1 N ⋅ B d = 2 0 F = ∑ y 0⟹N − A P−N = B 0 d = 1 14.5 in.PAd = 2 3.35 in. AB P=30 lbd = 1 14.5 in.d = 2 3.35 in. 30 lb⋅14.5 in.−N⋅ B 3.35 in.=0 N B 30⋅14.5=435 lb⋅in N = B ≈ 3.35 435 129.8507 lb N A N = A P+N B 4. Final Calculation: Rounding to one decimal place: ● Final Conclusion: The normal forces applied to the stud are 159.9 lb at point A (directed downward) and 129.9 lb at point B (directed upwa
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주어진 조건: 30 lb, 1 N, 14.5 in, 3.35 in, 0 N, 435 lb
구하는 것: (a) Normal forces at A and B ● Calculation Process 1
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Engineering Mechanics: Statics · 9th Edition
저자: Meriam, Kraige & Bolton
출판사: Wiley
단원: Rigid Bodies: Equivalent Systems of Forces