Engineering Mechanics: Statics 9th Edition · Statics of Particles · Problem 2_136
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Meriam, Kraige & Bolton — Statics of Particles: Problem 2_136
⚡ Mecademy AIENG정역학 · ch2 Problem Statement 2/136 Represent the resultant of the force system acting on the pipe assembly by a single force R at A and a couple M. Problem 2/136 To represent the force system as a single force at point and a couple , we need to find the vector sum of all forces and the vector sum of all moments (including individual couples) about point . We will use a coordinate system with the origin at the leftmost bend, as shown in the diagram. 1. Identify Force Vectors and Their Positions From the diagram, we identify six forces and one couple. Let's list them along with their position vectors in meters (converting from mm): at at at at at (this is point ) at Couple 2. Calculate Resultant Force The resultant force is the vector sum of all applied forces: RAM AO P F = 1 −160i NP = 1 (0,0,0) m F = 2 −100k NP = 2 (0.2,0,0) m F = 3 160(sin25j− ∘ cos25k) N≈ ∘ (67.62j−145.01k) NP = 3 (0.2,0.15,0) m F = 4 −160k NP = 4 (0.2,0.25,0) m F = 5 180j NP = 5 (0.5,0.25,0) mA F = 6 −120i NP = 6 (0.5,0.25,0.25) m C=50k N⋅m R R=F = ∑ i F + 1 F + 2 F + 3 F + 4 F + 5 F 6 R = x −160+0+0+0+0−120=−280 N R = y 0+0+67.62+0+180+0=247.62 N R = z 0−100−145.01−160+0+0=−405.01 N ● Resultant Force: 3. Calculate Resultant Moment at Point The total moment at is the sum of the moments of all forces about plus any applied couples. The position vectors relative to are : (Force is at point ) Now calculate individual moments : 1. 2. 3. 4. 5. Summing all moments and the couple : ● Resultant Fo
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주어진 조건: 100k, 25j, 25k, 67.62j, 145.01k, 160k
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Engineering Mechanics: Statics · 9th Edition
저자: Meriam, Kraige & Bolton
출판사: Wiley
단원: Statics of Particles