Lecture_6-3D_Equilibrium-1.ppt

Lecture_6-3D_Equilibrium-1.ppt - Equilibrium of a Particle...

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Unformatted text preview: Equilibrium of a Particle Lecture 6-3-D Force System (3-dimensional force equilibrium) Learning Objectives Be able to draw a free body diagram (FBD). Be able to apply equations of equilibrium to solve a 3- D problem. Pre-requisite Knowledge Units of measurements. Trigonometry concepts. Vector concepts. Rectangular components concepts. Position vectors. 3-D Force Equilibrium F = ⋅ + ⋅ + ⋅ = ∑ ∑ ∑ ∑ k F j F i F z y x = ∑ x F = ∑ y F = ∑ z F 1. What is the equation for the equilibrium of the force in x-direction for the system? Hint: position vectors AB and AC A) F 1 – (F 2 + F 3 )*3/(3 2 + 4 2 + 8 2 ) ½ = 0 B) F 1 – F 2 – F 3 = 0 C) F 1 – F 2 – F 3 – 100 = 0 D) 100 – (F 2 + F 3 )*3/(3 2 + 4 2 + 8 2 ) ½ = 0 Concept Question-1 F 1 F 2 F 3 100 N Solution 1 F 1 F 2 F 3 100 N Coordinates of Point A, B and C: A (0,0,0) B (-3,-4,8) C (-3,4,8) Note: F 1 and 100 N do not need to be further resolved. Only F 2 and F 3 need to be resolved. Unit vector u AB and u AC : u AB = (-3 i -4 j +8 k)/(32 + 42 + 82) ½ u AC = (-3 i +4 j +8 k)/(32 + 42 + 82) ½ Summation of all force vectors: Σ F = 0 F 1 i – 100 N k + F 2 (-3 i -4 j +8 k)/(32 + 42 + 82) ½ + F 3 (-3 i +4 j +8 k)/(32 + 42 + 82) ½ = 0 i + 0 j + 0 k F 1 – (F 2 + F 3 )*3/(3 2 + 4 2 + 8 2 ) ½ = 0 2. What is the equation for the equilibrium of force in y-direction for 2....
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This note was uploaded on 06/14/2009 for the course CE 221 taught by Professor Buch during the Spring '08 term at Michigan State University.

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Lecture_6-3D_Equilibrium-1.ppt - Equilibrium of a Particle...

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