5.6p_F_M_E - ENGR 211 Lecture 5 ENGR 211 Lecture 5 PTS Help...

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ENGR 211 Lecture 5 Fall 2007 Civil Engineering ENGR 211 Lecture 5 ENGR 211 Lecture 5 Beavers! •PTS – Help Session •See Announcement •Office Hours Today •2 – 3 pm Civil Engineering 3 Cable Example 3 Cable Example – Find Tensions in Cables Find Tensions in Cables
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ENGR 211 Lecture 5 Fall 2007 Civil Engineering Free Body Diagram A Replace the cables with Force Vectors Different Magnitude Same Direction As Cables Civil Engineering A = (0,0,10) B = (-4,6,0) C = (8,6,0) D = (1,-8,0) BA = [(0-(-4)) i +(0-6) j +(10-0) k ]ft. CA = [(0-8) i +(0-6) j +(10-0) k ]ft. DA = [(0-1) i +(0-(-8)) j +(10-0) k ]ft . |BA| = sqrt(36+16+100)=12.3 ft. |CA| = sqrt(36+64+100)=14.1 ft. |DA| = sqrt(64+1+100)=12.8 ft. A } Coordinates } Cable Vectors } Cable Magnitudes
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ENGR 211 Lecture 5 Fall 2007 Civil Engineering e BA = [0.32i-0.49j+0.81k] e CA = [-0.57i-0.43j+0.71k] e DA = [-0.08i+0.63j+0.78k] T BA = | T BA | e BA = | T BA | [0.32i-0.49j+0.81k] lbs. T CA = | T CA | e CA = | T CA | [-0.57i-0.43j+0.71k] lbs. T DA = | T DA | e DA = | T DA | [-0.08i+0.63j+0.78k] lbs. } Cable unit vectors Civil Engineering To be in static equilibrium Σ F x = Σ F y = Σ F z = 0 Σ F x = | T BA |(0.32)+| T CA |(-0.57)+| T DA |(-0.08) = 0 T BA = | T BA | e BA = | T BA | [0.32i-0.49j+0.81k] lbs. T CA = | T CA | e CA = | T CA | [-0.57i-0.43j+0.71k] lbs. T DA = | T DA | e DA = | T DA | [-0.08i+0.63j+0.78k] lbs.
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ENGR 211 Lecture 5 Fall 2007 Civil Engineering Σ F z = | T BA |(0.81)k+| T CA |(0.71)k+| T DA |(0.78)k - W = 0 Σ F y = | T BA |(-0.49)j+| T CA |(-0.43)j+| T DA |(0.63)j = 0 Equilibrium Equations: Σ F x = | T BA |(0.32)+| T CA |(-0.57)+| T
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5.6p_F_M_E - ENGR 211 Lecture 5 ENGR 211 Lecture 5 PTS Help...

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