University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2020
Views:

484

Pages:

2
Author:

FrextWhestembki

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 143. Free-Body Diagram: There is only one unknown of interest and, therefore only one equation is needed: ΣM AB = 0 Geometry:  1.05 m   = 16.2602°  3.5 m  θ = tan −1  xI = ( 2.50 m ) cos16.2602° = 2.4 m yI = 1.95 m − ( 2.50 m ) sin16.2602° = 1.25 m λ BA = uuur BA = BA −3.6i + 1.05j ( −3.6 ) 2 + (1.05 ) 2 =− 24 7 i+ j 25 25 continued COSMOS: Complete Online Solutions Manual Organization System rK/A = (1.8 m ) i − ( 0.525 m ) j + ( 0.225 m ) k rI/A = ( 2.4 m ) i − ( 0.7 m ) j + ( 0.45 m ) k ( ) W = − ( mg ) j = − ( 25 kg ) 9.81 m/s 2 j = − ( 245.25 kg ) j PJI uur JI = PJI = PJI JI −0.025i + 1.25j − 0.25k ( −0.025)2 + (1.25)2 + ( −0.25)2 50 10   1 = PJI  − i + j + k 51 51   51 Now, ΣM BA = 0: ( ) ( ) λ BA ⋅ rK / A × W + λ BA ⋅ rI / A × PJI = 0 − 24 − 24 7 7 0 0 1 PJI + 12.4 − 0.7 0.45 =0 1.8 − 0.525 0.225 25 25 51) ( )( −1 50 −10 0 − 245.25 0 − 1324.35 536.85 P =0 + 25 ( 25)( 51) JI Therefore: PJI = 125.811 N or PJI = 125.8 N

COSMOS Chapter 4 Solution 143

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