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

2019
Views:

141

Pages:

2
Author:

diamundgalx45p

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 142. 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  xG = (1.25 m ) cos16.2602° = 1.2 m yG = 1.95 m − (1.25 m ) sin16.2602° = 1.6 m λ BA uuur BA = = BA −3.6i + 1.05 j ( −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 rG/A = (1.2 m ) i − (1.95 m − 1.6 m ) j + ( 0.45 m ) k = (1.2 m ) i − ( 0.35 m ) j + ( 0.45 m ) k ( ) W = − ( mg ) j = − ( 25 kg ) 9.81 m/s 2 j = − ( 245.25 kg ) j PHG = PHG uuuur HG = PHG HG −0.05i + 1.6 j − 0.4 k ( −0.05) 2 2 + (1.6 ) + ( −0.4 ) 2 32 8   1 j− k = PHG  − i + 33 33   33 Now, ΣM BA = 0: ( ) ( ) λ BA ⋅ rK / A × W + λ BA ⋅ rG/ A × PHG = 0 7 0 7 0 − 24 − 24 1 PHG 1.8 − 0.525 0.225 + 1.2 − 0.35 0.45 =0 25 25 )( 33) ( 0 − 245.25 0 32 −1 −8 − 1324.35 342.45 P =0 + 25 25 ( )( 35) HG Therefore: PHG = 127.620 N or PHG = 127.6 N

COSMOS Chapter 4 Solution 142

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