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

2024
Views:

410

Pages:

2
Author:

Turfmorrybk9

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 110. (a) Have ΣF = R R =  − 0.52 j − 1.05j − 2.1( sin α i + cos α j) − 0.64i  kips = − 0.64 kip + ( 2.1 kips )( sin α ) i  − 1.57 kips + ( 2.1 kips ) cos α  j Then tan α = Rx 0.64 + 2.1sin α = Ry 1.57 + 2.1cos α 1.57 tan α + 2.1sin α = 0.64 + 2.1sin α tan α = 0.64 1.57 α = 22.178° or (b) From α = 22.178° Rx = − 0.64 kip − ( 2.1 kips ) sin 22.178° = −1.43272 kips Ry = −1.57 kips − ( 2.1 kips ) cos 22.178° = − 3.5146 kips . α = 22.2° COSMOS: Complete Online Solutions Manual Organization System R= ( −1.43272 )2 + ( − 3.5146 )2 = 3.7954 kips or Then R = 3.80 kips 67.8° M EF = ΣM EF Where M EF = ( 0.52 kip )( 3.6 in.) + (1.05 kips )(1.6 in.) − ( 2.1 kips )( 0.8 in.) − ( 0.64 kip ) (1.6 in.) sin 22.178° + 1.6 in. = 0.46146 kip ⋅ in. To obtain distance d left of EF, Have M EF = dRy = d ( − 3.5146 kips ) d = 0.46146 kip ⋅ in. − 3.5146 kips = − 0.131298 in. or . d = 0.1313 in. left of EF

COSMOS Chapter 3 Solution 110

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