COSMOS Chapter 3 Solution 97 | Answer Key - Edubirdie

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

2021
Views:

484

Pages:

2
Author:

Madelyn Bradshaw

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 97. ΣM Bz : M 2 z = 0 (a) Have ( ) k ⋅ rH /B × F1 + M1z = 0 (1) rH /B = ( 31 in.) i − ( 2 in.) j where F1 = λ EH F1 = = ( 6 in.) i + ( 6 in.) j − ( 7 in.) k 11.0 in. ( 20 lb ) 20 lb ( 6i + 6 j − 7k ) 11.0 M1z = k ⋅ M1 M1 = λ EJ M1 = −di + ( 3 in.) j − ( 7 in.) k d 2 + 58 in. ( 480 lb ⋅in.) Then from Equation (1), 0 0 1 31 −2 0 6 6 −7 20 lb ⋅ in. ( −7 )( 480 lb ⋅ in.) + =0 11.0 d 2 + 58 continued . COSMOS: Complete Online Solutions Manual Organization System Solving for d, Equation (1) reduces to 20 lb ⋅ in. 3360 lb ⋅ in. =0 (186 + 12 ) − 2 11.0 d + 58 d = 5.3955 in. From which or d = 5.40 in. (b) F2 = F1 = 20 lb ( 6i + 6 j − 7k ) 11.0 = (10.9091i + 10.9091j − 12.7273k ) lb or F2 = (10.91 lb ) i + (10.91 lb ) j − (12.73 lb ) k M 2 = rH /B × F1 + M1 i j k 20 lb ⋅ in. ( − 5.3955 ) i + 3j − 7k = 31 − 2 0 + ( 480 lb ⋅ in.) 11.0 9.3333 6 6 −7 = ( 25.455i + 394.55j + 360k ) lb ⋅ in. + ( − 277.48i + 154.285j − 360k ) lb ⋅ in. M 2 = − ( 252.03 lb ⋅ in.) i + ( 548.84 lb ⋅ in.) j or M 2 = − ( 21.0 lb ⋅ ft ) i + ( 45.7 lb ⋅ ft ) j .

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