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

2019
Views:

333

Pages:

2
Author:

SataTalluctft9

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 133. ( ) First replace the given couples with an equivalent force-couple system R, M OB at the origin. ΣF : ΣM O : R = − ( 35 lb ) i − (12 lb ) k M OR = − ( 200 lb ⋅ in.) i + ( 8 in.) j + ( 8 in.) k  ×  − ( 35 lb ) i  − (140 lb ⋅ in.) k + (10 in.) i + ( 4 in.) j ×  − (12 lb ) k  = ( − 200 − 48 ) i + ( − 280 + 120 ) j + ( 280 − 140 ) k = − ( 248 lb ⋅ in.) i − (1600 lb ⋅ in.) j + (140 lb ⋅ in.) k Now R= ( − 35 lb )2 + ( −12 lb )2 = 37 lb Then λ axis = 1 ( − 35 i − 12 k ) 37 R = − ( 35.0 lb ) i − (12.00 ) k (a) M1 = λ axis ⋅ M OR (b) Then = 1 ( − 35 i − 12 k ) ⋅ ( − 248 i − 160 j + 140 k )( lb ⋅ in.) 37 = 1 ( 35 × 248 − 12 × 140 )( lb ⋅ in.) 37 = 7000 lb ⋅ in. 37 M P= 1 = R 7000 lb ⋅ in. 37 37 lb or P = 5.11 in. . COSMOS: Complete Online Solutions Manual Organization System (c) Have M1 = λ axis M1 = 7000 lb ⋅ in. ( − 35 i − 12 k ) 37 2 Then M OR = M1 + M 2 or M 2 = ( − 248 i − 160 j + 140 k ) − 7000 ( − 35 i − 12 k ) 37 2 = − ( 69.037 lb ⋅ in.) i − (160 lb ⋅ in.) j + ( 201.36 lb ⋅ in.) k M z = rO/P × R Require i j k or − 69.037i − 160 j + 201.36k = 0 y z −35 0 −12 j: −160 = − 35 z z = 4.57 in. or k: 201.36 = 35 y or y = 5.75 in. ∴ The point of intersection is defined by y = 5.75 in. z = 4.57 in. .