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

2021
Views:

382

Pages:

2
Author:

greenteafrappl97

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 135. (a) First reduce the given force system to a force-couple at the origin. ΣF : Pλ BA + Pλ DC + Pλ DE = R Have  4 3  3 4  −9 4 12  R = P  j − k  +  i − j +  i − j+ k  5  5 5   25 5 25   5 3P ( 2i − 20 j − k ) 25 ∴ R = R 3P 25 + (1) 2 27 5 P 25 Have ΣM : Σ ( rO × P ) = M OR −4 P 3P  4P  4P 12 P   3P  −9 P j− k  + ( 20a ) j ×  i− j  + ( 20a ) j ×  i− j+ k  = M OR 5  5  5 25   5  5  25 ( 24a ) j ×  ∴ M OR = (b) Have M 1 = λ R ⋅M OR where λ R = . 24 Pa ( −i − k ) 5 R 3P 25 1 = = ( 2i − 20 j − k ) ( 2i − 20 j − k ) R 25 27 5 P 9 5 COSMOS: Complete Online Solutions Manual Organization System Then M1 = and pitch 1 9 5 ( 2i − 20 j − k ) ⋅ p= 24 Pa −8Pa ( −i − k ) = 5 15 5 M1 −8Pa  25  −8a =  = R 81 15 5  27 5 P  or p = −0.0988a (c) M1 = M 1λ R = −8Pa  1  8Pa ( −2i + 20 j + k )   ( 2i − 20 j − k ) = 675 15 5  9 5  Then M 2 = M OR − M1 = 24Pa 8Pa 8Pa ( −i − k ) − ( −2i + 20 j + k ) = ( −403i − 20 j − 406k ) 5 675 675 M 2 = rQ/O × R Require  8Pa   3P    ( −403i − 20 j − 406k ) = ( xi + zk ) ×   ( 2i − 20 j − k )  675   25   3P  =   20 zi + ( x + 2 z ) j − 20 xk   25  From i: 8 ( −403) Pa  3P  = 20 z   675  25  ∴ z = −1.99012a From k: 8 ( −406 ) Pa  3P  = −20 x   675  25  ∴ x = 2.0049a ∴ The axis of the wrench intersects the xz-plane at x = 2.00a, z = −1.990a .