COSMOS Chapter 4 Solution 102 | Michigan State University - Edubirdie

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

2023
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2
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COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 102. Free-Body Diagram: ( ) The weight W is W = mg = (170 kg ) 9.81 m/s 2 = 1667.7 N ΣM C = 0: or rCA × N A + rCB × N B + rCG × W = 0 ( −0.3 m ) i + (1.2 m ) k  × N A j + (1.8 m ) i + ( 0.9 m ) k  × N B j + ( 0.6 m ) i + ( 0.6 m ) k  × ( −W ) j = 0 or − ( 0.3 m ) N Ak − (1.2 m ) N Ai + (1.8 m ) N Bk − ( 0.9 m ) N Bi − ( 0.6 m )Wk + ( 0.6 m )Wi = 0 Equating the coefficients of the unit vectors to zero: i: −1.2 N A − 0.9 N B + 0.6W = 0 4 N A + 3 N B 0 = 2W j: (1) − 0.3 N A + 1.8 N B − 0.6W = 0 − N A + 6 N B 0 = 2W − 2 ×  Eq. (1)  + Eq. ( 2 ) (2) gives −9 N A = 2W NA = 2 W = 370.60 N 9 continued COSMOS: Complete Online Solutions Manual Organization System Now (2) gives NB = 1 2  10 W  2W + W  = 6 9  27 N B = 617.67 N, and from (1) ΣFy = 0: N A + N B + NC − W = 0 370.60 N + 617.67 N + NC − 1667.7 N = 0 NC = 679.43 N Therefore the forces on the blocks are: N A = 371 N N B = 618 N NC = 679 N

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