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

2024
Views:

134

Pages:

2
Author:

salcignofi10

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 109. Free-Body Diagram: (a) ΣFy = 0: 3 ( R ) − 135 N = 0 R = 45.0 N ΣM A = 0: rAB × R B + rAC × R C + rAG × FW = 0 or ( 0.9 m ) i + l k  × R j + l i + ( 0.9 m ) k  × R j + ( 0.450 m ) i + ( 0.450 m ) k  × ( − FW j) = 0 or ( 0.9 m ) R k − lR i + lR k − ( 0.9 m ) R i − ( 0.450 m ) FW k + ( 0.450 m ) FW i = 0 Equating the coefficients of the i unit vector to zero: i: − lR − ( 0.9 m ) R + ( 0.45 m ) FW = 0 Using that FW = 3R − l − ( 0.9 m ) + ( 0.45 m )( 3) = 0 l = 0.450 m or l = 450 mm continued COSMOS: Complete Online Solutions Manual Organization System (b) Free-Body Diagram: ΣM A = 0: rAB × R B + rAC × R C + rAG × FW = 0 or ( 0.9 m ) i + ( 0.5 m ) k  × R j + ( 0.5 m ) i + ( 0.9 m ) k  × R j + ( 0.450 m ) i + ( 0.450 m ) k  × ( −135 N ) j = 0 ( 0.9 m ) R k − ( 0.5 m ) R i + ( 0.5 m ) R k − ( 0.9 m ) R i − ( 0.450 m ) FW k + ( 0.450 m ) FW i = 0 Equating the coefficients of the unit vectors to zero i: − 0.5RB − 0.9 RC + 60.75 = 0 (1) k: 0.9 RB + 0.5RC − 60.75 = 0 (2) 0.5 × [ Eq. (1) ] + 0.9 × [ Eq. (2) ] gives  − 0.5 ( 0.5 ) + ( 0.9 )( 0.9 )  RB + ( 0.5 − 0.9 ) 60.75 = 0 RB = 43.393 N Now using (1) − 0.5 ( 43.393) − 0.9 RC + 60.67 = 0 RC = 43.393 N ΣFy = 0: RA + 43.393 + 43.393 − 135 = 0 RA = 48.2 N RB = 43.4 N RC = 43.4 N

COSMOS Chapter 4 Solution 109

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