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

2020
Views:

434

Pages:

2
Author:

garciavelascogr12

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 112 Free-Body Diagram: Express all forces in terms of rectangular components: rE = ( 3 ft ) i + ( 3 ft ) j rB = ( 3 ft ) sin 30° j + ( 3 ft ) cos 30°k = (1.5 ft ) j + ( 2.598 ft ) k rD = − ( 3 ft ) i + ( 3 ft ) j or or rA = (10 ft ) sin 30° j − (10 ft ) cos 30°k = ( 5 ft ) j + (8.66 ft ) k uuur BE = rE − rB = ( 3 ft ) i + ( 3 ft ) j − (1.5 ft ) j − ( 2.598 ft ) k uuur BE = ( 3 ft ) i + (1.5 ft ) j − ( 2.598 ft ) k, and BE = 4.243 ft uuur BD = rD − rB = − ( 3 ft ) i + ( 3 ft ) j − (1.5 ft ) j − ( 2.598 ft ) k uuur BD = − ( 3 ft ) i + (1.5 ft ) j − ( 2.598 ft ) k, and BD = 4.243 ft Then uuur ur BD = TBD ( − 0.707i + 0.3535j − 0.6123k ) T BD = TBD BD uuur ur BE T BE = TBE = TBE ( 0.707i + 0.3535j − 0.6123k ) BE continued COSMOS: Complete Online Solutions Manual Organization System rB × TBD + rB × TBE + ( 5 ft ) j + ( 8.66 ft ) k  × ( − 75 lb ) j ΣM C = 0: i j k i j k 0 1.5 2.598 TBD + 0 1.5 2.598 + 649.5 i = 0 or − 0.707 0.3535 − 0.6123 0.707 0.3535 − 0.6123 Equating the coefficients of the unit vectors to zero: j: −1.837 TBD + 1.837 TBE = 0 i: −1.837 TBD + 1.837 TBE + 649.5 lb = 0 TBD = 176.8 lb TBE = 176.8 lb Force equations: C x + (176.8 )( − 0.707 ) + (176.8 )( 0.707 ) = 0, or Cx = 0 C y + (176.8 )( 0.3535 ) + (176.8 )( 0.3535 ) − 75 lb = 0, or C z + (176.8 )( − 0.6123) + (176.8 )( − 0.6123) = 0, C z = 216.5 lb or C y = − 50 lb C = − ( 50 lb ) j + ( 216.5 lb ) k