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

2024
Views:

348

Pages:

2
Author:

behend6abs

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 147. Free-Body Diagram: Express forces in terms of their rectangular components: W1 = W2 = − ( 30 lb ) j uuuur − 6i + y j − 6k BH =T T=T BH ( −6 )2 + y 2 + ( − 6 )2 uuur AF 6i − 3 j − 6k 2 1 2 λ AF = = = i− j− k AF ( 6 ) 2 + ( − 3)2 + ( − 6 ) 2 3 3 3 rG/ A = ( 3 ft ) i rB/ A = ( 6 ft ) i rI / A = ( 6 ft ) i − ( 3 ft ) k continued COSMOS: Complete Online Solutions Manual Organization System Now, ( ΣM AF = 0: ) ( ) ( ) λ AF ⋅ rG/ A × W1 + λ AF ⋅ rB/ A × T + λ AF ⋅ rI / A × W2 = 0 2 −1 − 2 2 −1 − 2 1 3 0 0  + 6 0 0 3 −6 y −6 0 − 30 0 60 + ( − 36 − 12 y ) ( −6 )2 T 3 ( − 6 )2 + 2 −1 − 2 1 + 6 0 −3   = 0  3 + y 2 + 36 0 − 30 0 T 3 y 2 + 36 + 60 = 0 Solving for T : T = 30 30 + y T = 30 3+ y ( −6 )2 + y 2 + 36 y 2 + 72 Now to find the minimum of T, differentiate T with respect to y and equate the derivative to zero. dTm dy  1    ( 3 + y ) 72 + y 2  2  =  ( ) − 1 2 ( 2 y ) − ( 72 + (3 + y ) 2 1 2 2 y )  (1) 30  =0 Setting the numerator equal to zero and simplifying: ( 3 + y ) y − y 2 − 72 = 0, or y = 24 ft x = 0, y = 24 ft (a) Minimum occurs at: (b) Using the expression for T: Tmin 30 3 + 24 )2 + 72 28.284 lb or Tmin = 28.3 lb

COSMOS Chapter 4 Solution 147

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