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

2019
Views:

312

Pages:

2
Author:

marthamanayay9qr10

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 146. Free-Body Diagram: Express forces in terms of their rectangular components: W1 = W2 = − ( 30 lb ) j uuuur ( x − 6) i + y j − 6 k BH =T T=T BH ( x − 6 )2 + y 2 + ( − 6 )2 λ AF = uuur AF = AF 6i − 3 j − 6k ( 6) 2 2 + ( − 3) + ( − 6 ) 2 = 2 1 2 i− j− k 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 −1 − 2 2 −1 − 2 2 1 3 0 0  + 6 0 0  3 0 − 30 0 ( x − 6) y − 6 60 + ( −36 − 12 y ) ( x − 6 )2 T 3 ( x − 6) 2 2 2 −1 − 2 1 + 6 0 −3   = 0 3 + y 2 + 36 0 − 30 0 T 3 + y + 36 + 60 = 0 Solving for T : T = 30 30 + y ( x − 6 )2 + y 2 + 36 It is thus clear that for a given y, T will have its minimum value when x = 6 ft. Denoting this minimum by Tm: Tm = 30 3+ y y 2 + 36 Now to find the minimum of Tm, differentiate Tm with respect to y and equate the derivative to zero. dTm dy  1    ( 3 + y ) 36 + y 2  2  =  ( ) − 1 2 ( ( 2 y ) − 36 + y 2 (3 + y ) 2 ) 1 2  (1) 30  =0 Setting the numerator equal to zero and simplifying: ( 3 + y ) y − y 2 − 36 = 0 y = 12 ft x = 6 ft, y = 12 ft (a) Minimum occurs at: (b) Using the expression for T: Tmin = 30 3 + 12 ( 6 − 6 )2 + (12 )2 + 36 = 26.833 lb or Tmin = 26.8 lb