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

2019
Views:

91

Pages:

2
Author:

crawniadxfo

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 104. Free-Body Diagram: WAB = ( 5 lb/ft )( 2 ft ) = 10 lb First note WBC = ( 5 lb/ft )( 4 ft ) = 20 lb W = WAB + WBC = 30 lb To locate the equivalent force of the pipe assembly weight rG/B × W = Σ ( ri × Wi ) = rG ( AB ) × WAB + rG ( BC ) × WBC ( xGi + zGk ) × ( −30 lb ) j = (1 ft ) k × ( −10 lb ) j + ( 2 ft ) i × ( −20 lb ) j or ∴ − ( 30 lb ) xGk + ( 30 lb ) zG i = (10 lb ⋅ ft ) i − ( 40 lb ⋅ ft ) k From i-coefficient k-coefficient zG = 10 lb ⋅ ft 1 = ft 30 lb 3 xG = 40 lb ⋅ ft 1 = 1 ft 30 lb 3 From free-body diagram. of piping ΣM x = 0: W ( zG ) − TA ( 2 ft ) = 0 1  1  ∴ TA =  ft  30 lb  ft  = 5 lb 2  3  ΣFy = 0: or TA = 5.00 lb 5 lb + TD + TC − 30 lb = 0 ∴ TD + TC = 25 lb (1) continued COSMOS: Complete Online Solutions Manual Organization System ΣM z = 0: 4  TD (1.25 ft ) + TC ( 4 ft ) − 30 lb  ft  = 0 3  ∴ 1.25TD + 4TC = 40 lb ⋅ ft −4  Equation (1)  −4TD − 4TC = −100 ∴ TD = 21.818 lb Results: (3) −2.75TD = −60 Equation (2) + Equation (3) From Equation (1) (2) or TD = 21.8 lb TC = 25 − 21.818 = 3.1818 lb or TC = 3.18 lb TA = 5.00 lb TC = 3.18 lb TD = 21.8 lb