Practice Final - F10 - Solution

Practice Final - F10 - Solution - PROBLEM 1 Free-Body...

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PROBLEM 1 Free-Body Diagram: 1 2 / / / 4 2 4 6ft 1 (2 2 ) 3 2 4 2 4 = - - = = - - = - = - - = i j k λ i j k r i j r i j k r i uuur AF G A G A B A AF AF / 2/ / 0: ( ( 12 ) ( ( 12 )) ( ) 0 Σ = × - + × - + × = λ r j λ r j λ r AF AF G A AF G A AF B A M T / 2 1 2 2 1 2 1 1 2 1 0 4 1 2 ( ) 0 3 3 0 12 0 0 12 0 - - - - - + - - + × = - - λ r T AF B A / 1 1 (2 2 12) ( 2 2 12 2 4 12) ( ) 0 3 3 AF B A × × + - × × + × × + × = λ r T / ( ) 32 × = - λ r T AF B A
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PROBLEM 1 (CONTINUED) Since min T has no i component, wire BH is parallel to the yz plane, and 4 ft. x = ( a ) 4.00 ft; 8.00 ft x y = = t ( b ) min 10.73 lb T = t Because of symmetry of loading, 1 (Load) 2 = O 4.48 kN = O T We pass a section through KM , LM , LN , and use free body shown 3.84 0: (3.68 m) 4 (4.48 kN 0.6 kN)(3.6 m) 0 M LN M F Σ = + - = 3.954 kN LN F = - 3.95 kN LN F C = t 0: (4.80 m) (1.24 kN)(3.84 m) (4.48 kN 0.6 kN)(7.44 m) 0 L KM M F Σ = - - + - = 5.022 kN KM F = + 5.02 kN KM F T = t 4.80 1.12 0:
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Practice Final - F10 - Solution - PROBLEM 1 Free-Body...

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