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HW5-Solution - HW5 Solution Terry Cooke Cal Poly San Luis...

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HW5 - Solution Terry Cooke: Cal Poly San Luis Obispo
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Chapter 3, Solution 23 First note 2 2 2 ( 6) (2.4) ( 4) 7.6 m BC d   Then 2.5 kN ( 6 2.4 4 ) 7.6 BC T i j k We have / A B A BC M r T where / (6 m) B A r i Then 2.5 kN (6 m) ( 6 2.4 4 ) 7.6 A M i i j k or (7.89 kN m) (4.74 kN m) A M j k
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Chapter 3, Solution 25 ( a ) We have / A E A DE M r T where / 2 2 2 (2.3 m) (0.6 m) (3.3 m) (3 m) (810 N) (0.6) (3.3) (3) m (108 N) (594 N) (540 N) E A DE DE DE T r j T λ i j k i j k 0 2.3 0 N m 108 594 540 (1242 N m) (248.4 N m) A   i j k M i k or (1242 N m) (248 N m) A   M i k ( b ) We have / A G A CG M r T where / 2 2 2 (2.7 m) (2.3 m) (.6 m) (3.3 m) (3 m) (810 N) (.6) (3.3) (3) m (108 N) (594 N) (540 N) G A CG CG CG T   r i j T i j k i j k 2.7 2.3 0 N m 108 594 540 (1242 N m) (1458 N m) (1852 N m) A   i j k M i j k or (1242 N m) (1458 N m) (1852 N m) A   M i j k
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Chapter 3, Solution 26 We have 2 A AB AD R F F where (82 lb) AB   F j and 6 7.75 3 (82 lb) 10.25 (48 lb) (62 lb) (24 lb) AD AD AD AD AD i j k F F F i j k  Thus 2 (48 lb) (226 lb) (24 lb) A AB AD R F F i j k Also /
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