HW7-Solution

HW7-Solution - B M a 63.600 in. a F (.25 lb)(cos 25 sin 25...

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HW7 - Solution Terry Cooke: Cal Poly San Luis Obispo
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Chapter 3, Solution 85 ( a ) Equivalence requires : or 250 N F F P F 60° : (0.3 m)(250 N) 75 N m B M     M The equivalent force-couple system at B is 250 N F 60° 75.0 N m M ( b ) Require Equivalence then requires : 0 cos cos x A B F F F or cos 0 A B F F   : 250 sin sin y A B F F F   Now if 250 0 A B F F     reject cos 0 or 90 and 250 A B F F Also : (0.3 m)(250 N) (0.2m) B A M F or 375 N A F   and 625 N B F 375 N A F 60 625 N B F 60
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Chapter 3, Solution 89 Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-couple. Have 2.9 lb 2.65 lb 0.25 lb, B F where the 2.65 lb force be part of the couple. Combining the two parallel forces, couple (2.65 lb)[(3.2 in. 2.8 in.)cos25 ] 14.4103 lb in. M and couple 14.4103 lb in. M A single equivalent force will be located in the negative z -direction Based on : 14.4103 lb in. [(.25 lb)cos 25 ]( )
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Unformatted text preview: B M a 63.600 in. a F (.25 lb)(cos 25 sin 25 ) i k F (0.227 lb) (0.1057 lb) i k and is applied on an extension of handle BD at a distance of 63.6 in. to the right of B Chapter 3, Solution 97 Equivalence requires : (150 N)( cos 35 sin 35 ) (122.873 N) (86.036 N) F F j k j k / : A D A M M r F where / (0.18 m) (0.12 m) (0.1 m) D A r i j k Then 0.18 0.12 0.1 N m 122.873 86.036 [( 0.12)( 86.036) (0.1)( 122.873)] [ (0.18)( 86.036)] [(0.18)( 122.873)] (22.6 N m) (15.49 N m) (22.1 N m) i j k M i j k i j k The equivalent force-couple system at A is (122.9 N) (86.0 N) F j k (22.6 N m) (15.49 N m) (22.1 N m) M i j k...
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HW7-Solution - B M a 63.600 in. a F (.25 lb)(cos 25 sin 25...

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