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Sol Cap 02 - Edicion 8

COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 24. We compute the following distances: Then: 204 lb Force: 212 lb Force: ( ) ( ) 2 2 OA = 48 + 90 = 102 in. ( ) ( ) 2 2 OB = 56 + 90 = 106 in. ( ) ( ) 2 2 OC = 80 + 60 = 100 in. ( ) 48 F x =− 204 lb , F x =−96.0 lb ⊳ 102 ( ) 90 F y =+ 204 lb , F y = 180.0 lb ⊳ 102 ( ) 56 F x =+ 212 lb , F x = 112.0 lb ⊳ 106 400 lb Force: ( ) 90 F y =+ 212 lb , F y = 180.0 lb ⊳ 106 ( ) 80 F x =− 400 lb , F x =−320 lb ⊳ 100 ( ) 60 F y =− 400 lb , F y =−240 lb ⊳ 100 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 25. (a) P = = P y sin 35 ° 960 N sin 35° or P = 1674 N ⊳ (b) P = x = P y tan35° 960 N tan 35° or P x = 1371 N ⊳ Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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