Solucioﾌ］ al ejercicio 1
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This document summarizes the solution to an exercise involving free-body diagrams of levers. It includes: (1) Drawing free-body diagrams of levers BCD and calculating the forces and torques on them. (2) Using the free-body diagram equations to solve for the magnitude and direction of the force on lever C, finding it to be 449 N at an angle of 32.3 degrees. (3) Noting that the solution is from Chapter 4, Solution 19 of the textbook "Vector Mechanics for Engineers: Statics and Dynamics, 8/e".
Solucioﾌ］ al ejercicio 1
- 1. Soluciﾃｳn al ejercicio 1 Ahora en el nodo B Vector Mechanicsfor Engineers: Staticsand Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell ﾂｩ 2007 TheMcGraw-Hill Companies. Free-Body Diagram: (a) From free-body diagram of lever BCD ( ) ( )0: 50 mm 200 N 75 mm 0C ABM TS = - = 300ABT = (b) From free-body diagram of lever BCD ( )0: 200 N 0.6 300 N 0x xF CS = + + = 380 N or 380 Nx xC = - =C ( )0: 0.8 300 N 0y yF CS = + = N240orN240 =-= yyC C Then ( ) ( ) 2 22 2 380 240 449.44 Nx yC C C= + = + = and ﾂｰ= - - == -- 276.32 380 240 tantan 11 x y C C q or 449 N=C 32.3ﾂｰ COSMOS: Complete Online Solutions Manual Organization System Vector Mechanicsfor Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Chapter 4, Solution 19. Free-Body Diagram: (a) From free-body diagram of lever BCD ( ) ( )0: 50 mm 200 N 75 mm 0C ABM TS = - = 300ABT = (b) From free-body diagram of lever BCD ( )0: 200 N 0.6 300 N 0x xF CS = + + = 380 N or 380 Nx xC = - =C ( )0: 0.8 300 N 0y yF CS = + = N240orN240 =-= yyC C Then ( ) ( )2 22 2 380 240 449.44 Nx yC C C= + = + = and ﾂｰ= - - == -- 276.32 380 240 tantan 11 x y C C q or 449 N=C 32.3ﾂｰ COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 19. Free-Body Diagram: (a) From free-body diagram of lever BCD ( ) ( )0: 50 mm 200 N 75 mm 0C ABM TS = - = 300ABT = (b) From free-body diagram of lever BCD ( )0: 200 N 0.6 300 N 0x xF CS = + + = 380 N or 380 Nx xC = - =C ( )0: 0.8 300 N 0y yF CS = + = N240orN240 =-= yyC C Then ( ) ( )2 22 2 380 240 449.44 Nx yC C C= + = + = and ﾂｰ= - - == -- 276.32 380 240 tantan 11 x y C C q or 449 N=C 32.3ﾂｰ
- 2. Hallemos Hallemos Ahora en el nodo C Por ultimo hallemos