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• 1.     Método  de  los  nodos:     Primero  dibujamos  el  diagrama  de  cuerpo  libre:     945!!"! 32.3° ▹ ∴ TAB = 300 !! or C = 449 N ! ⎛ Cy ⎞ − 240 ⎞ ⎟ = tan −1 ⎛ ⎟ = 32.276° ⎜ ⎟ ⎝ − 380 ⎠ ⎝ Cx ⎠ ( 380 )2 + ( 240 )2 = 449.44 N ! or ΣFy = 0: C y + 0.8 ( 300 N ) = 0 C y = 240 N C x = 380 N !!! or (a) From free-body diagram of lever BCD Free-Body  Diagram: (b) From free-body diagram of lever BCD Chapter 4, Solution 19. ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0 ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 !!! !! COSMOS: Complete Online Solutions Manual Organization System !! !!!   and θ = tan −1 ⎜ ⎜ C = ΣFy = 0: C y + 0.8 ( 300 N ) = 0 𝐹! = 0      , 𝑅! +   𝑅! − 945   =  0     ∴ C y!= −240 ! N C y = 240 N or 2 2 𝑅C! =   Cx !+= y  945          (𝐼)   + ( 240 ) = 449.44 N ! + 𝑅! C = ( 380 ) and Then 2 2 Cx + C y = ∴ C y = −240 N ΣFx = 0: 0      , N + Cx + = 0  300 N ) = 0 𝐹! = 200 𝑅!! 0.6 ( ∴ C x = −380 N or C x = 380 N Then   2 2 ⎛ Cy ⎞ − 240 ⎞ ⎟ = tan −1 ⎛ ⎟ = 32.276° ⎜ ⎟ ⎝ − 380 ⎠ ⎝ Cx ⎠ θ = tan −1 ⎜ ⎜ or C = 449 N ram: tion 19. nline Solutions Manual Organization System   ∴ C x = −380 N (a) From free-body quilibrio  para  calcular  las  reacciones  en  B   Ahora  aplicando  las  ecuaciones  de  ediagram of lever BCD y  C  tenemos:   ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0   ∴ TAB = 300   (b) From free-body diagram of lever BCD   32.3° ▹
• 2. Chapter 4, Solution 19. Free-Body Diagram:   (a) From free-body diagram of lever BCD ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 𝑀! = 0:         (b) From free-body diagram of lever BCD ΣFx 3,75 N C + −12 945 + 12 += 0: 200𝑅!! + =x  0  0.6 ( 300 N ) = 0 ∴ C x = −380 N   𝑅!! =     Ahora  de  (II)    en    (I)  tenemos:   Then     ∴ TAB = 300 and C x = 380 N or 12 945 = 0: C + 0.8 300 N = 0 ΣFy = 720  𝑙𝑏        (𝐼𝐼)   ( ) y 12 + 3,75 ∴ C y = −240 N C = 2 2 Cx + C y = C y = 240 N or ( 380 )2 + ( 240 )2 = 449.44 N 𝑅!! +   𝑅!! = ⎛  945   Cy ⎞ −1 − 240 ⎞ ⎟ = tan −1 ⎛ ⎟ = 32.276° ⎜ ⎟ ⎝ − 380 ⎠ ⎝ Cx ⎠ θ = tan ⎜ ⎜ 𝑅!! +  720 =  945     or C = 449 N 32.3° ▹ 𝑅!! =  945 − 720 = 225  𝑙𝑏     Se  aplica  el  método  de  los  nodos  para  determinar  las  fuerzas  internas  en  los   elementos  BA,  AC  y  BC,  y  conocer  si  están  en  tracción  o  compresión       Ahora  en  el  nodo  B   !!" 36,87° !! !!" 225!!"!     tan 𝛼 =   𝐶𝑂 9 3 = =     𝐶𝐴 12 4 𝛼 =   tan!! 3 = 36,87°   4   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.
• 3. ⎛ Cy ⎞ − 240 ⎞ ⎟ = tan −1 ⎛ ⎟=3 ⎜ Cx ⎟ ⎝ − 380 ⎠ ⎠ ⎝ ( 380 )2 + ( 24 or ΣFy = 0: C y + 0.8 ( 300 N ) = 0 or ΣFx = 0: 200 N + Cx + 0.6 ( 300 (b) From free-body diagram of lever BCD ΣM C = 0: TAB ( 50 mm ) − 200 N ( (a) From free-body diagram of lever BCD ∴ C x = −380 N ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 2 2 Cx + C y = ∴ C y = −240 N (a) From free-body diagram of lever BCD -Body Diagram: ∴ TAB = 300 Ahora  aplicando  las  ecuaciones  de  equilibrio:   (b) From free-body diagram of lever BCD   θ = tan −1 ⎜ ⎜ C = 𝐹! = 0      , 225   +   𝐹!" sin 36,87 = 0   ∴ C y = −240 N C = C y = 240 N or 2 2 𝐹 = ( 36,87 = ) = 449.44 N C x + C y!" sin380 ) + ( 240−225   and   Then ΣFy = 0: C y + 0.8 ( 300 N ) = 0 Then   2 2 ⎛ Cy ⎞ −1 ⎛ − 240 ⎟ = tan−225 ⎞ = 32.276° ⎟ ⎜ ⎟ = !" ⎝ − 380 =   −375  𝑙𝑏,     ⎝ C x ⎠ sin 36,87 ⎠ θ = tan −1 ⎜ 𝐹 ⎜ and   or C = 449 N Free-Body Diagram: Chapter 4, Solution 19. COSMOS: Complete Online Solutions Manual Organization System   ΣFx = 0: 0      ,         − Cx +   0.6 ( 𝐹 cos = 0 𝐹! = 200 N + 𝐹!" +   300 N ) 36,87 = 0            (𝐼)   !" ∴ C x = −380 N or C x = 380 N 32.3° ▹ 𝐹!" = 375  𝑙𝑏                     𝐼𝐼  ,                    𝑪      𝑒𝑙  𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜  𝐵𝐴  𝑒𝑠𝑡𝑎  𝑒𝑛  𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖ó𝑛     Ahora  de  (II)    en    (I)  tenemos:     −𝐹!"   +   𝐹!" cos 36,87 = 0     −𝐹!"   +   375 cos 36,87 = 0     𝐹!" = 375 cos 36,87 =  300  𝑙𝑏     𝐹!" = 300  𝑙𝑏  ,                    𝑪      𝑒𝑙  𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜  𝐵𝐶  𝑒𝑠𝑡𝑎  𝑒𝑛  𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖ó𝑛       Ahora  en  el  nodo  C     !!" 67,38° !!" !! 720!!"! 𝐶𝑂 9 12 Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell 𝛽 =   tan Johnston, Jr.,= =     Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 𝐶𝐴 3,75 5 The McGraw-Hill Companies.   12 𝛽 =   tan!! = 67,38°   5
• 4. Free-Body Diagram: (a) From free-body diagram of lever BCD ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 ∴ TAB = 300 Ahora  aplicando  las  ecuaciones  de  equilibrio:   (b) From free-body diagram of lever BCD   ΣFx = 0: 0      ,        𝐹 C  x−   0.6 ( 300 N ) = 0 = 0   𝐹! = 200 N + + 𝐹!" cos 67,38 !" ∴ C x = −380 N or C x = 380 N   ΣFy = 0:𝐹!" cos 0.8 ( 300 N ) = 𝐹!"   C y + 67,38 =   0   or C y = 240 N 300 = 780  𝑙𝑏   cos2 + ( 240 2 2 2 C = cos 67,38 ( 380 ) 67,38 ) = 449.44 N Cx + C y = 𝐹!" =             Then ∴ 𝐹C y = −240 N !" =   ⎛ Cy ⎞ − 240 ⎞ ⎟ = tan −1 ⎛ ⎟ = 32.276° ⎜ ⎟ 𝐹!" = 780  𝑙𝑏  ,                    𝑻      𝑒𝑙  𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜  𝐶𝐴  𝑒𝑠𝑡𝑎  𝑒𝑛  𝑡𝑟𝑎𝑐𝑐𝑖ó𝑛   ⎝ − 380 ⎠ ⎝ Cx ⎠ and θ = tan −1 ⎜ ⎜ ector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., lliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. or C = 449 N 32.3° ▹